The Circular Orbital Velocity Relative to a Stationary Observer in Schwarzschild Geometry

Question

Relative to a stationary observer (constantly accelerating to maintain a fixed radius) what would be the velocity measured of an object in circular orbit. I've tried using J B Hartle (9.47) and (9.48) which is essentially an energy conservation but cannot get the standard solution.

Answer 1

A sketch of the full proof is as follows:

Take the effective potential in terms of the specific angular momentum $L/m$ and differentiate it with respect to $r$. Equate this to zero to find that the minimum (or maximum) occurs when $$ \left(\frac{L}{m}\right)^2 = \frac{c^2\,r^2\,r_s}{2r - 3r_s}\, , \tag*{(1)}$$ where $r_s$ is the Schwarzschild radius. The stable circular orbit, which occurs where the potential is a minimum, will have this value of specific angular momentum.

For a shell observer at fixed $(r, \theta, \phi$, the observed orbital speed of something in a circular orbit at the same $r$ is $v_{\rm shell} = r\, d\phi/dt_{\rm shell}$. In the shell observer's local inertial frame $dt_{\rm shell}$ is just related to the proper time of the orbiting object $d\tau$ by $dt_{\rm shell} = \gamma\, d\tau$, where $\gamma$ is the usual $(1 - v_{\rm shell}^2/c^2)^{-1/2}$.

Now, knowing that $(L/m) = r^2 d\phi/d\tau$ (by definition), we can write $$ v_{\rm shell} = r\frac{d\phi}{dt_{\rm shell}} = r\frac{d\phi}{d\tau}\frac{d\tau}{dt_{\rm shell}} = \frac{L}{mr\gamma} = \frac{L}{mr}\left(1- \frac{v_{\rm shell}^2}{c^2}\right)^{1/2}\, , $$ then solve this for $v_{\rm shell}$ and substitute for $L/m$ from equation (1) to get $$ v_{\rm shell} = c\left[ \frac{r_s}{2(r-r_s)}\right]^{1/2}\ . $$

Answer 2

We can sidle up to this result very easily. For the observer at infinity the orbital velocity is exactly the same as the Newtonian result:

$$ v = \sqrt{\frac{GM}{r}} $$

For a proof of this see ProfRob's answer here. It's a surprising result, and it's due to the way the Schwarzschild $r$ coordinate is defined.

Anyhow to find the velocity measured by the observer hovering at a fixed radius $r$ just divide by the time dilation factor $\sqrt{1 - r_s/r}$.

Answer 3

The tangential circular velocity relative to a local stationary observer is $\rm v_{\leftarrow} = \sqrt{G M/(r-r_s)}$, which is as John Rennie pointed out the shapirodelayed velocity observed at infinity divided by $\surd g_{\rm tt}=\sqrt{1-\rm r_s/r}$.

In the radial direction however you also have to multiply with $\surd g_{\rm rr}$ since the Schwarzschild coordinates only have proper circumference, but radial expansion, so in that case divide by $\rm 1-r_s/r$ without the square root.

In that direction it is not the shapirodelayed but the local escape velocity $\rm v_{\uparrow}=\sqrt{2 G M/r} \ $that is the same as under Newton.

For more details on the subject see Cole Miller, ASTR 498, page 4