If I shoot a light ray radially, assuming the Schwarzschild metric, what vector would I plug in to the metric tensor to get a relationship between the $t$ and $r$ coordinates? That is what vector do I plugin to get $ds^2 =0$?
2 Answers
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Since the photon has a null vector set $\rm ds^2=0$ and solve for $\rm dr/dt$
then you get $\rm dr= \pm \ dt \ c \ (r-r_s)/r$ which you integrate to get
$\rm \pm \ c \ t = r_2-r_1-r_s \ ln \ (r_1-r_s)+r_s \ ln \ (r_2-r_s):$
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$\begingroup$ What I am confused about is, a metric tensor takes in two inputs to give a number, what vectors did we input into the schwarschild metric to get 0? $\endgroup$ Commented Oct 14, 2023 at 2:12
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For a radial path, I think you mean the row vector $dx^{\mu} \equiv(c\ dt, dr, 0, 0)$ and its transpose. So $g_{\mu \nu}dx^{\mu}dx^{\nu}=0$.