General relativity: How is the 4-velocity and momentum defined in GR? I know that in SR, the 4-velovity $$ u^\mu = (d t/d \tau,d x/d \tau,d y/d \tau,d z/d \tau ) $$ and $$p^\mu = m u^\mu.$$ How do these generalize to GR? I imagine there are new complications, particularly by what we let $p^0$ be. And are these only defined for geodesics, or in general?
 A: Are you mentioning $p^0$ because you think of it as the energy?
If that is where your question is coming from, then perhaps this answer may be of use. Energy is something that an observer measures about an object. Let me explain. 
The observer has a 4-velocity, $u^\mu_{\text{obs}}$, and the object has a 4-momentum, $p^\mu_{\text{obj}}$. The energy that the observer measures for the object is the projection of the object's 4-momentum on the observer's 4-velocity: 
$$E = p^\text{obj}_\mu u^\mu_\text{obs}=p^\mu_\text{obj}g_{\mu\nu}u^\nu_\text{obs}$$ 
Here we see the metric tensor coming into play, irregardless if we are in a flat spacetime, or in a curved one.
If we think of ourself as the observer, then ${\bf u}=(1,0,0,0)$. The energy that we measure an object to have will be $E = g_{00}p^0_\text{obj}+g_{i0}p^i_\text{obj}$. If the time-space components of the metric vanish locally, the energy is $E = g_{00}p^0_\text{obj}$. In flat spacetime $g_{00} = 1$, and we get back to the initial idea that energy is the time component of the 4-momentum. 
Hope that helps. 
A: On a general Lorentzian manifold with a time-positive metric, $d\tau^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$. In SR $g_{\mu\nu} = \eta_{\mu\nu}$ which gives the familiar relation:
$$
d\tau^2 = dt^2 - dx_idx^i
$$
In GR the Einstein Field Equations tell you what $g_{\mu\nu}$ is so you can compute the altered line element.
Otherwise, the form of the 4-velocity and 4-momentum is the same in GR for a massive particle.
