# What is the relativistic energy of a bounded static particle?

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, $$m_e = U$$?

• 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. – Cryo Jun 5 '19 at 15:44
• @Cryo Please post that as an answer, rather than a comment. – rob Jun 5 '19 at 16:44
• In addition an electron will not stay at rest in an electric field (electromagnetic field is light is radiation). – anna v Jun 5 '19 at 16:47

$$(mc,\mathbf p)$$ 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.