Graduate Dirac "GTR" Eq. 27.11 -- how to show that a boundary term vanishes?

[Kostik]

TL;DR

In Dirac's "General Theory of Relativity", Dirac derives Einstein's field equations and the geodesic equation from the variation $\delta(I_g+I_m)=0$ of the actions for gravity and matter. The two dynamical variables in the variation are $g_{\mu\nu}$ and $p^\mu$ which satisfies $\delta p^\mu = (p^\nu b^\mu - p^\mu b^\nu)_{,\nu}$ where $b^\mu$ is an arbitrary displacements of an element of matter. He discards a boundary term during partial integration that is not easily (?) justified!

In Dirac's "General Theory of Relativity", p. 53, eq. (27.11), Dirac is deriving Einstein's field equations and the geodesic equation from the variation $\delta(I_g+I_m)=0$ of the actions for gravity and matter. Here $p^\mu=\rho v^\mu \sqrt{-g}$ is the momentum of an element of matter. He makes arbitrary displacements of an element of matter $b^\mu$. The two dynamical variables in the variation are $g_{\mu\nu}$ and $p^\mu$ which satisfies $\delta p^\mu = (p^\nu b^\mu - p^\mu b^\nu)_{,\nu}=0$.

The action is over a 4-dimensional volume $M$. Dirac integrates by parts: $$\int_M v_\mu (p^\nu b^\mu - p^\mu b^\nu)_{,\nu} \, d^4 x
= \int_M [v_\mu (p^\nu b^\mu - p^\mu b^\nu)]_{,\nu} \, d^4 x
- \int_M v_{\mu,\nu} (p^\nu b^\mu - p^\mu b^\nu) \, d^4 x .$$ Remember: on the boundary $\partial M$: $$(p^\nu b^\mu - p^\mu b^\nu)_{,\nu}=0.$$ (It's unusual for a divergence to vanish on a boundary, but remember this is actually a variation $\delta p^\mu = (p^\nu b^\mu - p^\mu b^\nu)_{,\nu}$. As $p^\mu$ is one of the dynamical variables, it is kept constant on the boundary. Hence, the variation $\delta p^\mu=0$ on the boundary.)

Dirac assumes the boundary term vanishes. How do we show this? $$\int_M [v_\mu (p^\nu b^\mu - p^\mu b^\nu)]_{,\nu} \, d^4 x = \int_{\partial M} v_\mu (p^\nu b^\mu - p^\mu b^\nu) \, dS_\nu = 0 \quad (\text{Show!})$$ where $dS_\nu$ is the oriented hypersurface element in 3D space.

If $v_\mu (p^\nu b^\mu - p^\mu b^\nu)$ were constant on the boundary, the result would be trivial. But this is not the case!

One can also write this integral: $$\int_M [v_\mu (p^\nu b^\mu - p^\mu b^\nu)]_{,\nu} \, d^4 x
= \int_{\partial M} v_\mu n_\nu (p^\nu b^\mu - p^\mu b^\nu) \, \sqrt{|h|} \, d^3 y$$ where $n_\nu$ is the "oriented unit normal vector" on $\partial M$. (Here I am supposing that the boundary $\partial M$ can be parameterized by $x^\nu = x^\nu (y^m)$, $m=1,2,3$ and $h$ is the determinant of the matrix $h_{mn}=g_{\mu\nu}\frac{\partial x^\mu}{\partial y^m}\frac{\partial x^\nu}{\partial y^n}$ ... but this is just linear algebra / change of variable stuff, and isn't important to the problem at hand.)

To repeat, the key fact that I have to work with is that, on the boundary $\partial M$:
$$(p^\nu b^\mu - p^\mu b^\nu)_{,\nu}=0.$$

 

[ergospherical]

I don't think it needs to be any more complicated than the fact that, for any physical (bounded) matter distribution, the vector field $p$ is going to vanish sufficiently fast toward infinity, i.e. $p{|}_{\partial} = 0$ and the boundary term $\int_{\partial} dS \ n_{\nu} v_{\mu} (p^{\nu} b^{\mu} - p^{\mu} b^{\nu}) = 0$

 

[Kostik]

I thought of this, too: since $p^\mu = \rho v^\mu \sqrt{-g}$, one can always consider a 4-dim. "cylinder" of spacetime between two time coordinates $T_1$ and $T_2$, and the ball $r=R$, and let $R$ be sufficiently large so as to encompass all the matter in existence. Then $\rho=0$ on the boundary, hence $p^\mu=0$.

However, I prefer to solve these boundary term issues with compact volumes. First, consider that this method does not work when trying to vary the Hilbert-Einstein action -- that's why the Gibbons-Hawking-York boundary term is usually added. Second, can I be sure that $\rho=0$ sufficiently far away? What if the matter-energy density of a (closed?) universe never vanishes?

 

[ergospherical]

Take any physical vector field $X$ (e.g. your term involving $p^{\mu}$) and integrate it over the $n-1$ sphere $\lim_{r\rightarrow \infty} S_r$$$\int dS \ n_{\mu} X^{\mu} = \int dS \ r^{n-1} \hat{n}_{\mu} X^{\mu}$$so the integral vanishes if $X = O(r^{-n})$, which is what we mean by assuming the field vanishes sufficiently quickly at spatial infinity.

 

[Kostik]

OK, I got it. The solution is physics, not math. I believe the integral, in general, need not vanish. However, going back to the variation $\delta(I_g+I_m)=0$, one has two integrals, and integration by parts turns the second one into two more. One can argue that all three integrals must vanish separately. Hence, the boundary term integral goes away, and the other two are left to yield the EFE and geodesic equation.