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Premise: The speed of light is set $c = 1$.

Let's consider an electron in an external electromagnetic field. Its four-momentum will be $$p^{\mu} = (E, \bar p) = (\gamma m_e, \gamma m_e \bar v),$$ where $m_e$ is the mass of the electron and $\bar v$ is its classical speed in a given frame.
Let's then choose a frame where $\bar v = 0$; the four-momentum will become $$p^{\mu} = (m_e, 0)$$ and, according to mass-shell condition, the electron energy will be $$\sqrt{p_{\mu}p^{\mu}} = E = m.$$ However we know that classically (and in this frame we are working in non-relativistic approximation, since the particle speed is zero) the energy of the static electron would be $$E = U,$$ where $U$ is the potential energy associated to the external electromagnetic field evaluated at the point in space time where the electron is located in the zero-speed frame. My question is thus the following: is it correct to say that, in the zero-speed frame, $m_e = U$?

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    $\begingroup$ No. Potential energy is rather arbitary. I can always define potential energy to be zero at that specific point&time where your particle is. What matters is the difference in potential energy, but that difference requires sampling at two different points in space-time, which will lead to complications once you start boosting into different frames of reference. $\endgroup$ – Cryo Jun 5 at 15:44
  • $\begingroup$ @Cryo Please post that as an answer, rather than a comment. $\endgroup$ – rob Jun 5 at 16:44
  • $\begingroup$ In addition an electron will not stay at rest in an electric field (electromagnetic field is light is radiation). $\endgroup$ – anna v Jun 5 at 16:47
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No. Potential energy is rather arbitary. I can always define potential energy to be zero at that specific point&time where your particle is. What matters is the difference in potential energy, but that difference requires sampling at two different points in space-time, which will lead to complications once you start boosting into different frames of reference.

PS: As suggested by @rob, I am making my comment an answer (does it help indexing?).

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Potential energy associated with external EM field is external energy, it does not contribute to the four-vector

$$ (mc,\mathbf p) $$ in any way. It does not change rest mass of the particle $m$. Only change in internal energy of the particle (change of internal composition) can change rest mass of the particle. For example, radioactive decay.

The external field will change momentum of the particle, but not its rest mass.

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