# Canonical or mechanical 4-momentum to measure energy?

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?

• 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. – Rudyard Aug 4 at 12:23
• 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? – Eli Aug 4 at 15:42