I Conservation of Energy in GR: A-B System Analysis

[Superposed_Cat]

Assume you have a two particle system, A, which has a mass and gravitational pull of g,
and B, an object with low mass,

The system starts at time 0 with the distance between A and B being 0, A being at rest and B having enough kinetic energy to move it a distance r away from A, until time t all it's kinetic energy is converted and stored as gravitational potential.

How does the curvature change between 0 and t? Does B's stress-energy tensor distort
the metric equally (create an equal gravitational pull) at 0 and t?

 

[PeroK]

This seems a very confused question.

 

[PeterDonis]

Moderator's note: Thread level changed to "I".

 

[PeterDonis]

Assume you have a two particle system, A, which has a mass and gravitational pull of g,
and B, an object with low mass,

What do you mean by "low mass"?

If you mean negligible mass, i.e., you are treating B as a test object moving in the spacetime geometry created by A's mass, then you don't have a two-particle system; you have a single object with mass and a test object moving in its spacetime geometry. This case has a known exact solution in GR, which seems to be the one you have in mind.

However, if B has non-negligible mass, so you have a true two-body system, there is no known exact solution and your questions don't make sense as you ask them.

How does the curvature change between 0 and t?

In the first case above (B is a test object), it doesn't change at all. The only curvature is due to A's mass and A does not change.

In the second case above, we don't have a known exact solution so the question cannot be answered except by a detailed numerical simulation.

Does B's stress-energy tensor distort
the metric equally (create an equal gravitational pull) at 0 and t?

In the first case above (B is a test object), B has no stress-energy tensor and does not change the metric at all.

In the second case above, your question doesn't make sense because there is no way to define "the amount of distortion in the metric" to make the question well-defined.

 

[Superposed_Cat]

What do you mean by "low mass"?

If you mean negligible mass, i.e., you are treating B as a test object moving in the spacetime geometry created by A's mass, then you don't have a two-particle system; you have a single object with mass and a test object moving in its spacetime geometry. This case has a known exact solution in GR, which seems to be the one you have in mind.

However, if B has non-negligible mass, so you have a true two-body system, there is no known exact solution and your questions don't make sense as you ask them.In the first case above (B is a test object), it doesn't change at all. The only curvature is due to A's mass and A does not change.

In the second case above, we don't have a known exact solution so the question cannot be answered except by a detailed numerical simulation.In the first case above (B is a test object), B has no stress-energy tensor and does not change the metric at all.

In the second case above, your question doesn't make sense because there is no way to define "the amount of distortion in the metric" to make the question well-defined.

You're right, it wasn't well defined, I meant the second case, I guess my more accurate question was whether everything was conserved in GR when kinetic energy was changed to gravitational potential, as with electric potential, elastic tension, and all the other ways to store it, there was a clear way to write it in the SE tensor, whereas when energy is stored gravitationally, I don't see where it goes in a recoverable way in the field equation

 

[PeterDonis]

I guess my more accurate question was whether everything was conserved in GR when kinetic energy was changed to gravitational potential

It is in a special class of spacetimes, stationary spacetimes (heuristically these are spacetimes that you can view as describing a system, like a central mass, that doesn't change with time). In a stationary spacetime there are two conserved "energy" quantities of interest:

(1) Energy at infinity: this is a constant of geodesic (free-fall) motion, and it is what is conserved when, as you describe, kinetic energy and gravitational potential energy are converted back and forth as an object follows a free-fall orbit. The name can be somewhat confusing since, for example, an object in a bound orbit is never at infinity; but you can think of it as the sum of kinetic and gravitational potential energy that works more or less the same way as it does in Newtonian physics (the only complication in GR is that you have to work in the rest frame of the central mass, since that is the proper frame for defining the kinetic and potential energy in the right way).

(2) Komar energy: Also called Komar mass, this is the GR version of the intuitive notion of obtaining the total mass of the central body by adding up the contributions of the stress-energy tensor of all the parts of the body. Note, however, that this is a different quantity from energy at infinity above; energy at infinity has nothing to do with the stress-energy tensor at all, and is almost always made use of in the vacuum region outside the central mass, since that's where objects will be in free-fall orbits--and in the vacuum region the stress-energy tensor is zero. The Komar mass is obtained by an integral over the interior of the central mass, and doesn't involve kinetic or potential energy; the key complications in GR are that the geometry of space inside the mass is not Euclidean, that the stress-energy tensor inside the mass includes more than just its energy density (for a static object the key additional contribution is from pressure), and that gravitational binding energy makes a negative contribution to the total mass. The Komar mass integral incorporates all of this.

, as with electric potential, elastic tension, and all the other ways to store it, there was a clear way to write it in the SE tensor, whereas when energy is stored gravitationally, I don't see where it goes in a recoverable way in the field equation

See the discussion of Komar mass above.