If we use the Schwartzschild metric with Eddington-Finkelstein coordinates what is the equation for a radial in-going null geodesic in this new coordinate system? The metric is given as

$$\textrm{d}s^2= \left(1-\frac{2m}{r}\right)\textrm{d}T^2-\left(1+\frac{2m}{r} \right)\textrm{d}r^2-\frac{4m}{r}\textrm{d}T\textrm{d}r=0 \\ \textrm{d}\theta=\textrm{d}\phi=0$$

Is it the line element that's the equation or is it the geodesic equation that is the equation for a radial in-going null geodesic?

My guess is that I want something similar to an in-going radial null geodesics for the standard schwartzschild metric.



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  • $\begingroup$ You are looking for the radial component of the geodesic equation. The Schwarzschild solution Wikipedia is helpful for this. Also note that if you find the radial geodesic difficult to manage there is also the first integral given by $\epsilon = g_{\mu \nu } \frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}$. $\endgroup$ – Rumplestillskin Jan 13 '17 at 11:09
  • $\begingroup$ So I'm looking for the radial component $r$ or $dr$ in other words? $\endgroup$ – Turbotanten Jan 13 '17 at 11:14
  • 1
    $\begingroup$ Please see my answer below $\endgroup$ – Rumplestillskin Jan 13 '17 at 11:30

You are looking to determine the radial component of the geodesic equation given by $$\frac{d^2x^\alpha}{d\tau^2} + \Gamma^{\alpha}_{\beta \gamma}\frac{dx^\beta}{d\tau}\frac{dx^\gamma}{d\tau}=0.$$ As a side note you may find the following interesting also:

If you haven't calculated the connection coefficients (Christoffel symbols) you can look for symmetries in the spacetime. We are dealing with a spherically symmetric solution so we can work in the equatorial plane where $\theta=\pi/2$ and hence $d\theta=0$. Furthermore we can identify $t,\phi$ as cyclic coordinates (ignorable) and hence indicate conserved quantities and can be determined using Killing vectors or Lagrangian mechanics (or other ways I'm sure too). With this we can work out $dt/d\tau$ and $d\phi/d\tau$. Then use the fact that $$ \epsilon = g_{\mu\nu} \frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}$$ is a first integral and bam! you can determine $dr/d\tau$ also.

  • $\begingroup$ Thanks for the answer but I still don't get it. Is it the equation of motion for the radial component that is the equation for a radial in going null geodesic? $\endgroup$ – Turbotanten Jan 13 '17 at 14:31
  • $\begingroup$ Yes it is! How did you go with it? $\endgroup$ – Rumplestillskin Jan 13 '17 at 22:48
  • $\begingroup$ Just to be clear you're looking to determine $dr/d\tau$ $\endgroup$ – Rumplestillskin Jan 14 '17 at 0:31

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