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

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?

• I have the feeling this is a duplicate of Inner product on space of mode functions for a scalar field. – AccidentalFourierTransform Jul 14 '18 at 22:09
• You’re proven that the integral over a closed surface is zero. That implies that the integrals over two different spacelike hypersurfaces are the same. – knzhou Jul 14 '18 at 22:11
• It’s the same proof that a force being conservative implies work is path independent. – knzhou Jul 14 '18 at 22:11

## 1 Answer

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.

• That's fantastic! Thanks very much. I'm curious about the relative minus sign in $\int_{\Sigma_{1}}-\int_{\Sigma_{2}}$. The orientation of the entire volume bounded by the two sheets stays the same because of this `-' sign right? – Greg.Paul Jul 14 '18 at 22:46
• And one more thing - the Gauss theorem usually assumes that the region $U$ is compact (and so bounded). Is the additional assumption that the integral at spatial infinity vanishes essentially fixing the fact that the volume bounded by the two sheets $\Sigma_{1}$ and $\Sigma_{2}$ is obviously not bounded? – Greg.Paul Jul 14 '18 at 22:48
• As for your first question, the orientation in Stokes' theorem has to be taken as "all outward". So on the top surface, the element $d\Sigma_\mu$ points upward, whereas on the bottom surface it points down. You want to compare the integrals where $d\Sigma_\mu$ points in the same direction on both slices, that's why I added a relative sign. – TempAccount2020 Jul 15 '18 at 1:43
• want to remove this comment – TempAccount2020 Jul 15 '18 at 1:45