From the flat spacetime metric, we can see that the line element corresponding to null geodesic predicts that photon travel at speed $c$, but when we make a generalized coordinate transformation can we still define the speed of objects on a null geodesic by comparing the coefficients of $(dt)^2$ and $(dr)^2$? What I meant is that can we say anything about the speed of objects moving on null geodesic by looking at the line element in a particular coordinate system.


Speed $c$ is always equivalent to $ds^2=0$.

Suppose we work in a $+-\cdots$ convention with $c=1$, so in Minkowski spacetime $ds^2=dt^2-\sum_i dx_i^2$ with speed $\dfrac{\sqrt{\sum_i dx_i^2}}{dt}$. If we consider a general diagonal matrix, the convention obtains $ds^2=\sum_\mu g_{\mu\mu}(dx^\mu)^2$ with $g_{00}>0>g_{ii}$ and the speed is $\dfrac{\sqrt{\sum_i g_{ii}(dx^i)^2}}{dt}$. For a more general metric in which $g_{0i}=0$ but $g_{ij}$ might be non-zero even if $i\ne j$, the speed is $\dfrac{\sqrt{g_{ij}dx^idx^j}}{dt}$ (summing over repeated indices implicit). We can apply this result to a completely general metric $ds^2=g_{\mu\nu}dx^\mu dx^\nu$, provided we first use the transformation $d\tilde{t}:=dx^0+\dfrac{g_{0i}}{g_{00}}dx^i$ to obtain $g_{\tilde{t}i}=0$.

  • $\begingroup$ If you do the same calculation in Schwarzschild metric, then you find that this corresponds to speed to be different at different values of r. For r=2m, the speed is zero, for r<2m the speed is negative, for r>2m the speed is positive. $\endgroup$ – Khushal Sep 1 '17 at 12:19
  • $\begingroup$ @Aniket Thanks; to be careful about signs we need to use $|dt |$ in denominators. $\endgroup$ – J.G. Sep 1 '17 at 14:26

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