Circular orbit: Schwarzschild vs Newton

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Box 10.3; Problems 10.10, 10.11.

For a circular orbit, the Schwarzschild angular speed is obtained from the angular momentum:

\displaystyle \omega=\frac{d\phi}{d\tau} \displaystyle = \displaystyle \frac{l}{r^{2}}\ \ \ \ \ (1)
\displaystyle \displaystyle = \displaystyle \frac{1}{r^{2}}\sqrt{\frac{r^{2}GM}{r-3GM}} \ \ \ \ \ (2)

We can write this as

\displaystyle \omega^{2}=\frac{GM}{r^{2}\left(r-3GM\right)} \ \ \ \ \ (3)

Comparing this with the angular speed measured at infinity {\Omega=\frac{d\phi}{dt}} we have

\displaystyle \Omega^{2}=\frac{GM}{r^{3}} \ \ \ \ \ (4)

Thus {\omega>\Omega}.

The relation between {r} and {l} for a circular orbit is

\displaystyle r=\frac{6GM}{1\mp\sqrt{1-12\left(GM/l\right)^{2}}} \ \ \ \ \ (5)

 

In the Newtonian case, we can use Kepler’s law at all distances, so we have

\displaystyle \Omega^{2} \displaystyle = \displaystyle \frac{l^{2}}{r^{4}}\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle \frac{GM}{r^{3}}\ \ \ \ \ (7)
\displaystyle r \displaystyle = \displaystyle \frac{l^{2}}{GM} \ \ \ \ \ (8)

Note that we can get the same result by defining the Newtonian energy

\displaystyle E_{N} \displaystyle = \displaystyle \frac{1}{2}\left(\frac{dr}{dt}\right)^{2}+V_{N}\ \ \ \ \ (9)
\displaystyle V_{N} \displaystyle = \displaystyle \frac{1}{2}\frac{l^{2}}{r^{2}}-\frac{GM}{r} \ \ \ \ \ (10)

and then solving

\displaystyle \frac{dV_{N}}{dr}=-\frac{l^{2}}{r^{3}}+\frac{GM}{r^{2}}=0 \ \ \ \ \ (11)

In the Newtonian case, the potential energy {V_{N}} has only one minimum.

For large {l}, we can expand 5 to get (using the minus sign)

\displaystyle r \displaystyle \rightarrow \displaystyle \frac{6GM}{1-\left(1-\frac{1}{2}\frac{12\left(GM\right)^{2}}{l^{2}}\right)}\ \ \ \ \ (12)
\displaystyle \displaystyle = \displaystyle \frac{l^{2}}{GM} \ \ \ \ \ (13)

Thus the Schwarzschild case reduces to the Newtonian case for large {l}.

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