# I Mass of potential energy in GR

1. Nov 9, 2017

### zonde

There is something that has been bothering me for some time about binding energy (and respective mass difference) but I was not sure how to formulate the question. Now it feels like I can ask it meaningfully enough.

In GR energy produces gravity just like mass. But how potential energy is accounted for in GR?

To illustrate my reasoning I will use scenario and it is supposed to be weak field limit so that concepts of Newtonian gravity are meaningful enough.
Lets say we have configuration of objects that are bound by being in potential well of gravity. We add from outside some energy $E_o$. So the mass of the configuration as seen form outside has increased by the amount $m=\frac{E_o}{c^2}$. According to virial theorem twice that amount of kinetic energy of the configuration will be converted into potential energy of configuration. Now if we compare theoretical "inside" descriptions of two configurations, mass of the objects stay the same while at the same time the kinetic energy has been reduced by amount $E_o$. So to get the correct change in mass as seen from outside we have to add the difference in potential energy too.
So how this mass increase due to potential energy shows up in GR formalism?

2. Nov 9, 2017

### Staff: Mentor

In the same way it does in Newtonian gravity. For the particular case you are describing, the spacetime is stationary, so a meaningful "gravitational potential energy" can be defined, and in the weak field limit it will be equal to the Newtonian gravitational potential energy. And as long as you work in the center of mass frame of the system as a whole, the kinetic energy will be equal to the Newtonian kinetic energy. So everything works exactly the same as it does in Newtonian theory. The only difference is the mass-energy equivalence, which lets us say that the externally measured mass of the system does actually increase by $E_o / c^2$ (which we could not say in Newtonian gravity).

3. Nov 10, 2017

### zonde

I think your own insights article (Does Gravity Gravitate? Part 2) gives a better answer: The difference between $M_0$ and $M$ is sometimes referred to as “gravitational binding energy”
Currently may take on this is something like this: as going along curved spacetime in direction of more curvature is responsible for appearance of kinetic energy so the change in curvature should account for potential energy correction as the gravity goes in the opposite direction so to say. But this is rather loose reasoning so I should think it over.

4. Nov 10, 2017

### Staff: Mentor

This statement referred to a static gravitating body like the Earth. It doesn't really apply to the scenario we are discussing here; the virial theorem doesn't apply to a static object like the Earth whose internal parts are under pressure (if you work out the actual stress-energy tensor components in the integrals for $M_0$ and $M$ in that article, you will see that they contain nonzero pressure and stress), and the corresponding integrals for such a system would look quite different (because the stress-energy is in a bunch of little pieces that can be idealized as point masses, not a single continuous object).

5. Nov 10, 2017

### zonde

Virial theorem applies to "stable system" with pressure present. It seems static enough to me that the reasoning of "static object" with pressure present should apply.

6. Nov 10, 2017

### Staff: Mentor

No, it doesn't. For a simple example, consider a small piece of the Earth at the surface (or more precisely a non-rotating planet). It has zero kinetic energy (we are working in the center of mass frame of the planet as a whole, which is the frame in which the virial theorem would apply if it applied), but its gravitational potential energy is negative (since it is sitting at rest at a finite altitude, not at infinity). In order for that piece of the Earth to remain static in the position it occupies, it must have nonzero proper acceleration; that nonzero proper acceleration is supplied by the pressure from the rest of the Earth below it.

The virial theorem only applies to objects in free-fall orbits. The objects can be orbiting something like a planet or star, but in that case the virial theorem only applies to the orbiting objects, not the planet or star they are orbiting.

7. Nov 10, 2017

### pervect

Staff Emeritus