Klein-Gordon Inner product being independent of the choice of spacelike hypersurface $\Sigma$ used to integrate it

Question

Here I'll work in flat 4-dimensional Minkwoski space, but using arbitrary coordinates (described by some metric $g_{\mu\nu}$).

Suppose we've got two complex-valued scalar functions $f$ and $g$ which solve the Klein-Gordon equation (ie. $(\Box_{x} + m^2) f(x)=(\Box_{x} + m^2) g(x) = 0$). Then we define the following current: $$ J^{\mu}(f,g)(x) = - i \left[ f^{\ast}(x) \frac{\partial g(x)}{\partial x} - \frac{\partial f^{\ast}(x)}{\partial x} g(x)\right] $$

This is a conserved current, in the sense that it solves the continuity equation $J^{\mu}(f,g)_{;\mu} =0$, or in arbitrary coordinates: $$ \frac{1}{\sqrt{ - \det(g) }} \frac{\partial}{\partial x^\mu} \bigg( \sqrt{ - \det(g) } J^{\mu}(f,g)(x) \bigg) \ = \ 0 $$

Using this we construct the Klein-Gordon inner product as: $$ \langle f,g\rangle = \int_{\Sigma} d^3\Sigma_{\mu}\ J^{\mu}(f,g) $$

Where $\Sigma$ is a space-like hypersurface and $d^{3}\Sigma_{\mu} = \frac{1}{2} \epsilon_{\mu\alpha\beta\gamma} dx^{\alpha} \wedge dx^{\beta} \wedge dx^{\gamma}$ being the 3-dimensional volume element ($\epsilon$ being the Levi-Cevita tensor).

The claim made in Takagi's `Vacuum noise and stress induced by uniform accelerator: Hawking-Unruh effect in Rindler manifold of arbitrary dimensions' is that the KG inner product is independent of the choice of space-like hypersurface $\Sigma$ used to integrate it.

The reason for this is that $J^{\mu}(f,g)$ is a conserved current which means Gauss' theorem may be applied.

My Question How do you prove that the KG inner product is independent of $\Sigma$?

Chapter 2.8 of Hawking and Ellis' book `Large-Scale Structure of Space-Time' has a bit on Gauss' theorem in arbitrary coordinates, where it's said that: $\int_{\partial U} d^3\Sigma_{\mu}\ X^{\mu} = \int_{U} d^4x\ X^{\mu}_{\ ;\mu}$ for a vector field $X^{\mu}$. But if we set $X^{\mu} \mapsto J^{\mu}(f,g)$, then the divergence is vanishing which would seem to imply the KG inner product being $0$! My diff geo is not great though and am probably mis-interpreting something - can somebody point out what is my error in this line of logic?

Answer 1

In your last paragraph you almost get it right. You definitely need to use that if $U$ is a region in spacetime, then $$ \int_{\partial U} d^3 \Sigma_\mu \, J^\mu = \int_U d^4 x \; \nabla_\mu J^\mu = 0 $$ since $J$ is a conserved current.

You want to prove the following: if $\Sigma_1$ and $\Sigma_2$ are two different timeslices, then the integral of $J^\mu$ over $\Sigma_1$ is equal to the integral over $\Sigma_2$. For instance, let $\Sigma_2$ be $\Sigma_1$ moved up by an arbitrary distance. Let $U$ be the volume in between the timeslices. Then on top the top/bottom the volume $U$ is bounded by $\Sigma_{1,2}$, and in the spatial directions $U$ is bounded by some 3d surface $V$. Now you assume that the integrand decays rapidly near spatial infinity, such that $$ \int_V d^3x \, J^\mu \stackrel{\text{assumption}}{=} 0. $$ Then the above Stokes theorem tells you that $$ 0 = \int_U d^4 x \; \nabla_\mu J^\mu = \left[ \int_{\Sigma_1} - \int_{\Sigma_2} \pm \int_V \right] J^\mu. $$ Using our assumption about the decay of $J$ at infinity, we find $$ \int_{\Sigma_1} J = \int_{\Sigma_2} J $$ as desired.