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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 $v$.

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I don't think the second equation is correct. Is it meant to be just the first equation divided by $c^2 dt^2$? If so, then the two terms on the right-hand side are both wrong. BTW, you might want to stop writing all the factors of $c$ when you do calculations in GR. They just make the equations look messy. Everybody in GR uses units with $c=1$. –  Ben Crowell Sep 29 '13 at 22:35
    
I was careless with pasting the second equation. I realise that $c=1$ is useful, but haven't been doing GR long enough to really feel at ease with it. –  Alyosha Sep 30 '13 at 18:33
    
In the revised version of the second equation, are you intending $v$ to be the transverse velocity? I don't really understand the point of this equation. What are you trying to do with it? In general, a quantity such as this $v$ is a coordinate velocity, which is of no particular interest and has no particular physical interpretation. –  Ben Crowell Sep 30 '13 at 19:02
    
I meant it to be transverse. Thank you, this was one of the problems I was having earlier. I'm trying to deduce the difference in the time signals between being sent by GPS satellites and Earthlings receiving them. –  Alyosha Sep 30 '13 at 19:44
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Ah, I see. Your calculation shows a sum of a gravitational blueshift and a kinematic redshift due to the transverse Doppler effect. GPS satellites are in circular orbits, so longitudinal Doppler shifts are reduced. However, the GPS user isn't at the center of the earth, so I think there will still be longitudinal Doppler shifts, and since longitudinal shifts go like $v$ rather than $v^2$, they are probably bigger than the effects represented by your equations. –  Ben Crowell Sep 30 '13 at 20:14

2 Answers 2

up vote 4 down vote accepted

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.

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