Klein-Gordon inner product with time derivative and "energy inner product"

Question

I’m currently working through Wald’s “Quantum Field Theory in Curved Spacetimes and Black Hole Thermodynamics” and am stuck at deriving equation (4.3.8) (which is claimed to be verifiable by “direct calculation”).

The situation is as follows. We’ve got a globally hyperbolic stationary spacetime with timelike Killing field $\xi$. We now consider (complex) classical solutions to the Klein-Gordon equation $\nabla^a\nabla_a\psi - m^2\psi = 0$ whose initial data (i.$\,$e. value and normal derivative) have compact support on some (and therefore any) Cauchy surface. For two such solutions $\psi_1,\psi_2$, we define $$B(\psi_1,\psi_2) := \int_\Sigma\left(\psi_2 n^a\nabla_a\overline{\psi_1} - \overline{\psi_1} n^a\nabla_a \psi_2\right)\sqrt{h}\mathrm{d}^3x,$$ which is, apart from a missing factor of $\mathrm{i}$, just the usual “Klein-Gordon inner product” (which is not positive definite on the space of all solutions), and $$\langle\psi_1,\psi_2\rangle := \int_\Sigma \xi^a n^b T_{ab}(\psi_1,\psi_2) \sqrt{h}\mathrm{d}^3x$$ with $$T_{ab}(\psi_1,\psi_2) := \nabla_{(a}\overline{\psi_1}\nabla_{b)}\psi_2 - \frac{1}{2}g_{ab}\left(\nabla^c\overline{\psi_1} \nabla_c\psi_2 + m^2 \overline{\psi_1} \psi_2\right),$$ where $\Sigma$ is any (smooth spacelike) Cauchy surface and $n$ its future-directed unit normal. Since $T_{ab}(\psi,\psi)$ is the energy-momentum tensor of the solution $\psi$, Wald calls $\langle\cdot,\cdot\rangle$ the “energy inner product” (which can be shown to be positive definite for $m\ne0$).

Equation (4.3.8), which I want to derive, claims, up to notational differences, that $$B(\psi_1,\xi^a \nabla_a \psi_2) = 2 \langle\psi_1,\psi_2\rangle$$ for all $\psi_1,\psi_2$. After a few manipulations of the definitions, I get (using that $\psi_2$ solves the KG equation and $\xi$ is Killing) $$B(\psi_1,\xi^a \nabla_a \psi_2) - 2 \langle\psi_1,\psi_2\rangle = \int_\Sigma n^b\nabla^a\left(\overline{\psi_1}\xi_b\nabla_a\psi_2 - \overline{\psi_1}\xi_a\nabla_b\psi_2\right)\sqrt{h}\mathrm{d}^3x.$$ But why does this expression vanish? Maybe I’m missing something really easy here, but I just don’t see it.

EDIT: Contrary to what I wrote first, boundary terms we would get by applying the divergence theorem / Gauß’ theorem do vanish, because either (a) $\Sigma$ is non-compact and $\psi_i,n\cdot\nabla\psi_i$ have compact support or (b) $\Sigma$ is compact and has no boundary.

Answer 1

I’ve now found a way to do the calculation, or, in fact, two.

The first method is to just “brute-force” decompose all tensors in $\int_\Sigma n^b\nabla^a\left(\overline{\psi_1}\xi_b\nabla_a\psi_2 - \overline{\psi_1}\xi_a\nabla_b\psi_2\right)\sqrt{h}\mathrm{d}^3x$ into normal and tangential components, and, after some (somewhat messy) calculation using Gauß’ theorem for $\Sigma$, one ends up with 0.

The second method is shorter and, in my opinion, a lot more elegant, since it uses coordinate-free notation (arguably, abstract index notation is also coordinate-free, so maybe I ought to call it “index-free notation”). Let $\star$ denote the Hodge star operator on the spacetime; for any vector field $X$, let $X^\flat := g(X,\cdot)$ be the associated one-form and $i_X\eta := \eta(X,\cdot)$ the interior product with any differential form $\eta$. The volume form of the spacetime is $\star1$, thus the volume form of $\Sigma$ is $\sqrt{h}\mathrm{d}^3x = \pm i_n{\star}1$ (here and in the following, $\pm$ will stand for signs we don’t care about and which can vary from term to term). Furthermore, we have the relation \begin{equation} \label{eq_1} \tag{1} i_X {\star}\eta = (-1)^{d-1} {\star}(\eta\wedge X^\flat) \end{equation} for any vector field $X$ and form $\eta$. For our integral, we get \begin{align*} &\int_\Sigma n^b\nabla^a\left(\overline{\psi_1}\xi_b\nabla_a\psi_2 - \overline{\psi_1}\xi_a\nabla_b\psi_2\right)\sqrt{h}\mathrm{d}^3x \\ &= \pm \int_\Sigma \left(i_n{\star}\mathrm{d}{\star}\left(\xi^\flat \wedge \overline{\psi_1} \mathrm{d}\psi_2\right)\right) i_n{\star}1 \\ &\overset{\eqref{eq_1}}{=} \pm \int_\Sigma {\star}\left(\mathrm{d}{\star}\left(\xi^\flat \wedge \overline{\psi_1} \mathrm{d}\psi_2\right) \wedge n^\flat\right) i_n{\star}1 \\ &= \pm \int_\Sigma i_n{\star}{\star}\left(\mathrm{d}{\star}\left(\xi^\flat \wedge \overline{\psi_1} \mathrm{d}\psi_2\right) \wedge n^\flat\right) \\ &= \pm \int_\Sigma i_n\left(\mathrm{d}{\star}\left(\xi^\flat \wedge \overline{\psi_1} \mathrm{d}\psi_2\right) \wedge n^\flat\right) \\ &= \pm \int_\Sigma i_n\left(\mathrm{d}{\star}\left(\xi^\flat \wedge \overline{\psi_1} \mathrm{d}\psi_2\right)\right)\wedge n^\flat \pm n^\flat(n) \int_\Sigma \mathrm{d}{\star}\left(\xi^\flat \wedge \overline{\psi_1} \mathrm{d}\psi_2\right) \end{align*} The first term vanishes since $n$ is normal to $\Sigma$ (and thus $n^\flat$ is 0 as a form on $\Sigma$), and the second one vanishes due to Stokes’ theorem, since $\psi_1$ has compact support and $\partial\Sigma = \emptyset$.