The energy-momentum relation of a free particle is (in SI Units):

$$ m^2c^4 =- c^2 \vec{p}^2 + E^2 $$ Minimal coupling is a way to fix a gauge freedom for the choice of canonical momentum (which I can in special relativity give as $ p_{\mu} = \left( \begin{matrix} \frac{E}{c} \\ p_x \\ p_y \\ p_z \end{matrix} \right)$. One way to write the minimal coupling would be by writing the canonical momentum now as:

$$ p_{\mu} = \left( \begin{matrix} \frac{E - e \Phi}{c} \\ p_x + A_x \\ p_y + A_x \\ p_z + A_z \end{matrix} \right)$$ (Here I emply kind of a relativistic hamiltonian formalism, where the movement of a particle in space time, parameterised by $s$, is given by the hamiltonian $\lambda (m^2 + p^{\mu} p_{\mu})$, with $\lambda$ being an abitrary positive function of s, that describes how fast the path in space time is gone through).

Does this mean that I can generalize the energy-momentum relation to: $$ m^2c^4 = - c^2 (\vec{p} + \vec{A})^2 + (E-e\Phi)^2~? $$

If I solve this for $E$ for $\vec{A} = 0$, it resolves to: $$ E = e \Phi + \sqrt{m^2 c^4 + c^2 p^2} $$ which to me seems plausible, and that's why I'm asking.

Edit: the question is not only wether the modified relation holds, but also wether the reasoning behind is right.


1 Answer 1


Yes, indeed it is the expression of the energy of a relativistic massive charged particle in the presence of an electric potential. Consistently, if the expression is expanded in the limit $cp\ll mc^2$, it reduces to the known non-relativistic formula $$ E = \sqrt{m^2c^4+c^2p^2} + e\Phi \simeq mc^2+ \frac{p^2}{2m}+e\Phi $$ with the addition of the rest mass term.


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