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Using the Schwarzchild metric for a body circularly orbiting a nonspinning black hole (i.e. $dr=0$), the relation between $d\tau$, the time between two light pulses sent out infinitesimally close together, as measured by the object, and $dt$, the time between the pusles as measured by the observer far away from the black hole who recieves these pulses, is $$c^2 d\tau^2=\frac{c^2dt^2 }{1+\frac{r_s}{r}}-r^2d\theta^2$$

$$\left(\frac{d\tau}{dt}\right)^2=\frac{1 }{1+\frac{r_s}{r}}-\left(\frac{r \dot{\theta}}{c}\right)^2=\frac{1 }{1+\frac{r_s}{r}}-\left(\frac{v}{c}\right)^2$$ where $r$ is the reduced radius.

However, which observer measures $d\theta$, and why? This will have measurable consequences for the value of $v$.

Meow
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2 Answers2

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GR doesn't have global frames of reference. No global or local frame of reference is needed in order to define coordinates in GR. Picking a local inertial frame is one possible way of defining coordinates locally, but it isn't the only way, and it doesn't typically suffice as a way of defining coordinates globally. In GR, coordinates are just labels for events, nothing more and nothing less.

The fact that the Schwarzschild metric, expressed in Schwarzschild coordinates, has an $r^2d\theta^2$ term in it is typically taken as the definition of the Schwarzschild $r$ coordinate. This is because the $\theta$ coordinate is simple to define: on a spherical shell, we just define a great circle as $2\pi$ of angle. Alternatively, $r$ can be defined as the radius of curvature of the shell. The reason that it's not so trivial to define $r$ is that we can't necessarily measure a distance from the center -- for a black hole, the center is a point that's missing from spacetime, and inside the event horizon the $r$ coordinate is timelike rather than spacelike.

This may be helpful: http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html#Section7.2

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You are expressing the metric in the coordinates of the Schwarzschild observer, i.e. the scientist watching from an (effectively) infinite distance. So the $\theta$ coordinate is the one used by the observer at infinity, just like $t$, $r$ and $\phi$.

Having said that, I think (I don't have my books to hand, so I could be remembering this wrongly) if you transform into a shell frame the $\theta$ coordinate stays the same. I'm not sure about an orbiting frame.

John Rennie
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