# General Relativity - gravity self coupling

1. Apr 8, 2010

### JustinLevy

I realize that energy is a touchy subject in GR in dynamical situations, and people argued for a long time whether or not gravitational waves carried energy. But it sounds like there has been agreement for quite awhile now that they do carry energy.

Since in weak gravity situations, we can write equations that make gravity look similar to the fields of electromagnetism, and on top of this
1) gravity waves carry energy
2) gravity should "self couple" according to semi-classical treatments of gravity
I don't understand why the Schwarzschild solution has a zero stress energy tensor. Shouldn't we include some energy for the 'gravitational field' itself?

Sort of like the charged black hole solution. Where energy is included in the electric field.

I wonder how this would affect planetary motion if we were indeed supposed to include such a term. Since the "gravitational field energy" would be such a small contribution to the sun's mass in this idea, it might not be measurable. Although it would make the mass of the sun appear to get slightly larger as you moved away from it (which bizarrely is what the pioneer anomaly saw, but is probably just a coincidence).

Last edited: Apr 8, 2010
2. Apr 8, 2010

### bcrowell

Staff Emeritus
Gravity does self-couple. That's why the Einstein field equations are nonlinear, rather than linear like Maxwell's equations. The nonlinearity of the equations already describes the self-coupling. If you added a term to the stress-energy tensor to describe the energy of the gravitational field itself, you'd be counting the effect twice.

Another reason why the stress-energy tensor can't include a contribution from the field itself is that it would break general covariance. By the equivalence principle, the gravitational field can have any value you like at any point in space, including zero. Therefore if you try to write down an equation for the energy density in terms of the gravitational field, it's guaranteed not to be generally covariant.

Computations of energy densities of gravitational waves are non-generally covariant approximations to the underlying generally covariant theory.

3. Apr 8, 2010

### JustinLevy

If this were true, we could identify a piece that is the contribution to the stress energy tensor due to the gravitational field. However none of the terms in Einstein's field equations can be considered this, since every single term is zero in vaccuum.

Instead it appears the equations are non-linear for the more mundane reason that they involve the curvature, which is necessarily non-linear in the metric components.

So yes, there is "self coupling" already in the sense that the equations are non-linear. But this is not due to a contribution to the stress energy from the gravitational field. And since there appears to be energy in the gravitational field, I wonder what it would look like if one included a term explicitly for this.

In the weak field limit, we have (consider g_uv = minkowski metric + h_uv, where h_uv is a small perturbation to the minkowski metric):
$$h^{0\mu}$$
acting analogous to the four-vector of electrodynamics.
So I'd naively expect at least some "gravitational stress energy" terms like:
$$F_{\mu\nu} = {h^{0\nu}}_{;\mu} - {h^{0\mu}}_{;\nu}$$
$$T^{\mu\nu} = -\frac{1}{\mu_0}[ F^{\mu \alpha}F_{\alpha}{}^{\nu} + \frac{1}{4} g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}]$$

Since I'm by hand selecting the time components, there must be some way to extend that for all components, but that at least gives the dominating main contributions and hints at what the form should be.

Hmm... I don't agree it would break the equivalence principle.
Adding a term to the stress energy tensor would not affect any experiment except a gravitational one. And the equivalence principle is a local rule, so doesn't seem affected.

The gravitational field in GR already provides a curvature: the spacetime in an accelerating rocket in flat background has no curvature, while sitting at rest on a gravitating body the spacetime clearly has a curvature. This doesn't violate the equivalence principle, even though gravity is non-linear. Adding a term to the stress energy tensor wouldn't change anything any more than an additional particle field that only acts gravitationally would. So it shouldn't violate the equivalence principle.

And I don't see how general covariance must be broken. Heck, even Newton's gravity can be written in a generally covariant formulation. As long as the "gravitational stress energy" we put in transforms like a tensor, the equations are still generally covariant.

There are difficulties discussing energy in dynamics in GR. But for things like a static spherically symmetric source, we can talk about energy just fine. In the weak field limit there appears to be energy in the fields. So why not include that in a stress energy tensor and see what happens?

As already mentioned, qualitatively the mass of the sun should appear to get larger as one gets farther away from it. It may turn out that this idea is already experimentally ruled out. I'm just curious:
1) what the modification would look like and do
and
2) why, if gravitational fields can carry energy in GR, why there isn't already a stress energy term for them in GR

4. Apr 8, 2010

### Stingray

This "mundane" nonlinearity can be viewed as including exactly what you suggest. If you choose a Minkowski background metric and write things in terms of its (ordinary partial) derivative operator, you get
$$\partial_{\alpha \beta} H^{\mu\alpha\nu\beta} = 16 \pi (-g) (T^{\mu\nu} + t^{\mu\nu}_{LL}),$$
where
$$H^{\mu \alpha \nu \beta } = \sqrt{-g} ( g^{\mu\nu} g^{\alpha \beta} - g^{\mu\alpha} g^{\nu\beta} ).$$
$t^{\mu\nu}_{LL}$ is the Landau-Lifgarbagez pseudotensor, and looks something like a gravitational stress-energy. It is quadratic in metric gradients. The right-hand side of the first equation is divergence-free with respect to $\partial_\mu$.

These equations are equivalent to the usual form of Einstein's equation. Your "gravitational self-coupling" is already there. This form is sometimes useful - especially in post-Newtonian approximations - but everything depends very strongly on the choice of background coordinate system. The Landau-Lifgarbagez pseudotensor may vanish in one coordinate system while being finite in another. This is a fundamental problem with any local notion of energy or momentum density of the gravitational field. A gravitational field can only be meaningfully defined to have energy over a finite region (sometimes!).

5. Apr 8, 2010

### JustinLevy

I had never seen that before. Thank you very much!

http://en.wikipedia.org/wiki/Landau-Lifgarbagez_pseudotensor [Broken]
They don't explain in any detail why this formulation is objected to. If the formulation is valid, and you get to have a conserved energy and momentum in GR now, why wouldn't this be embraced more?

Last edited by a moderator: May 4, 2017
6. Apr 8, 2010

### Stingray

People don't object to this per se. It is actually the starting point for most versions of post-Newtonian theory (which is one of the largest and most active areas of analytical research in GR).

Taking the LL pseudotensor seriously as an effective stress-energy tensor is a different matter. Questions regarding the (say) "gravitational energy" inside a particular volume depend on things external to the theory. They are, in a sense, subjective. This is very different from what happens with ordinary stress-energy tensors.

7. Apr 9, 2010

### JustinLevy

I've read through a few things now, and I'm still confused.
Is the Landau-Lifgarbagez formulation only true in certain coordinate systems? If so, how do I know if I'm using a "valid" coordinate system? If no, then in what sense is it not generally covariant?

Also, in the wiki page it says one of the requirements they imposed when searching for this pseudotensor was that:
"that it vanish locally in an inertial frame of reference (which requires that it only contains first and not second or higher derivatives of the metric)"
But there are only two terms, and BOTH have second derivatives of the metric in it. So what did they mean to say there?

8. Apr 9, 2010

### Phrak

It is not zero. The stress energy tensor exterior to the mass does not have all zero entires.

9. Apr 9, 2010

### JustinLevy

Are you referring to the stress energy "tensor" including the pseudotensor Stringray and I were talking about above? In that case, yes.

But the normal, actual stress energy tensor is identically zero exterior to the mass in the Schwarzschild solution. The cosmological constant is zero, and the Ricci curvature tensor is zero, so by the einstein field equations, the stress energy tensor must be zero in the Schwarzschild solution. It is a vacuum solution after all (with the "spherical mass" boundary condition).

10. Apr 9, 2010

### Stingray

That Wiki page is not very good. The LL pseudotensor does not involve any second derivatives of the metric if it is expanded properly. If you scroll down, the last two expressions on that page (before the Einstein pseudotensor) make this clear. I'm not going to guarantee that those equations are correct, so you might also want to look at Eq 2.7 of http://arxiv.org/abs/gr-qc/0007087 (for example).

Anyway, the LL formalism works in any coordinate system. No terms transform like tensors, however. The equations as a whole are covariant, but this could not be guessed by casual inspection (unless you knew everything was properly deduced from a covariant starting point).

11. Apr 9, 2010

### Phrak

Ekk. I was referring to the metric. nebermind