Is EM Radiation Present with Free vs Bound Electrons?

[vanhees71]

I argued that they don't, because the power radiated is zero. The power is proportional to the dipole moment (actually, its square), which is zero in this case.

One way to look at radiation is that the thing that actually produces radiation is a time-varying multipole moment. Accelerated charges radiate (in most cases) because they produce a time-varying multipole moment (in most cases). This is neither more right nor more wrong than the more conventional description, as it is more akin to changing coordinate systems, but often symmetries that are hard to see in the more conventional description are apparent when thinking in multipoles. Like this case.

If you really want to think in terms of the individual electrons, you can calculate the radiation from each and will discover that it is 180 degrees out of phase. In the radiation zone (r >> separation between the electrons) the fields (other than the overall E field from -2 units of charge) are zero.

I'm very surprised that you think that two electrons in interaction don't radiate. Of course the bremsstrahlung cross section is not 0 in QED!

Of course you are right, if you consider the dipole approximation there are the well-known dipole selection rules for one-photon emission.

 

[Vanadium 50]

@cmb, this is not the situation in the original question, nor is it classical physics, which is the section this is in.

 

[Vanadium 50]

I'm very surprised that you think that two electrons in interaction don't radiate.

I didn't say that. I said that there is no radiation in the situation described, not that two electrons can never radiate. See message #15.

...in QED!

And QED is not classical physics.

 

[vanhees71]

I think I simply don't understand your argument then. Why shouldn't two classical electrons accelerated in back-to-back motion due to their repulsion radiate?

Maybe I have to do the calculation to understand this. From a QT point of view it seems plausible though, because of the dipole selection rule. That's why I thought you argue within QT...

 

[sophiecentaur]

Why shouldn't two classical electrons accelerated in back-to-back motion due to their repulsion radiate?

This has already been discussed above. If you treat them a two classical charged particles in the absence of any other charges then you would have an elastic collision. The net EM energy that's transferred between will be zero - or they wouldn't be following Classical laws.
But electrons are not classical particles so the possibility of radiation does exist.

 

[Vanadium 50]

Why shouldn't two classical electrons accelerated in back-to-back motion due to their repulsion radiate?

Because the power radiated is proportional to the square of the change of their dipole moment, and the back to back electrons have a constant dipole moment (of zero). For this setup. Other setups would of course have different answers.

 

[vanhees71]

I have to do the calculation to see this. Both electrons are accelerated. So why don't they radiate within classical physics? Only because they are moving on a straight line? Do you know a textbook, where this calculation is done?

 

[Vanadium 50]

This exact problem? No. But the power radiated by a multipole? Lots - pretty much all of them. If you look at one, you will see a factor of p (two, actually), the dipole moment. This is always zero for this problem.

 

[vanhees71]

Of course, I know the expression of the radiation power in terms of the multipole expansion. Each multipole mode adds to the radiation power
$$P_{\ell,m}=\frac{c}{8 \pi k^2} |a_{\ell m}|^2.$$
See Jackson, Eq. (16.78), using Gaussian units, and indeed the $|a_{\ell m}|^2$ are proportional to the corresponding multipoles of the charge-current distribution.

So how do I see that all these vanish only because a charge moves (accelerated) along a straight line?

There are radiation losses in linear colliders though they are negligible even for electrons for all practical purposes. See, e.g.,

http://farside.ph.utexas.edu/teaching/em/lectures/node131.html

Though this doesn't make use of the multipole expansion but just of the Larmor formula for an accelerated point charge.

 

[jbriggs444]

Whatever the details, if there's an elastic collision, there can't be any radiation.

There cannot be any emitted radiation unless there was incident radiation. [Which, since arranging for the right radiation to be incident would be difficult, amounts to pretty much the same thing].

 

[jartsa]

OK, can I be a bit more specific then with a few of scenario questions I have?

Sorry, I can't answer any of those questions.
@Vanadium 50 What if two identical charged balls far away from each other are given simultaneous, sharp, equal pushes towards each other? Surely they would radiate? I mean pushed by a hand or something, not by an electric field.

 

[Vanadium 50]

This has been kind of annoying - responses to "the radiation in this case is zero" seem to fall into two categories: "I haven't done the calculation myself, but I just don't believe it" and "yes, but this other system has radiation!". (I wish my professors would haven fallen for this: "The answer was A. You wrote D." "Yeah, but D is the answer to another problem! I should get full credit")

@vanhees71, I am unable to get at my copy of Jackson, but somewhere in chapter 9 there is an expression for the power radiated from a dipole that looks something like dP/d\Omega \sim k^4 {\bf p}^2 \sin^2{\theta}, where p is the dipole moment vector (at maximum extent). (I am sure Jackson has all the c's and 1/12π's in the right spot, but there is no way I am going to get the leading constants right.) Since p = 0, dP/d\Omega = 0. No radiation.

One could say, "yes, but this system does not have a constant wavenumber k" which is true, but if I sum this up wavenumber by wavenumber, it's just summing up zeros.

Also, intgrating over angles to get P won't fix anything: if the power per unit angle is zero everywhere, integrating won't help.

Another way to look at this (which I suspect you will really hate) is if you have two electrons separated by a distance a (or 2a, if you prefer), then a's can only appear in radiation terms accompanied by q's. In the far field zone, the length a can be neglected because r >> a (this is the definition of the far field or radiation zone) but the dipole moment (qa) cannot, because it describes the strength of the radiation source.

Since all that matters in the radiation zone is (qa), which is how p is defined (up to a factor of 2 or 4), I can replace one of the electrons by an equal positive charge moving in the opposite direction and it will not - indeed, cannot - change what is happening in the radiation zone. But in that case, the field is obviously zero at large r.

This, by the way, is the same argument as saying that the radiation from electron 2 is 180 degrees out of phase with the radiation from electron 1. One might argue that this is not exact, which is true. It is only true to order (a/r). But in the radiation zone we neglect such terms because they are part of the definition for what it means to be near field. (They will also fall off as 1/r² and not 1/r)

I should point out that I am treating an "electron" as a classical object: an infinitesemally small charged ball. A physical electron also has spin, and a spin flip would of course radiate.

 

[vanhees71]

But we don't have a dipole here. It's much simpler to argue with the Larmor formula for accelerated point particles and there I don't see, why the particles shouldn't radiate when being accelerated by their mutual repulsive interaction forces. Of course you can solve this only approximately, i.e., full radiation reaction cannot be taken into account. If I find the time, I try to do the calculation, using the Larmor formula and see whether the fields are such that the Poynting vector vanishes for some reason at infinity (radiation part).

 

[Vanadium 50]

But we don't have a dipole here.

Exactly! It's zero.

 

[Delta2]

Exactly! It's zero.

No the dipole moment is not zero, check post #18 please. However it is not time varying as i claim it to be because after more careful consideration $r_1+r_2$ remains constant if the electrons are moving only due to mutual repulsion.

 

[hutchphd]

But we don't have a dipole here. It's much simpler to argue with the Larmor formula for accelerated point particles and there I don't see, why the particles shouldn't radiate when being accelerated by their mutual repulsive interaction forces. Of course you can solve this only approximately, i.e., full radiation reaction cannot be taken into account. If I find the time, I try to do the calculation, using the Larmor formula and see whether the fields are such that the Poynting vector vanishes for some reason at infinity (radiation part).

Exactly! It's zero.

If I remember correctly the radiation from the Larmor formula for a point charge is preferentially emitted in the "forward" direction. How can this possibly lead to exact cancellation for equal and opposite velocities? Can you provide the highlights of your calculation ? Or a reference?

 

[vanhees71]

Exactly! It's zero.

I don't know, how you are so sure that in this situation the classical electrons don't radiate. Here is a paper, proving the opposite (though it's not the interaction of two point particles but the radial motion (as well as other solutions) for a charged particle in an external Coulomb potential):

https://journals.aps.org/pr/abstract/10.1103/PhysRev.124.616

I don't claim to know the definitive answer, but I don't understand your argument with the dipole. Why should I use a multipole expansion in this case to begin with? The currents are not restricted to a finite region. So why should the dipole approximation be correct?

The solution of the retarded potential for the motion given by the accelerated charges when solving the radial Coulomb problem neglecting the radiation loss and then plugging this solution into the formula for the retarded (Lienart-Wiechert) potential is for sure tedious to find. I'm not sure, whether it's doable in closed form.

It's anyway a subtle point whether charges in unbound trajectories radiate or not. E.g., for decades people (including Pauli in his famous encyclopedia article on relativity) thought that a particle in hyperbolic motion doesn't radiate, but that's well known to be errorneous, and it was already known for a long time by a calculation by Sommerfeld, which however had been forgotten.

 

[vanhees71]

If I remember correctly the radiation from the Larmor formula for a point charge is preferentially emitted in the "forward" direction. How can this possibly lead to exact cancellation for equal and opposite velocities? Can you provide the highlights of your calculation ? Or a reference?

As I said in my previous posting, I don't want to claim anything without having done the calculation. I'm not sure whether it's doable at all. Here's an alternative argument for radiation, using the Abraham-Lorentz-Dirac equation, but not for exactly the case discussed here (see my previous posting):

https://journals.aps.org/pr/abstract/10.1103/PhysRev.124.616

 

[Vanadium 50]

No the dipole moment is not zero, check post #18 please.

I disagree with that definition, which leads to a position dependence of dipole moment (we don't wave or want one with monopole moment, usually called "charge" do we). I am used to seeing it without the r vector. But...

it is not time varying

So the power will be zero. The k⁴ comes about through time derivatives.

 

[Vanadium 50]

but not for exactly the case discussed here

I wish I had you as a professor. You'd give me credit for answering C instead of A because C was the answer to some other question, right?

 

[hutchphd]

Looking at my bookshelf I notice Reitz and Milford derive the zero radiation result (sec16-10 in 1st edition) but need to assume the distance between charges is small compared to wavelength. In that circumstance I can believe @Vanadium 50 argument . Otherwise not so much..can we see an outline?

 

[Vanadium 50]

but need to assume the distance between charges is small compared to wavelength

That is the definition of far field/radiation zone is it not?

 

[vanhees71]

Another source is Landau and Lifshitz: Of course the dipole approximation gives 0 radiation but that doesn't mean that the charges don't radiate. In such a case you have to consider the higher orders of the multipole expansion, and indeed already the quadrupole radiation is non-zero (Landau and Lifshitz, vol. 2, Paragraph 71; the problem to this section exactly discusses your case, and of course the system of two identical particles radiate).

 

[hutchphd]

I don't think so. Far field is a constraint on viewing distance...not on source vs wavelength.

 

[Vanadium 50]

@hutchphd, my copy of Reitz/Milford (and Christy) for me is not far from Jackson. But you got me thinking. I did make a mistake, but nobody caught it.

First, the power from the dipole is zero. Despite everyone arguing that it's not.

Second, there is a quadrupole moment. "Quadrupole?" you say? "How can there be a quadrupole when you only have two poles?" It comes from the piece you subtract off, the xx, yy and zz parts. One of those doesn't vanish. So this will radiate in the E2 mode, albeit weakly.

 

[vanhees71]

No, I did not argue that the dipole is zero. It is of course zero in this case (this is a one-liner). I argued against the wrong claim that this implies that there is no radiation. It's resolved in Landau&Lifshitz vol. 2 (see my above posting #53).

 

[Vanadium 50]

Far field is a constraint on viewing distance...not on object vs wavelength.

I see. I misread it. But that's a common assumption in these calculations. If you don't put that in, you have an ultraviolet catastrophe. (See the k⁴ term in the dipole) Even classically.

I see that @vanhees71 spotted the flaw in my reasoning as I was writing.

 

[cmb]

I wish I had you as a professor. You'd give me credit for answering C instead of A because C was the answer to some other question, right?

Might I please say that I am very confused now.

If I understand you, which I might not, are you saying 'this is the answer according to classic physics, and this is the question I am answering [because it is in the 'classic physics' section]', even though answering 'that' question give the wrong answer to the real world?

I don't think the OP cares if the correct place is in 'classic physics' or elsewhere. If someone assumes a thing is classic physics and the 'correct, real-world' answer is actually quantum physics, surely the thread should simply be moved?

 

[vanhees71]

I think it was just a misunderstanding of my argument or that I communicated in a not clear enough way. The error was to assume that there's no radiation if there's none within the dipole approximation, though one learns in QM1 and also in classical electrodynamics that if the dipole approx. gives no radiation you have to go to higher orders. I think many physicists have only the dipole approximation in mind and forget that it's only an approximation.