How is the approximate gravitational wave stress energy momentum tensor not 0?

Question

In Section 35.7 of Misner, Thorne, and Wheeler, p. 955, an "effective" stress energy momentum tensor for gravitational waves is defined:

$$T^{\text{GW}}_{\mu \nu} = \frac{1}{32 \pi} \left< \bar{h}_{\alpha \beta, \mu} \bar{h}^{\alpha \beta}_{\space\space,\nu} - \frac{1}{2}\bar{h}_{,\mu}\bar{h}_{\nu} - \bar{h}^{\alpha \beta}_{\space \space ,\beta} \bar{h}_{\alpha \mu, \nu} - \bar{h}^{\alpha \beta}_{\space \space, \beta} \bar{h}_{\alpha \nu,\mu} \right>.$$

The brackets indicate an averaging over a region of space much larger than a wavelength of the wave. The text then says that on a background space time with Einstein tensor $G^{\text{B}}_{\mu \nu}$,

$$G^{\text{B}}_{\mu \nu} = 8 \pi \left(T^{\text{GW}}_{\mu \nu} + T^{\text{other fields}}_{\mu \nu} \right).$$

But for a gravitational wave propagating through empty space, with $R^{\text{B}}_{\mu \nu} = 0$, wouldn't this then imply that $T^{\text{GW}}$ is $0$? What am I missing?

Answer 1

The treatment of this in section 35.7 of MTW says they're only presenting the result, whereas the actual derivation is given in section 35.13, p. 964. The basic idea is as follows. First you take the vacuum field equation $R_{\mu\nu}=0$, and expand it in powers of the wave's amplitude $A$. They argue that the term linear in $A$ must vanish, because any effect of gravitational waves on the background curvature must be a second-order effect. Turning next to the $A^2$ term, they break it up into a fluctuating part and a smooth background. Setting the sum of these equal to zero, we get something that looks like the ordinary Einstein field equations, for the background, with the wave term interpretable as a source, i.e., as an effective stress-energy.

[EDIT] In a comment, the OP asks:

Thank you! But how does that address my question, exactly? Wouldn't this still mean that a gravitational wave propagating in vacuum carries no energy/momentum?

Locally (i.e., in any experiment on scales smaller than a wavelength), there is no energy-momentum, because the stress-energy tensor is zero. But in GR it is not true that you can simply integrate the stress-energy tensor over some volume and find the total energy-momentum inside. This is why, for example, we say that a Schwarzschild black hole has mass, even though the stress-energy tensor is zero everywhere. Integration of a tensor with rank>0 fails because of the ambiguities introduced by the path-dependence of parallel transport.

Answer 2

A previous answer contains some correct statements but does not clearly answer the question IMO, so here goes.

In the equation

$$G^{\text{B}}_{\mu \nu} = 8 \pi \left(T^{\text{GW}}_{\mu \nu} + T^{\text{other fields}}_{\mu \nu} \right)$$

$G^{\rm B}_{\mu\nu}$ is not the whole Einstein tensor, it is only a part of it, but it is the largest part. In the case of the linearized (weak field) theory it is the first-order term. In a vacuum we have indeed that $T^{\text{other fields}}_{\mu \nu} = 0$. However, we don't then have $G^{\rm B}_{\mu\nu} = 0.$ Rather, we have $G^{\rm B}_{\mu\nu} + G^\text{the rest}_{\mu\nu} = 0$ where the second term is the rest of the Einstein tensor. It is precisely this term which has been shifted over to the other side of the equation and made to play the role of 'stress energy tensor' $T^{GW}_{\mu\nu}$.

That's the answer to the question. It may help to note also why the waves are being handled like this. The logic is that the Einstein field equation deals with these issues through its non-linearity in a rather opaque way, so we introduce a way to clarify what is going in. Here we are unpacking that non-linear behaviour so as to draw attention to the way the gravity field can cause matter to take up or give off energy. $T^{GW}_{\mu\nu}$ is defined so as to match the flux of energy and momentum gained and lost by the matter by its gravitational interactions.

Answer 3

Space is not empty if there is a gravitational wave, so that actually $R^B_{\mu \nu} \ne 0$.