Geodesic equation and spatial variation of time

Question

I am trying understand the interpretation of geodesic equations. For simplicity, let us take a metric $$ds^2 = g_{00}(x)dt^2 + a(x,y,z)(dx^2 + dy^2 + dz^2).$$

I interpret the metric to be a spacetime, where time, $t$, varies with $x$. Now how does $t$ vary with $x$? We know that $dt = \frac{d\tau}{\sqrt{g_{00}}}$, where $\tau$ is proper time. Since $\tau$ is an invariant quantity, is it right to say that time $t$ varies with $x$ as $t = \frac{t_0}{\sqrt{g_{00}}}$, where $t_0$ is some constant? or should one use the geodesic equation to determine the spatial variation of $t$?

The geodesic equation would obviously depend on $a(x,y,z)$, and hence the result should be very different from giving the time variation as $t \approx \frac{t_0}{\sqrt{g_{00}}}$.

If the geodesic equation does not provide the spatial variation of $t$, what would it signify here?

Answer 1

Your metric is indipendent on time. So, there is a Killing vector $\xi_\mu=(1,0,0,0)$ and you have the conserved quantity ($c=1$) $$ \xi_\mu\frac{dx^\mu}{d\tau}=g_{00}(x)\frac{dt}{d\tau}=\frac{E}{m}. $$ Also, from the metric you get $$ 1=g_{00}(x)\left(\frac{dt}{d\tau}\right)^2+a(x,y,z)\left[\left(\frac{dx}{d\tau}\right)^2+\left(\frac{dy}{d\tau}\right)^2+\left(\frac{dz}{d\tau}\right)^2\right] $$ and using the conserved quantity it yields $$ 1=g_{00}^{-1}(x)\left(\frac{E}{m}\right)^2+a(x,y,z)\left[\left(\frac{dx}{d\tau}\right)^2+\left(\frac{dy}{d\tau}\right)^2+\left(\frac{dz}{d\tau}\right)^2\right]. $$ Therefore, you also need the geodesic equation to solve for $t=t(x,y,z)$ and $\tau=\tau(x,y,z)$ as independent variables.

You can check this through the example of the Schwarzschild metric.