I've often seen that the energy of a particle $p$ with 4-momentum $p^\mu$ measured by an observer $O$ with 4-velocity $u^\mu = dx^\mu/d\tau |_O$ is given by $$E=-p_\mu u^\mu.$$ In general, what exactly is this $p^\mu$? Is it really mechanical momentum $p^\mu = m\:d x^\mu/d\tau |_p$ or canonical momentum $p_\mu = \partial L/\partial\dot{x}^\mu$, where $L$ is a suitably defined Lagrangian?

In the example of EM, canonical momentum is $p_\mu = g_{\mu \nu} \dot{x}^\nu+qA_\mu$. For a static observer at some asymptotic infinity we have $u^\mu = (1,\vec{0})$; so do we want them to measure $\dot{x}^0$ or $\dot{x}^0 + qA_0$ as energy?

  • $\begingroup$ Looking back at my question, it seems to me that $p^\mu$ should definitely be mechanical momentum. $\dot{x}^0$ is, in a sense, energy of the particle by definition. $\endgroup$ – Rudyard Aug 4 at 12:23
  • $\begingroup$ If I calculate the equation $E=-p_\mu\,u^\mu$ with $p^\mu=m\frac{d x^\mu}{d\tau}$ i get $E=-g_{\mu\nu} p^\nu\,u^\mu=-m\,c^2$ so this is not what you are look for? $\endgroup$ – Eli Aug 4 at 15:42

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