- #1

WannabeNewton

Science Advisor

- 5,803

- 530

## Main Question or Discussion Point

Hi guys! I have a rather brief question regarding circular free fall orbits in the Schwarzschild geometry. Consider an observer in a circular orbit in the equatorial plane at some allowed ##r = R##. The angular velocity as measured by an observer at infinity is given by ##\omega^2 = \frac{M}{R^3}## and the 4-velocity of the observer in Schwarzschild coordinates can then be written as ##u = (1 - \frac{2M}{R} - R^2\omega^2)^{-1/2}(\partial_t + \omega \partial_{\phi})##.

Consider now a static observer at ##r = R## whose worldline intersects that of the circularly orbiting observer at some event; the static observer has a Lorentz frame ##\{e_t,..,e_{\phi}\}##. We boost in the ##e_{\phi}## direction to write the 4-velocity of the orbiting observer relative to the static observer's frame: ##u = \gamma e_t + \gamma v e_{\phi}##. We know ##(1 - \frac{2M}{R} - R^2\omega^2)^{-1/2}\partial_t = \frac{(1 - \frac{2M}{R})^{1/2}}{(1 - \frac{2M}{R} - R^2\omega^2)^{1/2}}e_0 = \frac{e_0}{(1 - \frac{\omega^2R^2}{1 - \frac{2M}{R}})^{1/2}}## so ##v = (1 - \frac{2M}{R})^{-1/2}\omega R##. Now if ##T_R## is the period of the circular orbit as measured by the static observer (between events on the static observer's worldline) then it is related to the period measured at infinity by ##T_R = (1 - \frac{2M}{R})^{1/2}T## so the angular velocity as measured by the static observer is ##\omega_R = (1 - \frac{2M}{R})^{-1/2}\omega## hence ##v = \omega_R R##.

Why is the result ##v = \omega_R R##, an otherwise flat space-time relation, still holding in this curved space-time for the 3-velocity of the observer in circular orbit on the equatorial plane relative to a local static observer hovering at the same altitude as the orbit? Thanks in advance!

Consider now a static observer at ##r = R## whose worldline intersects that of the circularly orbiting observer at some event; the static observer has a Lorentz frame ##\{e_t,..,e_{\phi}\}##. We boost in the ##e_{\phi}## direction to write the 4-velocity of the orbiting observer relative to the static observer's frame: ##u = \gamma e_t + \gamma v e_{\phi}##. We know ##(1 - \frac{2M}{R} - R^2\omega^2)^{-1/2}\partial_t = \frac{(1 - \frac{2M}{R})^{1/2}}{(1 - \frac{2M}{R} - R^2\omega^2)^{1/2}}e_0 = \frac{e_0}{(1 - \frac{\omega^2R^2}{1 - \frac{2M}{R}})^{1/2}}## so ##v = (1 - \frac{2M}{R})^{-1/2}\omega R##. Now if ##T_R## is the period of the circular orbit as measured by the static observer (between events on the static observer's worldline) then it is related to the period measured at infinity by ##T_R = (1 - \frac{2M}{R})^{1/2}T## so the angular velocity as measured by the static observer is ##\omega_R = (1 - \frac{2M}{R})^{-1/2}\omega## hence ##v = \omega_R R##.

Why is the result ##v = \omega_R R##, an otherwise flat space-time relation, still holding in this curved space-time for the 3-velocity of the observer in circular orbit on the equatorial plane relative to a local static observer hovering at the same altitude as the orbit? Thanks in advance!

Last edited: