# Stress-Energy Tensor for Gravitational Field

1. Jul 6, 2012

### Alexrey

I'm trying to understand the way that the stress-energy tensor for a gravitational field is derived and I've run into a few problems. It seems that there are two main avenues which are kind of similar. One derivation involves looking at $$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$ where $$\eta_{\mu\nu}$$ is the standard Minkowski metric, where we then allow second-order terms in the perturbation $$h_{\mu\nu}$$ instead of ignoring them like the linearized theory of gravitational radiation did. We then define the Einstein tensor as (here c and G equal 1) $$G_{\mu\nu}=8\pi(T_{\mu\nu}+t_{\mu\nu})$$ where $$t_{\mu\nu}$$ represents the stress-energy tensor for a gravitational field. Then we simply find the connection coefficients, Ricci scalar, Ricci tensor and Einstein tensor to second-order in $$h_{\mu\nu}$$ and through lots of tedious algebra we then find $$t_{\mu\nu}$$ in terms of $$h_{\mu\nu}$$ and voila, we have our stress-energy tensor for a gravitational field! The other derivation seems to look at a more general metric $$g_{\mu\nu}=\tilde{g}_{\mu\nu}+h_{\mu\nu}$$ where $$\tilde{g}_{\mu\nu}$$ is now an arbitrary curved background metric instead of being flat like the first derivation, and $$h_{\mu\nu}$$ is again just a metric perturbation. But this derivation imposes a restriction that the scale of curvature $$R$$ of the curved background metric is far greater than the wavelength $$\lambda$$ of the metric perturbation or equivalently that the frequency $$f$$ of the metric perturbation is far greater than that of the curved background metric's frequency $$f_B$$, thereby allowing us to clearly separate the two and think of the metric perturbation as a gravitational wave riding on top of the curved background. Once this is done we then just do what we did in the first derivation and thus find $$t_{\mu\nu}$$ in terms of $$h_{\mu\nu}$$. Now, the first derivation said nothing about frequencies, wavelengths or background curvature, so I feel as though I am missing something very important. If anyone could enlighten me as to the proper way of deriving the stress-energy tensor of a gravitational field, and the reasons behind that derivation I would greatly appreciate it. Thanks! BTW sorry about the weird vertical layout, I don't know how to get it to display nicely!

2. Jul 6, 2012

### Staff: Mentor

An important clarification of terminology: there is no stress-energy tensor for the gravitational field. The things you're looking at are *pseudo* tensors. The key error is here:

No, that is *not* the Einstein tensor on the LHS of this equation. The correct equation with the Einstein tensor on the LHS is the standard Einstein Field Equation:

$$G_{\mu\nu}=8\pi T_{\mu\nu}$$

The stress-energy tensor on the RHS of this equation has *no* contribution from the gravitational field. The reason this is important is that the Einstein tensor on the LHS of this equation obeys the Bianchi identities, i.e., its covariant divergence is identically zero; therefore, the covariant divergence of the RHS is also identically zero. This is how local conservation of energy is expressed in GR, and it *only* works with the equation written the way I wrote it, with no contribution to the stress-energy tensor from gravity.

The various pseudo-tensors describing "energy in the gravitational field" are derived by extracting some piece of the LHS of the above equation, moving it to the RHS, and calling it $t_{\mu\nu}$. But there is no one unique way to do that, because there is no one unique way to split up the true Einstein tensor into a "background" part that stays on the LHS, and an "energy in the field" part that goes to the RHS. You describe two possible ways of doing it, taking either the Minkowski metric as the "background" or some more general curved metric (such as the Schwarzschild metric) as the "background" on which small-scale perturbations (such as gravitational waves) are superimposed. There are others. Which method you use depends on what you are trying to do with it; there is no one "right" answer.

You might want to check out this page from the Usenet Physics FAQ:

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

http://en.wikipedia.org/wiki/Stressâ€“energyâ€“momentum_pseudotensor

3. Jul 7, 2012

### Alexrey

Thanks very much, that makes a lot more sense now. I always wondered why the Bianchi identities were so important.

If I go the route of $$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$ and expand it to second order in the perturbation do I still need to have curvature and frequency restrictions? I thought not but then read that since the stress-energy pseudotensor takes an average of the metric perturbation of a finite volume, it would also retain some information about the metric perturbation. Would I thus still have to apply the restrictions?

4. Jul 7, 2012

### Staff: Mentor

Yes, because you're assuming that the perturbation is much smaller than the Minkowski metric term. Basically this is the same as the second scheme you described, with $\tilde g_{\mu\nu} = \eta_{\mu\nu}$.

A technical point: the "curvature" of the Minkowski metric is zero, so its "scale of curvature" is infinite, which means you can't formulate the "curvature and frequency restrictions" exactly the same as you described them for the case of a generic curved background metric. However, the curvature and frequency restrictions are really just a way of restating the real restriction, which is that the perturbation, $h_{\mu\nu}$, must be much smaller than the background metric, $\tilde g_{\mu\nu}$. The latter way of formulating the restriction works just fine with the Minkowski metric as the "background" metric.