Four-velocity in General-relativistic geodesic equation

I'm writing this question since I haven't find exhaustive enough answers searching on the Net. I'm writing an application for simulating General Relativity orbits calculating the acceleration from Christoffel's symbols, which are already derived from the Schwarzschild metric in isotropic coordinates:

$$\frac{d^2x^\alpha}{d\tau^2}=-\Gamma{^\alpha}{_\mu}{_\nu} \frac{dx^\nu}{d\tau} \frac{dx^\mu}{d\tau} \hspace{7.1em} (1)$$

My problem is: how should $\frac{dx^0}{d\tau}$ be defined from arbitrary spacial components of the four-velocity?

$$u^i=\frac{dx^i}{d\tau}=\text{arbitrary quantity} \hspace{4.4em} (2)$$

In special relativity, the time component of four-velocity, in terms of the spacial components, is:

$$u^0=\frac{dx^0}{d\tau}=\frac{cdt}{d\tau}=\sqrt{c^2+|u^i|^2} \hspace{4em} (3)$$

where $|u^i|^2 = (u^1)^2+(u^2)^2+(u^3)^2$ is the magnitude squared of the spacial components of the four-velocity. This comes from the Minkowskian metric. Then I derived a similar equation from the Schwarzschild metric in isotropic coordinates:

$$u^0=\frac{cdt}{d\tau}=\sqrt{\frac{B}{A}(c^2+B^2|u^i|^2)} \hspace{4em} (4)$$

where $A=(1-\frac{GM}{2R})^2$ and $B=(1+\frac{GM}{2R})^2$. If this is correct, it's quite straightforward to find that the latter definition gives a greater value than the one from special relativity, since $B > 1$ and $B > A$. Does this mean that the four-velocity magnitude is not constant in General Relativity, as it depends also on radial distance? (I knew that the four-velocity always equals the speed of light)

• No, the magnitude of the four-velocity is always the same. What changes is the metric, which affects how magnitudes should be computed. In the case of the geodesic equation it actually doesn't matter what magnitude you choose, as the equation is invariant under scaling/shifting of the parameter. Aug 10, 2016 at 1:16

The 4-velocity $\mathbf u$ of a particle with mass is always normalised so that $g_{\mu\nu}u^\mu u^\nu=-1$ (I am using a -+++ metric signature, you want $+1$ otherwise). From this you can express the $u^0$-component in terms of the other components, up to sign. Then pick the option which has $\mathbf u$ pointing forward in time.