# Effective stress-energy tensor for a gravitational wave, compared to static semi-Newtonian case

There is a calculation that I had been thinking for a long time of working out to my own satisfaction, both because of its intrinsic importance and because it seemed like it would be fun. This was to calculate the intensity of a gravitational wave for a given amplitude. E.g., for LIGO events we have amplitudes expressed as a fractional change in the metric, and these can be related to the mass-energy released in the event that created the waves. Misner, Thorne, and Wheeler have the following expression (I've changed the notation slightly) in section 36.7:

$$T_{cd}=\frac{1}{32\pi}\langle g_{ab,c}g^{ab}{}_{,d} \rangle ,$$

where $g_{\mu\nu}-\eta_{\mu\nu} \ll 1$. This is for the effective stress-energy tensor for a gravitational wave, in the transverse traceless (TT) gauge, averaged over several wavelengths.

MTW's derivation of this is complex and spread around in little chunks in different places in the book. I thought I would go through the features of this equation and see how many of them I could understand or give heuristic derivations for. Dimensionally, it makes sense, and this occurs iff we have two derivatives. It depends on the square of the amplitude, which makes sense in the small-amplitude limit for any wave. The derivatives are partial derivatives, not covariant derivatives, because the covariant derivative of the metric is zero. The average is required because by the equivalence principle, we can never have a local expression for the energy density of the gravitational field. Next I started worrying about the factor of $1/32\pi$, wondering if there was some heuristic way to produce it.

What occurred to me was to work out an expression, in similar notation, for the energy density of the gravitational field in Newtonian mechanics. The energy density is $-(1/8\pi)\textbf{g}^2=-(1/8\pi)(\nabla\phi)^2$, where the units are such that $G=1$, and $\nabla$ is a 3-gradient. In the static case, in the semi-newtonian limit, the metric is $g_{tt}=1+2\phi$. So translating the above expression for the energy density into GR-style notation, we have

$$\textbf{g}=-\frac{1}{2}\nabla g_{tt}$$

$$T_{tt} = -\frac{1}{32\pi}g_{tt,\mu}g^{tt,\mu},$$

where $\mu$ runs only over the spacelike coordinates.

So these two expressions have some obvious similarities, but they are also different in certain ways. It's nice that the $1/32\pi$ is the same in the two cases, although the sign is opposite. However, we're contracting over different indices. I'm also not very convinced by my own comparison of these expressions because the one for a gravitational wave is specific to a certain gauge, and I never actually used that assumption.

Is it possible to give a semi-heuristic derivation -- one that's more convincing than mine -- for the expression for the effective stress-energy tensor of a gravitational wave?

• Your equation shows negative energy density, which seems to be fiercely rejected by some theoreticians and firmly accepted by others. It is a confusing issue. See [physics.stackexchange.com/questions/390130/…, containing a quote from Steve Carlip. – S. McGrew Apr 19 '18 at 16:53
• @S.McGrew: No, the negative energy density for the gravitational field in Newtonian gravity is totally noncontroversial. The quote in the linked question doesn't make sense, and is presumably garbled or taken out of context, since a professional relativist like Carlip wouldn't be that confused about freshman physics. – user4552 Apr 19 '18 at 21:39
• The context was a conversation with Steve Carlip about general relativity and is definitely not garbled. He said: "To make gravity attractive in such a [vector-like] theory, you must require that the gravitational field has negative energy, which (apart from the obvious instabilities) would drastically disagree with binary pulsar observations". So it's a strong assertion that the energy density of the gravitational field is positive and that gravity is therefore not vector-like. – S. McGrew Apr 19 '18 at 22:02
• A further quote from Steve Carlip (emailMarch 6, 2018): "In a vector theory of gravity, where like masses attract, a gravitational wave would have to carry off negative energy. This would mean that a binary pulsar system would increase in energy as it emitted gravitational waves, which contradicts observation." So is there something wrong with the way the semi-Newtonian limit is taken? Am I misunderstanding your equation, or what negative energy density looks like? – S. McGrew Apr 19 '18 at 22:16
• In another post you said, " In GR, the equivalence principle makes the gravitational field (g, the thing that's 9.8 m/s2 on earth) unobservable, so we don't have an energy density expressible in terms of g." Could you expand on that? If the gravitational field isn't observable, what is the thing that's 9.8 m/s^2 on earth? What are we measuring when we measure the weight of a mass or the acceleration of a falling object relative to e.g., the Tower of Pisa? Not arguing, just asking-- – S. McGrew Apr 20 '18 at 3:56