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In Carroll's Spacetime and geometry section 7.5, it is deriving metric $\bar h_{\mu \nu}$ with Lorenz gauge in terms of quadrupole moment of energy density.

In Eq.[7.135],

\begin{equation} \int d^3y \tilde T^{ij} = \int \partial_k(y^i T^{kj}) d^3y - \int y^i (\partial_k T^{kj}) d^3y \end{equation}

It says "The first term is a surface integral which will vanish since the source is isolated". But I could not see the form of surface integral for the first term. Is it divergence theorem? I know divergence is $\partial_k T^k$, how does it relate to $\partial_k(y^i T^{kj})$? It seems to me that $y^i$ is redundant. If it is $\partial_k(T^{kj})$, it works and converts as $\int T^{kj}n^kd^2y$.

Moreover, to my understanding, the source is emitting energy through gravitational wave. How could it said to be isolated?

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Yes, it is the divergence theorem. The term is simply $$\int\partial_k(y^iT^{kj})d^3y=\int y^iT^{kj}n_k d^2y$$ where the integral is over the surface at infinity and $\boldsymbol{n}$ is the unit vector away from the surface. We presume that $T^{kj}$ vanishes at infinity (since the source is isolated) and hence the integral is zero.

You are right that the source is emitting energy through the wave, but here we are dealing with linearised gravity, where $T^{kj}$ only includes the contribution from ordinary matter, and not from the wave itself. This matter is isolated and so $T^{kj}$ vanishes at infinity.

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  • $\begingroup$ Thanks for the explanation. For the second part about stress tensor, is T non-zero for second order of h? Because energy is transmitted through $h^2$. $\endgroup$
    – Simon219
    Apr 8 at 13:51
  • $\begingroup$ If you are asking whether there is a contribution to the stress tensor from the wave itself (so that it is non zero at infinity), then yes, there is. This is a slightly dodgy way of looking at it though, and really it should be considered part of the Einstein tensor $G^{\mu\nu}$. I think this answer explains it quite well. $\endgroup$ Apr 8 at 14:32
  • $\begingroup$ Thanks. I get it now. $\endgroup$
    – Simon219
    Apr 8 at 14:55

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