Newtonian description of Gravitational Potential Energy

1. Apr 13, 2010

particlemania

I'm still on special relativity, and haven't yet reached general relativity, but had a question in mind...

does the Newtonian description of Gravitational Potential Energy

$$U= -G \frac{M m}{R}$$

fail for very strong gravitational fields?

or it only fails where curvature of space becomes evident?

2. Apr 13, 2010

bcrowell

Staff Emeritus
Not sure if you mean curvature of space or curvature of spacetime here. In GR, even weak fields are described using curved spacetime.

The equation U=-GMm/R isn't useful for strong fields. Actually, the whole concept of gravitational potential energy U isn't even useful for strong fields. Newtonian mechanics describes gravity as an instantaneous action at a distance, and this point of view is built into the whole idea of having such a function U(R). GR doesn't describe gravity as an instantaneous action at a distance.

3. Apr 13, 2010

particlemania

So does that mean that PE loses its meaning in General Relativity.

If that is so, there must be some equivalent, something that tells us about the energy contained in a body for keeping it static with respect to a g-field...

4. Apr 13, 2010

bapowell

Not at all. In the "weak field" limit, GR recovers the Newtonian potential (for a static source). In fact, this is an important feature of any gravitational theory -- it must make contact with non-relativistic physics in the appropriate limit. This is where the factor of 8 pi comes from in Einstein's equations -- it comes from matching the Einstein Eqs to the Poisson equation for the Newtonian potential in the weak static field limit.

In GR, the Newtonian potential is used extensively in cosmological perturbation theory (as you might expect). Just to whet your appetite:

The gravitational potential in GR is the metric tensor, $$g_{\mu \nu}$$. You can think of it as a 4x4 matrix with $$\mu$$ and $$\nu$$ indexing the entries. The 0-0 component couples to energy density. In perturbation theory, one writes:

$$g_{00} \simeq -(1 + 2\phi)$$

where I think you can guess what $$\phi$$ is...

5. Apr 13, 2010

utesfan100

If one tries to examine a potential energy in GR they would find that it goes to infinity if the density gets too high. They would also find that when the density is close to being too high in the neighborhood of our observer the potential would have to depend on the velocity of the object, with significant changes with changes in the direction of the velocity.

Any definition of a gravitational potential could be made to go to zero at any point, and this restricts the ability to describe a gravitational potential in terms of a tensor, as GR would like to do. The Landau-Lipgarbagez pseudotensor is a good start, if you want to see what the gravitational potential might look like in GR.

http://en.wikipedia.org/wiki/Stress-energy-momentum_pseudotensor

6. Apr 13, 2010

bapowell

You should read my post above.

7. Apr 13, 2010

utesfan100

I did. My learning style here is to think out loud and watch how the hammers bang out my thoughts :)

Your post was on how the GR theory becomes like the Newtonian theory. Mine was more on how the Newtonian theory breaks from what we know to happen under GR.

8. Apr 13, 2010

bapowell

Sorry, misunderstood. I thought you were suggesting that it was difficult to define a scalar potential in GR because of its tensorial nature. Apologies.

9. Apr 17, 2010

particlemania

Well as I am still on Special Relativity, I am still not very comfortable with tensors being used as potentials.

Also I couldn't guess what that phi $$\phi$$ was for....

Although I can imagine how, in strong fields, the velocity and direction of motion will have great influence on potential energy.

10. Apr 17, 2010

bapowell

Sorry. The $$\phi$$ is the Newtonian potential. It is defined perturbatively in GR. You shouldn't be too uncomfortable with tensors being potentials in GR. That is, if you are comfortable with the vector potential in ED, this is just one step further...