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.
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Related: https://physics.stackexchange.com/q/29082/2451 and links therein. – Qmechanic Sep 01 '17 at 12:02
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1You should note that the formal way of writing down the metric is coordinate independent, which means your ability to state the speed of objects moving along null geodesics does not change in different coordinate systems. – Jim Sep 01 '17 at 12:06
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Related and possibly a duplicate: GR. Einstein's 1911 Paper: On the Influence of Gravitation on the Propagation of Light – John Rennie Sep 01 '17 at 14:30
1 Answers
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 $g_{\tilde{t}i}=0$.

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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. – Khushal Sep 01 '17 at 12:19
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@Aniket Thanks; to be careful about signs we need to use $|dt |$ in denominators. – J.G. Sep 01 '17 at 14:26