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## Homework Statement

Find the linear velocity of a particle in a circular orbit of radius [itex]R[/itex] in the Schwarzschild geometry as would be measured as by a stationary observer stationed at one point on the orbit. (It's problem 10 in chapter 9 of Hartle, if that helps)

## Homework Equations

The Scwarzschild metric, obviously,

[itex]ds^2=- \left (1-\frac{2M}{r} \right )dt^2 + \left (1-\frac{2M}{r} \right )^{-1} dr^2 +r^2(d \theta^2 + \sin ^2 \theta d \phi ^2) [/itex]

And in the book they derive the equation (for a circular orbit of radius R)

[itex]\Omega \equiv \frac{d \phi}{d t}= \frac{M}{R^3}[/itex]

## The Attempt at a Solution

OK, so in this problem, obviously [itex]\frac{dr}{d \bar{t}}=\frac{d\theta}{d \bar{t}}=0[/itex], where [itex]\bar{t}[/itex] is the time measured in the stationary observer's frame. Ok so we know the angular velocity measured from the far away observer's frame:

[itex]\Omega \equiv \frac{d \phi}{d t}= \frac{M}{R^3}[/itex]

since the stationary observer is, by definition not moving (w.r.t. to our far-away observer), we can easily find [itex]dt/d\bar{t}[/itex] from the metric. We then have [itex]\frac{d \phi}{d \bar{t}}.[/itex] My question is, after we get the velocity 3-vector (i.e. the velocity w.r.t. to the stationary orbit), what do we do and why?