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It is well known that verifying momentum and energy conservation in the presence of electromagnetic fields requires care as the fields themselves carry energy and momentum (see Griffiths chapter 8 for instance). However, I have a thought experiment where I cannot seem to 'balance out' Newton's third law as I don't see how to give the electromagnetic field any compensating momentum. Thoughts and answers welcome.

Suppose I have a uniformly charged infinite plane placed on the plane $z=0$. We can take the charge per unit area to be $\sigma$. This gives a uniform electric field $$ |E_z| = \frac{\sigma}{2} ~,$$ pointing away from the plate. Now lets consider a massless positively charged particle approaching the plate from the negative $z$ direction with instantaneous position

$$ \vec r_p = (0,0, c t) ~,$$

such that it will cross the plate at time $t=0$.

According to Jackson 3rd edition problem 11.18 the electric and magnetic fields of the massless particle exist only in the x-y plane passing through the instantaneous location of the particle and are given by

$$ \vec E = 2q \frac{ \vec r_\perp }{|r_\perp|^2} \delta(ct-z)~, $$

and

$$ \vec B = 2q \frac{ \vec v \times \vec r_\perp }{|r_\perp|^2} \delta(ct-z) ~,$$ where v is the 3-velocity of the particle so in our case $(0,0,1)$ and $\vec r_\perp$ is the location of the observation point in the $z=ct$ plane.

Thus, while there is a force on the massless particle when approaching the x-y plane, there is no force on the charged plate. Like I said before, in electromagnetism forces between charges do not have to cancel as the field can carry momentum but in this case the Poynting vector is independent of time so its not clear how to restore momentum conservation.

For completeness let me write down the Lorentz force for massless particles. It is simply $$ \frac{d p_\alpha}{d t} = q F_{\alpha \beta} \frac{d x^\beta}{dt}~, $$ where the derivative can be taken with respect to any affine parameter so I take it to be with respect to the lab frame. In our case this simply reduces to $$ \frac{d \mathcal E}{dt} = -q E ~,$$ where $\mathcal E$ is the lab frame energy of the massless particle.

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  • $\begingroup$ How, exactly, is this question different from How is energy conserved in this electromagnetic scattering?? (i.e. make it clear in the text, and link to the previous question.) $\endgroup$ – Emilio Pisanty Feb 18 '16 at 12:40
  • $\begingroup$ Its related but due to symmetry cleaner. Also for some thinking in terms of momentum conservation is easier than energy conservation. $\endgroup$ – Borun Chowdhury Feb 18 '16 at 13:12

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