Free fall from rest at a particular radius to the central singularity

[happyparticle]

Homework Statement

Proper time from rest at a particular radius to the central singularity

Relevant Equations

Eddington–Finkelstein coordinates
$ds^2 = - (1-\frac{2M}{r}) dt^2 + 2dtdr +r^2 d\Omega^2$

I have this problem:

First of all, I would like to have an equation for the proper time.
I was using the Eddington-Finkelstein coordinates in the Schwarzschild metric : $$ds^2 = - (1-\frac{2M}{r}) dt^2 + 2dtdr +r^2 d\Omega^2$$
Knowing that the observer falls radially, thus $d\phi = d\theta = 0$.

Also, $ds^2 = -d\tau^2$ and after a little algebra we get:
$$1 = (1-\frac{2M}{r}) (\frac{dt}{d\tau})^2 - 2 \frac{dt dr}{d\tau^2}$$

I'm pretty stuck here. I found here and there that the equation for the proper time in this situation should be $\tau = \frac{\pi r_0^{3/2}}{2\sqrt{2M}}$.

I guess I need to find the value for $\frac{dt}{d\tau}$.

 

[TSny]

I was using the Eddington-Finkelstein coordinates in the Schwarzschild metric : $$ds^2 = - (1-\frac{2M}{r}) dt^2 + 2dtdr +r^2 d\Omega^2$$
Knowing that the observer falls radially, thus $d\phi = d\theta = 0$.

Also, $ds^2 = -d\tau^2$ and after a little algebra we get:
$$1 = (1-\frac{2M}{r}) (\frac{dt}{d\tau})^2 - 2 \frac{dt dr}{d\tau^2}$$

OK

I'm pretty stuck here.

I guess I need to find the value for $\frac{dt}{d\tau}$.

You can get an expression for $\frac{dt}{d\tau}$ by noting that free fall is motion along a geodesic. The equations of motion for a geodesic can be found using a variational principle. For example, see the first page of these notes. Here, they are working in Schwarzschild coordinates and arrive at the bottom of the page with an expression for $\dot t \equiv \dfrac {dt}{d \tau}$. You can apply the same method when using Eddington-Finkelstein coordinates.