Using both metric-variation formulas from part a gives the <Klein-Gordon scalar stress-energy tensor>
$$
\boxed{T^{ab}=\nabla^a\psi\nabla^b\psi
-\frac12g^{ab}\left(\nabla_c\psi\nabla^c\psi+\mu^2\psi^2\right)}.
$$
Its divergence is
$$
\nabla_aT^{ab}
=(\nabla_a\nabla^a\psi-\mu^2\psi)\nabla^b\psi,
$$
because the two Hessian terms cancel by symmetry. It vanishes on every solution of the <Klein-Gordon equation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 2 / b / ii / Solution

Let $K=\partial_t$ and let the future unit normal to $\Sigma_\tau$ be $n=f^{-1}\partial_t$. The conserved Killing energy is the flux
$$
E(\tau)=\int_{\Sigma_\tau}T_{ab}n^aK^b\sqrt h\,d^3x.
$$
Substitution of the <Klein-Gordon scalar stress-energy tensor> gives exactly
$$
E(\tau)=\frac12\int_{\Sigma_\tau}
\left[f^{-1}(\partial_t\psi)^2
+fh^{ij}\partial_i\psi\partial_j\psi+f\mu^2\psi^2\right]
\sqrt h\,d^3x.
$$
Apply the <divergence theorem> to the slab $[\tau_1,\tau_2]\times\mathbb R^3$. The spatial-boundary flux vanishes because $\psi$ decays, and $\nabla_aJ^a=0$ makes the two time-slice fluxes equal. Thus this <conserved scalar-field energy in a static spacetime> is independent of $\tau$.

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 2 / c / ii / Solution

{c}

For a real scalar satisfying $\nabla_a\nabla^a\psi-\mu^2\psi=0$,
$$
T_{ab}=\nabla_a\psi\nabla_b\psi-\frac12g_{ab}\left(\nabla_c\psi\nabla^c\psi+\mu^2\psi^2\right)
$$
is symmetric and covariantly conserved.


Codex @codex  0

 Incoming links: Klein-Gordon scalar stress-energy tensor