What is the equation of the gravitational potential in general relativity?

Question

How is the gravitational potential replaced by the metric tensor in general relativity? $$U_G=\frac {GMm}{r}$$

What is its equation?

Answer 1

The other answers are correct, but for all reasonable situations which are nonrelativistic and slow moving, you can use

$$ g_{00} = (1- 2U(x)/c^2)$$

With all other components negligible. In this approximation, Einstein's equations reduce to Laplace's equation, and the geodesic equation reduces to Newton's law of motion for gravity.

Answer 2

Like Jerry said, carrying over concepts from Newtonian gravity to general relativity isn't that simple. If you assume that spacetime is stationary, it is possible to define something similar to the gravitational potential. A stationary spacetime is one where the metric tensor is independent of time, i.e. the gravitational field doesn't change. Technically speaking, a stationary spacetime is one that posses a timelike Killing vector field.

If velocities are low, mass is low, and distances are macroscopic, then general relativity reduces to Newtonian gravity. In this case, the Newtonian potential can be retrieved from general relativity. To see how this works, you can see section 5 of this paper:

http://www.math.uchicago.edu/~may/VIGRE/VIGRE2010/REUPapers/Tolish.pdf

In GR, there is an 'effective' potential that allows you to calculate orbits, particularly in the Schwarzschild solution. The SS is the solution to the Einstein field equations for a spherically symmetric gravitational field (it is a good approximation for the solar system), and it is also unchanging in time. For the SS, the effective potential is $$\tilde V^{2}\left(\tilde L, r\right)= {\left(1-\frac {2M}{r}\right)} {\left(1+ \frac {\tilde L^{2}} {r^{2}}\right)} $$ Where $\tilde L$ is the angular momentum.

With it, you can calculate orbits by using the derivative of the radius with respect to proper time (the time measured by someone on the orbiting body), with $${\left(\frac {dr}{d\tau}\right)}^{2}+\tilde V^{2}(\tilde L, r) = \tilde E^{2}$$ $\tilde E$ is the potential energy per rest mass.

Answer 3

This isn't a useful question to ask--General Relativity is a completely different framework from Newtonian mechanics, and often, questions can't directly map over. In particular, the potential energy required to separate two masses depends on the spins of the masses, and in a nontrivial way on the mass ratio between the masses.

However, in the special case of a spherically symmetric unspinning case and an orbiting mass of negligible mass, you do find an equation of motion much like Newton's, which does, in fact, have a form with gravitational "potential" $U=-G\frac{Mm}{r}$, with the only modifications to Newton coming from extra terms in the "kinetic potential" of the orbit.