In general relativity, the potential energy is given by


Solving $V^{\prime}(r)=0$, there are two points where circular orbits can occur. given by


Now, what is the gravitational binding energy of the each orbit?

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    $\begingroup$ I think there might be an issue here Anonymous. Gravity converts potential energy into kinetic energy, and the orbits article is talking about a combination of potential energy and kinetic energy. $\endgroup$ Commented Jun 26, 2015 at 6:57

2 Answers 2


I agree with everything John Rennie said but let me just take a slightly different direction.

Note that Schwarzschild space-time has a time-like Killing field $\xi^{\mu}$. In a stationary space-time such as this, one can define the Newtonian analog of gravitational potential by $\phi \equiv \frac{1}{2}\log(-\xi_{\mu}\xi^{\mu})$. One can then easily show that the acceleration of an observer at rest with respect to the asymptotic Lorentz frame is given by $a^{\mu} = \nabla^{\mu}\phi$, just as in Newtonian mechanics (c.f. Wald exercise 6.4).

But this is for observers at rest with respect to spatial infinity. What about observers in circular orbit? They also follow orbits of a time-like Killing field; specifically, their worldlines are tangent to the Killing field $\eta^{\mu} = \xi^{\mu} + \Omega \psi^{\mu}$ where $\Omega$ is the angular velocity with respect to spatial infinity and $\psi^{\mu}$ is the axial Killing field. We can then define just as before $\varphi = \frac{1}{2}\log (-\eta_{\mu}\eta^{\mu})$ and perform the exact same calculation to show that $a^{\mu} =\nabla^{\mu}\varphi$ for these orbiting observers. Thus $\varphi$ can be interpreted as the "gravitational potential" for said observers.

What we've done is go to a coordinate system rigidly rotating with the observer, so that said observer would now be at rest, and looked at the "gravitational potential" in this corotating system since the motion of stationary observers is entirely determined by such a potential.

I have here and above put "gravitational potential" in quotes because it includes also the centrifugal potential. This is simply due to the fact that what we call gravitational and non-gravitational forces depends on the coordinate system. As John Rennie said, $\varphi$ is gauge dependent.

Explicitly, $\varphi = \frac{1}{2}\log(1 - \frac{2M}{R} - \omega^2 R^2)$. Note this also works in e.g. Kerr space-time but there we have also a gravitomagnetic vector potential.

A good reference for all this would be http://link.springer.com/article/10.1007%2FBF00757816

  • $\begingroup$ +1 I think this is a better answer than mine :-) $\endgroup$ Commented Jun 27, 2015 at 5:01

Your question doesn't have an answer because it isn't possible to split energy into a potential part and a kinetic part, or at least not in an observer independant way.

The Schwarzschild metric is time independant and spherically symmetric, and these symmetries mean there are two conserved quantities that you can think of as total energy, $E$, and angular momentum, $L$. We can use these symmetries to derive an equation for the dependance of the $r$ coordinate on proper time:

$$ \left(\frac{dr}{d\tau}\right)^2 = E^2 - \left(1 - \frac{2M}{r}\right) \left(1 + \frac{L^2}{r^2}\right) $$

and this looks like an equation of motion with an effective potential $V_{eff}$ given by:

$$ V_{eff}^2 = \left(1 - \frac{2M}{r}\right) \left(1 + \frac{L^2}{r^2}\right) $$

This is the equation you give (you've subtracted off the $1$, but that makes no difference to the working), but although it has the form of an effective potential this isn't a potential energy.

A graph of $V_{eff}^2$ against $r$ looks like:


and as you say, we get two circular orbits at the turning points (marked with arrows).

If you wished you could solve the equation for $V_{eff}$ at these points, but the result is not a potential energy in any useful sensed of the word.

You don't say what texts you're using to learn GR. If you're mainly driven by curiousity I strongly recommend A first course in general relativity by Schutz. The material above is dealt with in Section 11.1.


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