Understanding the Stress-Energy Tensor & Solar Mass in General Relativity

[empdee4]

TL;DR

In tests of GR, stress-energy tensor is set to 0 in Schwarzschild solution, then is curvature caused by the sun, or by the 0 stress-energy?

In the test of General Relativity by perihelion motion of mercury, the stress-energy tensor is set to 0 in Schwarzschild solution. Then, is the curvature caused by solar mass, or by the 0 stress-energy? Or, do we consider solar mass as the gravitating mass? Or the 0 stress-energy the gravitating mass (material)? Explanation greatly appreciated.

 

[Ibix]

By Birkhoff's theorem, the spacetime outside a spherical mass is Schwarzschild spacetime. You just have to match it to an appropriate interior solution at the surface of the Sun. So the source of curvature is the Sun's stress-energy, yes. It just turns out that (to the extent that you can model the Sun as a non-rotating sphere) details apart from the total mass don't matter outside the Sun, just as they don't in Newtonian gravity.

 

[empdee4]

Assume two bodies of masses m and x•m are interacting with each other. In Newtonian gravitation, the force between two bodies are the same no matter which is considered gravitating or gravitated. That is, whether mgravitating = m and mgravitated = x•m , or mgravitating = x•m and mgravitated = m, the gravitational force is the same,
F = k mgravitating • mgravitated / r2 = x • m • m = k mgravitating • mgravitated / r2 = m • x • m
But in Einstein gravitation, mgravitated disappeared because equivalence between inertial and gravitation masses. The gravitation field is caused only by mgravitating , then the acceleration from gravitation would be different whether m or x • m is gravitating.
a = k mgravitating / r2
= k m/ r2 (if m is gravitating)

= k x • m / r2 (if x • m is gravitating)
The acceleration would be x times different between the two results. Obviously, this cannot be right because there can only be one answer in Nature.

 

[Nugatory]

The gravitation field is caused only by mgravitating ,

Careful... that is not generally true of solutions to Einstein’s field equations. It is true only in one particular case, namely when the mass of the smaller object is completely negligible compared with the larger mass so we can use the Schwarzschild solution.

Newtonian gravity gives the same result for this case.

 

[PeterDonis]

Assume two bodies of masses m and x•m are interacting with each other.

There is no such thing in GR; gravity is not a force in GR, and massive bodies that are separated from each other do not "interact" directly. They each affect the overall spacetime geometry in which both of them move, and that spacetime geometry in turn affects how each one moves.

In other words, "gravity" in GR does not work the same as Newtonian gravity. Your questions assume it does, so they are based on a misconception.

 

[anuttarasammyak]

Summary:: In tests of GR, stress-energy tensor is set to 0 in Schwarzschild solution, then is curvature caused by the sun, or by the 0 stress-energy?

Then, is the curvature caused by solar mass, or by the 0 stress-energy?

R_{\mu\nu}=0
in the "empty" space where mercury moves. There curvature R=0 also but $R_{\mu\nu\xi\rho} \neq 0$ in general.

 

[Ibix]

But in Einstein gravitation, mgravitated disappeared because equivalence between inertial and gravitation masses. The gravitation field is caused only by mgravitating , then the acceleration from gravitation would be different whether m or x • m is gravitating.

No. Strictly, the Schwarzschild spacetime is a black hole in an otherwise empty universe, which is why no other stress-energy appears - there is none by definition. It turns out also, as I said, to be an accurate model of spacetime outside a spherically symmetric mass - again, there must be no other stress-energy anywhere outside the spherically symmetric mass.

But Schwarzschild spacetime is also a good approximation for a situation where there is only negligible mass outside a central spherical object. Thus it's a decent approximation for the solar system, and Einstein's calculation was based on this approximation. It's analogous to modelling the Sun as a mass fixed at the origin, which is common in Newtonian calculations. We all know it's not exactly right, but for any mass smaller than another star the errors usually don't matter much.

If you want to model multiple mutually gravitating bodies then you can do so. That's how the gravitational wave signatures LIGO searches for are generated. But serious computational capacity is needed (even by modern standards) since none of the symmetries that make the Schwarzschild solution analytically tractable are present and, to the kind of precision Einstein needed, the resulting path of Mercury wouldn't be different.

 

[PeroK]

The acceleration would be x times different between the two results. Obviously, this cannot be right

There are two types of students here.

The first student looks at this problem as says: I've just started learning about GR. How does GR handle the case of the two body problem - i.e. when both masses are large enough to affect the other? What am I missing?

The second student says: I've just started learning about GR and I don't understand how it can handle the two-body problem. GR must be wrong.

If every subject was considered "obviously" wrong whenever a beginner made a mistake, there wouldn't be much science left!

 

[vanhees71]

There is no such thing in GR; gravity is not a force in GR, and massive bodies that are separated from each other do not "interact" directly. They each affect the overall spacetime geometry in which both of them move, and that spacetime geometry in turn affects how each one moves.

In other words, "gravity" in GR does not work the same as Newtonian gravity. Your questions assume it does, so they are based on a misconception.

In other words: The gravitational interaction (and it is an interaction after all, which however can be reinterpreted as dynamics of the geometry of spacetime or "geometro dynamics" as Wheeler dubbed it) is described within GR as a non-linear theory in contradistinction to the Newtonian approximation, which is about very weak gravitational fields which can be described in the linear approximation of GR.

 

[empdee4]

Thanks very much for explanations.

 

[empdee4]

More questions. I understand when none of the masses is small and negligible, they both contribute to the stress-energy. There is no difference between gravitating and gravitated masses, they are both gravitating.
My question is: Can the Einstein equation with such a stress-energy be reduced to Newtonian gravitation? If yes, how?
Thanks.

 

[PeroK]

My question is: Can the Einstein equation with such a stress-energy be reduced to Newtonian gravitation? If yes, how?

The short answer is that, like everything else in GR, it's far from easy!

Here are a few things on this:

https://en.wikipedia.org/wiki/Two-b..._relativity#Beyond_the_Schwarzschild_solution

https://arxiv.org/abs/gr-qc/9709026

https://phys.org/news/2013-04-einstein-gravity-theory-toughest-bizarre.html

 

[Ibix]

See also the PPN formalism: https://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism.

 

[PeterDonis]

My question is: Can the Einstein equation with such a stress-energy be reduced to Newtonian gravitation?

The even shorter answer than the ones @PeroK and @Ibix gave is: no.

You asked about the perihelion precession of Mercury in the OP. That is a phenomenon that is not predicted by Newtonian gravitation. The two other classic tests of GR, gravitational time dilation and bending of light by the Sun, are also not predicted by Newtonian gravitation. (Technically, one can sort of handwave a prediction of light bending by the Sun from Newtonian gravity, but even if this is considered acceptable, it still gives a numerical value for the bending that is only half of the GR value.)

One can use Newtonian gravity as a sort of zeroth-order approximation to GR for this specific scenario (a roughly spherical gravitating massive body surrounded by vacuum), and then get gradually more accurate predictions by adding terms of higher order. This is what the PPN formalism that @Ibix referred to does. But this still is not the same as reducing the Einstein equation to Newtonian gravitation. In fact it is the opposite, it requires admitting that Newtonian gravity by itself is not the same as GR and applying corrections accordingly.

If you were actually asking about a scenario where there are two or more gravitating bodies that contribute significantly to the overall spacetime geometry, then as long as the bodies are still isolated--i.e., there is vacuum except where the bodies are located and the size of the bodies is much smaller than their separation distances--there is in fact a sort of "post-Newtonian" approximation for this case as well. It is called the Einstein-Infeld-Hoffman equations:

https://en.wikipedia.org/wiki/Einstein–Infeld–Hoffmann_equations

As noted in that Wikipedia article, in the limit $c \rightarrow \infty$, these equations reduce to the Newtonian equations for a many-body system in which gravity is the only force acting. But again, this is not the same "reducing" GR to Newtonian gravity; it's the opposite.

 

[empdee4]

Thanks very much.