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Assume that an electron is accelerated along the +x-axis.

The electron will radiate electromagnetic energy and momentum in every direction.

But it seems to me that the EM momentum it radiates in each direction is balanced by an equal amount of EM momentum radiated in the opposite direction. Is that true?

Therefore how can we account for the radiation reaction force using Newton's 3rd law?

P.S. The standard way to derive the reaction force using classical electromagnetism is to first derive the Larmor formula for the total radiated power by integrating Poynting vectors over all directions. By assuming an oscillating motion for the charge one can then deduce the radiation reaction force that one must work against to produce this power.

This standard argument seems to me a bit of an indirect method that does not give a lot of physical understanding.

Can another classical EM argument give a more intuitive picture involving conservation of momentum or do we have to resort to quantum field theory?

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  • $\begingroup$ Classically speaking, the answer is yes, that is true. One nitpick - the electron does not radiate any energy in the direction of acceleration or the opposite direction, so not technically every direction. $\endgroup$ – Brionius Jan 13 '15 at 10:20
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But it seems to me that the EM momentum it radiates in each direction is balanced by an equal amount of EM momentum radiated in the opposite direction. Is that true?

It is true in the frame where the particle is at rest when it produces the retarded radiation. In other frames, particle moves and its angular pattern of Poynting intensity of radiation is deflected in the direction given by the velocity.

Therefore how can we account for the radiation reaction force using Newton's 3rd law?

Newton's 3rd law does not apply in electrodynamics. Thus we cannot.

P.S. The standard way ... This standard argument seems to me a bit of an indirect method that does not give a lot of physical understanding. Can another classical EM argument give a more intuitive picture involving conservation of momentum or do we have to resort to quantum field theory?

The "derivation" of the LAD term via the periodic motion is a very misconceived procedure. It derives one possible expression for the self-force that is necessary to make the model consistent with the Poynting theorem. But when this force is put into the equation of motion, the equation leads to unphysical behaviour.

This is because the LAD force is only approximate expression for the self-force on the charged sphere. There is infinity of other terms that are neglected.

The LAD force was derived with convincing basis originally by Lorentz. He calculated approximately forces of parts of charged sphere on each other and summed these forces. In contrast to non-relativistic mechanics, the EM interaction leads to modification of effective inertial mass, a force term that is proportional to $d\mathbf a/dt$ and to infinity of other terms with higher derivatives of velocity.

This derivation of self-force for charged sphere is well-based in EM theory, but is only approximate since the calculation is very hard and cannot be done exactly without explicit model of the sphere (in relativity, extended charged object cannot be rigid). So far such tractable model of charged fluid forming a compact body is lacking.

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It is not necessarily the case that

the EM momentum it radiates in each direction is balanced by an equal amount of EM momentum radiated in the opposite direction.

Say that the electron is being accelerated in the $+z$ direction, so that the external force is performing additional work, in the $+z$ direction, to overcome the radiation reaction, and therefore the electron is radiating a net momentum in the $+z$ direction into the field.

This momentum will be radiated in essentially all directions, mostly concentrated on the $x,y$ plane, and with a density going to zero on the $z$ axis in both directions. The key thing to note is that the radiated momentum density is a vector along the $+z$ direction regardless of which direction it's being radiated in. Thus the EM momentum radiated in direction $\hat{\mathbf{n}}$ and in the opposite directions are both pointing in the same direction and they do not cancel out.

(Well, mostly. Some components may cancel out, but the $z$ components will not.)

I'm not quite sure how to answer the second half of your question (and, indeed, I'm not very sure what the question actually is), except to tell you that even within classical electrodynamics, the reaction force isn't nearly as well understood as one would like. For a nice review, see Marjan Ribarič and Luka Šušteršič's Conservation laws and open questions of classical electrodynamics (World Scientific, Singapore, 1990), or their paper The basic open question of classical electrodynamics (arXiv:1005.3943).

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  • $\begingroup$ I don't understand - Surely the momentum density points radially away from the accelerating charge? $\endgroup$ – John Eastmond Jan 13 '15 at 18:42
  • $\begingroup$ That's precisely the thing - it doesn't. The radiated EM momentum density is in general a complicated vector field, but it tends to go in the direction of acceleration. Why would it need to point radially away? (Note also that 'radially away' doesn't make that much sense. Away from the current position, or the retarded position, or what would be the retarded position had there been no acceleration?) $\endgroup$ – Emilio Pisanty Jan 13 '15 at 19:38
  • $\begingroup$ "the radiated momentum density is a vector along the +z direction regardless of which direction it's being radiated in." In the frame in which the particle is at rest when it radiates, the radiated field has both $\mathbf E$ and $\mathbf B$ perpendicular to the radial direction. Since the momentum density direction is given by vector product $\mathbf E\times \mathbf B$, it points radially from the particle, not in the $z$ direction. $\endgroup$ – Ján Lalinský May 11 '15 at 20:20
  • $\begingroup$ @JánLalinský But those components will cancel out (in terms of momentum). It is the subleading corrections, which depend on the past history of the motion, which will not. $\endgroup$ – Emilio Pisanty May 11 '15 at 21:08
  • $\begingroup$ The retarded field from the point charged particle was at rest is mirror symmetric with the plane of symmetry containing the point and depends on the trajectory of the particle at that single point. I do not see why you mention past history of the motion - retarded field does not depend on the whole past, it only depends on one past point. $\endgroup$ – Ján Lalinský May 11 '15 at 22:30

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