Questions & Answers

Each e-mailed question will be posted to this page, followed by the professor’s answer. When e-mailing questions, the students should specify whether or not they would like to remain anonymous.

 

Monday, October 21, 2024

Q1) I also have a question about Problem 17, Chapter 4, and how the problem simply states rho_bound = 0. I get that (rho_bound)^(e) = 0 because there is no polarization field, but (rho_bound)^(m) appears to have a solution. It seems each problem is not specifying whether it is (e) or (m) but just assumed to be (e). Is there a reason we are ignoring rho_bound and J_bound for (m)? Also, are (rho_bound)^(e) and (rho_bound)^(m) ever combined to make (rho_bound)^(e&m)? I’ve been searching through the book but can only find section 2.15 and 2.16 which define them, but don’t seem to go much further. Do we explore these more in a later chapter?

A1) Since we have reduced Maxwell’s equations to contain only E and B, the only charges & currents that appear in the (reduced) equations are electric charges & currents.

        If you manipulate Maxwell’s equations such that the only remaining fields are E and H, you will end up with electric charge & current densities in addition to magnetic charge & current densities.

        You cannot combine electric charge and magnetic charge, although both could appear in Maxwell’s reduced equations. Recall that the electric charges come from electrons and protons, whereas magnetic charges come from magnetic north & south monopoles. Also, electric and magnetic charges have different dimensions (and different units). These entities are physically different and do not combine, even though they can both be present in some of the equations.

Q2) Why is it that in your solution to Problem 17, Chapter 4, you do not perform a Fourier Transform on t ? Also is there a reason you first solved for J_bound and then performed a Fourier Transform? Because it seems like performing a Fourier Transform on M and then solving for J in the Fourier domain is simpler. What am I missing?

A2) i) When the time-dependence factor is simple, say, cos(w₀t), the Fourier transform in the time domain gives us a pair of delta-functions (localized at plus or minus w₀), then the inverse Fourier transform (in the w domain) returns us back to cos(w₀t). As long as we remember to invoke the sifting property of these two delta-functions (i.e., replacing w everywhere with either plus or minus w₀, accordingly), there is no need to keep writing everything in detail. I just keep cos(w₀t) as it is, and proceed to do the Fourier transformation only in three-dimensional space – always remembering that, somewhere along the way, w’s must turn into either w₀ or – w₀.

ii) We can do the conversion of M to the bound-current-density J_bound and the Fourier transformation in any order; whichever method is easier to do should be done. I probably thought the way it’s done in the solutions was easier. If you think the other way is easier, then you can do it your way; the end result will be the same.

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Sunday, October 20, 2024

Q) I was working through parts (b) and (c) of Problem #1 from last year’s 2nd midterm exam, and I was having some trouble understanding your solutions. You mention the formation of a cube of dimensions Δ * Δ * Δ. I however thought that due to the existence of a triangle function the upper and lower volumes would be like tetrahedra, with the combined volumes being an octahedron with a square midsection centered at origin on the x-y plane. I appreciate the clarification and can also come by if a pen paper approach is preferred. 

A) The function specified in this problem is the material polarization P(r); there are no pyramids or octahedra here. The polarization P is defined inside an elongated cube. The base of this elongated cube is a delta x delta square in the xy-plane; the height of the cube is 2*delta. The polarization P is uniform in any plane that is parallel to the xy-plane at z = 0. The magnitude of P, however, declines linearly as one moves either upward from the xy-plane at z = 0 (i.e., moving up along the +z axis), or downward from the xy-plane at z = 0 (i.e., moving down along the –z axis). The corresponding bound charge-density is uniform and positive inside the upper cube whose volume is (delta)³. Similarly, the bound charge-density is uniform and negative inside the lower cube, whose volume is also (delta)³.

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Sunday, October 20, 2024

Q) In your solution, when solving for A(r), you skip showing the dk_z portion and skip to that solution because of the delta(k_z). The problem I am having is with the (k_//)^2 + (k_z)^2 in the denominator. When I go through the steps of solving the k_z integral, I get (k_//)^2 + (1)^2 in the denominator, but you apparently get k_z² = 0, so you get just (k_//)^2 in the denominator. How did you do this or is this a mistake?

A) It’s based on the sifting property of delta(k_z). Since the delta-function is centered at k_z = 0, whenever k_z appears in the integrand, it must be set to zero.

Student’s response) Right. It’s the delta(k_z) that goes to one, the rest go to zero. I don’t know why I forgot that. I’ve been working on this homework all weekend with very few breaks, so maybe that is why. Many of these integrals are so difficult and time-consuming. Took me six hours on Saturday to get through the first problem and part (a) of the second. I’ve almost completed the first four today (Sunday). I am hoping to get to the remaining three before end of day Monday.

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Thursday, October 17, 2024

Q) I am really struggling with the OPTI 501 class. I believe that I understand conceptually what we are studying, in a broad sense, but applying the material is proving difficult. I am especially struggling with the integrations. Even trying to understand what is happening between example steps is frustrating. Are there any good resources for these integrations that you could recommend — resources that really show individual steps and perhaps explain why? Alternatively, are there any tutors that you could recommend?

A) If you’ve watched the Friday, 10/18, lecture, you should know by now how to deal with those difficult integrals: Ignore them! The important thing is to go to the end of each example (or the end of the posted solution to each homework problem) and try to understand the physical meaning of the final result. I’ve shown all those integrals in the book (and in the homework solutions) so that the students gain confidence that the final results are rigorously derivable. Once you’ve understood (in a physical way), the meaning of the final results, you may want to go back and check the integrals and derivatives, etc., to see if they are done correctly.

      Some integrals are easy: they can be done by a change of variable or by the method of integration by parts. Some other integrals are more complicated, and one has to resort to advanced methods such as complex-plane integration or just look them up in a handbook. (Also, Mathematica has a symbolic language that enables one to evaluate certain integrals analytically.) The point of this course is not to make you an expert in doing integrals; the goal is to make the students understand and appreciate electrodynamics.

       If an integral shows up in the exams, it will be because it is fairly easy to evaluate. I also provide hints within the exam problems to guide the students through the integration process.

      The textbook contains many problems at the end of Chapters 1 and 3 (these are primarily mathematical chapters) to guide the students through mathematical methods – some of these problems involve integration and differentiation. You may want to try solving some of those problems (or just go over their solutions) if you feel you need to learn how to evaluate complicated integrals.

      Our TA, Paola, has already offered you to help with tutoring (or answering your questions) during her office hours. I am also always available (by email, phone, or zoom) to discuss any difficulties you may have with the lectures, textbook, homework solutions, etc. Please let me know how you’d like to proceed.

Student’s response) Thank you both for your responses. For each problem, I’ll focus on the end goal and the path to get there. In general, I’ll try to keep the substance of the learning material in focus and minimize the amount of time I spend on the minutia. Perhaps I will at least provide an outline or synopsis of how I would solve a problem if I can’t solve it mathematically. 

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Friday, October 11, 2024

Q) In regards to Coulomb and Lorenz gauges, is it correct that we use Coulomb for electrostatic conditions and Lorenz for dynamic EM field propagation scenarios? These “gauges” (why are they called gauges?”) separate out the vector and scalar potentials (page 99 of your book). You have psi and A shown in (1a) and (1b). Do we separate out the scalar potential to understand if the propagation decreases / increases over time? 

A) So far, I have not talked in the class about scalar and vector potentials, nor about gauge; I’ll discuss these topics in the coming week. In the meantime, it seems that you have started reading Chapter 4, hence the questions about Lorenz and Coulomb gauges.

        There is no good explanation as to exactly what the word “gauge” is supposed to mean as used in modern physics. Its roots are in French, where it means “measure.” The word is sometimes used to mean “setting the scale.” Be it as it may, the best way to understand it in classical electrodynamics is to forget about the precise meaning of the word and to lean, instead, on the physical properties of the two potentials (i.e., the scalar potential psi and the vector potential A). In classical electrodynamics, the component of the vector potential A that is parallel to the local k-vector is what determines the gauge. If this parallel component of A is set to zero, we say that we are working in the Coulomb gauge. And if the parallel component of A is chosen such that, in the Fourier domain, k. A = (w/c ²)psi, then we say that we’re working in the Lorenz gauge.

         In general, electromagnetic fields have both a scalar potential psi(r,t) and a vector potential A(r,t). We need both of these potentials to describe the electromagnetic field in static as well as dynamic situations. One good feature of the Lorenz gauge is that it separates the contribution of the (total) charge-density rho(r,t) from the contribution of the (total) current-density J(r,t). The scalar potential (in the Lorenz gauge) is fully determined by rho(r,t), whereas the vector potential (again in the Lorenz gauge) is fully determined by J(r,t). The Lorenz gauge has additional advantages when one works with electrodynamic problems in the context of special relativity. In contrast, the Coulomb gauge has advantages in terms of simplifying the treatment of certain problems in non-relativistic quantum electrodynamics.

        In purely electrostatic problems, J = 0 and, therefore, the vector potential A vanishes. Similarly, in purely magneto-static problems, rho = 0 and, consequently, the scalar potential psi vanishes. In general, however, we must compute both psi and A from the knowledge of rho and J. Once psi and A are found, we evaluate the B-field as the curl of A; that is B(r, t) = curl of A(r, t). The E-field is found from both psi and A; that is, E(r, t) = minus gradient of psi(r, t) – time-derivative of A(r, t). These last formulas are universal, namely, they do not depend on the chosen gauge. Also, no matter which gauge you choose to work in (i.e., Lorenz, Coulomb, etc.) the final expressions that you will find for the E and B (or E and H) fields will be gauge-independent. In other words, the choice of gauge might simplify the mathematical solution of the problem under consideration, but it does not affect the final values obtained for the actual (i.e., physical) fields E and B (or E and H).

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Thursday, October 10, 2024

Q) Why is the analogy of white light used in the fourier domain for temporal frequencies (time dependence of individual colors) and single-frequency wavelength used in the spatial frequency? I understand that white light contains different frequencies of light that oscillate at different time frequencies but doesn’t single frequency light also contain both spatial and temporal? Is it mutually exclusive or does it overlap? 

A) Yes, white light contains many frequencies (i.e., colors), so a prism separates those colors and allows us to measure the Fourier transform of the white light, namely, the amplitude and phase associated with each color. Of course, single frequency light (or what we call monochromatic light) has a sinusoidal time dependence, but in the frequency domain (i.e., Fourier domain) it only has one frequency. In other words, it is a delta-function in the Fourier domain. The location of the delta-function on the frequency axis is the frequency (or color) of the monochromatic light.

        I used the example of white light and its spreading out by a prism to show what Fourier transformation in the time domain looks like. The white light in that example consisted of plane-waves all propagating in the same direction. Therefore, there was no spatial frequency variations in that example; only temporal frequencies (or colors) were involved.

       Later, when I wanted to give a clear example of spatial frequencies, I used monochromatic light, so that there was only one temporal frequency (i.e., the color of that incident monochrome light). By shining the monochrome light on a transparency that had a picture printed on it, I argued that the incident beam breaks up into plane-waves that propagate in different directions (after passing through the transparency). These were examples of spatial frequencies (i.e., the direction of propagation in space specifies the spatial frequencies of a plane-wave).

        Now, if you wanted to use white light in the above example and shine it on a transparency (with a picture printed on it), you could do that too. But now you will have each color (or temporal frequency of the incident light) breaking up into multiple directions (i.e., multiple spatial frequencies emerge out of the transparency). Each color of the incident white light now breaks up into many plane-waves going in different directions. This would be an example in which both temporal and spatial frequencies come into play.

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Thursday, October 26, 2024

Q1) In Problem 3, first midterm exam, we have a magnetic dipole, yet it is claimed that curl (H) = 0. How can this be when there is no time dependence on anything, so dD/dt = 0, but we have a dipole, which can be treated as a loop of current, so J_free should not be zero. Yet the question claims it is. How is there no J_free when we have a magnetic dipole?

A1) The curl of H is zero everywhere in the surrounding space of the point dipole, because the associated loop of current is only confined to the immediate vicinity of the origin. Around the origin itself, the loop integral of H is zero because the current goes into the loop on one side of the origin and comes out on the other side. To observe the current that gives rise to the magnetic dipole, you’ll have to do the loop integral of H on one side of the origin and then take the loop through the center of the dipole (located at the origin of coordinates). This requires knowledge of the H field on the z-axis at the center of the dipole, which is not easy to find. Later, in Chapters 4 and 5, I’ll show you how to address such problems. (Basically, you’ll need to enlarge the magnetic dipole a little bit and make it into a small sphere. Then the fields in and around the spherical dipole will not go to infinity, and you can answer all such questions to your heart’s satisfaction.)

Q2) I have a question regarding Maxwell’s equations. Are they universally numbered? As I was practicing for the exam, I noticed that the order of Maxwell’s equations differs between certain textbooks and your course materials, which has left me a little confused. In some of the questions, you ask us to find a solution using the 2^nd or 3^rd Maxwell equation, and I am unsure which equations you are referring to, as I have seen them ordered differently in various sources. Could you please clarify how you would like us to interpret the numbering of Maxwell’s equations?

A2) Indeed, the numbering of Maxwell’s equation is different in different textbooks. I always refer to them the way they are numbered in my own book, which is the course’s only textbook. That’s how we’ve been using them in the class and in the problem sets throughout the semester.

Q3) In Problem 3, midterm exam 1, there is a reference to the singularity of the curl and divergence operators at the origin of the spherical coordinate system. What do you mean by “singularity” in this context?

A3) Here, singularity means that the mathematical formula for the operator does not yield meaningful results at the point of singularity. A simple example is provided by the function f(x) = 1/x, which is well-defined for all values of x, except at x =0, where f(x) goes to infinity. We say the function f(x) has a “singularity” at x =0.

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Wednesday, September 25, 2024

Q1) When we look at the Poynting vector, we take cross-product of the real E field and real H field, leaving out the imaginary parts because the quantity is real and measurable. In class you mentioned that the absorption of material is tied to an imaginary quantity. Does this mean that absorption of a material isn’t directly measurable? Glass like Germanium and other IR materials do state rough absorption numbers in units of [cm^-1] which makes me believe it’s a measurable quantity, but I can’t think of a way to measure an individual piece of glass’s absorption coefficient without making some assumptions about its atomic structure. 

A1) The Poynting vector is always real-valued, but this does not mean that the absorption of light cannot be measured. In an absorptive medium, the divergence of the time-averaged S(r, t ) – a real entity — is nonzero, which information can be used to compute the local absorption rate. There are simple, standard ways to measure absorption of material media in the lab; no need to resort to the knowledge of their atomic structure.

       When we get to Chapter 6 of the textbook, we’ll discuss the origins of the refractive index and explain why, in general, the refractive index n is treated as a complex entity (i.e., n = n’ + in”). The imaginary part n” of the refractive index is related to the rate of absorption of electromagnetic energy inside the material. Even though the refractive index n is, in general, complex-valued, and even though we use complex functions and complex numbers to represent the E-field, the H-field, etc. — but not the Poynting vector, for reasons that have to do with the non-linearity of the cross-multiplication operation involved in defining S(r, t) in terms of its constituents, the E and H fields — you must remember that the physical entities that are measurable in practice are always real-valued. When we measure the refraction and absorption of light in material media, we use techniques that measure the real part of n and the imaginary part of n separately. So, in the real world, we always measure entities that are real-valued, e.g., energy, linear & angular momenta, force, torque, distance, time interval, velocity, acceleration, oscillation frequency, etc.

Q2) Not so related to Chapter 2 (but something I’ve thought about a lot), I remember listening to Feynman talk about the probability of reflected light on a glass surface in a recorded video. You discussed the window with real refractive index that transmits light and reflects some light (Transmission + reflection = 1). The reflected light is around 4^ish percent w/o AR coatings, but I was curious if you had a rough conceptual explanation on what dictates the amount of light reflected back for a real indexed material. Or, what about the incident light is non-uniform that 4% is being reflected in the opposite direction of the transmitted 96%?

A2) Indeed the reflected optical power from a well-polished, flat surface of a dielectric medium whose (real-valued) refractive index is 1.5 equals 4% of the incident light power at normal incidence. The general formula for reflectivity (at normal incidence) is R = |1 – n|^2 / |1 + n|^2. This equation is valid for real- as well as complex-valued n. (Please note that the formula for R involves absolute values, which is how the end-result turns out to be real-valued.)

       Once we get to Chapter 7, we’ll have a lot more to say about reflection and transmission from various material media at normal incidence, at oblique incidence, the dependence of R and T on the (complex) refractive index, on the polarization state, and also on the angle of incidence of the light beam, etc.

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Monday, September 23, 2024

Q) After reading the section about Maxwell’s boundary conditions and working on the homework, I found myself confused: If the H-field vector and B-field vector are perpendicular, then the M-source would need to vector in a direction such that when added to the H-field vector (multiplied by mu₀), the sum would equal the B-field vector, right? Also, all of the magnetic vectors would be coplanar, right? The same would apply for the analogous electric terms, too. If all of that is correct, what happens without dipoles (i.e.- the M-sources or P-sources)? How can the H [or E]-field vector (multiplied by mu₀ [or epsilon₀]) be equal to the B [or D]-field vector?

A) I don’t know how you got the idea that the B and H fields must be mutually perpendicular. This is not the case, nor is it required that the E and D fields be perpendicular to each other. You must be thinking about Maxwell’s boundary conditions, where we’ve described the continuity of the component of E that is parallel to a surface (or interface), and also the discontinuity of the component of D that is perpendicular to a surface (or interface) in the presence of surface charges. Same goes for the component of B that is perpendicular to a surface (this one is continuous), and the component of H that is parallel to a surface (this one may be continuous or discontinuous, depending on whether a surface current exists at the surface, and also depending on the direction of that surface current).

       In general, B, H, and M are co-planar at any given point in spacetime, but there is no requirement that any pair of them be perpendicular to each other. If no magnetization M is present at the point of interest (i.e., at r, t), then B is simply equal to mu₀*H, which means that the B and H fields are parallel to each other — albeit with different magnitudes, as dictated by the value of mu₀. The same goes for E, D, and P. In the absence of material polarization, i.e., when P(r, t) = 0, the displacement field D(r, t) will be parallel to E(r, t) — with a proportionality coefficient of epsilon₀, that is. When P is present, we just add P(r, t) (vectorially) to epsilon₀*E(r, t) in order to obtain D(r, t). That is all.

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Friday, September 20, 2024

Q1) What is the difference between “q” and “Q”? Is one the charge of a particle, and the other total charge in a volume?

A1) Yes, I’ve used q for the charge of a small particle (or point-particle), and Q for the total charge enclosed within a volume (i.e., integral of charge density rho over the entire volume). There is no standard notation for these, but that is how I like to represent them.

Q2) I did not quite follow the derivation of the energy density of the E field during lecture. In particular, why the factor of ½ exists alongside the dot produce of E and E.

A2) The derivative of the product of two functions, say, f (x) and g(x), is f ’g + f g’. From here, we find that the derivative of the product of f (x) and f (x) is f f ’ + f ’f  = 2f f ’. By the same token, d/dt of E.E is 2E.(dE/dt). And this is where he factor of 2 comes from.

Q3) I always kept a mental model of energy as atoms/molecules moving (vibrating as heat or bulk motion as velocity). This atomic motion can be “absorbed” by material objects, for example, microscopic structure change (material deformation, change in entropy), and transfer of electrons (chemical reactions). Is there an analogy for a physical object that “carries” the energy of an electric/magnetic field like how the random/bulk motion of particles carries kinetic energy?

A3) I don’t know of any good models for the energies of the E and H fields. All we know is that these fields — whether static or dynamic, whether inside material media or outside — carry electromagnetic (EM) energy, whose respective densities are epsilon₀*E² /2 and mu₀*H²/2. These energies, when absorbed by material media, could turn into kinetic or potential or thermal or chemical energy of the material media, and those energies are precisely as you’ve always understood them to be.

Q4) Problem 2-18: How do we define effective field? In this case, is it the total contribution of the field on the surface charges, half of which comes from outside the sphere, the other half from inside? If the sphere was in a uniform E field, would the effective field and regular field be equal?

A4) The effective E-field acting on the surface charges is the average of the E-field immediately above the surface and the E-field immediately below the surface. The E-field below the surface is zero and, therefore, the effective E-field is one-half the E-field immediately above the surface.

      Think of each little lump of charge on the surface as a tiny little ball of charge q. One half of the little ball (charge q/2) facing outward sees the external field, and therefore, the force acting on it is qE/2. The other half of the charge (again, q/2) that faces inward sees no field at all and, therefore, the force it experiences is zero. All in all, the force of the E-field experienced by this little ball is qE/2. And that is why we say that the effective E-field that acts on the little ball in its entirety is E/2.

       Note that the E-field we’re considering here is produced by all the charges on the sphere and, therefore, the E-fields immediately above and immediately below the surface are always present in the aforementioned way. Now, if you bring in a new E-field from the outside and immerse the entire sphere in that E-field, you’ll have to add (vectorially) the new E-field to the self-field of the sphere. Therefore, the total E-field will always have the same discontinuity when going from immediately above to immediately below the surface. The discontinuity is equal to the surface-charge-density (sigma₀₎ divided by epsilon₀. Even though the new field brought from the outside is continuous at the sphere’s surface, the contribution of the self-field to the total field will retain its discontinuity.

Q5) Problem 2-18: Why does shrinking the shell conserve total charge Q, but not surface charge density σ₀? Is this a matter of conservation of energy? I would have assumed that the surface charge density would remain constant — which would decrease Q with decreasing R.

A5) The total charge within a closed system is always conserved. This is another one of those laws of nature, just like the laws of conservation of energy, linear momentum, and angular momentum. (We have discussed conservation of charge in Sec. 2-11, in the context of the charge-current continuity equation.) So, when one shrinks the sphere, the total charge Q remains the same; it is the surface-charge-density sigma₀ that increases.

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Thursday, September 19, 2024

Q1) Equation (27) of chapter 2 defines the momentum density. Is there no presence of relative permittivity or permeability in the equation because these values would already be accounted for in the Poynting vector?

A1) Equation (27) of Chapter 2 describes the electromagnetic momentum density. It applies to inside as well as outside material media. The E and H fields, of course, must be computed at the location where the momentum density S/c² is being evaluated (i.e., inside or outside material media), but the formula yields only the electromagnetic part of the momentum. There usually exists a mechanical momentum, induced by the electromagnetic field, associated with the material media (in addition to the electromagnetic momentum). In free space, where no material medium resides, the only momentum is electromagnetic. Inside material media, however, we generally have both electromagnetic and mechanical momenta. The force law of electrodynamics can be used to determine the mechanical part of the momentum.

Q2) So the mechanical momentum would be dependant upon the medium through which the wave is traveling. But the electromagnetic momentum for a wave is always based on free space?

A2) Yes and No. The mechanical momentum is dependent on the material properties and also on the E and H fields of the electromagnetic field present at various locations inside the material. The EM momentum density is given by S(r, t)/c², where S = E x H is the Poynting vector inside the material medium at the location of interest. The values of E and H at (r, t) inside the material, will, of course, depend on the overall properties of the material medium – and on the external source(s) of the E and H fields.

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Tuesday, September 17, 2024

Q) While the real component of the fields represents the physically measurable values associated with the field, does the imaginary component represent anything other than helping describe phase relationships? Like how the imaginary component of impedance is the reactance, or the imaginary component of the refractive index is the extinction coefficient .

A) You’re right about the imaginary parts of the E-field, B-field, etc. There is no physical significance to these imaginary parts; simply put, the imaginary parts are there to simplify the math. In the case of refractive index or impedance of electric circuits, both real and imaginary parts do have physical meaning — and are significant in their own rights. In the case of complex representations of the E and B fields, there is no comparable significance to the imaginary parts.

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Monday, September 16, 2024

Q) I’m finding it a little difficult to understand homework 2, problem 13. Perhaps this is because the lectures are not in sync with HW assignment due dates? Anyhow, I do read all chapters you assign before trying to complete the HW but I find some gaps in understanding that hopefully can be bridged by lectures and the wonderful questions student are asking during class.

Since I’m a distance student, I could really benefit if you elaborate regarding to Maxwell boundary conditions in your book 2.17. I’m a little confused in terms of what is continuous and what is discontinuous, especially when there is current vs no current. I’ve been trying to understand your reference in terms of pillboxes and rectangle loops. I’m hoping you can give some more visual insights into how the H-vector flows regarding to J vector. I’m also curious to understand how this applies in the real world. What are some situations these things are important so that we don’t make errors in our thought process? I wish I could go to your office hours so I can frame these questions better; I’m sorry. Hopefully when I have a better understanding I can ask better questions. 

A) The boundary conditions of Sec. 2.17 have real-world importance when it comes time to understand the behavior of things like optical fibers, multilayer thin film stacks, resonant cavities, etc. We will talk about some of these issues in Chapter 7, where many of the real-world applications of Maxwell’s equations will be discussed. However, we cannot begin to talk about such applications until the students understand the meaning and significance of Maxwell’s equations. Whenever there exist two different material media adjacent to each other, the Maxwell equations can be solved in both media separately, but the solutions remain disjointed at the boundary between those two media. That is when we invoke Maxwell’s boundary conditions to stitch the two solutions together — and finally come up with a complete and consistent solution to the physical problem of light (or other EM waves) propagating from one homogeneous medium to a second (adjacent) homogeneous medium.

       With regard to the geometry of one of those Maxwell’s boundary conditions where the electric current flows on a surface (i.e., a thin sheet of surface-current-density J_s exists on the surface), here is how the argument goes: If you look immediately above the surface, you’ll see a magnetic H-field component that is parallel to the surface and perpendicular to the direction of flow of the current (i.e., direction of J_s). Similarly, if you look immediately beneath the surface, you’ll find a different H-field component that is also parallel to the surface and perpendicular to J_s. These two components of the H-field are not the same. The difference between the H-field immediately above and the H-field immediately beneath the surface is exactly equal to the magnitude of J_s. In this sense, there is a discontinuity in the parallel component of the H-field that, while parallel to the surface, is perpendicular to the direction of J_s.

       Of course, there is nothing special about this property of H and J_s at the aforementioned surface; it is a direct consequence of Maxwell’s second equation, as Sec. 2.17 attempts to explain with the aid of a thin rectangular loop. The other equations of Maxwell lead to similar conclusions with regard to the continuity or discontinuity of the various fields (i.e., E parallel, B perpendicular, D perpendicular). The important point is to try to specialize Maxwell’s equations to the boundary between two adjacent material media, where surface-charges and surface-currents may or may not exist. One then relates the various field components on one side of the surface (i.e., immediately above) to the same component of the field on the other side of the surface (i.e., immediately beneath).

       These have practical applications as we’ll show in later Chapters — especially Chapter 7. What I want the students to take away at this point is that, not only are Maxwell’s equations the embodiment of powerful laws of nature, but also that they prove to be very useful in practical applications. One can solve these equations, for instance, to find out what happens to light (or other forms of electromagnetic radiation) inside various homogeneous material media, but also the same equations can be employed to tell us how to stitch solutions obtained for adjacent media into a complete and consistent solution that would work for the entire system.

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Monday, September 16, 2024

A) I think by introducing multiple spheres you’re making the problem more complicated than it needs to be. Just think of a single closed surface – i.e., that of a balloon, not necessarily having a spherical shape. For as long as the charge q stays inside the balloon, the total integral of epsioln0*E over the surface of the balloon will be equal to q. The instant that the charge leaves the balloon, the integral goes to zero, as there is no longer any charge inside the balloon.

       It is true that, if the balloon is very large, it takes quite some time for the information about the location of the charge (i.e., inside or outside the balloon) to reach the distant parts of the balloon’s surface. However, in the immediate vicinity of the location where the charge exits the balloon, the E-field is very large; it also gets larger and larger as the point-charge q approaches the surface from within. As soon as the charge gets out, the E-field in the immediate vicinity of the charge (which is still very close to the surface of the balloon at the point where it exited, but now on the outside) will reverse its direction. Before exiting the surface, the local E-field was pointing one way (that is, on the nearby parts of the balloon’s surface), but after exiting the surface, the E-field will be pointing in the opposite direction (again, on the nearby parts of the balloon’s surface). This is the reason why the overall integral suddenly drops from q to zero at the very instant the charge gets out of the balloon.

       It is the contribution to the surface integral by the E-field in the immediate vicinity of the exit point that makes the biggest difference. All the far-away points on the balloon’s surface continue to contribute their share, of course, but the contribution of that part of the balloon’s surface that is very close to the exit point will flip sign immediately upon the particle’s exit and, by doing so, it cancels out the contributions from all the far away points on the balloon’s surface.

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Wednesday, September 2, 2015
 
Q) I’m struggling to understand problem 30 (Chapter 1) on the homework and the best way to approach it. Since it is half the magnitude of C0′ with angular velocity of + and -, I halved C0′ * cos(wt) with one -w and +w which equates to cos(0) =1.

1. A) Let me suggest a couple of things you might want to try. If you’re still stuck, it’ll be okay. Hand in your HW tomorrow, then check the solutions that will be posted later in the week. 
   
   1) One way to think of the vector C0’ * cos(wt) is to imagine it as the sum of two vectors of length ½ C0 that are rotating in a plane, one clockwise, the other counterclockwise. (The plane of these two vectors is the same as the plane defined by the vectors C0’ and C0”.) 
   
   2) The vector C0” * sin(wt) can likewise be decomposed into two rotating vectors. This time, however, the initial position of the two half-vectors (i.e., their position at t = 0) is *not* the same as the direction of C0”. You must rotate these half-vectors by 90 degrees (one of them by +90 deg., the other one by -90 deg.) in order to get the correct behavior at t = 0. 
   
   3) You now have 4 vectors instead of the original two. However, two of these vectors are rotating clockwise (which you can combine into a single vector — simply by adding them together), and the other two are rotating counterclockwise (which you can also combine into a single vector).
   
   4) The two counter-rotating vectors found above, when added together and followed in time, will describe an ellipse. Try to express the major and minor axes of this ellipse in terms of the two (rotating) vectors. 

Friday, February 20, 2015

Q1) I was just asked a question about the physical mechanisms behind Brewster’s angle. Specifically, they asked why does only s-polarization reflect completely, but not p-polarization. Fresnel equations aside, I thought it was because, when the light hits the boundary between the two media, some of it is absorbed and re-radiated by the electric dipoles in the media (so, reflecting the s-direction in free space). But what happens to the dipoles that refract p-light?

A1) I’ll explain it in detail if you come to my office hours sometime next week. It’s not easy to give a clear explanation by e-mail.

Q2) Thank you for clarifying the physical phenomenon behind Brewster’s angle yesterday. I had one last question: Usually when the topic of Brewster’s angle is explained, the incident light is unpolarized. However, the reflected light carries s-polarization. Does that mean that the dipoles in the dielectric medium are preferentially polarized (i.e. their dipole moments are already arranged such that s-polarization comes out no matter what)?

A2) Unpolarized light is a mixture of p- and s-polarized light. Since p- and s-light do not interact with each other, you may consider the contribution of each component (that is p-component and s-component) to the reflected/refracted light separately and independently. In other words, at Brewster’s angle, the p-polarized light that is contained in an unpolarized beam goes through the glass medium without being reflected, while the s-polarized light is partially reflected from the surface and partially refracted into the glass. The two components of polarization contained in an unpolarized beam of light behave exactly as they would if they were alone (i.e. shining on the glass separately).

Friday, January 9, 2015

Q) I am trying to get a better understanding of Gaussian beam propagation, which is given as the paraxial solution to the helmholtz equation: http://en.wikipedia.org/wiki/Helmholtz_equation#Paraxial_approximation. The first few lines of the Wikipedia article state that it is necessary to assume that spatial dependence and time dependence of a given solution to the wave equation are separable: U(r,t) = A(r)T(t). Why are we able to separate the time and spatial dependence of this solution?

A) This is a general technique for solving partial differential equations (such as the Helmholtz equation). You first separate the variables, then solve for these separable “eigen-functions” of the equation. In the end, any general solution is found by a superposition of these eigen-function solutions. 

Wednesday, December 17, 2014

Q) I wanted to take an opportunity, now having finished the class officially, to tell you how much I appreciated the course. The final two chapters in the course on the Lorentz Oscillator model and Plane Electromagnetic Waves in LHI Media have been especially interesting as well as more directly useful for the analysis I do on a day to day basis. Having had undergraduate experience in topics such as refractive index, polarization, etc., I feel as though I have a much stronger fundamental understanding of what dictates these properties.
     As a sophomore in the undergraduate program I worked with Dr. Moloney and Dr. Brio on a small project that involved a FDTD simulation of EM waves and I was studying Maxwell’s equations, PML and boundary conditions, and reflection/ transmission of thin film stacks. I recently read Chapter 8 in the textbook because of the very close correlation of topics. It is rewarding to acknowledge that while at the time, I was in way over my head, I can look back and understand the model I was working with and the problems we were analyzing.

A) Thank you for the kind remarks. I am delighted to know that you’ve enjoyed Opti 501. I hope there’ll be an opportunity to see you in another class.

Tuesday, December 16, 2014

Q) I guess I finally figured out why you express EM waves with complex-valued parameters. Last time when I saw you, I drew a picture of EM wave. So if the wave is actually spinning with respect to time, it forms a circle if we see it from the cross-section. Separating it into two components stops the spinning and simplifies the analysis. Is it correct? If that makes sense for E, the complex H is the same.
     Mathematically, I can tell k” indicates the direction of decaying. Does it mean the EM wave travels in a different direction k’ while decays in k” (as a result of inhomogeneous medium)?
     And for Poynting’s vector, since the very first I saw it, I thought I should have the direction of k vector. Expression 7-14 on your book seems certified my assumption. It expresses both the magnitude and direction of energy movement while k only denotes a vector and the XYZ coordinates.

     So the whole idea is to represent realistic EM waves instead of the simplest thing I drew last time. Am I right? I got my ideas from the attached graph (see the attached file “December 17, 2014 Question.png” at the bottom of this page.)

1. A) You’re correct. I’ve written the book very carefully, so that the equations contain all the important features of electromagnetic waves. Once you understand the complex scalar/vector field notation, you can just look at the equations and understand their physical meaning; it is not complicated at all. The equations are a complete and compact way of expressing the physical reality of electromagnetic fields (and their sources).

Monday, December 15, 2014

Q) I have been reviewing my notes and have some questions that I would like to ask. In lecture 24 we solved Maxwell’s Equations in LHI media. Is there ever a case when E'(0) is perpendicular to E”(0) and H'(0) is perpendicular to H”(0) such that E'(0) is parallel to H”(0) and E”(0) is parallel to H'(0)? What sort of field would this represent?
    We saw that for plane waves at non-normal incidence to a LHI material that k’ and k” are not parallel or perpendicular to each other.  k” travels along the direction or propagation and k’ travels at some angle. My question is: what determines the direction of propagation of k’ and k” for a plane wave at non-normal incidence to a LHI material?
     Lecture 26: For the portion of the lecture on time-averaged pointing vector. Is there any physical interpretation/reason why we analyze the conjugate of H, H*? What mathematical principal is satisfied by analyzing the real part of H as H(r,t) = Re{H(r)e^-iwt} = [H(r)e^-iwt + H*(r)e^+iwt]?
     You mentioned in lecture that once we add a tiny bit of imaginary part to mu(w) and epsilon(w) that we now have complex values for these parameters and the material absorbs. Can you give some practical values of the complex values used in different applications, like fiber communications, and say to what extent we can ignore any absorption and treat the material as transparent?

A) Below are the answers to your questions:

1) When the beam is elliptically polarized, it is always possible to decompose the E-field into two perpendicular components, E0’ and E0”; these will represent the major and minor axes of the ellipse of polarization. Since H0 is given by k x E0, it should be clear that a real-valued k will result in H0’ and H0” to be similarly perpendicular to each other, and behave as you described in your question.

2) Usually, the incidence medium is transparent and the incident k-vector is real. Then, upon transmission into a second medium, k can acquire both real and imaginary components, k’ and k”. The Snell’s law specifies the x and y components of k, which must be the same as those assigned to the incident beam. However, kz is determined by the dispersion relation, and here part of kz may become real and part of it imaginary. When you put kx, ky, and kz back together (for the transmitted beam), you’ll find that the k-vector has both real and imaginary parts, and these are what we call k’ and k”.

3) The physical E and H fields are ALWAYS real-valued. We use complex notation because it makes mathematical manipulations easier, but in reality all fields are real, and, in the end, we must take the real part of whatever complex entity we have associated with the field in order to obtain the actual (i.e., physical ) field values. When we derived the time-averaged Poynting vector, this is exactly what we did; we wrote the real part of E and cross-multiplied it into the real part of H. An alternative way to express the real part of a complex entity is to write it as ½ the sum of the complex field and its conjugate. That is how ½ (H + H*) came about.

4) In general, all media have an absorption coefficient, but sometimes the absorption coefficient is small enough that it can be ignored to a good approximation. If n = 1.5 + 0.1*i, for example, then the imaginary part is too large to be ignored. You can put this value of n into the complex exponential expression for the E-field (or H-field), that is, exp[i(k.r – wt)], and see that, within a few wavelengths of propagation, the field amplitude drops by 90% or so. However, if n = 1.5 + 0.00001*i, then it takes a long propagation distance (perhaps hundreds of meters or kilometers, depending on the wavelength) for the field amplitudes to drop by a significant amount (e.g., by 10%, or 20%, or more). All of this, of course, can be easily checked by putting the complex value of n into the complex exponential factor and see for yourself.

     In the case of fiber optical waveguides, the imaginary part of n could be as small as 0.00001; however, since the light needs to propagate for many kilometers in the fiber, this small absorption coefficient will eventually have significant effects on the field amplitude. In contrast, if the beam were to propagate inside a block of pure glass material having n = 1.5 + 0.00001*i for only 100 meters or so, then the attenuation would be so small that you wouldn’t have to worry about it — hence the neglect of the absorption coefficient under such circumstances.
     You see that whether or not one is allowed to ignore the imaginary part of the refractive index is application dependent. It depends on the wavelength of the light and on how far the light is going to propagate in the medium, and also what level of attenuation would be acceptable in that particular application. (You may search the internet for the complex refractive indices of many materials, then use the aforementioned exponential factor to determine how much the field amplitude drops over a given propagation distance.)

Monday, December 15, 2014

Q) I hope all is well with you. I’m not sure if you remember me immediately, I was a PhD student at the college of optical sciences and worked with Robert Norwood, I took your first 501 course way back and although it has been a long time, I wanted to contact you to thank you for being a key professor on that particular course which helped me get through most of the optical sciences program. Actually after I took your 501 course I realized that thanks to the many office hours I bothered you with, I managed to understand concepts that other students that were previously ahead of me couldn’t.
     I succesfully passed my final oral exam for my PhD and now I’m moving on with the rest of my life here in Mexico. I went back to Tucson for two days for the defense this semester after being on a leave of absence, and unfortunately didn’t have time to stop by your office, hence this email as a poor substitute. I’m now in my hometown and I have two daughters, and working part time as a professor at my alma mater university here in Guadalajara.

A) I was delighted to receive your e-mail and to learn that you’ve now settled back in your country, with your husband and children, and started to teach in your Alma Mater in Guadalajara. Many thanks for your kind remarks about Opti 501. When you took my Opti 501 class, I still had handwritten notes. Since then, however, my book on the subject has been published. I am attaching to this e-mail a pdf copy of the book for your personal use. I wish you the best in your personal and professional life, and hope to have an opportunity to see you again in the near future.

Saturday, December 06, 2014

Q) I’m trying to figure out why Eqs.(17a) and (17b) are valid in Chapter 7. From Maxwell’s equations we determined that the parallel components of the electric and magnetic fields must be continuous, so Ex and kx must be the same above and below. If this is the case, then the reflection will always be the same since the reflected and incident Ex0 are equal. This means that the electric field components must NOT be equal. However, from my understanding, if the kx vectors are the same at the interface, this will result in the Ex values also being the same since z = 0 at the interface and ky = 0 for p-polarized light. I must be missing something fundamental, can you tell what it is that I am getting wrong?

A) Continuity of a parallel component of E, such as Ex, means that Ex^(incident) plus Ex^(reflected) must be equal to Ex^(transmitted). In other words, the total Ex on one side of the interface (say, immediately above the interface) must be equal to the total Ex on the other side of the interface (immediately below). This condition does NOT imply that Ex^(incident) is equal to Ex^(reflected), as you seem to be assuming. You must take Ex^(transmitted) into account as well. 

Sunday, November 30, 2014

Q) In section 7.5 of the textbook, the rule Re(alpha) = 1/2 (alpha + alpha*) is used to expand and solve the S(r,t) function. I have a few questions regarding that:

  (i) Is there any reason this rule is applied/used on H(r,t) and not E(r,t)?

  (ii) What does alpha* physically represent? Once we get to <S(r,t)> = 1/2 Re[E(r) x H*(r)] I’m unsure what H*(r) actually represents?

  (ii) Why does the sign of the exp(iwt) swap after use of the above rule, s.t. H(r,t) becomes 1/2[H(r)exp(-iwt)+H*(r)exp(+iwt)]?

1. A) (i) You may use the conjugation either on E or on H; the final result will be the same, because E x H* and E* x H are conjugates of each other and, therefore, their real parts (which is what <S> turns out to be) will be the same.
   
   (ii) In complex notation, H = |H| exp(i.phi). The conjugate of H is, therefore, H* = |H| exp(-i.phi). Thus the conjugate of the H-field, has the sign of its phase reversed.

(iii) We are conjugating the entire H-field, including its time-dependent factor exp(-i.w.t). When you conjugate something like A.B.C, you’ll get A*.B*.C*. In the present case, both H(r) and exp(-i.w.t) must be conjugated; that is why we get H*(r). exp(+i.w.t).

Thursday, November 20, 2014

Q) I remember there is one method for solving EM problems in the Jackson’s book, called “Green’s Function,” but you don’t mention this method in your book. Do you think this method is too mathematical to solve EM problems? Or is it useful in other physical field?

A) The Green’s function approach to solving differential equations has its own advantages, but I don’t use it in my book because I try to use the simplest possible (and also most general) mathematical tools to get to physical answers. The only place where I have mentioned Green’s functions in my book is in problems 40 and 41 at the end of Chapter 1.

Saturday, November 15, 2014 

Q) In lecture 21 you mentioned that the penetration or skin depth is described as the depth at which the oscillations, caused by the incident plane wave, of the atoms extinguishes to zero. From the lecture I gathered that the skin depth is a function of the amplitude and phase of the incident plane wave, the wavelength of the incident wave, and the dielectric properties of the mirror material. If I were an optical engineer who was designing a fiber optic transmission system and wanted to pick a core and cladding combination which minimized the skin depth would there be a formula I should consider which describes the behaviour of the skin depth?
     Just out of curiosity, are there any tricks you can play by altering the polarization while not compromising TIR. One thing that initially comes to mind, would be to pick a core and cladding combination which has the same TIR and Brewster’s angle. Would such a technique be beneficial to isolate one polarization state in order to increase the transmission length along the fiber of another polarization state?

7. A) You’ll see the answer to your question once we get through the material in Chapter 7. The penetration depth (or skin depth) and the depth of the evanescent field that resides in the cladding of a fiber are intimately related.
   As for your second question, the TIR angle does not depend on the state of polarization, although the field amplitudes in the evanescent field are polarization-dependent. You’ll also see in Chapter 7 the relation between the Brewster angle and the critical angle of TIR. The Brewster angle is generally smaller than the critical TIR angle, so it is not possible to get rid of one polarization while allowing the other polarization to propagate through the fiber via TIR.
   It’s best if we don’t discuss these things before learning how to solve Maxwell’s equations in material media — which will be done in Chapter 7.

Sunday, November 09, 2014

Q1) During lecture 21, I had a couple of questions regarding the physics of a PEC mirror vs normal mirrors:

(i) Is the reflected light only a function of radiated plane waves from e- dipole oscillations? If so, in a PEC mirror, would the reflected light be only 50% of the incident beam’s intensity?

(ii) In a normal mirror, do the radiative plane waves propagating to the right of the surface (through material) trigger additional e- dipole oscillations in the next atomic layer? If so, does the radiated plane wave from the second layer back in the direction towards 1st layer, cause additional e- dipole oscillations in that 1st layer? Ultimately, I’m not sure how we ever get something like 95% reflectivity of a mirror, as it seems like most of the radiative energy is lost to heat in the material. I’d love to know that inner working a bit more.

A1) You will learn the inner workings of the perfectly electrically conducting (PEC) mirrors as well as those of normal mirrors in a few weeks, after we go through Chapters 6 and 7. In the meantime, let me answer your questions as best I can — without the tools to be developed in Chapters 6 and 7.
     (i) Yes, the reflected light is only due to the oscillating conduction electrons, which, together with the stationary nuclei of the atoms, can be considered as oscillating dipoles (for more on this see the Drude model of conduction electrons in Chapter 6). The reflected light’s amplitude, and also its intensity, are 100% of those of the incident light; no energy is lost in the process.
     (ii) The answer to most of these questions is yes. The dipoles in each atomic layer are excited by the incident beam as well as by the light radiated by all the other dipoles (that is, dipoles within the same layer, as well as those in other layers). You have to think in terms of interference among different waves (incident wave and radiated waves by various dipole layers). Destructive interference cancels out the fields in many places. In the first few atomic layers of the mirror, the fields survive and excite the atomic dipoles (or the conduction electrons). These dipoles absorb a few percent of the incident beam and radiate the remaining fraction of the incident optical energy in the backward direction (i.e., toward the source of the incident light).

Q2) Couple of follow up questions to (i) and (ii): In your description of the PEC mirror during lecture, it was described that the surface conducting e- oscillations cause plane waves in both directions (for conducting sheet), with the waves into the material destructively combining with incident light (due to H-field phase shift) to effectively yield no propagation. I guess I’m thinking about this like a balloon with a source hole at the top, and two holes on each side of the balloon. If the incident air flow then “leaks out” each hole equally, but we consider the hole to the right to be producing air that is null, the stream on the left would be 50% that of incident. I can see the logical incongruence in my metaphor, but I’m not sure I see how to properly think of the mirror afterall. For (ii), it seems that my lack of knowing in (i) is constipating my understanding of (ii). I’m hoping you might have the prescription for that.

A) You seem to be ignoring the effects of “interference.” Your balloon metaphor fails to take into account the interference between the incoming wave and one of the outgoing waves. The radiated field that propagates in the same direction as the incident wave has both its E and H fields oppositely oriented to the E and H fields of the incident wave. Therefore, these two waves completely cancel out. Their energies are *not* added together but rather subtracted from each other. As I mentioned in response to a student’s question in the class, the Poynting vectors are not additive (because Maxwell’s equations are linear in E and H, but not in E x H). You must think in terms of superposition of fields, which allows for constructive as well as destructive interference. Your analogy with the air particles getting in and out of the balloon is misleading precisely on this point, because it cannot account for interference between two waves.

Thursday, November 06, 2014

Q) You mentioned in the lecture today that a lag in phase of susceptibility means the material is lossy, whereas a lead in phase of susceptibility means there is a gain in the material. If the susceptibility lags by a phase of 3pi/2, could this be rewritten as leading by a phase of pi/2? If that is the case, how can we distinguish between gainy and lossy materials? Do absorptive materials never have a susceptibility phase lag of greater than pi?

A) The phase angle should be limited to the interval (-pi, pi]. Ultimately both epsilon and mu (permittivity and permeability) must work together to determine if the material as a whole is gainy or lossy. The imaginary part of the refractive index n (which is defined as the square root of epsilon times mu) will be either positive or negative. When the imaginary part of n is positive, the material will be lossy, and when the imaginary part of n is negative, the material will exhibit gain. The phase angles of the electric and magnetic susceptibilities will eventually end up determining the sign of the imaginary part of the (complex) refractive index n.

Monday, October 27, 2014

Q) I have a question about Bessel functions of the second kind Y(x). In page 73 of the book, Eq.(32), I can say the Bessel function of first kind J(x) is either even function or odd function which depends on the number n, but how about the Y(x)? Is it neither even function nor odd function because it has an imaginary part?
    My second question is about the equation appearing on the 4th line from the top of page 117 of your book. Why is the coefficient (2π)^3 not (2π)^-3 in the integral? Although I know if it is (2π)^-3, the answer will be wrong!

A) The way the function Yn(x) is defined it is neither odd nor even. The appearance of the imaginary “i” in Eq.(33) of Chapter 3 has something to do with the term ln(x/2) in the functional form of Yn(x) given in Eq.(26) of the same chapter.
     Concerning your second question, I don’t see why you are being confused about the factor (2*pi)^-3. If you write (2*pi)^-4 appearing in front of the equation as (2*pi)^-1 (2*pi)^-3, then move (2*pi)^-3 inside the first integral and place it before the second integral, you’ll see that everything is exactly correct. There should be no reason for confusion here.

Saturday, October 25, 2014

Q1) In looking at the posted solution to Chap.4, Prob. 4, I’m seeing incongruity with how we described point charges during lecture 15 and 16. Is this because rho is only a function of r in the homework but is a function of r,t in the lecture example?

A1) After you work with the Fourier method for a while, you recognize that sometimes it is easy to handle the time-dependence directly. For instance, when the charge or current distributions are time-independent, there is no point to introduce delta(omega) in the calculations, then remove it in the end. You’re essentially working at one frequency, omega = 0, and this can be taken into account directly, without going through the trouble of dealing with delta(omega). This is what’s going on in HW 4, Prob. 4. 

      In some other examples and problems of Chapter 4, the time-dependence factor is cos(omega_0*t). This means that we only have two frequencies, +omega_0 and –omega_0. We then avoid using the time part of the Fourier transform, concentrate on the space part, and insert omega = + or –omega_0 into the equations by hand. This is just a short-cut that comes from familiarity with the Fourier method; the essential approach to solving the equations using the method of Fourier transform has not changed; we’ve just become lazy.

Q2) Hopefully one day this will become more familiar to me as well. At the moment, I’m feeling quite defeated by the math in front of me. Concepts and inter-relationships between elements (e.g., how scalar potential is related to charge density, etc.) is making sense in both domains to me, but inevitably I get to a point in every problem where I hit a wall mathematically. I’ve been going over your posted solutions to problems I couldn’t compute and finding, even with your solution, I more often than not have no idea what is being done, but the end result makes sense.

Of the fundamental math elements, I see great value in having quick and comfortable knowledge of the following topics/tools:

1) Chain rule (especially with PDE fns);
2) Integration by parts;
3) Conversion to alternative coordinate system and consequences to intergrand elements (changes in limits, changes in volume elements, relationship between x,y,z and related coordinate system coordinates);
4) Trigonometric identities;
5) Common Fourier transformation equations and resultants.

    Are there any other topics you could recommend I put some time into? I have a feeling this midterm might be pretty rough for me, but I’m trying my best to persevere through this, even if the light bulb doesn’t go off until far after the course is complete.

A2) To your list I’d add vector algebra and a little bit of complex number theory (mainly complex algebra, Euler’s identity, conversion from “Real and Imaginary” representation to “Magnitude and Phase” representation, etc.). At the end of Chapters 1 and 3 (the math chapters), there are many problems that deal with all the math tools that you need in this course. You may want to go over the solutions to some of those problems once again.
      The math required for the exam is not going to be nearly as difficult as in the text and homework problems. If you’re getting the physical concepts, you should be able to do the exam problems with some elementary knowledge of the algebraic methods.

Saturday, October 25, 2014

Q) I wrote some comment and have two questions about Bessel function in the attachment.

A) Your understanding of the energy flow in your diagram depicted at the top of your attached page is correct.
     Your understanding of the Bessel-function-relations throughout your attached page is incorrect. For the correct relations, please see Chapter 3, page 73, Eqs.(32) and (33). These equations, in turn, may be derived from Eqs.(25) and (26) on page 72 (Chapter 3).

Saturday, October 25, 2014

Q1) In example 5.5.2, would it have been valid to set the observation point to r=(0,0,0) with r’=(ρ,0,z) instead of r=(ρ,0,0) with r’=(0,0,z)? In this problem it seems like we would still get the right answer, but I am just wondering if this is a valid general approach.

A1) No, I’m afraid you cannot assign r and r’ arbitrarily; for instance, r’ must be a point within the source. In Example 5.5.2, the source is a long, thin wire located on the z-axis. The only choice you have for r’ is x’ = 0, y’ = 0, and z’ anywhere from –infinity to +infinity. There are no current sources anywhere else in the system, and one must not deviate from the above choice for r’.
     As for the observation point r, because of the symmetry of the problem, the choices for phi and z are irrelevant, that is why we chose them both to be zero. (You can choose any other fixed values for phi and z without affecting the outcome of the calculation.) The only coordinate for the observation point whose choice matters is its distance from the current-carrying wire, which is defined as rho in the cylindrical coordinate system. That is why we chose for the observation point r = (rho, 0, 0,).
     I didn’t understand how you came up with your alternative choice for r and r’. Please let me know if the above answer is not clear.

Q2) Your answer makes sense, thank you. The solution for problem 26a in Chapter 5 is given in one line but I am having trouble getting to the right answer. For the polarization vector, I have: P=δ(x)δ(y)*I0*d*sin(ωt), leading to: Jtotal=ω*δ(x)δ(y)*I0*d*cos(ωt) (both in the z direction). Are these correct so far? I am having trouble computing the integral to calculate the vector potential field A, So I’m wondering if there is a trivial way to solve this that I’m not thinking of.

A2) The current is already given to you; it is I0*sin(omega*t). Let’s say the cross sectional area of the little cylinder is A. Its height is given as d. Then the volume is A*d. This is a small volume by definition. So there is no integral to take; in other words, your dr’ = dx’dy’dz’ is simply A*d. To obtain the current density J, divide the current by A. However, when you multiply the current density by the volume element (i.e., dr’ = A*d), the cross-sectional area A cancels out. The end result is the one given in a single line in the solution to this extremely simple problem. There is no integral to take, because the current occupies only an infinitesimal volume.
     You also seem to be confusing the polarization P(t) with the current I(t). Recall that the bound-current-density is given by J(t) = dP/dt (partial derivative). The current density is I(t)/A. To obtain P(t) you must take the integral of I(t)/A.

Saturday, October 25, 2014

Q) On page 95 of your book, why do you say the time-averaged rate of the Poynting vector in Eq.(52) is (1/4)Z0(Jso^2), instead of (1/8)Z0(Jso^2)? Why don’t we consider the< (cos(w0t))^2> term in Eq.(52)?

A) I “did” take into account that the time-average of cos^2 is ½. However, the energy is propagating on “both sides” of the current sheet, that is, on the +y-side as well as on the –y-side. Therefore, the result, after time-averaging, must be multiplied by 2 to give us the “total” rate of flow of energy out of the sheet. That way, the factor 1/8 is multiplied by 2 to give us ¼.

Thursday, October 23, 2014
 
Q1) On problem 9 (p78) from Chapter 3, does A(r) mean Ao exp(i (k dot r)) and B, Bo exp (i (k dot r)), with r = <x,y,z>, k = <kx,ky,kz>?

A1) In problem 9, Chapter 3, the A(r) and B(r) fields are *not* plane-waves. However, you can always express any field [such as A(r) and B(r)] as a superposition of plane-waves. In other words, you may express the fields as an integral of plane-waves over all possible vectors (kx, ky, kz). This is just saying that you should be working in the Fourier domain.

Q2) Thanks, so, taking the F.T. of A(r) we have integral of [ A(r) exp-i(k dot r) dr ] and I don’t have to worry about the actual form/value of A?

A2) Yes.

Sunday, October 19, 2014

Q) During this week’s lectures I noticed a mathematical principle being used repeatedly that I want to ensure I’m clear on. As we transition from space-time domain to Fourier domain (indicated below as =>), it seems that the following events occur:

Grad => i*k (e.g. -Grad(Psi(r,t) => -i*k*Psi(k,omega)

d/dt => -i*omega (e.g. dE(r,t)/dt => -i*omega*E(k,omega)

I’m not 100% clear as to why this can be employed and was hoping for some guidance.

A) Please take a look at the examples in Chapter 1. For the gradient, divergence, and curl operations, there are examples in the book that show what happens when you apply these operators to plane-waves. If, after seeing those examples, you still feel that you’d need help, please feel free to e-mail our TA or contact me again.

Sunday, October 19, 2014

Q) Starting in section 4.9, I noticed that the spatial fourier transform is frequently found without including the time domain. Why don’t we need to do the time domain transform as well here?

A) Good question; I’m glad you asked. When the time-dependence is written in the form of exp(i*omega0*t), as it is in that example, we already have it in the Fourier domain. There is no point in doing a Fourier transform of exp(i*omega0*t), and then an inverse transform, because you’re going to end up with the same exp(i*omega0*t) in the end. So, as a shortcut, I deleted this step in the calculations. I let the time-dependence stay as it is, namely, exp(i*omega0*t), and only worried about the spatial dependence of the function. That is why I Fourier-transformed P(r, t) in three-dimensional space only, and ended up with P(k, t). [Recall that we are dealing with different functions here, so the fact that the letter P is used to denote the function does not mean that P(r, t) and P(k, t) are the same functions. The argument of one being (r, t), and of the other being (k, t), helps to distinguish the two functions.] 
Friday, October 17, 2014

Q) I had a question on choosing to do a Fourier transform in 3 vs 4 dimensions (excluding the temporal fourier transform). Is there some way you can tell from the initial problem whether you need to do the 4 dimensional transform or not (like in example 4.12)?

A) Good question; I’m glad you asked. When the time-dependence is written in the form of exp(i*omega0*t), as it is in that example, we already have it in the Fourier domain. There is no point in doing a Fourier transform of exp(i*omega0*t), and then an inverse transform, because you’re going to end up with the same exp(i*omega0*t) in the end. So, as a short cut, I deleted this step in the calculations. I let the time-dependence stay as it is, namely, exp(i*omega0*t), and only worried about the spatial dependence of the function. That is why I Fourier-transformed P(r, t) in three-dimensional space only, and ended up with P(k, t). [Recall that we are dealing with different functions here, so the fact that the letter P is used to denote the function does not mean that P(r, t) and P(k, t) are the same functions. The argument of one being (r, t), and of the other being (k, t), helps to distinguish the two functions.]
Wednesday, October 15, 2014

Q) I have a math problem about Laplacian operator in cylindrical coordinates (or spherical coordinates?). It is difficult to type in the e-mail, so the problem is attached on the e-mail. Why there are two additional terms in E(r) and E(phi) in the red region? Why can’t I just write the bottom formula in the attachment.

A) We will discuss the Laplacian operator in Chapter 5. You’ll have to wait until next week to see how these terms in the Laplacian come about.

Wednesday, October 08, 2014

1. Q) I think the place I am struggling the most in this class is with the vector operations like we saw in problem 2 of the midterm. For 2A, I am trying to understand where you got k1 x k1 from the operations. Naturally, I understand if you can get to this point then the result is zero since a vector has zero orthogonality with itself. I’ve attached a scan of where I’ve gotten in my work, can you take a look at it and let me know if I’ve done anything wrong and where I could’ve gotten k1 x k1?
   
   A) Please take a look at Chapter 1, Example 1, Eq.(11), and then Example 3, Eq.(21). You should be able to see where k x k comes from.

Friday, September 26, 2014

Q) After thinking about the Lorentz for law and what it comes from I realize the solution from HW2#8 results in a B-field gradient along the radial axis between the two current containing wires. Is the resulting force a result of the violation of Maxwell-Ampere Law? The H-field gradient should only be possible if there is a surface current density present? If the wires are present in a vacuum or some medium that does not contain a free charge density then is this natures way of correcting the imbalance by trying to align field lines of equal magnitude? That’s how I’m interpreting this so please tell me if I’m wrong.

     So my question is if the (insulated) wires are inside a medium which contains free charge density such as a liquid NaCl solution or an ionized gas does the force disappear and current density appear in the B field gradient? Is the Lorentz Force Law only applicable in the absence of free charge carriers between the 2 current wires? or is there an appearance of M to compensate somehow?

A) I am sorry, but I have no idea what your question is. Are you talking about Problem 8, Chapter 2, here?
     Problem 8, Chapter 2, is about the magnetic field and magnetic flux between two current carrying wires. Where does the Lorentz force law come in? Why do you think the Maxwell-Ampere law is violated? Why do we need free charge-density in the system? I have no idea what you are thinking. Please break down your question into simpler ones, then ask me one question at a time.

     What we have in this problem is two current-carrying wires in vacuum. The B (or H) fields are the result of the superposition of the fields produced by two independent sources (i.e., the two wires). Each source produces a field that satisfies the Maxwell-Ampere equation. The “gradient” of the H-field does not appear in the Maxwell-Ampere law, rather it is the curl of the H field. In this problem the curl of the H-field is zero in the space between the two wires (it is nonzero only where the currents exist, in accordance with the Maxwell-Ampere law). Why should the Lorentz force law come into play in the context of this problem baffles me. Please clarify.

Wednesday, September 24, 2014

Q) I find the coefficient of curl(M) in Eq.(13b) of Chapter 2 difficult to understand. Then I realized your definition of H is slightly different from the book I am using. It’s Markus Zahn’s Electromagnetic Field Theory. I took two photos for you to understand what happened. Please help me to understand this.

1. 
   A) In the literature you find two different definitions of B. Some authors define it as B = mu_0*H + M (as I do in this course), while other authors define it as B = mu_0*(H +M). It is just a matter of how one chooses to define M. In our method, M is mu_0*I*A, where I is the current in the loop and A is the loop area. In the alternative definition used by other authors, M = I*A, without any mu_0. Both definitions work fine, and one has to always stay consistent in his use of M and B.

Wednesday, September 24, 2014

Q) I had question on problem 31 while studying for the exam tomorrow. When you solve for the voltage using the differential equation, you get a negative cosine. However, in conventional circuit theory when you solve for the voltage using phaser math you would get a positive cosine. Do you know why this is?

A) You may be thinking about current rather than voltage. The relation is always the same: I(t) = C dV/dt. It then depends on whether your current is sin(wt) or your voltage. It also depends on the direction you choose for the current, i.e., going into or out of the plate relative to which the voltage is being measured. In any case, what you get out of phasor analysis should be no different than what you get out of the above equation.
Wednesday, September 24, 2014

Q) I have question on midterm 1 (2013) for the last problem. We use maxwell’s 1st equation between R1 <R< R2 to find the sigma1. In this case we use the condition: E field inside the metallic conductor is zero. Then we found the sigma2 (not zero), according the shell is initially charged neutral. I think, if sigma2 has some value, then a new E field will induced by sigma2. As a result, the E field inside the metallic conductor will not be zero.

A) The field inside the metallic conductor is zero everywhere. Remember that a uniformly charged cylindrical shell does *not* have a field *inside* the cylinder, but there is a field outside. So, the inner cylinder surface at R1 produces an E-field in the metal which is cancelled out by the field produced by the central wire. The outer cylinder will produce an E-field only outside the outer cylinder, where R > R2, but it cannot create any fields inside the cylinder, where R < R2. You can prove this by using Gauss’s law.

Tuesday, September 23, 2014

Q) In Lecture 8 somebody asked about a 3 medium scenario about the parallel E-field components. It made me think of another scenario where you would see a discontinuity of the E-field in the same medium at the region where there is discontiguity from the other 2 media. Does this imply there is an accumulation of charge at this point due to the boundary condition of maxwell’s 1st equation. Could you take a look at the attached figure and maybe talk about it in today’s class?

A) This is a good question; you get a point for asking it. There is no need for a discontinuity of the E-field in medium 3 of your diagram. Since E||^1 and E||^2 must be the same, there will be a discontinuity between D||^1 and D||^2, which is accommodated by the appearance of a (free) surface-charge-density at the boundary between media 1 and 2. If free charge is not available in the system, then all the fields will have to be zero at that junction in order to satisfy the requisite boundary conditions.

Saturday, September 20, 2014

Q) I am doing problem 19 part (e) of the homework which asks us to find the angular momentum of the system. I found equation 34 in section 2.22 for the angular momentum density. I figured I would just integrate my result over the volume of the solenoid in which an electric field is present, but then I realized that the solenoid is infinitely long. Am I doing the problem correctly, or should I be looking for a different method?

A) Angular momentum-density is the angular momentum per unit volume. For this you do not need to know the total volume of the solenoid. Also, in general, one may calculate the angular momentum and torque for a unit length of the solenoid (i.e., angular momentum and torque per meter of the solenoid length).

Friday, September 19, 2014

Q1) I have a question about the EM momentum. In your book, you say the EM momentum is given by the Poynting vector divided by the square of the speed of light in vacuum and it is under any and all circumstances. But when I google the EM momentum in the material, there is a controversy, called Abraham-Minkowski controversy concerning this issue. So the question is, is the Eq.(27) in Chapter 2 applicable when the light is propagating in a material? If it is true, does that mean we have considered the E field and H field are in the material when we calculate the Poynting vector? But why there is a Abraham-Minkowski controversy? Is it resolved?

A1) Yes, the E and H fields must be computed in the material medium if a material medium resides at (r, t). Recall that everything is local; everything is calculated at (r, t), so if you want the momentum-density at that point, you must calculates the E and H fields at that point. Many things can reside in the SAME location in space-time. For example, you can have rho_free, J_free, P, M, E, H, etc., all located at the same point (r, t). In other words, fields and materials (i.e., sources) can be co-located.

     The momentum-density E x H/c^2 is known in the literature as the Abraham momentum-density. Presently, there is no controversy about the Abraham momentum being the correct “electromagnetic” (EM) momentum; just about every author who writes about these issues now agrees that the “electromagnetic” momentum is equal to the Abraham momentum. What has been controversial is the question of “total” momentum, which is the sum of the electromagnetic and mechanical momenta. In Opti 501 we don’t talk about total momentum, even though I believe the controversy in the case of total momentum has been resolved as well — however, other authors may not agree with me on this point.
      Nowadays, the controversy about total momentum can be translated into a related controversy about the electromagnetic force and torque equations, and whether the correct expressions for EM force and torque are those of Lorentz, or Minkowski, or Abraham, or Einstein-Laub, or Chu. Again, in my book I don’t go into any details about these controversies; the force and torque equations that I believe are correct are those of Einstein and Laub, which I have used in the book to the exclusion of all others.
      You can read about these controversies in the vast literature of the subject (see, for example, the references cited at the end of Chapters 2 and 10). I have written about these matters extensively during the past ten years; if you visit the arXiv on the internet (http://arxiv.org/), you can search under my name and find nearly 40 papers on this and related topics.

Q2) Concerning the previous letter, I have searched your paper published in Optics Communication detailing the A-H Controversy, and I will take some time to read your paper to learn more details about the controversy. Another question is, you have said a example about the EM momentum: A light passes through a glass which is coated a layer of anti-reflective material on both sides. When the light passes through the glass, the glass will have a small displacement. But why? If the incident light and transmission light have the same momentum, why the glass has a displacement? Does it violate the conservation of momentum?

A2) Okay, that paper of mine is a good place to start with; there are many references at the end of that paper, which you might consult as well.

     The displacement of the glass slab does NOT violate momentum conservation. By the time the light pulse has exited from the glass slab, the glass has stopped, so it does not have any momentum. (Remember that momentum is the product of mass and velocity; if the glass stops moving, its momentum will become zero; therefore, all the momentum is now back in the light pulse and, therefore, momentum is conserved.) When the light pulse is still inside the glass slab, it has LESS momentum than it does when it is entirely in free space. Therefore, the difference between the momentum of the light pulse in free-space and its momentum inside the glass will be picked up by the glass slab. As long as the pulse remains inside the glass, the glass has nonzero velocity and, therefore, a non-zero momentum. During this time (i.e., while the pulse is still inside the glass), the glass slab travels a little bit forward. That is the reason for its displacement at the end. But, in the end, since the glass slab has zero velocity, it has zero momentum. Consequently, the momentum conservation law is not violated. 

Thursday, September 18, 2014

Q) I’m not sure I understand the difference between the E field and the D field. If you measured an electric field, would you have any way of determining whether it was an E field or a D field? Is the only difference that the E field includes fields that rise from bound charge?

A) The E-field is the electric-field that you have in vacuum. If the field is measured at a location in space where some polarization P also exists, then the D-field at that location will be equal to epsilon_0*E plus the polarization P. It may not be easy or even possible to separately measure E and D. But, conceptually, you may always think of a field E that exists in some region of space, and whenever it so happens that there is also some material medium which has polarization P at a given point within that region, then the D-field will be the sum of the two things mentioned above, namely, D(r,t) = epsilon_0*E(r,t) + P(r,t). 

Thursday, September 18, 2014

Q) Wanted to pass on some information, in the event you have some students who want to read through the Feynman lectures, that the entire collection is available for free on http://feynmanlectures.info/. It’s in partnership with CalTech and contains all fixed errata.

A) Thanks; this is very useful information.

Monday, September 15, 2014

Q) I have a question about Problem (18C) on the 501 homework. The question asks for the effective field. I read through the chapter and could not find anything on the effective field. I did however find something on page 228. It says, “The effective E-field acting on this charge-density is the average of the E-fields immediately above and immediately below the surface.” Does this same definition apply to this problem?

A) Think of the layer of charge on the surface of the sphere as being very thin (but not of zero thickness). The E-field on one side of the layer is zero, while on the other side its magnitude is E. What is the average field acting on the charge layer? Answer: It should be half-way between E and zero, which is E/2.

Saturday, September 13, 2014

Q1) You mentioned that, although not yet found in nature, magnetic monopoles are theoretically possible. If a magnetic monopole was discovered, would Maxwell’s 4th equation be disproven?

A1) If monopoles were discovered, the corresponding magnetic charge-density must be added to the right-hand-side of Maxwell’s fourth equation. Also, the corresponding magnetic current-density must be added to the right-hand-side of Maxwell’s third equation. Equations 3 and 4 would then resemble equations 1 and 2. In other words, there will be symmetry between free charge and free current densities of electricity and magnetism.

Q2) I can see how Maxwell’s second equation yields a B field for a infinitely long wire with current that is inversely proportional to the distance to the wire [B=µ0I/(2pi*r)]. If this is the case, though, does the magnetic field approach infinity close to the wire? That doesn’t make sense to me.

A2) Any realistic wire must have a finite thickness. Therefore, you cannot get arbitrarily close to the center of the wire, and the field cannot go to infinity. As you enter the wire, the field begins to decline.

Wednesday, September 10, 2014

Q) I was curious about question number 6, chapter 2. You are asking for the charge densities on the top surfaces of the plates however, you are also giving us the surface-charge densities. Isn’t that redundant?

A) Each plate has two surfaces: top and bottom. So altogether there are four surfaces in this problem. The problem gives you the total charge-density on EACH plate. You now have to determine how that charge-density is divided between the top surface of the plate and its bottom surface.

Tuesday, September 09, 2014

Q) In the last lecture you mentioned how the equation describing the Electric Displacement D = (epsilon0)E + P implies that the E-field polarizes vacuum by creating an electron-positron dipole. Is there a similar phenomena happening with the corresponding H-field? Does the H-field magnetize vacuum by creating a pair of particle-antiparticle magnetic monopoles as a magnetic dipole in a vacuum. Is this the theoretical basis for the existence of magnetic monopoles and would it be a good place to look for such particles?

A) I was using the analogy between P and epsilon_0*E to indicate that, in the absence of P, one may interpret the D-field as though it were due to the polarization of the vacuum. One way to interpret this polarization is to associate it with electron-positron pairs, which supposedly reside in vacuum.
      A similar qualitative argument can be applied to the case of the magnetic field: There is an equivalent analogy between M and mu_0*H. Therefore, in the absence of magnetic materials (i.e., when M = 0), one could say that the B-field is produced by the magnetization of the vacuum in accordance with B = mu_0*H.
      No one has ever observed the breakdown of the vacuum into a magnetic monopole–anti-monopole pair, so whatever we say at this point is pure speculation. It could as well be that the H-field somehow induces tiny loops of electrical current in the vacuum. (Recall that tiny loops of electrical current are indistinguishable from magnetic dipoles formed by a pair of equal and opposite magnetic monopoles.)
    In the absence of experimental data, it’s best to treat these analogies as speculations only. Maxwell’s equations, of course, are correct, but what is beneath the surface of these equations is anybody’s guess.

Tuesday, September 09, 2014

Q) From the last lecture, we saw that magnetization can be obtained by taking the sum of magnetic dipole moments in a small volume and dividing by that same volume as it approaches zero. This seemed identical to the method for achieving electrical polarization except that magnetic dipole moment is replaced with electric dipole moment. Could we describe a magnetic dipole moment as a pair of opposite magnetic charges (in units of Webers) separated by some distance?

1. A) Yes, you may. As far as Maxwell’s equations are concerned, it makes no difference whether magnetic dipoles are represented as Amperian current loops, or as pairs of north and south poles separated by a short distance. The units of those magnetic poles, as you’ve indicated, would be Webers. In reality, of course, no one has ever seen magnetic monopoles, so all the natural dipoles that we know of are apparently made up of electric current loops.

Monday, September 08, 2014

Q1) I have a question regarding HW 2, Problem 2, parts (c) and (d). In my understanding, part (c) is asking about the continuity equation applied to the individual species and part (d) is asking about a single continuity equation that describes the superposition of both particle species.
    I think (c) should state something along the lines of: for each individual specie, the continuity equation still applies because the change in p over time could include space and time factors as well as surrounding charge influence. The continuity equation says that for any change in p over time there is an equal and opposite change in the gradient of the charge density and for each specie the continuity equation still applies. Am I on the right track? It seems to me like a matter of conservation of charge?
    If (c) is true, then (d) should be a straightforward combination of the two species in terms of p and J (from parts a and b) and then writing the continuity equation for the combined terms?

A1) The continuity equation does NOT apply to individual species. For example, suppose a positive charge comes into contact with a negative charge, resulting in their mutual annihilation. You will then have a change in the local charge density for each species without having a corresponding term for Div.J1 or Div.J2.
    The sum of the two charges, however, is always conserved; that is why the continuity equation is valid for the total charge.

Q2) I have a follow up question. Can the scenario of a charge coming into contact with another charge resulting in mutual anhilliation be thought of as a drastic change in p_free such that d/dt(p_free) does not exist (p_free is no longer a smooth and continuous function at the point) and thus the continuity equation does not hold for a single specie? 

A2) First, I think “specie” is not a word, at least not one that has the same meaning as the singular form of species. I have always seen “species” used in plural form.
    Second, the continuity equation is useful because it indicates that electric charge is conserved. When different species of charge annihilate each other, you no longer have conservation, and then the continuity equation loses its meaning. You may be able to use delta functions to model the annihilation (or creation) of individual species of charge, but then the divergence of J (i.e., div.J) may not play any useful role in your model. All in all, it is best to consider only the total charge-density and the corresponding total current density. When individual species of charge are involved, one will be able to treat them in certain mathematical ways, but one cannot apply the overall continuity equation to these individual species.

Thursday, September 04, 2014

Q) I have some concerns about my abilities to work through this week’s homework assignment. I’m definitely finding that my working knowledge of the mathematical topics at hand are insufficient, where this week’s homework took me nearly 10 hours to complete, and was unable to even complete questions 9, 23, or 30 due to a fundamental lack of understanding what the question was asking for and/or how to proceed through it.

      I’m wondering, in your opinion, if this is significant cause for concern or if this is par for the course? When other students in the past have struggled with these topics, have any study aids or methodologies helped them correct course (even something as dramatic as dropping 501 and going back to take or self study undergraduate material)? Looking through the course material, there is consistent use of these topics and so I want to ensure I have a firm grasp prior to moving forward, or else will never truly “get” the subject or the physical principles at hand.

1. A) In the past, some students have spent more than 20 hours on each HW. So, 10 hours is not bad at all!

     Problems at the end of Chapter 1 are meant to help you get up to speed with math. (More mathematics will be forthcoming in Chapter 3.) After you have practiced with the problems in Chapters 1 and 3, you should have sufficient knowledge of the required math for Opti 501.

     Please look at the posted solutions for problems at the end of Chapter 1. See if you understand them. If any difficulty, please call me or the TA to ask for guidance. You should ask the questions, and we’ll try to guide you through the solutions. You need to persevere; many students have had difficulty with this course in the beginning. You’ll learn how to handle the math. It’s not as bad as it looks at first.

Tuesday, September 02, 2014

Q) I was confused about the cross-product. Let A= (0,0,1), and B=(0,0,1) Do the cross-product of those two vectors. My question is that the result of the cross product is the original point or not exist? 

A) The cross-product exists; it is (0,0,0). When two vectors are parallel to each other, as it happens to be in this case, their cross-product is a vector of length zero. The vector (0,0,0) has zero length. When the length is zero, the “direction” of the vector cannot be defined.

Thursday, August 14, 2014 

Q) I am a first-year graduate student and this Fall I will take the Opti 501 class. Since I would like to prepare for the first week of classes, I ordered the class textbook few weeks ago but it seems that it will take more time to arrive. I noticed in the syllabus that you recommend reading Jackson’s book. Do you recommend this one or another book to start reading before classes start?

501. 
     A) Welcome to Opti 501. You probably have ordered a hardcopy of the textbook. It typically takes 6-8 weeks before it arrives, because they print on demand. Had you ordered the electronic version, you would have received it in a couple of days.

      I suggest to read The Feynman Lectures on Physics, Volume 2 (instead of Jackson). Also, if you used David Griffiths’ book for your undergraduate course in electromagnetics, it will be a good idea to review that book as well.

Saturday, December 07, 2013

1. Q) I have a question about Problem 1B on the 2012 Fall final. In equation 9 of the solution, you equate the transmission coefficient to 1 + rho. I thought rho + tau = 1. Is this not true in this case?
   
   A) rho + tau is not equal to 1. You are thinking of reflected and transmitted power (or Poynting vector). When we write 1 + rho = tau, it is a direct consequence of the continuity of the tangential component of the electric field at the interface. Just multiply both sides of the equation into Ex^(inc) to see that it becomes Ex^(inc) + Ex^(ref) = Ex^(trans).

Tuesday, December 3, 2013

1. 
   Q) In chapter 7, problem 21 parts a and b, when we find E and H fields from the scalar and the vector potentials, how can the fields be imaginary, when E and H are always said to be real?
   
   A) When c = c’ + ic”, then Im(c) = c”. In other words, the imaginary part of a complex number is always real. The fields in Problem 21 are therefore real; they are just written as the “imaginary parts” of complex functions.  

Wednesday, November 27, 2013

Q) In today’s lecture you mentioned that some of the energy goes into the evanescent wave, whose Poynting vector points along the boundary, and said that without frustrated TIR, the energy couples back into the prism into the reflected beam. I’m wondering how this happens if the Poynting vector is parallel to the boundary. I don’t think this is the case, but it seems that if one were to place a detector just beyond the prism, in the plane of the boundary, then the energy would be carried away from the prism to the detector. How does the energy return into the prism if the Poynting vector is not pointing back into the prism?

1. A)Insightful question; you get a point for asking it. What I said applies to a plane-wave. Remember that plane-waves are infinite in extent, so there are no leading and trailing edges to the beam.

      Now suppose the beam has a finite cross-section, so that it can be treated as a beam that has a finite-diameter footprint at the bottom of the prism. Let’s say that in the xz cross-sectional diagram the beam touches the bottom of the prism between points A on the left and B on the right (the x-axis is parallel to the bottom of the prism). The finite-diameter beam is no longer a single plane-wave, but a superposition of an infinite number of plane-waves, each having a slightly different k-vector. The Poynting vector of the evanescent wave now has a vertical component in the vicinity of the point A, which brings energy down into the free-space region below the prism, and another vertical component near the point B where it takes the energy back into the prism. In between the points A and B, the Poynting vector in the free space region below the prism is essentially horizontal.

    If you now imagine the beam getting wider and wider (to approximate a true plane-wave), the points A and B move to minus/plus infinity along the x-axis, respectively, and the vertical components of the Poynting vector move out of sight, leaving us with only the horizontal Poynting vector beneath the prism.

 Monday, November 25, 2013 

1. 
   Q) Do we represent quantities like D(r,t) = Re{εoε(ω) Eo exp[i(k⋅r −ω t)]}, as a standard mathematical notation or does it have something to do with the physical meaning like exponential decay or something similar?
2. A) This is a plane-wave of the D-field, like any other plane-wave. The E-field has its own plane-waves, so does the H-field, and so does the D-field. Because the medium is linear, isotropic, and homogeneous, we have written the polarization P as a plane-wave whose amplitude is proportional to the exciting E-field. When you combine epsilon_0*E with P, you get the D-field in the form D(r,t) = Re{εoε(ω)Eoexp[i(k⋅r −ω t)]}.

Sunday, November 17, 2013

6. Q) I have two questions about section 6.5 on the Clausius-Mossotti relation:

1. a) Is it true that this section is in order to get the relation of polarizability coefficient C(w) and electric susceptibility X(w), which stands for the relationship between P and E in microscopic and macroscopic worlds, respectively? In other words, the effective E-field and X(w) is actually for the microscopic world, while the local E and C(w) is for the macroscopic one?

1. b) Why does the model in this relation use the small sphere instead of small pillbox or other shape block? Is it just for mathematical calculate convenient or because that we have to?
   
   A) The Clausius-Mossotti relation is intended to relate the polarization P(r,t) to the “total” local E-field E(r,t). Before the Clausius-Mossotti correction is applied, P(r,t) is related to the E-field that drives (i.e., excites) the local dipoles; the proportionality constant in this case is C(w). After the correction, the E-field is the “total” local E-field, which includes the driving field as well as the “internal” field of the dipoles. Since the internal field has already been taken into account (in the form of the spring-constant alpha), the Clausius-Mossotti correction removes that contribution from the local E-field. In the end, the polarization P(r,t) is related to the “total” local E-field E(r,t) via the proportionality constant Chi(w).

      The spherical shape is purely for mathematical convenience. The field inside a spherical dipole is “uniform” and proportional to P(r,t). That is why Clausius and Mossotti chose this shape. If they had used a cube or some other shape, the internal E-field of the dipole would not have been proportional to P(r,t), in which case the calculations would have become very difficult (an non-local). The only way to get a simple relation between the driving E-field (i.e., the E-field that actually excites the dipoles) and the total local E-field E(r,t) is to choose a spherical shape for the local dipole.

Wednesday, November 13, 2013

Q1) In Problem 1 of chapter 6, I am not able to imagine how do we consider a volume in the solution, when the problem asks us to assume a surface element?

A1) Suppose there is a thin box to the left and also a thin box to the right of the surface element Delta_s shown in the picture. The box has the same surface area as Delta_s, but its thickness is small. There are many dipoles in each box, some of which cross the surface Delta_s, so that their positive charge ends up in one box, while their negative charge ends up in the other box.

Q2) I have just watched the last lecture where you spent some time discussing electronic transitions in the context of the Lorentz Oscillator model. I can see that the energy in the applied external electric field will cause some electrons to transition into higher energy orbitals and in my understanding, once an electron has made that transition, its properties are changed. In this model I suppose that would correspond to it having a different spring constant, resonance frequency and friction coefficient. Obviously, if the applied field is weak, not many electrons will make this transition and overall the material will behave ‘normally’. However, I am wondering what happens if the applied field is very strong and a large proportion of the electrons have been forced into a higher energy orbital. Does the model still apply, or does there come a point when the material deviates from its expected behaviour?

A2) I didn’t mean to confuse the students with two fundamentally different models. The model we are discussing in Chapter 6 is the Lorentz oscillator model. It is just a mass-and-spring model, and should not be confused with the quantum theory of transitions between different levels in atomic systems. (When a student asked me about the relation between the Lorentz oscillator model and different atomic transition frequencies, w_0k, I mentioned transitions between different energy levels of the atoms.) In general, the Lorentz model applies to material media (naturally composed of atoms and molecules) provided that the applied E-field is not too strong. When the field becomes strong, we enter the regime of nonlinear optics, at which point the Lorentz oscillator model is no longer satisfactory.

      It is also possible to pump some materials (e.g., gain media used in lasers) so that their atoms would reside primarily in the excited state (i.e., the upper energy level); we talk about population inversion in this case. One can apply the Lorentz oscillator model to such media as well, but I don’t want to distract you at this point. The main idea of the Lorentz oscillator is that a mass-and-spring model can describe the behavior of the atomic system when the exciting E-field is not too strong. Hendrik Lorentz invented his model before the electronic structure of atoms was known; we should not mix the two models.
Saturday, November 9, 2013

Q) After Thursdays lecture I have a couple of questions about material and refractive index: (1) Why does atomic/molecular density affect refractive index? If refractive index is a measure of how hard it is to oscillate an atom’s electrons, why would the spacing to the next atom affect the refractive index? I can see why it may introduce a retardation if there is delay in the process of transmission but I don’t see why it affects the electron “shaking”. (2) Is it possible for material to be linear in polarization and not magnetization or the other way around, and if so, is it still considered linear?

A) The density tells us how many dipoles per unit volume are oscillating at the same time. The more dipoles you have, the more radiation you’re going to get; it is as simple as that. There should be a difference in the strength of radiation coming out of a material with, say, one dipole per cubic nanometer and 100 dipoles per cubic nanometer. That’s why the refractive index (actually, the susceptibility) is proportional to both the strength of each dipole and the number of dipoles per unit volume. You may want to watch my next lecture (the one I gave on Friday 11/8 and is now available on the web; the link and password are given in the attached file) to see how the susceptibility formula is derived for the Lorentz oscillator model.
        With regard to your second question, the answer is that both P and M must be linearly related to the exciting electric and magnetic fields in order for the material to be considered linear. If either P or M fails to follow the exciting field (E or H) linearly, then the material will, in its interactions with the electromagnetic field, behave nonlinearly. 

Saturday, November 2, 2013

Q) In this week’s lecture, we again discussed how to force an integral to zero by multiplying it by an exponential with an imaginary component that is small. This forces the integral to zero at negative infinity even though it goes to infinity at positive infinity. My question is: does the converging direction matter? Would the integral be affected if we forced it to zero at positive infinity instead and to infinity at minus infinity?

1. 
   A) Very good question. It turns out that the integral converges both ways. You can force the integrand to go to zero either as t goes to minus infinity or as t goes to plus infinity. The physical interpretation is easier (i.e., more meaningful) when you force it to zero at  t = – infinity. However, mathematically speaking, both techniques work and are equivalent.
   Wednesday, October 23, 2013
   
   Q1) In Question 21(a), it is asked to find “d” such that the electromagnetic field remains confined between two plates. Does it mean that the Poynting vector has to be zero outside of plates? Or both electric field and magnetic field separately has to be zero?

      In part (b) Why the E-field has to be zero inside both plates? And if E is zero, how can it provide force to keep oscillating? How when E field is zero, therefore, oscillations sustain themselves?

A1) When the gap between the two plates is properly adjusted, both the E field and the H field outside the cavity go to zero. This means that the radiation from one plate is exactly out of phase with respect to the radiation from the other plate. Therefore, in the regions outside the cavity, the total E-field and the total H-field everywhere will be zero.

       The E-field inside “perfect” conductors must be zero. This is because perfect conductors, by definition, have no resistance and, therefore, do not need any E-field to sustain the oscillations of their conduction electrons. (If the metallic resistivity were finite but small, the net field inside the conductor would not be zero, but it would be small.) Of course there is always “radiation resistance,” which is the E-field produced by the “radiation” of each plate on itself (the so-called self-field). Since we have two plates, each plate drives the oscillation of the other plate by canceling the corresponding self-field. By driving here I mean: countering the effect of the self-field (i.e., radiation resistance) of each plate. In other words, one plate creates an E-field on itself that opposes the oscillation of the electrons, while the E-field produced by the other plate cancels that self-field.

Q2) Can you please attempt to shed some light onto a part of the 3D Fourier Transform that I cannot understand? For a problem with a random function f(x):

                 Fourier [f(x)] = F(k) = (Triple Integral) from -∞ to + ∞ ||| f(x) exp(-ik*r) dr,

where does the additional 2πr^2 sinθ exp(cosθ) originate from? An example of this from the problems is Chapter 3, Problem 4(a), between the first and second step in the solution. It goes from this triple integral of f(x) exp(-ik*r) dr  to the double integral | r=0 to ∞ | θ = 0 to π f(x) exp(-ikrcosθ) 2πr^2 sinθ dr dθ.

A2) The factor 2πr^2 sinθ dr dθ is the volume of the ring. The ring has radius r*sinθ; therefore its perimeter is 2πr*sinθ. In the theta direction, the width of the ring is r*dθ, while in the r direction its thickness is dr. When you multiply these three factors, you’ll get the volume of the ring as 2πr^2 sinθ dr dθ.

      The exponential factor is the same as exp(i*k . r). Since the angle between the vector k and the vector r is theta, the dot-product k . r in the exponent becomes k*r*cosθ.

Q3) I have a question from problem 2 of the 2012 midterm concerning the scalar potential and throwing out the infinite term. I see in part B you explain that the term can be ignored since it does not effect the E-field, but I don’t understand why it is ok to throw out the term earlier in part A when defining the scalar potential.

A3) The infinite term is essentially constant, because its gradient is negligible. One can always add or subtract constant terms to potentials without affecting their physical behavior, because their base values do not matter as far as the physical content of the potential is concerned. What is important about a potential is usually its gradient or curl, which does not depend on such (constant) additive terms. Any added constant (even when it is infinitely large) can be dropped from a potential function without affecting the things that matter, namely, the E and B fields.

     By the way, we have not yet covered the formulas for the potentials that are used in this 2012 exam; the subject will be covered in Chapter 5 next week.

Q4) 1) In the potentials for the Lorenz gauge, the denominator has k^2-(w/c)^2 factor. I guess I was under the impression that k = w/c, so wouldn’t the denominator just cancel out?

2) This may have been answered in class already (I don’t recall and I didn’t have it in my notes), but in what situations is it better to use the Coulomb gauge versus the Lorenz gauge?

3) When one is given the charges/currents and asked to find the fields, one has to take the Fourier transform of the charges/currents, find the potentials, relate the potentials to the field, then inverse transform to get the fields in space-time. I know that the FT of the charges/currents gives the amplitude of their corresponding planes, but what was the reason that the potentials couldn’t be related to the charges/currents in space-time directly?

A4) 1) In general, when sources are present, “k” is NOT equal to “w/c”; only in the absence of sources (i.e., in “empty” free space) is k equal to w/c. For example, when the sources are time-independent (i.e., static) w is equal to zero, but the various plane-waves (associated with charge, current, E and H fields, etc.) have non-zero k values.

2) In this course, for the most part we use the Lorentz gauge. People working in quantum optics use other gauges (usually the Coulomb gauge). It yields the same E and B (and H and D) fields, but in applications that involve the Schrodinger equation, for example, the use of the Coulomb gauge makes the calculations somewhat easier.

3) Potentials can be related directly to the charges and currents in space-time, as you suggest. This is what we will do next week when we discuss Chapter 5. In general, it is possible to solve electrodynamics problems both in the Fourier domain and in the space-time domain. We have employed the Fourier technique in Chapter 4, and will use the space-time method in Chapter 5. Generally speaking, it is good to be familiar with both methods. The Fourier method is much easier to grasp conceptually. Besides, the proof of the relations in the space-time domain is easiest when it’s done by inverse-transformation from the Fourier domain.

Q5) In problem 18, chapter 4, while we represent the magnetisation of a point dipole why do we not consider the distance “d” and we just simply represent as del(x)del(y)del(z)?

A5) The distance “d” between the poles of the dipole is already “encoded” in the magnitude m_0 of the dipole. As I said in the class (with regard to an electric dipole, p_0), the dipole moment is the product of the charge q and the distance d. If you let the distance d go to zero while the charge q goes to infinity, the product p_0 = q*d remains constant, while the dipole approaches a point-dipole. The same logic applies in the case of a magnetic dipole m_0, where the charge “q” now represents a magnetic monopole.

Q6) While solving the homework problem #13 and looking at some of the past exams, there are instances where we only consider functions of space and not of time. For example, M(r) = m*delta(r)*z^. What is the physical meaning (if any) of something that is not a function of time? Isn’t everything truly a function of time? If it is not changing with time, then it is a static function, but is still a function of time, just not dependent on it.

     Trying to solve these problems then only requires using the spatial Fourier transform and not both spatial and temporal frequencies. Looking through the book, we take the transform of a time-independent function and result in a delta function in temporal frequency space. For example, F.T.{M(r,t) = m*delta(r)*z^} is m*delta(omega)*z^. Why do we omit this temporal frequency from some problems? It makes it easier, but if we are looking to solve the full equation, then shouldn’t we take both spatial and temporal frequencies into consideration?

A6) When we say something is not a function of time, we mean that it “does not change with time.” In other words, it’s “static,” or “time-independent.” Different words, same meaning.

     When something is independent of time, its temporal Fourier transform has only one frequency (w=0). Rather than doing a Fourier transform and then an inverse Fourier transform, we just keep in mind that w is equal to zero for this system. We don’t need to go through the process of writing delta(w), then inverting it back to the time domain. The only thing we need for calculating the F.T. of the potential functions is this single value of w (i.e., w=0), which we use in the calculation process. We just don’t bother writing delta(w) then inverting it.

     The same goes for single-frequency time-dependent functions such as cos(w_0*t). We usually express cos(w_0*t) as the sum of two complex exponentials. Each one of these has a single frequency: one has the frequency w_0, the other has the (negative) frequency –w_0. As long as we keep track of each frequency and use the correct frequency in the denominator [i.e., where k^2 – (w/c)^2 appears], there is no need to Fourier transform cos(w_0*t) back and forth.

     This is just a way of reducing the amount of algebra that we need to do. If you are not comfortable with it, you may continue to do the full four-dimensional Fourier transforms.

Q7) I have a question about how you described light reflecting off a mirror or going through a concrete wall. As I understand that train of thought, the incident electric field passes through a material, and the material’s response to that electric field sends out an electric field in both directions. For the aluminum reflector, the electric field on the rear side is perfectly out of phase and thus the two cancel each other out leaving only the electric field that is observed as reflecting from the surface.

      So is this a mathematical construct in Maxwell’s equations to describe this behavior of light reflecting or transmitting through material, or are you saying this is the physical mechanism that takes place in reality? Does this idea extend to light described as a photon instead of a wave / e&m field?

A7) Maxwell’s equations are the best description we have of the nature of electromagnetism. Our physical understanding is based on these equations. So, if the equations say that is what’s happening, then that would be the way we believe EM waves behave in nature.

Tuesday, October 22, 2013

Q1) I think I’m a little bit stuck on problems 13, and 17. Getting the magnetic point dipole into Fourier space is straight forward, and results in m*z-hat. The bound current is equal to mu0 times the curl of the magnetism, so working in Fourier space, I believe it’s just a simple crossing into ik, which gives a k X z term. Then to get the potential and B field, it’s the curl of the potential, which crosses k into the k X z term from before, giving k X (k X z). Using the identity, I get the field as being: m z-hat – m(k-hat dot z-hat) k-hat, where I’ve cancelled the |k| with the k^2 in the denominator from the potential.
The H-field is the B field less the magnetization (time mu0^-1), which conveniently cancels out m z-hat from before, and leaves the second term. So it seems that the H-field in spatial space would be the inverse fourier transform of -m(k-hat dot z-hat) k-hat.

     When I change the integral coordinates using symmetry arguments, I replace k-hat with r-hat * cos(theta). Since there are two k-hats in that term, I end up with a cos^2(theta).
       -m cos^2(theta) (r-hat dot z-hat) r-hat 
Along with the sin(theta) from the coordinate change, the integral becomes quite tricky. 
         For problem 13, I’ve gotten to a point where I have three terms in the integral:

                       Integral[ksin(kr)/r – 2cos(kr)/r^2 – 2sin(kr)/kr^3] (k, 0 to inf) 

Am I on the right track, or have I done something wrong? It seems problem 17 is very similar.

A1) You may be better off trying to find the inverse Fourier transform of the vector potential A(k,w) at first, then using the curl operator in the space-time domain to determine the B and H fields from Curl x A(r,t). The H-field of the magnetic point-dipole should be analogous to the E-field of the electric point-dipole given as an example in Chapter 4. The difference is that one calculates the scalar potential in one case and the vector potential in the other, but the fields (E in one case, H in the other) should have the same mathematical form.

Q2) I am a little confused about the boundary condition in different situations. Could you please tell me the physical meaning of it? And in the midterm #2, 2008, Problem1(d), the answer gives a note that ψ is related to ρ. I don’t understand how it comes.

 

A2) The boundary conditions were described in Chapter 2. Please read Section 17 of Chapter 2 again.
In 2008 we covered different parts of the course in a different order. You will see the relation between rho and psi in Chapter 5.

 

Q3) I am going through the 2007 midterm 2 and I have a question about problem 2b, to find the surface current on the mirrors. My problem is that I don’t see what happened to the time-dependent sine terms on the H-field, as the answer just has the coefficients of the H-field as the value for J.

 

A3) The solution gives only the “magnitude” of the surface current. In other words, it’s only the “amplitude” of the oscillating current. The sinusoidal factor should be included in the expression (as you’ve suggested) if one is interested in the instantaneous values of the surface current.

Monday, October 21, 2013

Q1) I didn’t clearly understand how we represent (how cosine and sine terms gets into representation) of magnetization of a spinning dipole around z axis, using delta functions in chapter 4 problem 18. Can you please help me with this.

A1) The magnetic dipole moment is a vector in the xy-plane having an angle theta with the x-axis. Therefore, its projection on the x-axis is proportional to cos(theta), while its projection on the y-axis is proportional to sin(theta). If the dipole begins to rotate around the z-axis, the angle theta becomes a function of time. In this case theta = w0*t.

Q2) After reviewing my solution to homework problem 4.3 (assuming it is correct), I noticed that if I return the scalar potential to space-time, it involves a Laplacian operator, operating on the scalar potential (using the Coulomb gauge). This has not been covered explicitly in class, so I was wondering what the exact physical significance of the Laplacian operator is with respect to the scalar and vector potentials and if it can actually be extracted from other parts of the scalar and vector potential equations? Also, does this only appear in the Coulomb gauge?

A2) You are correct about the Laplacian, but this problem does not ask you to work in the space-time domain; the solution is wanted in the Fourier domain only. Next week we’ll start Chapter 5, where we will discuss the solution of Maxwell’s equations in the space-time domain. There you will see how the Laplacian operator comes about in a general setting, both for scalar and vector potentials.

Q3) I was listening to the lecture about finding the current density of spherical shell in the k space. When you are going to project r along k, you said it becomes k cos(theta). But as I understand it should be I r I* cos(theta)*k, and because I r I = 1, so we can write cos(theta) k. Am I correct?

    Also, why do you draw a ring around the k-vector? What is the purpose of that? We are going to find current density in the k space and we have to integrate in the whole of the space which can be in the spherical coordinate or cylindrical coordinate and you automatically reached that point ( the element of spherical coordinate) but you used a method that I can’t understand the purpose of that (using ring).

A3) I don’t recall what I said in the class; the same problem, however, is written in the book. Please check the book and see what I’ve said there. If you project only the magnitude of r on k^, then you should be getting r*cos(theta). However, if you are integrating the unit-vector r^ over the ring, because of the particular symmetry of this problem, you will end up with a vector along k^, in which case the contribution of each volume element to the integral will be cos(theta)*k^.

     With regard to your second question, the ring is necessary because the functions involved are symmetric with respect to the k-vector (not with respect to the z-axis). We have the function exp(ik.r), which may be written as exp[i*k*r*cos(theta)]. This function has a “constant value” over the ring, where k, r and theta are the same for all the points residing within the ring. We then integrate over theta (from zero to pi) and over r (from zero to infinity) to cover the entire 3D space. Because the axis of symmetry is now k (rather than z), the method of integration that I’ve used (i.e., the ring centered on the k-vector) is the correct method. If you draw the ring around the z-axis, the function exp(ik.r) will NOT be a constant within that ring, and this is enough to prevent you from carrying out the integration any further.
Thursday, October 17, 2013

Q1) In class you mentioned that in the spatial domain, according to Huygens’ principle, we find the value of a field at another position by adding up spherical waves. It hadn’t occurred to me that this is analogous to adding up plane waves in Fourier space. Are there other spaces with different shaped waves? If so, are they ever useful?

A1) Yes, if you have a radiator with cylindrical symmetry (such as a straight wire carrying a sinusoidal current), there will be cylindrical waves that can be superposed. In cylindrical coordinates, waves can be written as superposition of Bessel functions also. In general, there are many systems of orthogonal functions that can be used to decompose any function into a superposition of those orthogonal functions. Mathematicians over the centuries have created many sets of orthogonal functions for different applications in different geometries. The Fourier series and Fourier integral are just the tip of the iceberg.

Q2) I was reviewing example 4.7 in the text (the magnetostatics problem involving the hollow cylinder) and I noticed that your final solution involving the piecewise function does not explicitly state the condition for rho=R. You do state the magnetic-field discontinuity situation immediately following, but I was wondering why you chose not to include it in the resulting function? Also, according to G&R 6.512-3, for the condition of rho=R, the equation would be 1/(2b) with ‘b’ being equal to R. So wouldn’t that reduce the magnitude of the surface current density by 1/2 at rho=R while still maintaining the necessary boundary condition? I am unclear as to where the ‘1/2’ has gone?

A2) The value of a function at a “single point” does not have any physical significance. If you recall, when I first introduced the step function, I mentioned that some people like to define the value of the function at the point of discontinuity as ½, but the value at a single point does not have any physical significance. The same holds here. We have the H-field outside the cylinder and inside the cylinder. At the cylindrical surface itself (thickness = 0) the value of the H-field is 1/(2R), which is half-way between the fields immediately inside and immediately outside the cylinder. This value, however, does not enter the calculation of the discontinuity of the H-field at the cylinder surface.

     Imagine giving the wall a small thickness, then connecting the H-field outside the wall to that inside the wall with a steep, continuous line (similar to what I did with the step-function when we wanted to find its derivative). The value of the H-field in the middle of the wall-thickness will now be one half of the discontinuity (i.e., 1/2R). But this value does not determine the actual discontinuity of the H-field, which is the difference between the field immediately outside the wall and that immediately inside. That difference is still equal to 1/R (not 1/2R).

     In Example (4.7), I only wrote the solution for the H-field in the two continuous regions separated by the cylindrical surface. If you want, you can add to this the value of the H-field at the surface itself (G&R 6.512-3 gives the correct mathematical answer). However, physically speaking, the value of a function at a point of discontinuity has no significance. Any measuring device will have a finite physical width, and the value that it measures will be the average value at the point of discontinuity (i.e., half-way between the inside and outside values of the H-field).
Thursday, October 10, 2013

 

1. Q) What is the difference between group and phase velocity. What is the physical meaning of both and why exactly should we have to define two velocities?

 

6. A) We will discuss the group velocity in Chapter 6. For now you should only use the concept of phase velocity for a plane-wave, as was described in Chapter 1. In vacuum there is no difference between phase velocity and group velocity; they are exactly the same. When the refractive index of a material medium (e.g., glass) becomes a function of the frequency of the light, a pulse of light propagates in the medium with a velocity that differs from the phase velocity. That velocity is called the group velocity, because a pulse consists of a group of plane-waves. You must wait until we get to Chapter 6, where we will discuss the refractive index and its frequency dependence.

 Monday, October 7, 2013

 

1. Q) Does the sin() function need to have an argument with pi in it? I have seen sinc() defined as both sin(x)/x and sin(pi*x)/pi*x, so I am not clear as to when you would want to modify the argument and denominator to include pi and when you would not?

1. A) It is a matter of definition. The sinc function has been defined both ways. In my class, I’d like to stick with one notation, and that’s going to be sin(pi*x) / (pi*x). This is also the definition used by R. Bracewell in his book on Fourier Transform Theory.

Sunday, September 29, 2013

Q1) In the make up class on two Thursday ago, a guy asked you why there is no any free charge density in the EM energy? You would said because when E exerts energy on the charge and it doesn’t move there is no any work on it.I believe in this part your are right, but charge gets hot so there is energy and the E get charges hot, in this case no need to move them. If I am wrong please correct me whenever you have time.

P.S. In previous terms you have given some questions like imaging charge or something that you have not covered in this current semester, are we supposed to study these kinds of subject by ourselves?

A1) Charges do not get hot just because the E-field is acting on them. They get hot when they move and collide with the lattice (i.e., ionic cores of the atoms arranged in a lattice). There is no energy expenditure (kinetic, thermal, or otherwise) when there is no motion. Remember that “work” (in its technical sense) is given by F.delta_x. If delta_x is zero, there is no work done and no energy is being spent.
Previous exams are always good to study, even though each year I cover different things; so things that I covered in previous years may not appear in this year’s exam. But the HW problems and previous year’s exams are pretty fundamental. If you have time, it’s always a good idea to think about them and try to solve them with the knowledge that you already have. At the very least, you should look at the solutions and see if there is anything there that requires new knowledge.

Q2) I am a distance 501 student and I have a question about the glass bead example you mentioned in class on 9/26. I am familiar with Snells law, which states that light’s path bends as it changes from one medium to another when they have two different indicies of refraction, and that the resulting velocity changes. What I don’t understand is why the light bends at all, instead of just slowing down. I don’t mean in a crystal, but in an amorphous medium I don’t understand why light bends to a different direction.

A2) The short answer to your question is: Maxwell’s equations. If you take two plane-waves in two adjacent media having different refractive indices, it turns out that the only way to satisfy the boundary conditions (i.e., continuity of the E and H components parallel to the interface) is to have refraction according to Snell’s law at that interface — i.e., bending of the beam toward the surface normal in the medium that has the higher refractive index. We will discuss this in detail in Chapter 7.
      There are other ways to argue that the light must bend. Fermat’s principle, for instance, tells us that a light ray always takes a path that corresponds to the least time needed to go from point A in medium 1 to point B in medium 2. If the ray passes through point C at the interface (in going from A to B), you can calculate the time needed to go from A to C to B as a function of the position of the point C on the interface. Upon minimizing the travel time (by adjusting the position of C), you’ll find that C is always located at a place that makes the two rays AC and CB satisfy Snell’s law.

Saturday, September 28, 2013

Q1) I read a lot of examples on web trying to visualise the difference between energy and momentum, but still I am not able to get a clear picture. When two objects collide whether energy is exchanged or momentum?

A1) Both are exchanged. A particle of mass m moving with velocity V has kinetic energy equal to MV^2/2 (a scalar entity), and momentum equal to MV (a vector entity). These are two very different things.
     Suppose a particle of mass m is standing still, while another (identical) particle of the same mass m is moving with velocity V toward it. The first particle has no kinetic energy and no momentum. However, as soon as the moving particle hits it, the energy and momentum of the two particles will be exchanged completely. The first particle (originally stationary) will move with velocity V in the same direction as the other (moving) particle, whereas the second particle (originally moving) will come to a halt. This is an excellent example of both energy and momentum being fully exchanged in an elastic collision.

    Try solving the two equations (conservation of energy and conservation of momentum) for two particles having different masses (m1 and m2) and different velocities (V1 and V2), and see for yourself how in different situations different amounts of momentum and energy are exchanged between the two particles. 

Q2) To expand on your brief discussion of optical tweezers, as I understand it the linear momentum of the laser beam focused on the micro-bead will cause the bead to be displaced in the opposite direction of propagation, and if the beam is circularly polarized the angular momentum can keep the bead inside the focus to move it in a transverse direction. Is this correct?
      I am a little bit more familiar with optical vorticies using orbital angular momentum where the bead would travel around the donut beam focus like an orbit around the vortex. I know OAM is a product of the wavefront spiral phase instead of polarization, but one thing I have never been clear on is if the wavefront and polarization can be connected. We typically discuss light’s wavefront or polarization, but I cannot think about discussing them together. Are they two separate thoughts to fill in the whole picture, or can they be linked?

A2) 1) The linear momentum of the incident beam does both lifting up and keeping the particle inside trap. No circular polarization (or angular momentum) is involved here. It is all understandable in terms of ray-tracing. Just move the particle slightly away from its position of equilibrium, and you’ll see the momentum of the rays will change in such a way as to exert a force on the particle that wants to bring it back to its equilibrium position.

2) I actually talked about this at some length in Lecture #11. You can combine vorticity and circular polarization to create a circularly polarized vortex beam. A small particle placed in the path of such a beam will spin on its own axis (because of circular polarization) and will also orbit around the axis of the vortex (because of the orbital angular momentum of the vortex). The two things are not very different. Please watch Lecture #11 first, and if your question is not fully answered contact me again.

Q3)  I am going through the chapter 2 problems to help prepare for the midterm and I have a couple questions. I noticed a former test had problem 2.10 (magnetized disk), and some of the parts of the question involve using the circ and rect functions which have not been covered in class. These functions are not overly difficult so I could learn their properties if need-be, but do you think this or any other chapter 2 questions are off-limits from testing?
     Also in problem 2.11 (3 charges on vertices of equilateral triangle) you ask if the position where the fields cancel is a stable equilibrium position for a charged particle. Your solution says that for the position to be stable, div(E) must be positive or negative, but that since there are no particles at this point, the charge density rho at this position is 0, and thus div(E)=0. But if we place a charge there to test the stability of the position, won’t we now have a charge density and thus dev(E)=/=0?

A3) 1) Some students are familiar with the Circ and Rect functions, because they see them in their Fourier Optics class. Others have come to my office hours and asked about them. These functions are not difficult to understand; in fact you can Google them and find their definitions on the internet. I will formally introduce them in Chapter 3. But if you’re not comfortable with this problem, you may skip it for now. In previous years, I covered things in a different order compared to this year, so sometimes you see a problem in previous year exams (or at the end of chapters) that looks unfamiliar. When you encounter such problems, please ignore them for the time being; at the end of the semester, when you prepare for the final exam, most of these problems will begin to make sense.

2) The question of stability has to do with what other charges are going to do to a test charge. If you place a charge at the location where the E-field (produced by the other three particles) is zero, then the total force on the test particle will be zero. However, if you move the test particle away from that location, there is bound to be some position where the field (produced by the other three particles) will try to push the particle further away (rather than bringing it back to the original location). The divergence of E has two parts, one produced by the three particles located at the vertices of the triangle, the other produced by the test particle. However, the test particle does not exert a force on itself, so its own contribution to the divergence is irrelevant. What is important is the E-field produced by the other three particles, and that E-field cannot possibly push the test particle back to its equilibrium position under all circumstances, because the divergence of the E-field produced by the three particles is zero. This means that there are always going to be points (in the immediate vicinity of the point where the E-field vanishes) such that the E-field of the three particles will tend to pull the test particle further away — rather than pushing it back in.

Friday, September 27, 2013

1. Q) First, when I went through the Maxwell’s boundary conditions, it puzzled me that why H||(r1,t)-H||(r2,t) equals to Js(r0,t). I think it is due to that the flux of D-field through the loop vanishes when the area is close to zero, but why can’t we define a local surface-D-field-density Ds like Js? What or which law limits us to do so?

      Second, if we look at an infinitely long, thin, straight wire carries the constant current along one direction, we can directly calculate the H-field by using Maxwell’s second equation. My question is that when we look a certain point in the space, it is possible for us to draw an infinite number of circles all including this point. Is the final H-field in this point the sum of all these H components in all these circles, or is it just meaningless to do such add-up and we just calculate this H-field in a certain circle we want or according to the curl operation?

A) 1) Remember that the surface-current-density J_s has units of ampere/m, whereas the ordinary (volume) current-density has units of ampere/m^2. This means that, in the vicinity of the surface that contains J_s, the volume current-density is infinite. That’s why you can reduce the width of the rectangle to essentially zero and yet some current continues to flow through that rectangle. If there was a case in which the D-field at and near the surface could become infinite, then you would have to take account of D in the boundary conditions as well. However, such a situation never arises in practice. In other words, we have situations where the current is confined to a surface layer (which corresponds to an infinite volume current-density), but we don’t have situations where there is a thin layer containing an infinitely intense D-field at the surface.

2) I don’t see why you want to draw so many different circles. There is only one circle that passes through a given point, is centered on the wire, and has a plane that is perpendicular to the wire. This is the circle that we must choose, because the symmetry of the problem then allows us to apply Ampere’s law (i.e., Maxwell’s second equation) and obtain the H-field. You could choose some other circle (or any arbitrary loop for that matter), but you won’t have the symmetry to apply Ampere’s law in that case. There is nothing in the curl equation (i.e., Ampere’s law) that says that you have to take many circles and then add up their contributions. The curl law applies to a single loop only. You may take any loop you want, but you are not allowed to take many different loops and then add their contributions; this would violate the meaning of the curl operation.

Friday, September 27, 2013

Q) a. For relativization, Would the mass of an object change with different temperature?

1. If the thermal energy can be a part of the mass, how about the potential energy? If a pressed spring’s mass is bigger than it in original length? And a free electron’s mass is bigger than one in atom?

1. About the mass we talked in class today, is it the inertia mass or gravity mass or both?
   
   d. When come to the electric potential energy, would the mass of a electron change after accelerated by a electric field? Or just the electric field’s mass change?
2. 
   Since the total energy is different with different observer, is the total university’s energy change with different observer? If yes, is it possible to find two special observer which can make the total university’s energy min and max respectively?
   
   A) a. Yes, the heat content of an object contributes to its mass.
   
   b. The work done to compress a spring contributes to the mass of the spring. When two hydrogen atoms fuse to form a helium atom (e.g., at the center of the Sun) the difference in the binding energies of the electrons is released as pure energy. This is the internal energy production mechanism for the Sun as well as other stars.
   
   
3. Inertial mass and gravitational mass are always the same.
   
   d. When an electron is accelerated in an external electric field, it begins to radiate electromagnetic energy. The situation is not so simple to analyze in this case. But if a “massive” charged particle is accelerated under an E-field, it would be possible to ignore the radiation effects (because of the large mass, the acceleration will be insignificant). In that case the rest mass of the particle does not increase, but its mass in motion [i.e., m_0/sqrt(1-v^2/c^2)] goes up with increasing velocity. The increase in the energy of the particle then comes at the expense of the energy stored in the total electric field, that is, the E-field of the particle plus the applied external field.
   
   e. I don’t know anything about the total energy of the Universe, and whether it will change depending on the observer’s velocity. What I can tell you is that, for a closed system, the total energy and the total momentum (i.e., those of fields plus matter) form a 4-vector. The 4-vector transforms between different inertial frames (i.e., for different observers moving relative to each other at constant velocity) according to the Lorentz transformation rules. Thus the energy of the system seen by one observer could be different from the energy of the system seen by another observer.
   
   Wednesday, September 25, 2013
4. 
   Q) I’ve been having a few issues with the last few problems this week… Firstly I think there is a problem either with the way I am setting up my original E and H fields, or with the method I am using to get the subsequent terms, so that the corrected E and H fields are not in the format I expect them to be (and do not seem to satisfy all of Maxwell’s equations in 32). This is probably something I will need to look at the solutions for once they are released. But my main question is about the boundary conditions.

     In 31 e) the question seems to imply that I should be using the first and second approximations for Ez and H_phi to work out the surface charge and surface current density. Does this mean that the E or H field above the surface charge on the top plate, i.e. no longer in between the plates can be thought of as the 1st approximation and the field in between the plates as the 2nd approximation, where the interaction between the fields is strongest? I thought both estimations were for the form of the fields in between the plates, but surely if I am looking at the boundary condition across the charged surface I need to take one field from in between the plates and the other from the space outside the capacitor? Hope it is clear what I mean.

1. 
   A) When the solutions are posted tomorrow, please take a look and make sure you follow all the steps. It’s important to see how the various Maxwell equations interact to produce the correction terms to the static solutions.

     As for Problem 31(e), the E and H fields “inside” the metallic plates are zero at all times. What you find in parts (a)-(c) of this problem is the E and H fields “in between” the two plates. The boundary conditions then allow you to calculate the surface charge density and the surface current density on the lower facet of the top plate and also on the upper facet of the bottom plate. The perpendicular component of the D-field (including all the corrective terms) gives you the surface charge density, and the parallel component of the H-field (again including all the corrective terms) gives you the surface current density.

Tuesday, September 24, 2013

Q1) I’m looking through the solution to problem 1 of chapter 2, and it looks like I was almost there. On page 2 of the solution, first line, it’s noted that the free current density goes to zero at x goes to the infinities. I’m not sure I 100% understand that. Is it just that with a surface at infinity all charge will be enclosed, so none will be passing through the boundary?

A1) When we use the Fourier method of solving Maxwell’s equations, we must assume the Fourier transforms of the various functions exist. One such requirement is for the functions to go to zero as x, y, and z go to infinity. Essentially, we are assuming that charge, current, P and M have a finite extent in the xyz space. In reality, of course, nothing extends to infinity; everything is located within some finite region of space. Therefore, our solutions generally apply to physically realistic situations, not to hypothetical cases where sources and fields extend all the way to infinity without bound.

Q2) I’ve been banging my head against the wall trying to find the E-field in problem 32a, but to no avail. At first, I simply tried taking the gradient of the given voltage, but that would give me an E-field in the z direction when I clearly need an E-field in the rho direction. I then tried using Gauss’s law, but since the charge density is zero, I simply have the divergence of my E-field is zero, which I don’t believe helps. Is there something I’m missing or are my assumptions wrong in some way?

A2) Charge density on the cylinder surfaces is NOT zero. In fact, in part (c) of Problem 32, you’re asked to calculate the surface charge-density on both cylinders. What you need to do here is assume that the E-field is directed along rho, with its magnitude dropping as 1/rho — as expected from a static situation with uniform charge distribution on the outer surface of the inner cylinder. The integral of this E-field (along rho) should give you the voltage difference between the two cylinders. The unknown parameters are then obtained by ensuring that Maxwell’s equations are satisfied.

Sunday, September 22, 2013

Q) I am having some trouble with part B of problem 2.32 (the coaxial cable with varying current and voltage). The issue is proving the curl of H is equivalent to epsilon_0*dE/dt. My results would match if I could equate the coefficients of each side of Maxwell’s equation II.
   In other words, my curl of H term has I_0, and my epsilon_0*dE/dt term has epsilon_0*V_0*c/ln(b/a). Aside from confirming these 2 coefficients have the same units, I do not see a way to prove their are equal. But I am fairly certain I did the E and H fields correctly in part A; I used Gauss’ law for an assumed charge density to find E as a function of Q, then integrated to get V as a function of Q, and thus find E as a function of V to plug in the given V(t) in the problem. And the H field was a simple ampere’s law problem.
    Am I going about solving the fields incorrectly? Or is there some way to prove the equivalence of those coefficients?

A) You must set the coefficients on the two sides of the equation equal to each other in order to force the Maxwell equation to be satisfied. The same thing must be done with the other curl equation (curl x E = -dB/dt). This is how you’ll find the propagation velocity and the relation between the current and voltage in the coaxial cable.

Saturday, September 21, 2013

Q1) Did you derive this new (magnetic monopole) form of Gauss’ law for B-fields and the Maxwell-Faraday eqn, or is intuition from electric charges used to fill in the new terms? I can mostly see where the two new terms come from when I look at Eqns 1 and 2, but without some underlying law such as Coulombs law to derive Eqn1 I don’t see how the new forms of Eqns 3 and 4 are derived.

A1) There is no underlying law, because we don’t have experimental data to decide what the actual monopoles do. In the case of the Coulomb law, experimental data guided Coulomb in enunciating his law, but I wrote the laws of magnetic monopoles based purely on symmetry arguments and the knowledge of magnetic dipoles. Unless magnetic monopoles are discovered one day and these “new laws” are put to the test, we will never know if they are correct.

Q2) I would like to clarify my thoughts on how you related the emf, voltage difference V_a – V_b, and the Maxwell – Faraday Eqn. When you say that Integrate [E,{dL,a,b}] = V_a – V_b, shouldn’t this only be true for the static case when you said the Integral of E from A to B is equal to the Integral of E from B to A? In a purely mathematical term, Integrate [E,{dL,a,b}] = V_b – V_a = -(V_a – V_b). Over a closed loop as in the Maxwell – Faraday law this does not become an issue, but for an open loop I was thinking it would be an issue unless you strictly make the static assumption given above. Please clarify.

A2) I did “not” say the integral from A to B is equal to that from B to A. I said the integral from A to B is path independent. Indeed, if you follow my argument (in the absence of contributions from dB/dt), what I said is based on the fact that the integral from A to B is MINUS the integral from B to A. The integral from B to A can be taken on any path; in particular, it can be taken in the reverse of the same path from A to B. When you take these integrals, the direction of travel (i.e., how you get from one point to another) is very important and should not be ignored. When you reverse the path from A to B, the sign of the integral changes because every delta_L changes its sign.

Wednesday, September 18

1. 
   Q) I have been thinking a bit lately about where current in a thick wire would be located. Since there is no net charge, I don’t see any reason it would be on the outer surface of the wire. But now I’m considering the wire as a collection of very thin wires, which I believe would be attracted to one another through the Lorentz force law. Does this mean that the current travels more toward the center of the wire? This would seem to imply that there would be some sort of surface charge as the mobile charges are pulled inward, I guess until that force is balanced by the force from the surface charge on the outside, and the repellent force from the other mobile charges. Is this what’s going on? I feel like I’m missing something.
   
   A) Ohm’s law relates the current-density J to the local E-field, namely, J = sigma*E, where sigma is the electrical conductivity of the wire. There must be a small E-field along the direction of the current inside the wire. The E-field is essentially uniform throughout the wire, and so the current is uniformly distributed through the thickness of the wire. To produce this E-field, electrical charges must be distributed over the surface of the wire. The charge density (on the surface) must decrease slowly along the length of the wire; this is how the E-field is produced inside.
   The magnetic field that goes around the wire (both inside and outside) is in the azimuthal direction (phi). But the magnetic field cannot push the current too much toward the center (in fact not much at all), because the conduction electrons are like an incompressible fluid. Besides, the positive ions of the lattice would not allow the electrons to wander too far from them. 
   The surface charges produce a radial E-field outside the wire, and, together with the azimuthal magnetic field, they give rise to the Poynting vector outside the wire. The electromagnetic energy is in fact carried entirely outside the wire — not inside — via this Poynting vector. 
       Also, remember that this is an electrostatic and magnetostatic problem. Therefore, curl x E = 0 everywhere, and Div.E = 0 everywhere except on the surface of the wire, where the charge-density is nonzero. These requirements put severe limitations on how the E-field (and, consequently, the current-density J) must be distributed both inside and outside the wire. 
      The situation described above applies to DC and low-frequency currents only. At high frequencies, the current is confined to the skin-depth of the wire, which is a thin layer at the surface. Most of the interior of the wire does not carry any current at all.

Tuesday, September 17, 2013

 

Q1) In the ‘Field, Force, Energy and Momentum in Classical Electrodynamics’ book chapter 1, you mention the example of a Moibus strip, I actually followed your instructions with twisting a piece of paper with the surface normals defined, that helped understand the surface normal direction discontinuity. You then continue with this example by bringing the narrow ends close together but not touching, and raise the question of how does this modification affect Stokes’ theorem. I am not sure I understand what you mean by bringing the narrow ends together, could you please clarify that and relate it to the affect of Stokes’ theorem?
 
A1) When you make the Mobius strip, if you glue the narrow ends of the strip together, you will have a closed loop. However, the place where the edges are glued together, the two line integrals do not cancel each other, because they are in the same direction.
     If you do not glue the edges together, you will keep both edges alive, and they both contribute to the integral, as they should. That way Stokes’ theorem works as usual.
    So, the trick is to “not” eliminate those two edges, i.e., do “not” glue them together. Keep the loop open, and Stokes’ theorem will work without any problems. Glue the edges together (i.e., remove those edges from participating in the boundary integral) and the Stokes theorem will no longer apply.

Q2) Problem 2.4 in the last line is written I = j(x) dy = j(y) dx. But as far as I know and I understand it must be I = j(x) dy + j(y) dx otherwise if you consider a segment perpendicular to x axis so it will be zero.

A2) The problem is correct as stated. You may take a cross-section parallel to the x-axis, OR a cross-section parallel to the y-axis. The current going through both cross-sections must be the same. Remember, you can choose the cross-section in any direction that you want, and the current going through cannot depend on how you choose the cross-section.

Monday, September 16, 2013

Q) I have a question about obtaining the integral form of Maxwell’s 3rd equation (M3). I understand that one must first integrate both sides of M3 about the surface in question and apply Stoke’s Theorem to the Dell x E side of the equation, but I do not understand why the d/dt can be pulled off of the B-field and put out of the integral on the right side of the equation (pg 32, going from equation 10a to 10b). Could you please explain to me why this can be done?

A) The integral is over a surface, so the order of differentiation with respect to time doesn’t matter. You can find the contribution of dB/dt (partial d/dt here) to all the surface elements, then add up the results over those surface elements. Alternatively, you may first add up the dot-products of B into every surface element, then carry out the time derivative. The integration is with respect to spatial coordinates, while the differentiation is with respect to the time coordinate. Therefore, the order of the sequence is immaterial.

Monday, September 16, 2013

Q) Whilst doing the problem set this week, I am struggling with understanding the solution you provided to Chapt. 2, Problem 10. Firstly I’m not sure I fully understand what is meant by the notation Circ(r) and Rect(z). Then I do not see why you can reduce the curl of M to ∂Mr/∂z phi_hat or similarly the divergence of M to 1/r.∂(rMr)/∂r. I assume it is some sort of symmetry based argument, but I can’t quite grasp it. Also, in part (d), I can follow the argument as far as the expression for J(m)bound, but not how it relates to the curl of P. Essentially I would really like to see a worked solution with more details of all the steps involved, if that is possible.

     My background is actually in Chemistry, and although I’ve done a lot of Physical Chemistry, including Maxwell’s equations before, it has always been from a slightly different viewpoint to what we have been doing in this course. As a result I am a bit unfamiliar with some of the more Physics based concepts such as electrodynamics or magnetism. I was wondering whether there is a text you can recommend for someone in my position that is trying to bridge that gap.

1. 
   A) 1) The function Rect(z) is equal to 1 when |z| < ½, and it’s equal to zero otherwise. I did a Google search under Rect(x) and found it right away in Wikipedia. (The values of the function at z = ½ and z = -½ are also defined as ½, but these isolated points usually don’t make any contribution in a physics – as opposed to mathematics — problem.)
   
   2) The function Circ(r) is defined in cylindrical coordinates; it is equal to 1 when r < 1, and is zero otherwise. I did a Google search under “Circ function” and found it in a Wikipedia page on Fourier transforms; the Circ(.) function was defined in a footnote.
   
   3) If you look at Appendix B of the textbook, under cylindrical coordinates, you’ll see the expression for the curl operator. Since M(r,t) only has a component along r^, and since that component is a function of z (and not a function of phi), only one term in the curl expression is non-zero, and that term is (dM_r/dz) phi^. A similar argument applies to the divergence of M(r,t).
   
   4) Part (d) of the problem is for advanced students who are familiar with the theory of special relativity. In special relativity, a moving magnetic dipole produces a co-moving electric dipole, and vice-versa. The polarization P(r,t) depicted in Fig.(b) is, in fact, the relativistic companion of the M(r,t) shown in Fig.(a). Problem 10(d) shows how to compute the magnetic monopole current associated with P(r,t), and how to determine the strength P_o of the relativistically-induced polarization by setting the corresponding magnetic monopole current equal to that obtained directly in Part (c) of the problem.
   
   5) I have listed several books on the course website as supplementary reading material. You may want to take a look at Feynman’s lectures on Physics, Vol. II. Feynman describes electromagnetic theory in simple terms, using as little mathematics as possible.

Wednesday, September 11, 2013

1. Q) You’ve had a discussion in your lecture and in your book about calculating n x C” with n being perpendicular to the plane that contains C’ and C”. The plane that contains C’ and C” can have C’ and C” orthogonal to each other, but this is not necessary. I took their orthogonality to be the most simplistic configuration and made a right handed coordinate system, however after attempting to solve n x C”, the projected vector was along the orthogonal C’ direction, and therefore the length is C’. This did not lead to the correct solution of C”sin(wt). I was able to prove this solution after taking C’ and C” as non-orthogonal and n x C” to be in the same plane as C’ and C”, but not along the C’ axis. Can you elaborate on how to solve this by taking C’ orthogonal to C”?  Perhaps I do not have a clear understanding of the normal vector n and the cross product of n x C”.
   
   A) Look at the problem geometrically, not algebraically. C’ and C” form a plane. The unit vector n is perpendicular to this plane. When you cross n into C” you obtain a vector whose length is C” (because the length of n is equal to 1 and also because n and C” are perpendicular to each other); the cross-product n x C” lies in the plane defined by C’ and C” — and is perpendicular to C”. All these properties are direct consequences of the way the cross-product of two vectors is defined.
      In your special case, where C’ and C” are orthogonal to each other, your vector n x C” lies in the C’C” plane along the length of C’, but in the direction opposite to C’. Everything I said in the solution to this problem works in this special case just as it would if C’ and C” were not orthogonal to each other.

Friday, September 6, 2013

Q1) I am wondering if you can answer a few of my questions on some remarks you made in lecture for 501. My first question is in regards to today (Thursday, September 5th) lecture. When we were using Gauss’s Theorem to obtain the integral form of Maxwell’s first equation, we noted that Q was a function of t. I am wondering if we can exploit the charge conservation equation, and using Gauss’s theorem to obtain a relationship between D(r,t) and J_free. I tried doing this, and I got: d/dt (D(r,t))=J_free, but I’m not entirely convinced my math was correct.

A1) The relation you derived between D and J_free is not correct. The only relation one can obtain from a combination of Maxwell’s first and second equations is the charge-current continuity equation, Div.J_free + d(rho_free)/dt = 0. I’ll derive this in the class next week. 

Q2) My other question is in regards to the Fourier transform of an arbitrary wave. We mentioned in class that anytime we have a wave, we can decompose it into a summation of plane waves. Because plane waves have the nice property (are they considered eigenfunctions of the del operator?) where the divergence or time derivative is another plane wave, this becomes a useful procedure to solve difficult wave interactions and phenomenon. However, I am wondering (in 512, we discussed conformal maps) if instead of using a Fourier decomposition, we instead change the space in which we are operating, such that the original form of the wave now takes on one that is a plane wave in this space. Thus, rather than having a series of plane waves, we have only one. If this is a valid solution, does this simplify solving problems in the presence of strong gravitational fields, where the metric of space-time is given?

A2) We only discuss the Fourier transformation of the fields in Opti 501. We shall leave the space-time flat (i.e., we’ll work in the Minkowski space-time) and only Fourier transform the fields. What you are talking about sounds like “transformation optics,” which is a popular topic of research these days. There is a recent book by Ulf Leonhardt that describes the basic principles of transformation optics. The subject draws its inspiration from General Relativity and the distortions produced in the fabric of space-time by gravitational fields.

Thursday, September 5, 2013

1. 
   Q) On Tuesday you mentioned that we can describe an ellipse given any two vectors (such as A’ and A”), and showed that we can redefine the axes so that A’ and A” are orthogonal. I can see how the ellipse would be equivalent, but isn’t the phase different? It almost seems as if the speed is changing, since the angle from A’ to A” is not necessarily the same as from A”, say, to -A’, even though the phase difference between the two is equal. Is the angular speed changing? By changing axes to a perpendicular A’ to A”, is it really equivalent beyond the ellipse that it traces out?
   
   A) This is a good question; you’ll get a point for it.

     The two ellipses are equivalent in every respect, phase included. If you follow the solution to HW Problem 30 (Chapter 1), you will see how the major and minor axes of the ellipse relate to the Co’ and Co” vectors (these are the same as A’ and A” mentioned in your question). This HW problem shows that there is a unique way to pick the two orthogonal vectors (i.e., major and minor axes of the ellipse) for each choice of Co’ and Co”. It takes one quarter of one period for the tip of the sum vector (located on the ellipse) to go from Co’ to Co”, and another quarter of a period to go from Co” to –Co’, etc. The same statement applies to the major and minor axes of the ellipse. There is a strong connection between the vectors Co’ and Co” on the one hand, and the major and minor axes of the ellipse on the other. Not every choice for the axes is acceptable.

     Please consider these comments and analyze a few special cases for yourself. If you are still bothered by the phase problem, please let me know, so I can try to give you a more elaborate answer.

Tuesday, September 3, 2013

Q) In attempting the first three problems from the homework, I see that the function being used is f(r) and that it isn’t a function of time. Should I assume that this function is indeed a function of time or just of position? Also is f(r) a vector field or a scalar field? I don’t think it was stated one way or the other, but I could be wrong about that.

A) Divergence, curl, and gradient are operations on spatial coordinates. If the function happens to be time-dependent, we must freeze the time. In this case, the functions are not time-dependent, so there is nothing to worry about. (In other words, all operations will be the same, whether or not the function f is time-dependent). Also, in the case of the gradient operator, the function f is “not” boldface, which means that it is a scalar function of r. In the case of divergence and curl, the function f is written in boldface. This makes it a vector function of r.

 

Monday, December 5, 2012

Q) Does the interface between a material with a positive (n > 1) refractive index and a material with negative (n < 0) still result in a Brewster’s angle? Is there a Brewster’s angle for all linear, homogenous, isotropic interfaces? And, is the Brewster’s angle still frequency dependent? For example, at high frequencies where the difference in refractive index may be more pronounced, can this reflection-null be used for filtering applications, or gratings?

 

1. A) The answer to all your questions about the Brewster angle is contained in the Fresnel reflection coefficients. Whenever either rho_p or rho_s becomes zero, the incidence is at Brewster’s angle. Yes, it can happen for negative-index media; yes, it’s frequency-dependent; yes, it could be the s-light that exhibits the Brewster angle rather than the p-light, etc. Please take a look at Problem 16 (Chapter 7) for a discussion of some of these points.
   The details depend, of course, on the specific values of mu(w) and epsilon(w) for any pair of materials (what I call media “a” and “b” in the lectures). Without knowing the specific numbers for the two media, I cannot answer your questions, but the important point to remember is that, irrespective of their specific numerical values, mu and epsilon must be able to make either rho_p = 0 or rho_s = 0 at some angle of incidence. If this happens then you’ll have a Brewster’s angle; if it doesn’t then you won’t.

 

Thursday, December 1, 2012

Q) Few years ago, when I worked in the semiconductor fab, I saw the oxide on the silicon wafer have different color dependent on the thickness of the oxide. I wonder whether we can explain that using OPTI501 material?

A) Yes, you certainly can. The thickness of the oxide layer affects the Fresnel reflection coefficients; since the thickness (in units of wavelength) is different for different wavelengths of the white light, you get different reflectivities for different colors. (The refractive indices of the oxide and the silicon substrate are also wavelength dependent.) When red light, for example, is reflected more than green and blue, you get red coming back to your eyes — blue and green are absorbed in the underlying silicon substrate more than red. So, yes, the concepts we learned in Chapter 7 can definitely explain the phenomenon that you described.

 

Wednesday, November 30, 2012

Q1) I just watched lecture 28 and had a question on the last plot you draw on the board near the end of the lecture. In the plot illustrating the phase shift versus angle of incidence (beyond the critical angle) for S and P polarized light what is general scale on the y-axis? How much of a phase shift would we expect for visible light assuming material 1 is a common transparent glass type and material 2 is air? Are we talking about a small amount of phase shift or can the phase shift be substantial? In the lecture you mentioned that the beam’s lateral displacement at the interface during total internal reflection due to the Goos-Hanchen effect was very small. Does this imply that the resulting phase shift is very small as well? In the same plot the phase shift appears larger at the critical angle and decreases as the angle of incidence increases beyond the critical angle. Does this imply that the Goos-Hanchen lateral shift magnitude is also at a maximum at the critical angle and decreases beyond the critical angle?

A1) Good question; you get a point for asking it. The range of angles is zero to 180 degrees for both p- and s-light. You can easily obtain numerical values for the phase as a function of angle of incidence using the Fresnel reflection coefficients that I derived in class yesterday. You may also want to check Chapter 27 of my book, “Classical Optics and Its Applications” (2nd edition, Cambridge University Press, 2009) for a detailed discussion of total internal reflection and the Goos-Hanchen shift.

       What matters (as far as the Goos-Hanchen shift is concerned) is not the actual value of the phase but the “slope” of the phase as a function of k_x. The slope is largest immediately above the critical angle, and so is the Goos-Hanchen shift. Recall that plane-waves are infinitely wide and, therefore, their lateral shift is meaningless. To observe the Goos-Hanchen effect, one must use a finite-diameter beam with a small footprint at the glass-air interface. This means that the incident beam is not just one plane-wave but a superposition of many plane-waves, each arriving at a slightly different angle of incidence. The shift theorem of Fourier transform then tells us that a “linear” phase imparted to a function causes a displacement of its Fourier transform. The linear phase in this case is due to the slope of the phase of the Fresnel reflection coefficient as a function of the angle of incidence (assuming the range of incidence angles is sufficiently narrow to allow a linear approximation to the phase function).

Q2) Thank you for the insightful response. I have a general brief follow-up question. Are there existing technologies that utilize the phase shifting aspect of TIR to purposefully modulate phase and/or detect phase modulation? Given the large range of phase shifting potential it seems like it could be very useful in some technology applications.

A2) The phase of the Fresnel coefficient is directly responsible for the existence of modes within slab waveguides. (A similar relation exists for other waveguides, such as optical fibers.) So, the fact that the waveguides can be designed to have desired modes is indirectly related to the phase of the Fresnel reflection coefficients. People have used frustrated total internal reflection to make modulators (by adjusting the gap width or the refractive index of the material in the gap). There is also attenuated total internal reflection, where there is a mirror behind a prism with an adjustable gap in between (some of the light gets absorbed in the mirror, strongly so at the surface plasmon resonance, hence the “attenuated” terminology). There may be other applications of which I am not aware, but the phenomenon is certainly rich and has been known for a long time, so there must exist a lot of applications out there.

 

Q3) Regarding problem 22 on page 185, I’ve checked through Chapter 7 and couldn’t find where r_p and r_s are defined. Are these the reflection coefficients for p-polarized light and s-polarized light respectively, or are these the reflectivity coefficients defined on pg. 173?

A3) Sorry for the confusion. r_p and r_s are the same thing as rho_p and rho_s, respectively.

 

Q4) I’m attaching a picture of a demonstration of total internal reflection in an acrylic rod, from here:
http://en.wikipedia.org/wiki/File:Laser_in_fibre.jpg. Question — obviously we don’t have all knowledge about the experiment, but it appears to show the effect that you had drawn in lecture where the beam is slightly translated along the propagation direction and resulting in a “damped” reflection (as opposed to a crisp, ray traced reflection) — is this what I’m seeing?

 

A4) The beam seems to be coming from a laser pointer, and in any case it doesn’t broaden by diffraction as it propagates down the medium, so I assume the beam diameter is on the order of 1 mm. The “apparent” shift observed here is much greater than a millimeter and, therefore, cannot possibly be a Goos-Hanchen shift. It may be that the beam is following a spiral path within a cylindrical guide, and what you see as a shift is really the result of the projection of the spiral onto a flat plane. That’s my best guess, but I definitely think it is not Goos-Hanchen; the shift is much too large for that.

 

Wednesday, November 23, 2012

Q) I think i might have found an error in the equation on the bottom of page 168. There you write E(r) as E0 exp(i(k*r-wt)). You treat H(r) in a similar way. (Does that mean I found two errors?) Shouldn’t E(r) have no time dependence? To be fair, the very next step sees the time dependence vanish, but it seems a little unfair that it gets to vanish twice, once when taking the integral over one period, and another time when dealing with the conjugate.

A) Strictly speaking, you are correct, and I’ll give you a point for spotting this bit of inconsistency on page 168. The fact is that most books write the time-averaged Poynting vector as I’ve written on the last line of page 168. Because of conjugation, the time factor exp(-iwt) drops out and the final result remains the same, whether or not one includes the time factor in the expression.

Tuesday, November 22, 2012

Q1) In lecture, you found the critical angle of incidence for TIR between two transparent media. Is there an analogous situation for light incident on the boundary between media that are not transparent?

A1) When the media are not transparent there is always some absorption and, therefore, there is no possibility of 100% reflection. (The reflectance could be very high, but it will not be 100%.) So, strictly speaking, total internal reflection only happens at the boundary between two transparent media. You may want to take a look at Chap.7, Problem 9 for a more detailed discussion.

     There is also the possibility of 100% reflection from the surface of a plasma; that is because the refractive index of the plasma is purely imaginary (n = ik). The wave inside the plasma looks very much like an evanescent wave, but the reflectivity at the surface is 100% even at normal incidence from the vacuum! So again this does not count as total internal reflection, as there is no need for a medium of incidence (hence “internal” does not necessarily apply) and also there is no critical angle.

Q2) I think that, with gain- and loss-free media, you can physically recreate an incident wave by measuring the phase, intensity, and angle of the reflected and transmitted waves and reversing their directions of propagation through the media. I don’t think you’d need information about the propagation media.

      My question is: If a lossy medium is involved, can you still physically recreate an incident wave without more information about the medium? What, if any, information about the incident wave would be lost?

 

A2) I believe you are talking about time-reversal symmetry here. Time reversal does NOT imply that you don’t need to know the properties of the media involved; all it says is that, if you time-reverse the reflected and transmitted beams (by reflecting them from a phase-conjugate mirror, for instance), then they’ll propagate backward and will recombine at the original interface to reproduce the incident beam. I have written about this in my other book, “Classical Optics and Its Applications,” Cambridge University Press (2002), Chapter 14.

       Time-reversal symmetry does NOT work when either of the two media involved happens to be absorptive. The general condition for time-reversal symmetry to hold is that the dielectric tensors of the media must be real-valued.

Saturday, November 19, 2012

Q) I found a typo on page 164 in the last line. It reads “c1/c2 as (c1c2*)/(c2c2*),” and should read “c1/c2 as (c1c1*)/(c2c2*).”

A) The book is correct. Both the numerator and denominator must be multiplied by the same thing, which is c2* in this case.

Thursday, November 17, 2012

Q1) With regards to your discussion of group/phase velocity, you spoke also of phase-locked or mode-locked lasers where there is a discrete distribution of spectral lines, and periodicity in the space-time domain pulses — would the Discrete Fourier Series be appropriate for our analysis?

 

A1) Discrete Fourier series applies to situations where both a function of time, f(t), and its Fourier transform, F(s), are sampled at regular (or irregular) intervals. (In the case of Fast Fourier Transform (FFT) algorithm, the sampling intervals are regular.) In any case, when discussing phase-locked lasers, I mentioned that the spectrum becomes discrete, but the time-domain function, namely, the time-dependent E-field or H-field amplitudes are still continuous.

       And, by the way, discrete Fourier transform is nothing new or special; it is just a way to approximate continuous integrals (forward or reverse Fourier transforms) with discrete sums. In the case of phase-locked lasers, what we have is a number of discrete frequencies that are super-imposed to create a periodic function in the time domain. You may think of this as a comb function multiplied into the spectrum, resulting in the convolution of a comb function (the Fourier transform of a comb is also a comb) with the time domain signal. Such a convolution will make the time domain function periodic, hence the pulsed nature of the output from phase-locked lasers.

 

Follow-up to Q1) Hmmm — because we could look at the frequency spectrum in the phase-locked case as “discrete” as it’s a series of delta functions, then I would think that the DFS applies when we want to move back to continuous time if we’re working this out on paper. If we use a DFT, as you state, both the frequency spectrum and the time domain signal should be discrete and periodic. In Matlab, you would take the FFT of the pulse train and get a periodic frequency spectrum which would then be periodic, and the IFFT would result in a discrete time signal — I suppose it’s a moot point and implied that there’s some sort of lowpass filtering or that the sampling is high enough to be considered “continuous”… I don’t know, it’s a cool relationship and it looks nicer in plots than in words 🙂

Q2) Nonlinear optics often reference the chi^(3) (X^3) term — is that the higher order approximation that you spoke of last week Thursday (lecture 24) that causes spatio-temporal distortion in phase-locked pulses? 

 

A2) No, I was still talking about linear effects; chi_3 has to do with nonlinearities of the medium. Effects of chi_3 show up only at high intensities, whereas the pulse distortion due to absorption and/or dispersion that I was talking about have to do with the linear refractive index. When the spectrum of the pulse is wide enough that the first two terms in the Taylor series expansion of n(w), namely, the constant term and the term that is linear in w, do not give a good approximation to the refractive index within the spectral bandwidth of the pulse, then you’ll have to introduce higher-order terms in the Taylor series expansion of n(w). These higher-order terms produce distortion through dispersion, but have nothing to do with chi_3.

 

Friday, November 11, 2012

Q) I have a few questions about metamaterials with respect to the material we’ve covered so far. In lecture 24 you mentioned that the group velocity for metamaterials is negative. Does this mean that for a plane wave the wave vector k and Poynting vector S are in opposite directions? It seems like the right rule becomes the left hand rule.

      How are the boundary conditions for the electric field per Maxwell’s Equations 1 and 3 maintained at an interface between a conventional material and metamaterial?

      Is it possible to make a GRIN material that transitioned between a a negative index and positive index for a given wavelength? I realize that this would not be a linear isotropic material.

 

1. A) It is the phase velocity (not group velocity that you mention) that is negative in negative-index media. And yes, that makes the k-vector opposite in direction to the Poynting vector.
   The boundary conditions 1 and 3 remain the same as before, that is, perpendicular D-field will have a discontinuity equal to the interfacial electric charge-density, and the tangential E-field remains continuous at the boundary.
   I don’t see why a GRIN medium could not transition from a positive index to a negative index, so, yes, I think it’s possible. Just remember that it is the phase index that is negative; the group index is still positive in negative-index media.

 

Thursday, November 10, 2012

Q) Practically speaking, is plasma frequency somewhat fixed for a material? It’s dependent upon only number of oscillators, charge, and the mass — mostly material/density properties. In other words, for a given material we have 2 regions: (i) w < w_p for which the material is reflective; (ii) w > w_p for which the material is transparent. How much control is there in shifting w_p? Also, is there a temperature dependence?

A) The Lorentz oscillator model must be used with caution. For one thing, the charge q and the mass m are not the actual charge and mass of an electron; they represent some sort of “effective” charge and mass, which should be determined from quantum calculations. [Recall also that there is that oscillator strength parameter (f_k) in the formula for C_K(w), which could be said to account for the “effective” charge and mass of each electron.] For any material at any given frequency, one should add all the contributions to the susceptibility from various oscillators. So changing w from below w_p to above w_p may not be sufficient to get a transition from being reflective to becoming transparent. There are many other effects that the Lorentz oscillator model does not account for, for instance, interaction among different electrons, the band structure of the material, the impurities in the lattice, temperature dependence of the band structure, etc.

     I use the Lorentz oscillator as a good physical “toy model” to gain a rough insight into the behavior of real materials. When it comes to using it in numerical simulations (or for other exact calculations), I always check the experimentally measured values of n and k over some range of frequencies to see what parameters of the Lorentz model should be adjusted to get a good fit to the measured values.

Wednesday, November 9, 2012

Q) I just watched lecture 23 and have a quick question. In the previous lecture 22 you introduced the dimensionless electric and magnetic susceptibilities, Xi_e and Xi_m, and related them to the polarization and magnetization where in both cases the susceptibilities are functions of frequency. In lecture 23 you showed the differences in the treatment of the electric susceptibility between the Lorentz oscillator model for bound electrons versus the Lorentz oscillator model for conduction electrons via the Clausius-Mossotti relation and Drude model, respectively. My question is if there is an analogous oscillator model to describe a simple system that reacts under the influence of an oscillating magnetic field. If so, are there corresponding differences in the treatment of the magnetic susceptibility such as there are for the electric susceptibility in the Lorentz oscillator models?

A) Free charges (such as conduction electrons) do not have a magnetic counterpart, so there is no Drude model for magnetization. As for the Lorentz model and the corresponding Clausius-Mossotti correction, people sometimes use the same formulas for magnetic susceptibility, using w_o, gamma, and w_p as adjustable parameters to fit to experimental data over a desired range of frequencies. In general, however, the physical basis for magnetic susceptibility differs from that for dielectric susceptibility. Common materials exhibit various forms of magnetic behavior, named paramagnetic, diamagnetic, ferromagnetic, anti-ferromagnetic, and ferrimagnetic. Also, at the atomic scale, there are two different sources of magnetization, namely, spin and orbital angular momenta. The mass-and-spring model of Lorentz may be suitable for paramagnetic and diamagnetic response of materials to external magnetic fields, but it doesn’t quite fit the ferro- and ferri-magnetic response, where magnetic domains are present within the material medium. In response to a magnetic field, these magnetic domains expand or shrink, and the magnetic moment within individual domains may also rotate in different directions. Such domain-related behavior is usually slow and does not occur in response to high-frequency fields (e.g., at optical frequencies). In optics, when we deal with ordinary matter (as opposed to the recently discovered meta-materials), we always set mu(w) = 1 + Xi_m(w) = 1; in other words, ordinary magnetic materials simply do not respond to the optical H-field. At microwave frequencies, the response of magnetic matter to applied H-fields is usually dominated by changes in the orientation of magnetization or movements of domain walls, so the physical mechanism for such behavior is fundamentally different from that based on a mass-and-spring model. It is perhaps best to use experimentally obtained data for Xi_m(w) in such cases. Magnetic materials also exhibit hysteresis and nonlinearity, which makes the task of modeling their response even more complicated.

 

Tuesday, November 8, 2012

Q1) Does the n”(w) curve in example 6.7.4, figure 5 then relate to the gain-bandwidth of an active medium?

A1) To the extent that the medium satisfies the restrictions of a single-oscillator Lorentz model, I suppose you could infer the gain-bandwidth of the medium from the plot of n”(w) in Fig. 5.

Q2) Is there a particular advantage to defining the damping parameter in physical units? That is, it’s often defined in these 2nd order differential equations as “critically damped, gamma = 1”, “overdamped, gamma > 1”, and “underdamped, 0 < gamma < 1”, where instead gamma is normalized by 2*resonant frequency.

      Finally, is there a significance to the Q value for an electric or magnetic dipole oscillator?

A2) As you can see from the discussion of Sec. 6.8 [in particular, Eq.(20)] we are usually in the under-damped regime (that is gamma < 2*wo). Since we are not concerned about the system being over-damped or critically-damped, it makes sense to use a definition of the damping coefficient that has a physical significance. The units of gamma are the same as those of frequency (Hertz), and it usually gives a good estimate of the absorption line-width.

       As for the quality-factor Q of the oscillator, Q is usually defined as the ratio of useful energy (e.g., stored energy or radiated energy) to the wasted energy (e.g., lost to heat or leaked out of the cavity). Q can also be shown to be inversely proportional to the line-width of the oscillator/resonator. You can see in Chap.8, Sec. 7, a discussion of a Fabry-Perot type resonator. I am not sure how one can define a Q for the Lorentz oscillator model that could easily be related to radiated energy and lost energy. There may well be one such definition, I am just not aware of it.

 

Monday, November 7, 2012

Q1) With regards to the discussion of complex refractive index, n'(w) + i n”(w), you mentioned that ordinary, passive materials have an imaginary component n”(w)>=0, while active, gain media have n”(w)<0. Does this dependence then become more complicated than stated? I realize this isn’t a laser class, but most laser gain media are pumped to energize the gain atoms (i.e. to achieve population inversion). This operation is energy dependent and independent of our discussion — but, does the n”(w) change with the pumping operation?

A1) The value of n”(w) for a gain medium depends on the conditions under which the medium is operated. All I said in the class was that for gain media n”(w) is negative. The magnitude of this negative number will depend on how strongly the material is pumped, and also how rapidly the stimulated emission process depletes the inverted population.
      Another thing that is different between passive and active media is the sign of the oscillator strength (see Example 4, Sec. 6.7.4). In other words, the entire susceptibility chi_e(w) — not just its imaginary part — must be multiplied by a minus sign in the case of gain media. This ensures the causality of the response of the dipoles to the exciting field, which is another way of saying that a negative oscillator strength ensures that the Kramers-Kronig relations (Sec. 6.9) remain valid for gain media.

Q2) Is the electric susceptibility dependent upon a time delay, as the physical interactions are not instantaneous?

 

A2) The way we have defined susceptibility, chi_e, it is a function of the frequency w. Since single frequency oscillations span the entire time (from –inf. to +inf.), a time delay would appear in chi_e as a phase-factor, that is, chi_e must be a complex-valued function of w. You will see in Secs. 6.8 and 6.9 that such complex phase-factors are essential to ensure causality. This is the essence of the Kramers-Kronig relations.

Thursday, November 3, 2012

Q1) In Chapter 5, problem 39 should there be a delta function in the infinite sheet’s current density description? Should it read: Js(r’,t) = Jso delta(x) sin(wt-Ky’) ? It currently reads: Js(r’,t) = Jso sin(wt-Ky’). The example in section 5.5.3 has a delta function, delta(y), explicitly expressed in the infinite sheet’s current density description, although the sheet is oriented differently it’s still aligned at the origin as in problem 39. Problem 40 in chapter 5 also refers to the same current density as in problem 39. Should the delta function be expressed there as well? Same for problem 38.

A1) The distinction I have made in these problems is between “surface” current-density and ordinary current-density. If I write the current density as J_s (units: ampere/meter), then it is a surface current and no delta-function is needed to describe it. However, when I write them as ordinary current-densities (units: ampere/m^2), then they need the delta-function. Recall that these delta-functions have a dimension of 1/m, so I cannot use them together with the J_s notation, as they would modify the units.

      Also, with the surface currents, I have to use words to specify the plane to which the current is confined. In contrast, the use of delta-functions in conjunction with the ordinary current densities obviates the need for additional specifications.

Q2) I have another question on Chapter 5 problems 38 and 39… In problems 38/39…from a pure notational point of view, is there a reason that the current densities are expressed explicitly as functions of r’ instead of r ? As far as the retarded time goes the current density change at r’ is observed at r with a time lag due to propagation at the speed of light so this makes sense to me. Example 5.5.3 gives the current density description in terms of r, but the diagram shows it in terms of r’. In problem 39 the current density description is given in terms of r’ as well as shown in the diagram in terms of r’. In both the 5.5.3 example and problems 38/39 the observation point r is given. Personally, I like the fact that problems 38/39 explicitly express the current densities as functions of r’ , but I was confused at first based on the notation in example 5.5.3 which I read before I looked at problems 38/39. I hope I’m not missing anything fundamental here….

A2) I use r and r’ interchangeably, as these are dummy variables that describe the position of a point in space. Only when two different points appear in the same expression does one have to be careful as to which one is called r and which one r’. I usually refer to the source location as r’ and the observation point as r, but this is just an arbitrary choice. When there is no possibility of confusion, one can use either r or r’ to refer to a given point in space.

 

Q3) Does the Biot-Savart law really tell us anything new? It seems like Maxwell’s equations and the vector potential handle this case.

A3) As you can see from Prob. 20, Chap.5, the Biot-Savart law is a direct consequence of Maxwell’s equations; there is nothing fundamentally new in that law, although it sometimes helps in simplifying the calculations. For example, if you have a current ring and need to compute the H-field on the z-axis, the Biot-Savart law allows you to do an integral over the current profile of the loop and get the answer. If, however, you try to do the same calculation with the vector potential, you’ll need to find the vector potential not only on the z-axis, but also in the vicinity of the z-axis, so that you can compute its curl.

Tuesday, November 1, 2012

Q) Regarding the “method of images” used in chapter 5, problem 11 — a test charge is set at a distance from a sphere to evaluate the scalar potential of a sphere — can this method be used to evaluate the vector potential or magnetic field?

A) Good question; you get a point for it. The important feature of problems that can be solved by the method of images (e.g., Problems 6 and 11 in Chap. 5) is that there is a region of space where Laplace’s equation (i.e., wave equation in electrostatics) is satisfied, and also the boundary conditions for the original problem are the same as those for the problem in which some part of the system is replaced by image charges placed in proper locations “outside” the aforementioned region of space. The fact that a differential equation (Laplace’s in these examples) and its associated boundary conditions are identical for the original problem and for the modified problem guarantees the uniqueness of the solution, hence the validity of the method of images.

      The method of images can be used whenever one can identify an image for a source of radiation (or, in general, a source of electric and/or magnetic fields), which reproduces the boundary conditions for the region of interest “without” modifying the sources inside that region. The method of images is particularly useful in electrostatics because ordinary conductors can be considered equi-potential surfaces, and this fact simplifies the problem of identifying image charges that, together with the original source charges, reproduce the boundary conditions (e.g., constant potential on the surface of the conductor). One can do similar things in magnetostatics with high-permeability materials known as “mu metals” — for their large values of the permeability mu.

 

Monday, October 31, 2012

Q1) I have always seen the wave equation written in terms of the electric and magnetic fields rather than the scalar and vector potentials so it was interesting to see it in terms of the potentials. I followed the derivation in lecture 20 and chapter 5 to obtain the wave equations for the two potential fields and see the importance of using the Lorenz gauge to get the answer. In the past I’ve seen the derivation of the wave equation for the electric field from Maxwell’s equations, starting with taking the curl of both sides of Maxwell’s 3rd Eq (curl(curl(E))=curl(uoJ+dE/dt). This leads to: [Laplacian-(1/c^2) d^2/dt^2]E(r,t)=uo d/dt J(r,t)+(1/e0) grad p(r,t). In the derivation of the wave equation for the electric I don’t remember anything having to do specifically with the Lorenz gauge (attached is the derivation I’m referring to). Are the wave equations for the electric and magnetic fields truly independent of the choice of gauge? It seems as if the wave equations for the potentials (using any gauge) could be combined to get gauge independent wave equations for the electric and magnetic fields. Also, is it common to use the wave equations for the electric and magnetic fields to derive the wave equations for the potentials by imposing a particular gauge? I guess it’s sort of a chicken and egg question.

A1) Gauges are important only for the potentials. Once you derive the E and B fields from the potentials, the results will be independent of the chosen gauge. Likewise, the wave equation for the E-field is gauge-independent, as you can clearly see in the derivation of that wave equation: the gauge doesn’t appear anywhere in the derivation.

      The advantage of the Lorentz gauge is that it allows a neat separation of the contributions of charge and current densities; the former produces the scalar potential, the latter the vector potential. Work in any other gauge, and the contributions of charge and current will get mixed. In the end, both charge and current density distributions contribute to the wave equation for the E-field, no matter which gauge is used.

Q2) When we earlier discussed the scalar/vector potential functions, you mentioned that some physicist believed they exist in physicality… now that we are looking at them in space-time, their description is much more apparent as a “observational” or “measurement” field. And, equations (15) and (16) introduce a “retarded time”, (t-|r-r’|/c), or the time for “communicating” the state of the source, etc, etc. Could these instead be defined with, t+|r-r’|/c, for a “predictive time” or “anti causal time” ?

A2) Chapter 5 opens with a quote from Feynman; please take a look. You may also want to read Feynman’s entire discussion of the Aharonov-Bohm effect, where he argues that the vector potential should be granted the status of a real physical field. Not everyone agrees with him, of course, because the potentials are gauge-dependent, whereas real-fields are supposed to be uniquely defined at each point in space-time. I am inclined to accept Feynman’s point of view, because I also tend to believe in the “locality” of physical effects, and the Aharonov-Bohm effect can be made local only if the vector potential is accepted as a real field. The Lorenz gauge is the natural gauge for defining these potentials, as it allows the potentials to propagate from source to destination at the vacuum speed of light, c. However, even the Lorenz gauge is not unique, as you can see from Problem 31 at the end of Chapter 4. So, with this caveat, one may think of the potentials as real fields.

     In classical physics there are no experiments that require the potentials to act locally and, thereby, become visible; all observable effects can be associated with the local action of the E and B (or H) fields. In quantum physics, however, certain observations (e.g., the Aharonov-Bohm effect) provide a basis for speculation as to the nature of the potentials. What’s more, the basic formulation of quantum mechanics (in terms of the Hamiltonian for particles and fields) is impossible without the use of these potentials.

       Concerning the second part of your question, if you check the derivation of Eqs. (15) and (16) in Chapter 5, you’ll notice that we could as well have written the potentials in terms of advanced (rather than retarded) time. In other words, these expressions could as well be written with t + |r – r’|/c instead of t – |r – r’|/c. Both solutions will be equally acceptable (mathematically), but the “retarded” expressions appeal to our desire for causality, hence their widespread usage. Feynman and his thesis advisor (Wheeler) published a paper in the 1930s in which they argued that the potentials should be written as sums of two terms, with equal contributions from “retarded” and “Advanced” expressions. This particular formulation turned out to have some advantages in quantum electrodynamics. In classical physics, however, there is no advantage to working with “advanced” potentials, and everyone sticks to the retarded

Q3) I have another question about the material in lecture 20… When conceptualizing the scalar and vector potentials in space time per Eq 15 and Eq 16 in the book we must consider the retarded nature of an EM field which manifests from a point r’ and influences a point r, for which the time from manifestation to influence is |r-r’|/c. If we look at the electric and magnetic fields in terms of the potential fields in the forms of Eq 15 and Eq 16 using the time delay |r-r’|/c per E(r,t)=- grad(psi(r,t))-d/dt(A(r,t)) and B(r,t)=curl(A(r,t)) we also get a delayed cause and effect between the point r at which E and B are evaluated and the point r’ from which the charge and current densities are physically. For the case when d/dt(A(r,t)) is zero there can be a charge density and a current density, but no accelerating charge so the electric field is just: E(r,t)=- grad(psi(r,t)). In this case we do not have “light”, just an electric field that may or may not be static. My question has to do with the retarded nature of the cause and effect for a case when the source terms produce “light” versus when the source terms only produce EM fields that are not “light”. Do EM fields which are not “light” always travel at the same speed as EM fields that are “light”? In vacuum it seems like both types of EM fields would have travel at the same speed. Are there circumstances in which the speed of light in a particular medium is different than the speed of influence of a non “light” EM field? More than anything I want to make sure that I understand the differences between EM fields that are considered light and EM fields that are not considered light, particularly in the context of how they travel through different media.

A3) I don’t know what you mean by “non-light.” Either the fields are static or dynamic. If the E and H fields are time-independent (i.e., static), as would be the case when the sources are time-independent, then there is no meaning to the term “propagation speed.” The fields are constant everywhere, and you can’t talk about the speed with which they get from the source to the destination — in other words, the source has always been there, and any propagation has taken place (and ended) long ago. As soon as you have time-dependence (i.e., dynamic situation), you’ll have retardation, precisely as given in Eqs. (15) and (16) and, therefore, everything propagates with the vacuum speed of light, c. Note also that, according to Maxwell’s 2nd and 3rd equations, any time-dependence of E must produce a local H-field and vice versa, so you can’t have one without the other.

 

Q4) Thank you for the responses. By “non-light” I was trying to describe the mechanism in which a change in a field as observed at location r is relayed from a perturbation of a source at location r’ in the case where the source perturbation does not involve the emission of an electromagnetic wave. As I think about it more I cannot come up with a good example. As I understand things “light” can only be created by the following: Accelerating charge, charge annihilation, and energy transitions from a higher energy state to a lower energy state within an atom. So, even in the simplest case when a charge is moved from one static location to another static location there exists accelerated charge which creates the light that informs all locations within |r-r’| that the electric field is now different. When we start to consider the retarded time is it still OK to use plane waves for the description of light? It seems as if spherical waves are necessary such that all locations within a radius of |r-r’| get the news that something happened at the same time.

A4) In chapter 5 we are no longer using plane-waves. The plane-wave formulation of chapter 4 allowed us to go through the Fourier transform and inverse Fourier transform operations, so that we are now, in Chapter 5, dealing with the most general form of electromagnetic radiation. Each source point in Eqs. (15) and (16) is in fact radiating spherical waves!

       Note that these formulas allow for static fields as well, so both “light” and “non-light” — in your terminology — are encompassed by the formulas.

Q5) Another question on lecture 20… What happens to the retarded time dependence for the scalar and vector potentials in Chapter 5 Eq 15 and Eq 16 when the charge density and/or current density are moving at relativistic velocities with respect to a stationary observer? What if both the source terms (charge/current) and observer are moving relativistically? For example, if two electrons are both moving at relativistic velocities how do the potential fields (psi(r,t) and A(r,t)) from electron 1 influence electron 2?

A5) You have asked a couple of good questions today and I’m going to give you 2 points. Maxwell’s equations are fully compatible with special relativity and, therefore, they can be used in situations where the source, or the observer, or both are moving at relativistic speeds. The potentials that we have discussed in Chapters 4 and 5 are valid for any speed of the sources, no matter how close that speed may be to the vacuum speed of light, c.

      In this course, I have said from the beginning that we are going to stay in a single (inertial) frame of reference and not switch between reference frames, simply because I didn’t want to introduce the concepts of 4-vectors and the Lorentz transformation between various inertial frames. Switching between reference frames, however, would pose no problem for our formulation; all one needs to do is Lorentz-transform various 4-vectors such as the space-time coordinates (x, y, z, t), the charge-current densities (Jx, Jy, Jz, rho), the potentials (Ax, Ay, Az, psi), which must be in the Lorenz gauge for this purpose, the momentum-energy (px, py, pz, energy), etc., between the various inertial frames. In each frame then, one can obtain the fields and forces using the formulas we have derived in this course.

 

Sunday, October 30, 2012

Q1) Example 5.5.3 on page 120: You begin by saying “Figure 1 shows an infinite sheet of current J(r,t) = Jso * delta(y) * cos(wo*t – kappa*z)zhat. However, in Figure 1 itself, the current density is labeled as “J(r’,t)”. Why did you switch notation? I thought the r’ notation was reserved for the “source” point and the r notation was reserved for the “observation” point? The caption on Figure 1 also says “Js(r,t)” instead of “Js(r’,t).”

A1) r and r’ both represent the coordinates of a point in 3D space. I use them interchangeably, the meaning being understood from the context. The only time that one has to be careful is when “both” the observation point and the source point appear in the “same” equation. In such cases it is important to use r for one of them (usually the observation point) and r’ for the other (usually the source).

Q2) At the beginning of the lecture, we are looking at the static cases (B and E are time independent). In this case, shall the scalar and vector potential are also time independent? However, we later have the equation for scalar and vector potential with time dependent. Does that mean these equations are also applied for time dependent cases?

A2) The equations for scalar and vector potentials are general; they apply to both time-dependent and static cases. If the charges are static, then the scalar potential (psi) will be independent of time. And if the current distribution is time-independent, then the vector potential (A) will also be independent of time. Now, the E-field will just be -grad(psi), because dA/dt = 0. And the B-field, which is equal to curl of A, will be time-independent as well.

Thursday, October 27, 2012

Q) When we were working with the Fourier transform we saw that the potential functions were undefined when k^2 = (w/c)^2. Now in the space-time domain, we see that the potential functions are undefined when r’ = r. Can you please explain what this means?

A) Very good question; you get a point for asking it. The integrals have singularities when their denominators go to zero. For example, a charge located at a point r’ will produce a finite potential at every point r in space, except at its own location, i.e., when r = r’. The potential produced by a point-charge at its own location is infinite. When we evaluate these integrals, we essentially “exclude” the neighborhood of the point r’ from the domain of integration. That is how these singular integrals are evaluated — just like integration in the complex-plane when a pole is excluded from the domain of the integral.
     The same thing happens in the Fourier domain; when k = w/c, the denominator goes to zero and the integrand goes to infinity (provided that charge or current densities are non-zero at k = w/c). Again the way to evaluate these integrals is by “excluding” the singularity from the domain of integration.

     Later in chapter 7, we will see that plane-waves can exist in empty space when k = w/c; in other words, no sources (charge and/or current) are needed to sustain such plane-waves. Any charge/current distribution that has k = w/c will, therefore, produce a plane-electromagnetic wave with infinite magnitude. It is obvious that the only way to avoid such infinities is to exclude from the domain of integration the neighborhood of those points where k = w/c.

 

Sunday, October 23, 2012
 

Q1) I have a question on page 100. I don’t understand where the (1/2) coefficient on equation (68), “A(r,t)=(1/2)[A1(r,t)+A2(r,t)],” comes from. Comparing to the equation (58) on page 97, it is reasonable to put a (1/2) coefficient on the equation because the J1 and J2 on page 96 don’t have the (1/2) coefficient shown on equation (53). Could you explain the (1/2) coefficient for equation (68)?

A1) You are correct. You have found another mistake in the book, for which you will get a point. The factor 1/2 in the beginning of Eq.(68) should be dropped. The rest of the equation, however, is correct, and all the results that follow Eq.(68) are also correct. (I must have thought that J_1 and J_2 were not already multiplied by 1/2, similar to a previous example).

 

Q2) I’ve been going through the OPTI 501 midterms you posted from past semesters in preparation for the upcoming midterm. Problem 2 on the fall 2005 midterm asks about the limiting form of the Bessel functions Yo(x) and Jo(x) for small x values (near field). Chapter 4 problem 28 also references limiting forms of the Bessel functions, but for the far field. I see the results in the posted solutions. I didn’t see anything in terms of a derivation for the limiting forms of the Bessel functions, but they appear to be very useful. Is it possible to go over this briefly in class or could you point out a good reference for the limiting forms of the Bessel functions? Are there any other useful approximations to be made for the near and far fields for the (scalar and vector) potential fields when not in terms of Bessel functions?

A2) Please take a look at Chapter 3, the section on Bessel functions. I listed the useful approximations there, and gave general references at the end of the chapter. Also, the end of chapter problems (Chap. 3) and their posted solutions provide proofs for some of these approximations.

     There are many useful approximations in the theory of electrodynamics; many of these are discussed in Jackson’s book, many more in the book by Born and Wolf. You might say that the entire theory of (electromagnetic) diffraction is devoted to various approximations to Maxwell’s equations.

 

Saturday, October 22, 2012

Q1) Very interesting to talk about is happening when light is incident on ideal and non-ideal conductors in the context of sheet of current. What is happening when light is incident on a non-opaque insulator, like glass? Is the transmitted beam created by a current generated on the surface, or is it simply the non-attenuated incident beam? Is there only a surface current generated as much as the material has reflectivity? Thanks, Matthias Whitney.

A1) We get to these issues in Chapter 7. For now, let me just say that, when the light goes through a sheet of glass, all atoms get excited and begin to radiate; there is no difference between atoms at the front surface, those inside, and those at the back surface. These various layers oscillate in response to the incident radiation (and also in response to the radiation by other atoms within the same sheet). The net result is that some of the light appears to be reflected, some is transmitted, and the speed of light inside the glass slab is reduced by a factor which is commonly known as the refractive index.

 

Q2) In your Thursday lecture you used the rate of change of energy equation, but you left out a few terms from when the equation is in the book. In equations 2.23 and 2.24, you also have the terms for time-rate-of-change of the local energy density of the E- and H-fields. You do later say that you need to add to the time average 1/2*epsilon_o*E^2 and 1/2*mu_o*H^2, but this still does not have the d/dt term of the E^2 and H^2 terms as in the equations of the book. Why can we ignore these terms? Is it because these are plane waves oscillating at a constant frequency? Thanks, Matthias Whitney.

A2) Nothing has been ignored; the energy that is coming out of the sheet goes either into building up the E- and H-field energy densities in the surrounding space, or gets out of the two surfaces I drew on the board (to the right and left of the oscillating sheet). The d/dt terms simply indicate the rate of increase (or decrease) of the energy stored in the field. What I said in the class (or at least what I meant to say) is completely based on Eqs. 2.23 and 2.24 (and all other equations related to energy flow and energy storage discussed in Chap. 2). In the class, I don’t have enough time to go through exact derivations. But I point these things out so that the students would go and calculate everything in detail and see for themselves that things work out perfectly when various terms in the energy flow/storage equations are taken into account.

Q3) Another question from Thursday’s lecture. You mention in your example with two infinite current sheets that the standing wave will continue without any additional driving forces. What will happen if there is a continuing driving current? For example, like you explained in Tuesday’s lecture, if you have a positive sheet and a negative sheet that are continually moving against each other? Will this cause the amplitude of the standing wave to increase? Will there be any wave propagating outside of the sheets, or will it still be canceled out by the ever increasing standing wave? Thanks, Matthias Whitney

A3) Things get complicated when you raise the driving forces “continuously.” For example, the fields outside the sheets will no longer cancel out exactly because there is a delay between the time that the fields from the first sheet arrive at a given point, and the time of arrival of the fields from the second sheet. If, however, you simply raised the driving force and then kept it constant for a while, the transients will die down and you are back to the same steady state situation as before, albeit with a larger field now residing between the two sheets.
      Note also that the E-field produced by each sheet is now driving the oscillations of the other sheet; therefore, the person/mechanism that is shaking the sheets does not feel any opposing force (i.e., no radiation resistance). The external mechanism now does not need to exert any force of its own (nor spend energy) to shake the sheets; the oscillations have become self-sustaining.

 

Friday, October 21, 2012

Q1) With regards to your intuitive discussion of the principle of superposition and the boundary conditions of a mirror — what are frequency considerations for transmission and reflection? Something like aluminum will be opaque to visible frequencies, but not x-rays for an extreme example… perhaps there are better examples that could be applied towards a “non-thin film” dichroic mirror?

 

A1) The incident light (or x-ray, or any EM wave in general) must be able to excite the material in such a way that a strong sheet of current will be induced at the surface of the material. Metals at visible frequencies (and also at infra-red and microwave frequencies) have this property, because they have a very large susceptibility — meaning that a small E-field can give rise to huge currents in the material. Dielectrics (such as ordinary glass) do not have a large susceptibility, and that is way I couldn’t use the same argument for a piece of glass. The light excites more than just the surface atoms of the glass; that is why it is only partially reflected; most of the light goes through the glass and excites the atoms/molecules deep inside. A material such as silver acts like a metal in the visible range, but in the UV it becomes transparent. Recall that the point of departure for my argument was the presence of a thin sheet of oscillating current. This sheet is present in an aluminum mirror at visible (and many other) frequencies, but not at x-ray frequencies. A dielectric multilayer mirror also reflects light, but the principle of operation is very different: no surface currents here, but many layers of excited atoms deep inside the dielectric mirror.

        In Chapter 6 we will discuss the properties of materials and you’ll see under what circumstances the susceptibility of a material becomes large enough to act like the metallic mirror in the example I gave in the class. The response of each material is not just an intrinsic property of the material; it also depends on the excitation frequency.

 

Q2) I had a question following the discussion of incident reflection of EM waves off of ideal conductors in Lecture 17.  I’m aware that ferromagnetic material (iron, cobalt, nickel, ect.) can sustain a net internal magnetization in the absence of an external H-field. If the ideal conductor were ferromagnetic, would the net internal magnetization of the ferromagnet alter the polarization of the reflected H-field relative to the incident H-field at normal incidence?

 

A2) In ferromagnetic materials, the magnetization is constant in time; it does not oscillate with the frequency of the incident light. The surface current we talked about in the class yesterday is an “induced” current; it is induced by the incident laser beam and has the same frequency as the light beam. The two fields (one produced by the oscillating electrons at the front facet of the mirror, the other produced by the zero-frequency magnetization of the mirror) are independent and additive. There is, therefore, no effect on the reflectivity of the mirror due to the magnetization in this case — unless you wanted to dig deeper to find out how the light interacts with the magnetic material if it could penetrate it, in which case you would have the magneto-optical Kerr effect, which, depending on the direction of magnetization of the material relative to the surface normal, could result in the rotation of the polarization of the reflected beam.

        If, instead of a ferromagnetic material, you had a material that was magnetizable in the presence of the incident beam (i.e., one that had a non-zero magnetic susceptibility at the frequency of the incident light), then you could end up with effects that depend not only on the metal’s conduction electrons but also on its magnetic dipole moments. We will address these issues in Chapters 6 and 7. At optical frequencies, ordinary materials turn out to have essentially zero susceptibility, so the induced magnetization of the material is usually ignored; at microwave frequencies, however, the magnetic susceptibility becomes important and needs to be taken into account. You’ll have a better appreciation for these phenomena once we cover Chapters 6 and 7.

 

Q3) Hello Professor Mansuripur,

 

1) On page 88, example 4.5, you say that calculating the scalar potential will lead to a divergent integral.
a) What does it physically mean for the scalar potential to be divergent?
b) Are there methods to prevent this integral from diverging? Throughout the book, you always have mathematical tricks for us. We are always applying some form of the exp() function to get integrals to converge; for instance, we apply exp(w”t) in example 4.9 on page 91 to help with convergence.

 

2) On page 89, example 4.7. Similar as question 1 but this time it applies to the vector potential A.
a) What does it physically mean for A to be divergent, etc.?
b) Is encountering divergent integrals when trying to compute scalar or vector potentials a “limitation” of the Lorenz gauge? Would we have the same issue if we went with the Coulomb gauge?
c) So if we run into divergent integrals when trying to compute the scalar or vector potentials (as in these two examples), why bother computing them if we can ultimately use Equations 15 and 16 instead to arrive at the E and B fields?
d) Do the scalar or vector potentials (when they exist) simplify the process of computing the integrals? Is that why we use them?
e) How would one know ahead of time if the scalar or vector potentials will have a divergent integral? Do you have to “turn the crank” and “see what happens”? Or is there something special about the geometry of the problem that tells you to avoid using Equations 13 and 14 and just go straight to Equations 15 and 16?

 

3) The “gauges” are interesting to me. Is the Lorenz gauge THE standard/de-facto method in EM theory? What is its advantage over other gauges, say the Coulomb gauge? Is there a Mansuripur gauge?

 

A3) Let me answer your questions in a general way. Some of the systems I have chosen for analysis in Chapter 4 (also Chap. 5) are artificial, in the sense that they are infinitely large in 1 or 2 dimensions. In these cases, in the absence of oscillations, the potentials become infinitely large as one moves away from the source, although the fields (E or H) remain finite. In such situations, if you tried to compute the potentials, you will see that your integrals do not converge, but the integrals for the fields can be made to converge — using tricks such as introducing the exponentially-growing time-dependent factor. In one of the problems (Chap. 4, Prob. 10) I have asked you to start with a thin wire of finite length, L, find the potential, then let L go to infinity. This is the kind of trick that can be used, but the integrals are usually very complicated, and it is better to not bother with the potentials in such cases, and, instead, go straight for the fields.
       The systems used in practice are always finite in their dimensions and, therefore, there is no problem finding the potentials as well as the fields for them, but analytical solutions don’t usually exist for such systems and one will have to resort to numerical methods.
       In chapter 5, you will see another method of solving these problems; perhaps you’ll then get a better feel as to why the potentials sometimes go to infinity. If you take an infinitely large plate that is uniformly charged, for example, its electric field remains constant no matter how far away from the plate you go. Therefore, the scalar potential, which is just the integral of the E-field in the direction perpendicular to the plate, gets larger and larger as you move further and further away from the plate. It is an anomaly only because the plate is infinitely large. A finite-diameter plate will never exhibit such an anomaly. This is not a limitation of the Lorenz gauge; you will encounter similar problems in other gauges as well.
      Scalar and vector potentials usually simplify the solution of electrostatic, magnetostatic, and electrodynamic problems; they are conceptually interesting and powerful tools, and most importantly, they are needed for quantum treatment of electromagnetic systems. Strictly speaking, classical electrodynamic theory can do without these potentials, but in any text book that you care to look, these potentials are introduced and used effectively.
      I think someone previously asked your last question about the various gauges. Please take a look at the Q&A section of the website. (No gauge, I’m afraid to say, has ever been named after me!).

 

Follow up on Q3) Great explanation!  It is still interesting that Eq. 13 and 14 are numerically “unstable” for these artificial problems.  The minute you apply the gradient and curl operators to them (and subsequently get Eq. 15 and 16), then they become numerically “stable”.  I like the magical properties of the gradient and curl operators … math and physics are fascinating.

 

Thursday, October 20, 2012

Q1) For Example 4.9, you make two references to Imag(iwo*/c) = w”/c > 0 (on page 92 and 93). Do you actually mean Re(iwo*/c) = w”/c > 0? I get: iwo*/c = (1/c) (i ( w’ – iw”)) = (1/c) (iw’ + w”). So Re(iwo*/c) = w”/c and Imag(iwo*/c) = w’/c. Can you please explain?

A1) Yes, you are right; my mistake. I give you a point for finding an error in my book.

 

Q2) have another question about Example 4.9. On page 92: in going from line 3 to line 4 of the math at the top of the page, how did you “know” to split the one integral into three? In particular, what methodology did you use to get the last two integrals in the 4th line? That is, how did you go from line 3 to line 4? Did you use partial fractions?

A2) In line 3, if you take the term that contains k^2, then add and subtract (wo/c)^2 to k^2, that is, replace k^2 with k^2 – (wo/c)^2 + (wo/c)^2, you’ll get to line 4.

Q3) In examples 4.8 and 4.9, you have k.z where z is a unit vector. When shifting into spherical coordinates, you replace k.z with kcos(theta)*(r.z) where both r and z are unit vectors with the explanation “Azimuthal symmetry around r makes it possible to replace k with its projection onto the unit-vector r-hat (r unit vector)”
      In the homework problems, I find myself with similar terms inside the integral like k_y*x and k_x*y where x and y are unit vectors. Can I make similar arguments to replace them with kcos(theta)*(r.x) kcos(theta)*(r.y) respectively? How does it work when it is only a component of k? Maybe something like sqrt(k)cos(theta) instead? Thanks, Matthias Whitney.

A3) Matthias: We have done similar problems in a few examples in Chapters 3 and 4. You don’t need to treat k_x and k_y separately; the entire term in the exponent, k_x*x + k_y*y, can be written as k_|| * r_|| * cos(theta). Please see the diagram in Chapter 3, Example 5, and also those in Chapter 4, Secs. 4, 6, 7, 11 and 12. The other term that you have in some of these problems is “k cross z^.” Since z^ is a constant vector, you can take it outside the integral. Now you are left with k (a vector) under the integral sign. The diagrams mentioned above tell you that the only component of this vector (k) that matters is its projection onto the rho^ unit-vector (sometimes this same unit-vector is written as r_||^) — symmetry eliminates the projection of k in the direction perpendicular to rho^. You may thus replace the vector k with |k|*cos(theta)*rho^. Now, rho^ is a constant (because the observation point r is a fixed point), and, therefore, you can take rho^ out of the integral and combine it with z^ (which was previously taken out of the integral) to write the combination as “rho^ cross z^.” What is left of “k cross z^” under the integral sign will thus be |k|*cos(theta).

 

Wednesday, October 19, 2012

Q1) In the oscillating electric dipole example, the E-M fields appear to be linearly polarized (vertical with the alignment to z-direction). Would an electric dipole be able to produce other polarizations?

A1) Very good question; you get a point for it. Please see Problems 18 and 27 (Chapter 4) for rotating electric and magnetic dipoles; these create elliptical polarization with varying degrees of ellipticity, depending on the location of the observation point with respect to the dipole.

Q2) Could we have avoided the “trick” (defining omega as complex) in the oscillating electric dipole example by defining the dipole as causal, starting at t=0 ? Placing a step function in the definition seems to make this needlessly complicated (in comparison), but would that give the same result?

A2) First, causality does not mean that you start the excitation at t = 0; it means that, no matter where you start the excitation, the response of the system will occur “after” the excitation begins. As for using a step-function for the time-dependence of the dipole, i.e., Step(t)cos(wo*t), it will make the problem even harder. The Fourier transform of Step(t) is pi*delta(w) – i/w. This must then be convolved with the Fourier transform of cos(wo*t), namely, pi*delta(w–wo) + pi*delta(w+wo), to yield the Fourier transform of Step(t)cos(wo*t). The result is (pi^2)[delta(w–wo) + delta(w+wo)] – i*2*pi*w/(w^2–wo^2). Note that you still have the two “troublesome” delta functions, in addition to the new function w/(w^2–wo^2) to contend with.

     Considering the generality of the solution we obtained for the dipole radiator (i.e., exact solution everywhere in space, near-field as well as far-field), the mathematical trouble we went through is not all that bad. There is a simpler solution, however, which you’re going to see in Chapter 5, Sec. 5.5.1. But, to get there, we must first inverse-transform our general equations for scalar and vector potentials, and this will involve some more math.

Q3) I have couple questions about hw 5 problem 30:

1. Is there an equal sign for ρ >= R for H field?

2. Because Bessel function 2nd kind is going to infinity at ρ=0, does that mean E and H field are going to infinity also? if that is true, how do we explain for that?

A3) Here are the answers to your questions:

1. No, there is no equal sign (=) for the H-field when rho = R. The H-field is discontinuous at the boundary.

2. In none of the equations do you see Y0(0) or Y1(0). Take another look at the equations!

 

Tuesday, October 18, 2012

Q1) I would like to request that the second midterm exam scheduled for October 25th be pushed back to October 27th. The reason I request this slight delay is to account for the delay in our last assignment that set us all back a few days in approaching the new material that will be covered on this exam. I am finding the material extremely challenging and would benefit greatly from studying and practicing the material more and having additional time to meet with the TA or you during office hours. Additionally, this exam follows a 502 exam that takes place Thursday, October 20. Many of us will need to focus on studying the 502 material on Tuesday and Wednesday, which cuts our time short for completing this homework and studying for your test. Please consider these comments and the fact that I just want to be able to more fully understand these Fourier techniques before I am placed in a testing situation.

A1) I do understand all your concerns. Under normal conditions, I would have considered moving the exam by a couple of days. But I have planned this exam to overlap with my overseas trip next week. I’ll be leaving Tucson on Saturday and returning late night on Wednesday. We did not arrange for tape-ahead, so it is impossible to make up the Tuesday lecture if I were to postpone the exam to next Thursday. Besides, we have 32 students on campus and 9 distance students, all of whom have been told from the beginning of the semester the dates of the exams. In the case of distance students, they also have to arrange for proctors; changing the exam date will cause a lot of trouble for a lot of people.

     So, I am sorry that we have to stick to the schedule. You’ll have Friday, Saturday, Sunday and Monday to study the material for the exam. I’ll be here until Friday evening, in case you have questions. After that please send me e-mails, or talk to the TA. I’m sure you’ll be fine.

 

Q2) For our first few examples in Chap 4, we’re dealing with electrostatic cases where an E-field is present or magneto static cases where an H-field is present. In both cases, the Poynting vector S=ExH = 0. In the electrodynamic examples of Chap 4, both E and H are present, therefore S is non-zero. This makes sense as our moving charges/dipoles induce the H-field.
    What if we place an electrostatic source next to a magneto static source, say a point charge is place near a long thin wire? It seems this would become a dynamic system automatically, and the Poynting vector would describe the energy output from this system (conserving the energy required to “put together” this system). Firstly, is there such thing as a electro-magneto static system in free space? Secondly, does the Fourier description of the Poynting vector S(k,w) mean anything useful? It seems like a PSD-field.

A2) Yes, it is possible to have a Poynting vector in a static system where both E- and H-fields are present; see Chap. 2, Problem 19 for an example. Also, the practice problem I just posted to the website (under Exams & Solutions, Fall 2011) gives you another example of static electromagnetic systems.

       The Poynting vector is a vector-field and, as such, it has a Fourier transform. Usually, however, we don’t work with the Fourier transform of S(r, t). We find the E- and H-fields (by the Fourier method or by some other means), then cross-multiply them to get the Poynting vector distribution in space and time. This will give us a feel for how the energy is propagating through space-time.

 

Q3) Is a scalar potential tied to a specific source, or can it be a combination of sources? It seems to me the latter, as the point of this technique is to write out all the sources in terms of shifted circ/step/delta functions and let the Fourier transform do the combination for you…

P.S.: Why didn’t Feynman use Fourier analysis in the Volume II Lectures on Physics?

A3) All the Fourier operations we have discussed are linear. This means that you can find the potentials (scalar or vector) for each object separately, then add them all together (that is, add all the scalar potentials for various objects together, and/or add all the vector potentials for various objects together).
     Feynman was teaching the course to sophomores at Caltech. I guess they had not learned Fourier transforms yet, so it didn’t make sense for Feynman to spend a lot of time teaching Fourier transform theory before getting to electrodynamics. Besides, most treatments of electrodynamics stick to the space-time domain anyway, so the students wouldn’t miss much by learning it the standard way. We are also going to learn the standard way of doing things (i.e., the space-time method) in Chapter 5. The Fourier method is conceptually simple and technically powerful in its own right, of course, but one of the things we are going to do with it is to derive the space-time formulas, which will then be used in Chapter 5 to solve the same problems that we are now solving in the Fourier domain in Chapter 4.

Q4) In chapter 4 going from equation (17) to (18) on the second to the last step, I don’t understand how you change variables from 2sin(kr)/(kr) * dk to sin(x)/x * dx. I suppose you got rid of the 2 by recognizing that the sinc function is even and so you increase the limits of integration to –inf to inf. But is kr = x something fundamental that I’m missing? Also, the r in the denominator is pulled out of the integral so really it appears that the lines go from: sin(kr)/k * dk = sin(x)/x *dx. Maybe it works without the change of variable and so you just have the integral of sin(kr)/(kr) *dk from –inf to inf = pi/r ? Is that valid? Can you please clarify this step? (Matthias Whitney)

A4) Matthias: The variable was changed from k to x = kr. That is why dk was replaced by (1/r)dx. Also, as you noted, the function sin(x)/x is even; therefore, the factor of 2 is accounted for by extending the range of the integral from (0, inf) to (–inf, inf).

 

Monday, October 17, 2012

 

1. Q) I was going over my notes and the book on chapter 4 and have a question. How do the magnetic charge/current densities play into the scalar and vector potentials? Almost everything I’ve read thus far deals with the electric charge/current densities, which we used to derive the relations for the scalar and vector potentials. Problem 2 in chapter 4 relates the longitudinal components of the E and H fields to the electric and magnetic charge densities and the transverse components of the E and H fields to the electric and magnetic current densities, but doesn’t explicitly use the scalar and vector potentials.
   In problem 13, the total electric current density has a non-zero value while the total magnetic current density is zero due to the lack of a time dependence in M and because P does not exist. Thus, the vector potential with the electric current density case has a non zero value while the magnetic current density case forces the vector potential to zero. This is what makes me wonder about how the magnetic charge/current densities play into the scalar and vector potentials.

 

1. A) Excellent question; you’ll get a point for asking it. As you may recall from the discussions in Chap. 2, it is possible to write the contributions to Maxwell’s equations of P and M either in terms of bound electric charge/current densities, or in terms of bound magnetic charge/current densities. What we did in Chapter 4 was to write everything in terms of electrical charge/current densities, then proceeded to define the scalar and vector potentials accordingly. This is nice, because rho_free and J_free are already electrical (i.e., they do not have a magnetic equivalent). Consequently, all the charges are combined as a single charge source (rho_total), and all the currents are combined as a single current source (J_total). We then defined the potentials A and psi, and imposed the Lorenz gauge constraint so that rho_total gave rise to psi, while J_total gave rise to A.
   Now, suppose the system did not contain rho_free and J In that case we could as well define the total charge and current densities in terms of the magnetic entities (i.e., rho_total = -div.M and J_total = dM/dt – (1/epsilon_0)*curl x P ). Maxwell’s equations will then relate D and H to this rho_total and J_total, and they will resemble the equations used in Chap.4 as we began to define the potentials, but now it is the divergence of D that is zero, so the new vector potential (call it A’ ) must be defined such that curl x A’ = –D (the minus sign here will simplify things, although, strictly speaking, it is not necessary). Similarly, the new scalar potential (call it psi’) must be defined such that H = -grad(psi’) – dA’/dt. Substituting these expressions into Maxwell’s 3rd and 4th equations yields:
   
   psi’(k,w) = rho_total / {mu_0*[k^2 – (w /c)^2]},
   
   A’(k,w) = epsilon_0 * J_total / [k^2 – (w /c)^2],
   
   k.A’(k,w) – (w /c^2)*psi’(k,w) = 0        (Lorenz gauge).
   
   One could then proceed to solve for A’ and psi’, from which the solutions for D and H will follow. As I’ve said before, the textbooks usually do NOT use magnetic charge/current and the corresponding potentials, but, in principle, there is nothing wrong if one tried to solve Maxwell’s equations in this way. Needless to say, the final answers for the E, D, H, B fields will always turn out to be the same, no matter which potentials are used.

Thursday, October 13, 2012

Q1) Here’s a softball: what are the units of the scalar and vector potentials as we have defined? I’m getting psi in Volts and A in (Volt-seconds/meter). Are these definitions of scalar/vector potential affected by gauge, or is this straightforward as it seems?

A1) Your units are correct, and it is as straightforward as it seems. Since E = – grad(psi) – dA/dt, and since the units of E are volt/meter, then psi must be in volts (because grad is differentiation with respect to x, y, z), and A must be volt.sec/meter, because after differentiation with respect to time, the “sec” would disappear from the numerator. Of course you could use other relations, for instance, B = curl(A), would give A the units of weber/meter, but it is easy to show that weber/meter is the same as volt.sec/meter. This is because weber = henry.ampere = henry.coulomb/sec = henry.farad.volt/sec = sec^2.volt/sec = volt.sec. The change of gauge does not affect the units of the potentials, because the above relations between the potentials and the fields must remain valid in any gauge.

 

Q2) After watching Tuesday’s lecture and starting on the homework, it got me thinking about the Lorenz and Coulomb gauge transformations. How are the gauges affected from the point of view of relativity? Especially with regards to the Coulomb gauge, the equations look like if the charge density changes at one location, then the potential would respond immediately as measured from a distance much further away. Would this violate the rule that nothing can travel faster than the speed of light and if so, how can it be compensated for in the transformation?

 

A2) There was a similar question the other day; please see my answer posted to the website (under Questions & Answers) to the question dated Wednesday, October 12. The potentials do NOT have to satisfy the rules of causality if they are not “real” fields. In fact, most people treat A(r, t) and psi(r, t) as mathematical “scaffolding” to get to the actual fields (E and B), so the fact that causality is violated in the case of A and psi is not important — because they are not considered physical entities. If, however, you believe that the potentials are real physical entities (as, for instance, Feynman does), then you should stick to the Lorenz gauge, which ensures that, in response to changes in the source(s), the potentials change only after a delay associated with the finite speed of propagation of electromagnetic fields in vacuum (c).

 

Wednesday, October 12, 2012

Q) I just watched lecture 15 in which you went over the Lorenz gauge, where we define (curl of A) equal to B and then stipulate that the longitudinal component of A is equal to (w/c^2) times Psi. In chapter 4 problem 3 we are asked to use the Coulomb gauge for which we define (div of A) to be zero. How many other gauges are used in E&M with respect to the vector potential and in what cases would you be better off using a particular gauge versus another? The Lorenz gauge seems to make things really nice, under what conditions would one want to deviate from using the Lorenz gauge?

A) Excellent question; you get a point for asking it. There exists an infinite number of possible gauges; see Problem 31 at the end of Chapter 4 for a general discussion of various gauges. The Lorenz gauge has nice properties under the Lorentz transformation (i.e., when taking the potentials from one inertial frame to another using the relativistic coordinate transformations associated with the name of Hendrik Lorentz). You will also see in Chapter 5 that, in the Lorenz gauge, the relation between the potentials and the sources is causal, meaning that the potentials can change in response to a change of the sources only after a proper delay associated with the finite speed of light. This would be a necessary condition for the “physical reality” of the potentials, and there are people (e.g., Feynman) who argue that these potentials are in fact real physical fields. If that’s the case, then the Lorenz gauge would be the most natural gauge in which to define the potentials. However, another school of thought considers the potentials as merely convenient mathematical tools that help in the derivation of the actual (physical) fields, e.g., E and B. In that case, any gauge will be acceptable so long as the physical fields (E and B) remain unchanged, i.e., gauge-independent.
       Some of the gauges invented over the years (for mathematical convenience in solving specific problems) are associated with the names of their inventors, for example, Poincare, Goeppert-Mayer, Fock-Schwinger, etc. In this course, I’ll use the Lorenz gauge for the most part, mentioning the Coulomb gauge only occasionally, usually in homework problems, so that you become familiar at least with its existence. Later, when you take a course in quantum optics, you’ll see that they use the Coulomb gauge all the time, simply because it simplifies the mathematical treatment of the Schrodinger equation in the presence of electromagnetic fields. In relativistic quantum electrodynamics, however, the preferred gauge is the Lorenz gauge, for reasons that I mentioned earlier.

Monday, October 10, 2012

Q) Does the choice of basis function relate to the Fresnel/Fraunhofer diffraction regions? For example, if have a spherical charge distribution but I’m only interested in plane-waves, Fourier transforms are best. With that same spherical distribution, if I were interested in a “near field” propagation it would seem that a more “spheroidal” basis set would be appropriate — the concern is not in the frequency-domain description, but rather the ease of inverse transform…

PS: I didn’t see any notes of yours regarding “negative points” for questions…

 

5. A) There are certain relations to near-field and far-field, but at this point it is too early to worry about such things. We are going to Fourier transform charge and current distributions, then relate them to the field distributions (E and H fields) in the Fourier domain. Finally we are going to perform an inverse Fourier transform on the latter to obtain the field distributions in the entire space (that is everywhere, far field, near field, even inside the charge and current distributions). So, please be patient until we have covered chapters 4 and 5.  … And no, there are no negative points!

 

Saturday, October 8, 2012

Q1) It just dawned on me that when we use the Bessel function to “simplify” Fourier transforms in the cylindrical coordinate system, we’re really performing a transform with a new basis set, which I believe is also the Hankel transform. If this is true, is there an equivalent spherical function?

A1) We are not changing the basis set; the basis set is still plane-waves, exp[i(k.r-wt)], which are natural for the Cartesian coordinates. Just because Bessel functions appear in the integrals it doesn’t mean that we are expanding the original function into Bessel functions. Of course every function can be expanded into cylindrical waves or spherical waves, but the basis would then be different. Here we are doing only Fourier expansions in the standard basis of plane waves, exp[i(k.r-wt)].

 

Q2) In the book and in much of your lecture time, you write the Fourier integral with dr(vector)dt at the end. It appears that dr(vector) is really dv, the differential volume element, and it is in fact a scalar. I have to admit that as I was working on the homework this caused me quite a bit of confusion before you had lectured on the topic. Durring your lectures you seemed to make clear that integrating with respect to a differential volume element like (in the spherical examples you chose) 2*pi*r^2*sin(theta)*dr*dtheta is the way to go. I believe you’re very deliberate about what you write, so I wonder if there is some general meaning about dr(vector) that I am missing. Could you tell me if I’m missing something? Thanks, Jim Brookhyser.

A2) Jim: By drdt I mean dxdydzdt. Some authors like to write this as dVdt, or d^3rdt, but my preference is drdt, with r written as boldface (or with an arrow over it) to indicate its multi-dimensionality. Of course, dr is not limited to Cartesian coordinates, and can be evaluated in other coordinate systems as well. Similarly, when we integrate in the Fourier domain, the 4D volume element is written as dkdw (with k written as boldface or with an arrow over it). I am sorry if this has caused confusion; I should have explained my notation sooner.

 

Thursday, October 6, 2012

Q) With the understanding that the FT in 3D space or 4D space-time is essentially the superposition of plane waves weighted by our density distributions — we’re really examining all the frequencies in each direction present in the 3D/4D distribution, correct? So, is a propagating wave that originates from this source of plane waves really just a “sample” from the FT set of *all* possible emitted waves? In other words, does the FT act like a probability density function and any wave emitted from our source is a realization of this pdf?

 

5. A) We are not yet talking about waves emitted from the sources (sources are the charge and current distributions that we have Fourier transformed so far in our examples). Nothing is being emitted; we are only writing the source (charge or current distribution) as a superposition of plane-waves, that is all for now. There is no “probability” distribution involved. Every frequency component has a certain (complex-valued) amplitude, and when you add all these plane-waves you end up with the function that you started with (namely, the charge and current distributions in our examples so far). Next week I’ll talk about using these Fourier transforms to determine the electromagnetic fields emitted by the sources, but that requires the solution of Maxwell’s equations, which is the subject of Chapters 4 and 5.

 

Monday, October 3, 2012

Q) I just finished this week’s homework. I noticed that I ran into some very big integrals that were quite cumbersome to solve. I wonder if I missed some simple tricks. Anyway, I am all caught up in the class. I can’t wait to see the HW solutions … to see how you solve the problems.

 

1. A) The integrals could be long, but there are some tricks that I’ll go over in the class. Also, we’ll do so many of these integrals over the next few weeks that you’ll begin to see the pattern. Look at Appendix C for all the integrals that you’ll need in this course (and won’t be able to do them in any simple way). I take these integrals from the Gradshteyn and Rhyzhik Table of Integrals and use them in the course. Many of them can be done with the complex-plane techniques that you learned last semester, but not everyone in Opti 501 knows the complex-plane tricks. Since the goal here is to learn electrodynamics not mathematics, it is okay to use the Table of Integrals, but you’ll have to figure out how to reduce 3D and 4D integrals to 1D integrals, because 1D is all that the Table gives us.

 

Saturday, October 1, 2012

Q1) I have several questions I am hoping you can help me with to get me unstuck on HW #12 and 13:

a) In the solution to HW#14, why is the derivative of Sphere(r/R) equal to Delta(r-R)?
b) What is the derivative of Rect()? Should I multiply Rect() by exp(alpha*|x|) and let alpha -> 0?
c) What is the derivative of Circ()? I don’t really have a clue on this.

A1) The derivative of Rect is the easiest. Once you understand this, the other ones will be easy to grasp. Think of Rect(x) as the superposition of two step functions. What is the derivative of a step-function? Well, it is just delta(x), as I discussed in the class yesterday. Now, Rect(x) has two steps, one going up (from zero to 1) as you increase the value of x, the other going down (from 1 to zero). So, the first jump should give you a +delta function (upon differentiation), while the second should give a -delta function. But the jumps are not located at x =0, they are at x = 1/2 and -1/2. Therefore, the delta functions are correspondingly displaced, becoming, +delta(x+1/2) and -delta(x-1/2).

Q2) In the calculations involving Total Force, Momentum, Torque, and Angular Momentum, you define the integral as “for all space” and then use a pulse of light as an example. So, “all space” then just constitutes the volume occupied by this pulse. If dealing with a continuous wave laser or some constant source, do we still “chop” the EM wave in space-time and then merely integrate in time?

A2) No, you don’t chop off anything. I integrated over the volume of the pulse because that’s the only place where electromagnetic waves were present. In general, you have some source of light (say, a laser), some initial time at which the laser is turned on, and some volume of space which contains all the material media with which the laser beam comes into contact. Nothing should be assumed to come in from “infinity,” then chopped off. All the sources are in some finite location in space and start at some finite time. The rule is to make sure that the system is closed, i.e., nothing comes in from infinity and nothing disappears into infinity.

 

Friday, September 30, 2012

Q) a) In Chap 3, Example 4, how did you know r and k were separated by the spherical “coordinate” angle theta?
b) Similarly, in Chap 3, Example 5, how did you know r// and k// were separated by the cylindrical “coordinate” angle phi?

A) You are a bit ahead of the class; I haven’t covered this material yet. In any case, here is the answer to your questions:

a) Theta is one of my integration parameters. I pick an angle theta, use symmetry of the situation to write an expression for the desired Fourier kernel, then integrate over theta.

b) Same answer; phi is a parameter which, thanks to the symmetry of the situation, makes it easy to evaluate the integral. The integral is then taken over all possible values of phi.

Monday, September 26, 2012

Q1) Here’s a thought experiment I can’t resolve concerning the Poynting vector:  I can place two permanent magnets near each other in a vacuum to create an approximately constant H-field between them. Then I charge two plates with static electricity to create an approximately constant electric field between the plates and arrange them such that the E-field is perpendicular to the H-field. Something similar is done in a mass spectrometer if I’m not mistaken. It seems to me that I don’t need to have wires attached to the plates once they’re charged, and the static charges will stay put once in place. So I’m imagining a perfectly static situation where an E-field is perpendicular to an H-field, but the Poynting vector is clearly not zero, and so energy should be flowing through my volume. For the continuity of energy equation E*J is zero (no J), and everything is static so dD/dt and dB/dt are zero. So this means that div*S is also zero. OK, I agree with that; I don’t see any reason why my Poynting field must have divergence. But still, the presence of a Poynting field means energy is flowing through my volume. Where is it coming from and where is it going to? Why can’t I observe it?

 

A1) Great question; you’ll get a point for asking it. Yes, the divergence of the Poynting vector is going to be zero everywhere, because no electromagnetic energy is being accumulated, or given to material media, or taken away from material media. But the Poynting vector itself is not zero. You must consider not just the region between the capacitor plates and between the poles of the magnets, but also the regions outside; in particular pay special attention to the edges of the capacitor plates and the magnets. You’ll find that the Poynting vector is circulating in closed loops. In other words, the energy is circling around the plates and around the poles. The energy density at each point in space is constant, of course, but that energy density is circulating in a way that keeps the divergence of the Poynting vector equal to zero everywhere.
Now, the net momentum of such a field is zero, but there is a net angular momentum associated with the fields. In the process of setting up the system, e.g., moving the capacitor plates from infinitely far way to their final position, the Lorentz force acting on the charged plates exerts a torque on them; the integrated torque (with respect to time) is the mechanical angular momentum picked up by the plates and, in order to conserve the overall angular momentum of the system, the electromagnetic field must pick up an equal (albeit opposite in direction) angular momentum. This is the origin of the electromagnetic angular momentum present in the final configuration of the system that you described. Please take a look at Problem 19, Chapter 2 (and its posted solution) for a similar system for which it is easy to compute the angular momentum.

 

Saturday, September 24, 2012

Q1) I n the boundary condition example of an off-normal-axis incident plane wave reflecting off an aluminum mirror, you mentioned that the surface current oscillates at the same frequency as the incident beam. Can a surface current be applied to change (or filter) the frequency content of the reflected beam?

 

A1) We are not applying any driving force to the aluminum surface. The incident beam is driving the electrons and, naturally, it drives them at its own frequency. This is why the reflected beam has the same frequency as the incident beam. If the incident beam is strong enough to drive the mirror into the nonlinear regime, then other frequencies can be created as well. But as long as we deal with linear systems, the reflected beam will have the same frequency as the incident beam.

 

Q2) I have been working on some of the previous midterms on the website and I have a quick question regarding problem 1 on the midterm from Spring 2010. In part b you ask us to transform Maxwell’s equations into the Fourier domain. My main question is why are the Fourier Transforms of current and charge densities relatively unchanged (ie. no i, k, or w terms multiplying the densities). Is this just because these densities already factor in the volume of your material and thus would be a constant in the Fourier integral?
        Also, if we are given a problem like this on the midterm, would you want us to show the work for the Fourier Transform of each component of each term of each equation or would it be sufficient to show the work for the x component for a divergence and curl operator and it be understood that all the divergence and curl operators are computed in that fashion?

 

A2) The Fourier transform operation will be covered in Chapter 3, and we’ll begin to transform Maxwell’s equations in Chapter 4. For the first midterm, you won’t be asked any Fourier transform questions. (When I taught the course in Taiwan in the Spring of 2010, their lecture periods were much longer than in the U.S., so I covered a lot more material during that semester. Also, I only gave them one midterm, which meant that, by the time the midterm came along, a lot of material had been covered. Your class is going to have two midterms.)

 

Friday, September 23, 2012

Q1) Quick question — I recall from power engineering that complex power is expressed in volt-amp-reactive where the real part is the average power. Does taking the complex Poynting vector give us this same quantity and could this then be used for measuring transmission line losses?

A1) This is a good question; you’ll get a point for asking it. We will define a complex Poynting vector in Chapter 7, when we discuss monochromatic (i.e., single frequency) waves. The complex Poynting vector is then defined as (1/2)E x H*, where E and H are the complex E- and H-field amplitudes — their time-dependence being exp(-i*w*t). Here H* is the complex-conjugate of the H-field amplitude. We will see in Chap.7 that the real part of this complex Poynting vector represents the time-averaged (i.e., averaged over one period of oscillations) rate of flow of energy per unit area per unit time, just as our conventional definition, Re(E) x Re(H), would represent this same entity if we proceeded to take the average of Re(E) x Re(H) over one period of these monochromatic oscillations.
      Personally, I have never used the imaginary part of (1/2)E x H* to represent anything, but I suspect that it will behave very much like the reactive power does when you use the imaginary part of V(w) I(w)* in electric circuit theory. I can see, for instance, how in the case of a plane-wave propagating in a uniform, isotropic, homogeneous, linear medium described in terms of its dielectric function epsilon(w) and permeability mu(w), the imaginary part of the complex Poynting vector could be useful; it will have the same meaning as the reactive power has in electrical circuit theory, with inductive and capacitive energy being exchanged between the electric and magnetic dipoles of the medium. Whether the usage can be made more general and applied to situations involving diffraction and dispersion, for instance, or when nonlinearity and anisotropy are present in the medium, is something that I can’t comment on, simply because I have not looked into these matters in the past.

     You may want to check J. D. Jackson’s “Classical Electrodynamics” Chap. 6, Sec. 10, where he briefly discusses the complex Poynting vector. (I only have Jackson’s 2nd edition here in my office; the section number may be different in the 3rd edition of the book.) There is also the book by Fano, Chu and Adler, where they discuss the application of electromagnetic theory to circuits and transmission lines, which you might find helpful.

Q2) Is the ebook available now? I saw the problems in new homework assignment are in chapter 3, is it possible for you to post chapter 3? If not, can you just post the problems?

A2) The book is available from Bentham e-books: http://www.benthamscience.com/ebooks/9781608052530/. Have you tried to order it? As I have said before, I am not allowed to distribute the book, because I have transferred the copyright to Bentham e-Publishers. Only if you (or some other students) have placed the order and not received the book yet can we discuss the possibility of posting the chapter. Please let me know the status of your book order.

Q3) You mentioned the magnetization state of magnetic recording devices are generally “hard” magnetic materials, which means they require a large H field to be applied to change magnetic state (coercivity). Conversely, “soft” magnetic materials only require a small H field applied, and are used as a sort of temporary magnet (like shielding). That said, is this coercivity parameter important in “memristors” design, especially those that use the spin of electrons for a dynamic RAM replacement technology? Inherently, you’d want this to be less permanent (and lower power) so lean to the soft materials…

A3) Hard magnets have a “hysteresis loop,” which means that they (more or less) stay magnetized in one direction until the externally applied magnetic field reaches a certain threshold (known as coercivity), at which point the material switches to the opposite direction of magnetization. For some materials the coercivity is very large; in today’s hard disk magnetic drives, the coercivity is a few thousand oersteds. There are other materials, however, which have a very small coercivity, as small as a few oersteds. In contrast, soft magnetic materials do NOT have a hysteresis loop. The relation between their magnetization and the applied magnetic field is linear, meaning that if the field is zero, then the magnetization is zero, but if you applied a magnetic field to them then they would develop a magnetization parallel in direction and proportional in strength to the field.
     Now, the “memristor” that you mentioned — if I recall correctly, it’s an invention of Stan Williams of the Hewlett-Packard company — is non-magnetic. The electrical resistance of this device shows a hysteretic behavior in response to an applied voltage, which involves an internal motion of crystalline domain boundaries, but is not related to magnetism at all. What you’re thinking about is perhaps MRAM (for magnetic random access memory). These miniature storage elements indeed have a magnetic hysteresis loop; an electric current is used to reverse the direction of their magnetization. MRAM is a form of non-volatile memory — meaning that it will retain its state of magnetization (up or down) after the power is turned off. I don’t know the exact values of coercivity exhibited by MRAM devices, but I suspect that it won’t be too high, given the desirability of low-power switching as you’ve pointed out.

 

Thursday, September 22, 2012

Q1) Based on the EM linear and angular momentum equation, they don’t depend on the propagation velocity. Does that mean static EM also have momentum?

A1) Good question; you’ll get a point for it. Yes, indeed; static fields can have momentum and angular momentum as well. Look, for example, at Problem 19 at the end of Chapter 2.

Q2) I just watched the OPTI 501 Lecture 10 and have a quick question. Does a beam of light incident to glass at an angle introduce angular momentum to the glass or just linear momentum, or both? It seems like the refraction would yield angular momentum.

A2) Excellent question; you get a point for this. Indeed, when light is incident at an oblique angle onto a glass slab, it will transfer both linear and angular momenta to the slab. If the incident plane-wave is p-polarized and arrives at Brewster’s angle, there will be no reflection (at either entrance or exit facets); the beam then emerges with the same linear momentum as it had upon entrance, but it will be laterally displaced. The result is that angular momentum is transferred to the slab in this case, but no linear momentum.

      If you’re interested in this subject, please take a look at Chapter 10, Sec. 7 of the book. You may also want to take a look at Sec.7 of my 2004 paper:  Optics Express, Vol. 12, pp 5375-5401 (Nov. 2004).

Wednesday, September 21, 2012

Q1) I am working on HW 3 problem 32. Part a, I got H = I/(2*pi*r), and using the argument in the lecture for the capictor, I have I = 2pi*r*dD/dt to find E. I am not sure that that is right or not, but I can’t see how do start Part b and where v = 1/sqrt(mil*epsilon) come to play. Can you provide some hint?

A1)    Imagine some charge density, uniformly distributed on the outer surface of the inner cylinder. Use Gauss’s law to determine the E-field as a function of the radial distance r (or rho, as stated in the problem). At this point you can integrate the radial E-field from rho = a to rho = b; this will give you the voltage difference between the two conductors. Having found simple expressions for the E- and H-field distributions, you may now try putting them in the full set of Maxwell’s equations (including all the dynamic effects, i.e., terms that contain d/dt). You will see that the simple solutions are, in fact, exact solutions of Maxwell’s equations for this particular problem.
      To find the surface charge and current densities, use the boundary conditions that we discussed in the class last week.
     As for the last part of the problem (inductance and capacitance), in case you’re not sure how to define the L and C, please take a look at problems 26 and 27, as well as their posted solutions.

Monday, September 19, 2012

Q1) I have a question regarding lecture 8. The examples you went through for the boundary conditions regarding reflected light from an aluminum mirror assumed that the substrate had no internal E-field. I understand this in the sense that w/o an internal E-field the conduction electrons on the material surface can support oscillations from the incident light. What happens if there is an internal E-field in the material? Will the substrate only reflect light with high enough incident energy to overcome the internal E-field’s influence on the surface electrons? It seems like there could be some technological application for applying an E-Field internal to the mirror substrate. Possibly a tunable reflective filter, or tunable angle of incidence mirror…? What if the internal E-field was varied sinusoidally in conjucntion (or out of phase) with the incident light E-field (assuming the light is monochromatic)?

A1) What I said applies to perfectly conducting mirrors only (metals with infinite conductivity). Ordinary metals have a finite conductivity, albeit very large. Therefore, in the case of real metallic mirrors, the surface current is not truly on the surface, but confined to a thin layer, whose thickness is equal to the skin-depth of the material at the frequency of the incident light. For aluminum, this skin depth is about 20 nm for red light. For gold and silver, it is somewhat larger, but not greater than 100 nm, which is still a small fraction of the wavelength of the visible light.

       What happens is that the incident light excites these surface (or near-surface) electrons. The electrons radiate both forward (into the mirror) and backward (in the direction of the reflected beam). Within the bulk of the mirror, the forward radiated wave then “exactly cancels” the incident beam; that is why there is no field inside the mirror. That is how Maxwell’s equations work; in the case of conductors, they ensure that the interior of the metal is shielded from the fields (both electric and magnetic).

      There are, of course, other types of mirror that allow the light to penetrate deep inside. Multilayer dielectric stacks are examples of these second type of mirror. Here the materials are not conductive at all; they are described in terms of their refractive index, which is related to their polarizability — we will describe these materials in Chapter 6. Light penetrates deep into these dielectric mirrors, but in the end the entire incident beam is reflected from such dielectric stacks (even higher reflectivity than metallic mirrors can be achieved). Such mirrors can indeed be tailored to be frequency selective, or work at certain ranges of incidence angles, or both.

Q2) Regarding problem #29 of HW #3, we have a thin straight wire carrying a constant current and a nearby loop of wire. Do we assume that the magnetic B-field is uniform for the problem or that it is indeed radially dependent?

 

A2) The field is dependent on radial distance.

Friday, September 16, 2012

Q1)  In the presence of a uniform electric field, an electric dipole is initially oriented off-axis at an angle of theta. The Lorentz force pushes upon this dipole through the x-axis to -theta, at which point the dipole is torqued in the counterrotation direction for a nice little oscillating system. Given this scenario, the charges are moving and a bound-current is induced with a sinusoidal signature. Does Maxwell’s 2nd eqn imply that a magnetic field is also induced? I.e. do the equations only read left to right?

A1) Yes, the dipole oscillates under the conditions that you describe. A magnetic field is generated in accordance with Maxwell’s 2nd equation. But then this time-varing magnetic field will give rise to a time-varying electric field in accordance with Maxwell’s 3rd equation. That E-field will, in turn, modify the H-field according to Maxwell’s 2nd, and so on. In the end you will have an electromagnetic field radiating away from the oscillating dipole. The local (time-dependent) E-field thus created acts on the dipole in a way that counters the effect of your initial (static) E-field. In other words, the radiation will put an opposite torque on the dipole that will eventually stop the oscillations. The potential energy that was initially present in the system (because of the deviation of the dipole orientation from the direction of the static E-field) will be converted to radiated electromagnetic energy, and the dipole will come to rest in a position of minimum potential energy, namely, aligned with the static E-field.
      I don’t understand your comment about the Maxwell equation reading (working?) from left to right. There is no such thing as an equation reading left to right! It is only an equality between two entities. At any moment in time, the left side of the equation must be equal to the right side, and vice versa.

Follow-up on Q1) For posterity, here’s my diagram on the previous mentioned scenario:

t0
—-> +
—-> –
—->

t1
—->
—-> – +
—->

t2
—->
—-> –
—-> +

t3
—->
—-> – +
—->

repeat from t0…

      Thinking about this further, I can imagine that a current loop *is* created. If, for instance, there were no E-field and the + charge were simply rotating about the – charge, there would be a current loop and a magnetic dipole coming out of the plane. In the presence of the E-field, the same magnetic dipole is created, but is “switched” in direction at a certain frequency (the angular frequency over 2x the original angle above the plane it seems).

Did I answer my own question? If so, can I award myself points? 😉

Answer to the follow-up question) Your oscillating electric dipole is NOT a current loop. It is true that the positive charge and the negative charge are moving in opposite directions, but current is the product of charge and velocity, and, therefore, the two currents are always parallel to each other; so there is no current loop here.

      In principle, it is possible to have a current loop whose direction oscillates with time. This is called an oscillating magnetic dipole. It will radiate similarly to an oscillating electric dipole. We will see examples of these in Chapter 4.

Thursday, September 15, 2012

Q1) Is it true that a material with a surface charge density as applied in the “microscopic world” sense will always require an active current, electric field or army of ants? And, that in the “macroscopic sense”, this need not be true for a material(dielectric) with a bound surface charge density since it might have been “charged” or “energized” prior? Is there then typically a time constant associated with the bound surface charge density such as with capacitors?

A1) You can put charges, say, free electrons, on a good conductor (e.g., a piece of metal). These charges will then arrange themselves, as well as re-arrange the conduction electrons that already exist within the metal, in such a way that, after a short period of time, the “net” charge density inside the body of the metal will be zero, but there will be a charge distribution on the surface. The surface charges remain stationary as long as the external conditions do not change. The time constant associated with the redistribution of charges mentioned above is related to the conductivity of the material and also to its geometrical shape. The charges on the metallic plates of a capacitor are of the above kind.
       It is also possible to put free charges into dielectric materials. For example, a plastic comb gets electrically charged when you comb your hair. Or a glass substrate placed inside an electron microscope and bombarded with electrons gets negatively charged. These charges are not necessarily on the surface, because the material is a poor conductor of electricity, and the charges will sit where they are deposited (or perhaps move very slowly if the mobility is small but not zero). In all these examples the total (integrated) charge on the piece of material is either positive or negative, but not zero; in other words, the material has acquired a net charge.
      There is another kind of charge that is associated with polarized or polarizable materials. This is the charge associated with P(r, t). A piece of glass or plastic placed between the plates of a capacitor is not initially charged, nor are there any charges injected into it, or pulled out of it. When the capacitor is charged, the atoms or molecules of the dielectric material get polarized, in the sense that their electrons are pulled slightly to one side, leaving their positive charges on the opposite side. The net effect is that surface charges develop on the surfaces of the dielectric material. But the total (integrated) charge of the entire piece of material in this case is zero — because the atoms have not lost or gained any electrons; their charge distribution has simply shifted within the atom. There are also materials that are permanently polarized such as barium titanate (BaTiO3); these are called ferroelectric materials; their atoms are already polarized because of special arrangements within the crystal structure without the need for externally applied electric fields.
      The terms “microscopic” and “macroscopic” used in conjunction with Maxwell’s equations could be misleading. Electric and magnetic dipole moments, p and m, are every bit as microscopic as free-charge and free-current densities, rho and J. The former are associated with individual atoms and molecules, the latter with electrons and positive or negative ions, so in this sense they are both microscopic. Historically, however, people did not like to think of Polarization P(r, t) and Magnetization M(r, t) as continuous functions of space and time unless they imagined a small volume of material (say a 10 x 10 x 10 nm^3 cube) filled with atoms and molecules, then used a spatial averaging of dipole moments p and m within this small volume to define P(r, t) and M(r, t). This spatial averaging over a volume that is large compared with atomic volumes (but nevertheless small in comparison with optical wavelengths) is the origin of the word “macroscopic” in reference to Maxwell’s equations. As far as I am concerned, however, Maxwell’s equations are mathematically exact and self-consistent as they are; they do not need any such spatial averaging. One could say that the equations are mathematically exact, but that the real world obeys their predictions in an approximate way because in the real world atomic dipole moments need to be averaged over small volumes in order to produce the continuous and differentiable functions P(r, t) and M(r, t).

 

Wednesday, September 14, 2012

Q1)  Problem 7 of HW 2 got me to thinking. I was hoping your posted solution would answer my questions, but right at the beginning you said we could assume the conductor had no net charge. My question is how are electric dipoles treated when the charges are not equal and opposite? For instance two charges +q and -q a distance D apart produce the electric dipole moment q*D. But if the charges were 0 and 2q or q and 3q, the torque applied to such a structure by an E-field would be the same even though the linear force would change. Do the latter two situations exhibit any dipole moment? If so, how much?
      Also, my mind is really racing thinking what that cylindrical conductor would have done if it had some charge to it, under normal situations, that charge would be uniformly distributed around the surface, but then how would that affect the distribution of charges on the top and bottom surface in response to the moving current? Maybe that’s too in-depth a question.

A1) Dipoles “by definition” have equal and opposite charges separated by some distance d. The reason they are interesting to study is that atoms and molecules of real materials are ordinarily neutral (i.e., have zero charge); any imbalance of charge on one side of the atom will be compensated by an equal and opposite charge on the other side, hence the electric dipole. A stick with charge q on one end and charge 2q on the other is some other object; it is not a dipole. It doesn’t occur very often in nature and so people don’t devote a whole lot of time to studying it. You can make up lots of other things that are certainly possible, but not useful or interesting. For example, how about putting four charges of magnitude q, 7q, 18q and 1034q on the four-corners of a cross? That is a weird object that is possible but not very interesting.
       The cylinder could definitely be charged on its surface, but, if it is metallic, those charges will be redistributed until they reach equilibrium, i.e., until they stop moving around. Again, it is possible to solve problems like this, and in fact in electrostatics people solve Laplace’s equation with all kinds of charge distributions on various objects such as cylinders and spheres, but we are not going to address those problems in this course. My goal is to select the simplest possible systems that radiate, then solve the Maxwell equations for them and analyze the properties of the radiated fields. A neutral cylinder with an oscillating current along its axis (e.g., Problem 7, Chapter 2), or a sinusoidally oscillating electric point-dipole, or an oscillating magnetic point-dipole are examples of such systems that we will discuss during this semester.

Tuesday, September 13, 2012

Q1) Currently, there is only one midterm exam and solution posted for Fall and Spring 2010 on the OPTI 501 website. Do you plan on posting the 2nd midterms?

A1) In those two semesters I only gave one midterm exam.

Q2) I ordered the text book as soon as it became available. The publisher charged my credit card for the book nearly two weeks ago but I have not received anything yet. I tried to contact them to ask them about it but they have not responded to my e-mails. Are we going to need the next chapter of the book for the next homework? If so is there a reserve copy available somewhere that we can get to do the homework with?

A2)  I have e-mailed the editor at Bentham e-books twice in the past few days to find out what is going on; he has not responded yet. I’ll try to make sure that all the books ordered by the students are sent out as soon as possible. In the meantime, I have posted chapters 1 and 2 of the book (as well as the preface and the Table of Contents) to the course website (under textbooks), so you have all the information that you need for the first 3 homework assignments. If there are further delays by the publisher, I will post the third chapter next week. So, please be patient, while I try to figure out what is going on at the publishing house. This is my first experience with e-book publishers, and I am already regretting it!

Saturday, September 10, 2012

Q1) I have been having trouble with the third problem on the chapter 2 homework set. I understand how this system can be modeled as a delta function seeing as the only spike in the graph would be where q is at any given instance of time. To start I took the Fourier Transform of delta(x) to get exp(-i*f(t)*k) and similarly for the y and z components. But now I seem to have run into a road block on how to relate these delta functions to rho and J. I figure rho would be the sum of these transformed delta functions but I’m not sure how they relate to J. Any suggestions on where to go from here?

A1) You do not need to Fourier transform the delta-functions for this problem. You should just take the derivatives to come up with expressions for divergence of J and d(rho)/dt. Treat the delta-function as you would any other function of x, y, z and t, and use the chain rule when differentiating the delta function. If you are not familiar with the delta-function (and especially with its derivatives), you might want to look them up in Chapter 3. Otherwise, please wait until I post the solutions. (These solutions will probably make much more sense once we’ve covered the material in Chapter 3.)

Q2)  I have a question about HW2 Problem 6. Since we were not told explicitly that the charge densities are equal and opposite, I assume that we cannot make that simplification to the problem? Furthermore, is it reasonable to say that P(r,t) is zero, thus making D(r,t) = epsilon_0E(r,t)?

A2) You’re right on both questions. The charge densities are “not” equal and opposite, and P(r, t) is zero, because the space between the plates is filled with vacuum (not with any dielectric material, which would have contained P).

 

Friday, September 09, 2012

Q1) I am having a little trouble with question 1 part d of HW2, where we are taking the Fourier transform of the charge continuity equation. If you perform integration by parts on the Fourier transform (with respect to time) of the time derivative of charge density, you end up with two terms: one is the Fourier transform of charge density multiplied by iw, and the other is the charge density times E^(-iwt) evaluated at t -> positive and negative infinity.
     It would be nice if this second term went to zero; we would then have the continuity equation as “the divergence of the Fourier transform of current density equals -iw times the Fourier transform of charge density.” But I don’t see that we can assume that charge density goes to zero as time goes to positive or negative infinity, which I believe is the only way to make that problematic term disappear. Any suggestions for how to resolve this?

A1) Excellent question; you get a point for asking it. Functions that have a Fourier transform are usually those that go to zero at +/- infinity. This is not always the case, of course, as evidenced by the existence of Fourier transforms for functions such as Step(x), Sign(x), etc. However, in these exceptional cases, one usually has to “force” the function to zero at +/- infinity, then find the limiting Fourier transform as these imposed conditions (at infinity) are relaxed. You’ll learn about this technique in Chapter 3.

    In the case of the charge/current continuity equation, it is easy to see that J(r,t) and rho(r,t) can be assumed to have finite extent in the spatial (r) domain. The current density J(r,t) can also be assumed to approach zero as time (t) goes to +/- infinity. However, there is no reason, in general, to believe that rho(r,t) at each and every point r in space must have been zero at t = – infinity, or that it “must” end up being zero at t = +infinity. Nevertheless, charge distributions that remain finite as time approaches +/- infinity are not Fourier-transformable and, therefore, cannot be treated mathematically by techniques that require Fourier transformation.

    For purposes of Problem 1(d) of HW#2, you may assume that rho(r,t) goes to zero (for all values of r) as t approaches +/- infinity. A charge-neutral piece of material in the absence of excitations is a good example of something that can be assumed to have no charges and no currents at t = -infinity. It may be excited at later times, but eventually the excitations must die down and the material returned to its charge-neutral state as t goes to +infinity.

    Another way to deal with such problems is to assume that the space is filled with an initial charge distribution (which may be treated as the stationary background) plus a charge-neutral material superposed on the background. We will then keep the background stationary (i.e., time-independent) at all times and assume that all the changes occur in the charge-neutral matter, starting at some finite time, say, t_0. The static electric field of the background must, of course, be added to the dynamic solution obtained for the time-varying excitations. This dynamic solution can be computed from t = -infinity up to any desired (but finite) time, say, t_1, at which point one must force the neutral material to return to its neutral state once again. The solution thus obtained is valid in the time interval (t_0, t_1).

    We will use the Fourier transform method extensively in Chapter 4 to solve Maxwell’s equations. There you’ll see that the problems that we pick to solve by this technique are such that their various functions vanish (or can be forced to vanish) as x, y, z and t approach +/- infinity.

Wednesday, August 31, 2012

Q1) Another thing you had mentioned was that V’ and V” are independent of one another — I understand that you’re building up our model of plane waves and this is generic yet. But, if they are truly independent, then that must mean that we are dealing with unpolarized light. If we have circular, linear, or elliptical polarized light, we know a lot about our light (maybe we used a polarizer?) and they are highly correlated. So… what does the correlation between these two vectors mean? I see something on the web about DOP — degree of polarization — would that be a quality metric for a polarizing optic then?

A1) Vo’ and Vo” are independent of each other in the sense that any choice that you make for one will be independent of the choice that you make for the other. But remember that these vectors just represent the “amplitude” of the field, not its spatial and temporal variations. Once you have chosen Vo’ and Vo”, the polarization state of the beam (i.e., the plane-wave) is fixed throughout the entire space-time. That means that, if your choice resulted in linear polarization, then it will be the same polarization everywhere and at all times. Similarly, if your choice resulted in circular or elliptical polarization.
     If you send a polarized beam (and this term means linear, circular, or elliptical) through a polarizer or a half-wave plate or a Fresnel rhomb, etc., the optical element will change the relationship between Vo’ and Vo”, but once the beam comes out of the optical element, it will have a fixed, well-defined state of polarization again (albeit a different one than the incident state of polarization). Depolarized light “cannot” be constructed from a single plane-wave; you’ve got to add in a mix of different frequencies and different k-vectors to create depolarized (or partially-polarized) light. Degree of polarization is a term used in the context of partially-polarized light.

Tuesday, August 30, 2012

Q1) Last week Thursday, when explaining the real part of the plane wave equation, you made a comment, paraphrased: “if Vo’ = Vo” AND Vo’ is orthogonal to Vo”, then they are circularly polarized”. But, if Vo’ = Vo”, doesn’t that imply that they are orthogonal? Vo’ cos( w’t – k’ r’) + Vo” sin (w’t – k’ r’ ), so the cosine/sine arguments are equal at each “t.”

A1) Vo’ and Vo” are vectors. What I said was that their “lengths” are equal but their directions are perpendicular to each other. If the two “vectors” happen to be equal to each other, then not only will their lengths be equal, but also they will be parallel to each other (i.e., aligned in the same direction).

     When the two vectors are parallel to each other, the light will be linearly-polarized, irrespective of whether their lengths are equal or not.

     When Vo’ and Vo” happen to be perpendicular to each other, the light will be elliptically polarized AND the major and minor axes of the ellipse will be aligned with Vo’ and Vo”.

     Only when the lengths of the two vectors are equal AND the two vectors are perpendicular to each other do we get circularly-polarized light.

 

Monday, August 29, 2012

Q1) I am working through this week’s homework problems and I was wondering if there was a general system of equations that could calculate gradient (for example) for all curvilinear coordinate systems. It just seems like there would be a general case that would work for Cartesian, Cylindrical, and Spherical (as well as any others) if you implemented the proper scale factors for each coordinate system.

A1) No, there are no general formulae for gradient, divergence, and curl in various coordinate systems. Each one is unique and you’ll have to derive the corresponding formulae in each case. The only thing that these operations have in common — across the coordinate systems — is their definitions (e.g., for the divergence operator the definition says: integrate the vector field over a small closed surface, normalize by the volume, then let the volume approach zero). The definitions are coordinate-independent, but the mathematical formulae are coordinate-specific.
      Appendix B of my book lists the mathematical operations of gradient, divergence, and curl in various coordinate systems (Cartesian, cylindrical, spherical). Your first assignment asks you to derive these formulae. You may want to see the books by Feynman or Jackson (or any elementary EM book for that matter) to see the derivations in Cartesian coordinates. In cylindrical and spherical systems the procedures are similar to those in Cartesian, only the shape of the chosen volume (in the case of divergence) or the loop (in the case of curl) must correspond to a “natural” shape in that particular coordinate system.

Follow-up to Q1) I was doing a little more research on this when you mentioned the Appendices. After looking in my old E&M book, I found that Appendix A had something like what I was looking for for gradient, divergence, and curl. I was then able to use similar terminology to find it online (because I wasn’t sure if it was just an author specific convention) and found it at Wolfram Mathworld (Equation (8) http://mathworld.wolfram.com/Divergence.html). Looks to be just a general equation for divergence that can be derived for a general coordinate system u1, u2, u3 and then completed for any coordinate system with the proper scale factors h1, h2, h3 for that system. I’ve tried using it for Cartesian, Cylindrical, and Spherical systems and it seems to work. I just wanted to get your opinion on this.

 

Answer to follow-up) Alright, I see what you mean; I misunderstood your question at first. Eq. (8) on Wolfram’s website is fine. In fact, when you try to derive the divergence formula in cylindrical and spherical coordinates (Assignment 1), you’ll find that you are essentially “deriving” this formula on your own. My intention in this assignment is to get the students to think geometrically. Scale factors and partial differentiation are fine, but I want you to visualize a piece of a cylinder or a cut through a sphere and see for yourself the six surfaces of the resulting (small) volume, the contribution of each surface to the overall surface integral, the physical meaning behind various partial derivatives, etc.

 

Sunday, August 28, 2012

Q) Has the Z-transform be used in our studies of decaying/growing electromagnetic waves? In digital signal processing, this transform maps a discrete signal to a geometric series with the “z” variable and assigned a region-of-convergence (you’re probably more familiar with it than I, in which case these notes may just be for me). I would assume that the homework is merely showing the mathematics of convergence and that our work with fields will be mostly continuous time domain anyway.

A) We will not use Z-transforms in this course, only Fourier transforms. The first Assignment deals with basic mathematics, just to get you re-acquainted with stuff that you have already seen in your calculus and algebra courses. I look forward to having you in my class this semester.

 

Saturday, August 27, 2012

Q1) I was just wondering if all 31 homework problems were due on Sept 1st or if we are just turning in problems 22-24 and 29-31. I’m just a little worried about getting all of them done in time considering my work load this week and the fact that we haven’t gotten to some of the topics covered in the homework problems in class yet.

A1) Only problems 22-24 and 29-31 are due September 1. The others are for your own practice (solutions to the other problems are already posted to the website). Also, in the future, I won’t necessarily cover everything in the class before the HW is due; the book contains detailed explanations, and I expect the students to read the book. In the class, I only select a few topics and discuss in detail, but you are expected to learn the entire chapters 1-7 of the book.

Friday, August 26, 2012

Q1) I just watched the 2nd lecture for OPTI 501 and have a question…I really like the way you describe the physical meaning behind the mathematical features in the complex notation of the vector plane wave. When we decompose k, w, and Vo to:

k = k’ + ik”
w = w’ + iw”
Vo = Vo’ + Vo”

we get physical meaning for each individual term:

k’ –> wavelength
k” –> rate of growth/decay in space
w’ –> oscillation frequency
w” –> rate of growth/decay in time

For Vo’ and Vo” the explanation of the interaction between the two as a function of time to describe polarization (circular, elliptical, linear) is really intuitive. My question is whether or not there are any other physical distinctions between Vo’ and Vo” individually rather than as a time based interactive pair?

Maybe it’s not worth a point, but I’m interested in the answer 🙂

Hard to type vector notation in email (all k and Vo are vectors)

A1) Vo’ and Vo” are two real-valued (i.e., ordinary) vectors residing in 3D space. Their magnitudes and directions are the same everywhere in space, and remain the same at all instants in time. So, you see that there is really not much that one can say about these two vectors; pick a pair of vectors arbitrarily and call one of them Vo’, the other one Vo”, then attach this pair to each and every point in space. That’s all we’ve got to work with. Until you introduce the time- and space-dependent factor, exp[i(k.r-wt)], into the expression of a vector plane-wave nothing happens. As soon as you introduce the exponential factor, the vector Vo = Vo’ + i*Vo” becomes alive. If you fix the position and watch the vector as a function of time, you will see the ellipse of polarization. If you fix the time at t = to and watch the vector field at different points in space you’ll see a real-valued vector of varying length and orientation as you look at different points in space. There is really not much else to Vo’ and Vo” other than what happens to them in the presence of the complex-exponential factor exp[i(k.r-wt)].

 

Wednesday, August 24, 2012

Q1) You mentioned during the 1st lecture that the “complex” mathematical formalism sometimes using imaginary values is done as a means to perform the calculations, while the physical quantities involved are real valued. Are there alternative E&M mathematical formalisms in existence that use all real valued quantities in the calculations? I understand the utility of the standard formalism and think it’s great, I’m just wondering if other complete mathematical E&M formalisms exist.

A1) I have not come across any books/papers that avoid the use of complex notation. I guess you could refuse to use complex notation and, every time you see something written as a complex-valued entity, replace it with its real part. Pretty soon, however, you’ll find yourself submerged in real mess (no pun intended). My advice to you is: don’t even try; learn how to deal with complex values from the outset.

 

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