4-velocity in Schwarzschild metric Is 4-velocity always equal to light speed in any metric? In Schwarzschild metric, the light speed with respect to a remote observer is $1 - r_{S}/r$. Should that be the 4-velocity of a falling particle? If that is correct,; should that be the time component $dt/d\tau$ of the 4-velocity for a particle at rest, since all other components go zero?
 A: Remember that the norm of the four velocity is given by:
$$ V^2 = g_{\mu\nu} v^\mu v^\nu $$
If we have a stationary object at a distance $r$ then $dr = d\theta = d\phi = 0$ so the only non-zero component of the four velocity is:
$$ v^0 = \frac{dt}{d\tau} = \frac{1}{\sqrt{1 - r_s/r}} $$
Then the norm is:
$$ V^2 = g_{00}v^0v^0 = c^2 (1 - r_s/r) \left( \frac{1}{\sqrt{1 - r_s/r}} \right)^2  = c^2 $$
So the norm of the four velocity is indeed just $c$. The calculation is going to get more fiddly for a moving object, but the norm of the four velocity is always going to come out as $c$.
A: Given a set of coordinate $q^i$ to describe a point in the 4-dimensional spacetime, $\mathbf{X}(q^i)$, you can define the 4-velocity of a point as
$\mathbf{U} := \dfrac{d\mathbf{X}}{d \tau}(q^i(\tau)) = \dfrac{\partial \mathbf{X}}{\partial q^i} \dfrac{d q^i}{d \tau} = \dfrac{d q^i}{d \tau} \mathbf{q}_i$ ,
having defined $\mathbf{q}_i = \frac{\partial \mathbf{X}}{\partial q^i}$ as the vectors of the natural basis induced by the set of coordinates $q^i$, and having dropped the explicit dependence of the functions on their independent variables for brevity.
If a metric can be locally transformed to the Minkowski metric, with a change of coordinates, $q^i(Q^k)$ and the inverse transformation $Q^k(q^i)$, using only the law for derivatives of composite functions it's possible to write,
$\mathbf{U} = \dfrac{\partial \mathbf{X}}{\partial q^i} \dfrac{d q^i}{d \tau} = \dfrac{\partial \mathbf{X}}{\partial Q^k} \underbrace{\dfrac{\partial Q^k}{\partial q^i} \dfrac{\partial q^i}{\partial Q^j}}_{=\delta^k_j}\dfrac{d Q^j}{d \tau} = \dfrac{\partial \mathbf{X}}{\partial Q^k} \dfrac{d Q^k}{d \tau} = \dfrac{d Q^k}{d \tau} \mathbf{Q}_k$,
being $\mathbf{Q}_k$ a local basis of the spacetime with Minkowski metric $\mathbf{Q}_k \cdot \mathbf{Q}_j = \eta_{jk}$. Using this set of coordinates, you can interpret a variation of them as $(dQ^0, dQ^1, dQ^2, dQ^3) =  (c dt, d\mathbf{r})$, whose derivative w.r.t. the proper time reads
$\dfrac{dQ^0}{d \tau} = \dfrac{d (ct)}{d \tau} = \gamma(\mathbf{u}) c$
$\dfrac{dQ^i}{d \tau} \mathbf{Q}_i = \dfrac{dt}{d \tau} \dfrac{dQ^i}{d t} \mathbf{Q}_i = \gamma(\mathbf{u}) \mathbf{u}$,  for $i = 1:3$,
having introduced the definition $\gamma(\mathbf{u}) = \dfrac{d t}{d \tau} = \dfrac{1}{\sqrt{1 - \mathbf{u}^2/c^2}}$, and the 3-d velocity, $\mathbf{u} = \displaystyle \sum_{i=1}^{3} \dfrac{dQ^i}{d t} \mathbf{Q}_i$. Now, we can write the 4-velocity using basis $\mathbf{Q}_{\alpha}$, as
$\mathbf{U} = \gamma(\mathbf{u}) c \, \mathbf{Q}_0 + \gamma(\mathbf{u})\mathbf{u}$,
and evaluate
$\mathbf{U} \cdot \mathbf{U} = \left[ \gamma(\mathbf{u}) c \, \mathbf{Q}_0 + \gamma(\mathbf{u})\mathbf{u} \right] \cdot \left[ \gamma(\mathbf{u}) c \, \mathbf{Q}_0 + \gamma(\mathbf{u})\mathbf{u} \right] = \gamma^2 c^2 - \gamma^2 \mathbf{u}^2 = \gamma^2 \left( c^2 - \mathbf{u}^2 \right) = c^2$.
