In my post on gauges and gauge transformations in electromagnetics, I mentioned the full and complete solution for Maxwell’s equations, using the electric and magnetic (vector) potential Φ and A. Feynman frames it nicely, so I should print it and put it on the kitchen door, so I can look at it everyday. 🙂
I should print the wave equation we derived in our previous post too. Hmm… Stupid question, perhaps, but why is there no wave equation above? I mean: in the previous post, we said the wave equation was the solution for Maxwell’s equation, didn’t we? The answer is simple, of course: the wave equation is a solution for waves originating from some source and traveling through free space, so that’s a special case. Here we have everything. Those integrals ‘sweep’ all over space, and so that’s real space, which is full of moving charges and so there’s waves everywhere. So the solution above is far more general and captures it all: it’s the potential at every point in space, and at every point in time, taking into account whatever else is there, moving or not moving. In fact, it is the general solution of Maxwell’s equations.
How do we find it? Well… I could copy Feynman’s 21st Lecture but I won’t do that. The solution is based on the formula for Φ and A for a small blob of charge, and then the formulas above just integrate over all of space. That solution for a small blob of charge, i.e. a point charge really, was first deduced in 1898, by a French engineer: Alfred-Marie Liénard. However, his equations did not get much attention, apparently, because a German physicist, Emil Johann Wiechert, worked on the same thing and found the very same equations just two years later. That’s why they are referred to as the Liénard-Wiechert potentials, so they both get credit for it, even if both of them worked it out independently. These are the equations:
Now, you may wonder why I am mentioning them, and you may also wonder how we get those integrals above, i.e. our general solution for Maxwell’s equations, from them. You can find the answer to your second question in Feynman’s 21st Lecture. 🙂 As for the first question, I mention them because one can derive two other formulas for E and B from them. It’s the formulas that Feynman uses in his first Volume, when studying light:
Now you’ll probably wonder how we can get these two equations from the Liénard-Wiechert potentials. They don’t look very similar, do they? No, they don’t. Frankly, I would like to give you the same answer as above, i.e. check it in Feynman’s 21st Lecture, but the truth is that the derivation is so long and tedious that even Feynman says one needs “a lot of paper and a lot of time” for that. So… Well… I’d suggest we just use all of those formulas and not worry too much about where they come from. If we can agree on that, we’re actually sort of finished with electromagnetism. All the chapters that follow Feynman’s 21st Lecture are applications indeed, so they do not add all that much to the core of the classical theory of electromagnetism.
So why did I write this post? Well… I am not sure. I guess I just wanted to sum things up for myself, so I can print it all out and put it on the kitchen door indeed. 🙂 Oh, and now that I think of it, I should add one more formula, and that’s the formula for spherical waves (as opposed to the plane waves we discussed in my previous post). It’s a very simple formula, and entirely what you’d expect to see:
The S function is the source function, and you can see that the formula is a Coulomb-like potential, but with the retarded argument. You’ll wonder: what is ψ? Is it E or B or what? Well… You can just substitute: ψ can be anything. Indeed, Feynman gives a very general solution for any type of spherical wave here. 🙂
So… That’s it, folks. That’s all there is to it. I hope you enjoyed it. 🙂
Addendum: Feynman’s equation for electromagnetic radiation
I talked about Feynman’s formula for electromagnetic radiation before, but it’s probably good to quickly re-explain it here. Note that it talks about the electric field only, as the magnetic field is so tiny and, in any case, if we have E then we can find B. So the formula is:
The geometry of the situation is depicted below. We have some charge q that, we assume, is moving through space, and so it creates some field E at point P. The er‘ vector is the unit vector from P to Q, so it points at the charge. Well… It points to where the charge was at the time just a little while ago, i.e. at the time t – r‘/c. Why? Well… We don’t know where q is right now, because the field needs some time travel, we don’t know q right now, i.e. q at time t. It might be anywhere. Perhaps it followed some weird trajectory during the time r‘/c, like the trajectory below.
So our er‘ vector moves as the charge moves, and so it will also have velocity and, likely, some acceleration, but what we measure for its velocity and acceleration, i.e. the d(er‘)/dt and d2(er‘)/dt2 in that Feynman equation, is also the retarded velocity and the retarded acceleration. But look at the terms in the equation. The first two terms have a 1/r’2 in them, so these two effects diminish with the square of the distance. The first term is just Coulomb’s Law (note that the minus sign in front takes care of the fact that like charges repel and so the E vector will point in the other way). Well… It is and it isn’t, because of the retarded time argument, of course. And so we have the second term, which sort of compensates for that. Indeed, the d(er‘)/dt is the time rate of change of er‘ and, hence, if r‘/c = Δt, then (r‘/c)·d(er‘)/dt is a first-order approximation of Δer‘.
As Feynman puts it: “The second term is as though nature were trying to allow for the fact that the Coulomb effect is retarded, if we might put it very crudely. It suggests that we should calculate the delayed Coulomb field but add a correction to it, which is its rate of change times the time delay that we use. Nature seems to be attempting to guess what the field at the present time is going to be, by taking the rate of change and multiplying by the time that is delayed.” In short, the first two terms can be written as E = −(q/4πε0)/r‘2·[er‘ + Δer‘] and, hence, it’s a sort of modified Coulomb Law that sort of tries to guess what the electrostatic field at P should be based on (a) what it is right now, and (b) how q’s direction and velocity, as measured now, would change it.
Now, the third term has a 1/c2 factor in front but, unlike the other two terms, this effect does not fall off with distance. So the formula below fully describes electromagnetic radiation, indeed, because it’s the only important term when we get ‘far enough away’, with ‘far enough’ meaning that the parts that go as the square of the distance have fallen off so much that they’re no longer significant.
Of course, you’re smart, and so you’ll immediately note that, as r increases, that unit vector keeps wiggling but that effect will also diminish. You’re right. It does, but in a fairly complicated way. The acceleration of er‘ has two components indeed. One is the transverse or tangential piece, because the end of er‘ goes up and down, and the other is a radial piece because it stays on a sphere and so it changes direction. The radial piece is the smallest bit, and actually also varies as the inverse square of r when r is fairly large. The tangential piece, however, varies only inversely as the distance, so as 1/r. So, yes, the wigglings of er‘ look smaller and smaller, inversely as the distance, but the tangential piece is and remains significant, because it does not vary as 1/r2 but as 1/r only. That’s why you’ll usually see the law of radiation written in an even simpler way:
This law reduces the whole effect to the component of the acceleration that is perpendicular to the line of sight only. It assumes the distance is huge as compared to the distance over which the charge is moving and, therefore, that r‘ and r can be equated for all practical purposes. It also notes that the tangential piece is all that matters, and so it equates d2(er‘)/dt2 with ax/r. The whole thing is probably best illustrated as below: we have a generator driving charges up and down in G – so it’s an antenna really – and so we’ll measure a strong signal when putting the radiation detector D in position 1, but we’ll measure nothing in position 3. [The detector is, of course, another antenna, but with an amplifier for the signal.] But so here I am starting to talk about electromagnetic radiation once more, which was not what I wanted to do here, if only because Feynman does a much better job at that than I could ever do. 🙂
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:
https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 17, 2020 as a result of a DMCA takedown notice from Michael A. Gottlieb, Rudolf Pfeiffer, and The California Institute of Technology. You can learn more about the DMCA here:https://en.support.wordpress.com/copyright-and-the-dmca/
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How very unfortunate that CalTech is blocking a website that is trying to promote what they teach. A sad day for science, and Feynman would be embarassed.