Is EM Radiation Present with Free vs Bound Electrons?

[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.