1
$\begingroup$

In David J. Griffiths' Introduction to Electrodynamics, Third Edition, he says that radiation reaction is the recoil force exerted by the radiation emitted by an accelerating charge. In the next section, he says that the radiation reaction force is a net unbalanced force exerted on the charge due to each part of the charge applying force on the other parts that do not cancel out (the third law is not valid).

Aren't these two different explanations of the cause of radiation reaction force?

$\endgroup$
3
  • $\begingroup$ arxiv.org/pdf/2308.02628 $\endgroup$ Commented Aug 1 at 15:12
  • $\begingroup$ @JerroldFranklin 1) Links to the abstract are better appreciated than links to the whole article. 2) At least mention in your comment that this is your work. $\endgroup$ Commented Aug 1 at 15:38
  • $\begingroup$ I don't have a link to the abstract by itself. The arviv link includes the abstract and, incidentally, my name. I did not intend to hide anything. $\endgroup$ Commented Aug 2 at 1:16

1 Answer 1

0
$\begingroup$

In classical electrodynamics, properties of point charges have to be inferred from extended charge distributions, in the limit of zero size.

Moreover, classical electrodynamics is fully compatible with special relativity. That being the case, the "real" explanation should not involve action at a distance. Instead, each part of an extended charge distribution interacts with the field in its immediate vicinity, disturbances in the field propagate through space, and in this way, charges are able to act on other charges.

If the amount of momentum taken up by the field when it interacts with one part of an extended charge distribution is not equal and opposite to the amount of momentum taken up by the field when it interacts with some other part of the extended charge distribution (or the same part at a later time) then there will be a net transfer of momentum from the charge distribution to the field—the radiation reaction.

Since the field is the means by which charges exert forces on other charges, the aforementioned description of the phenomenon is not physically different from the one in which the forces exerted by different parts of the charge distribution on each other are calculated directly without mentioning the fields.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.