Varying the scalar and integrating by parts gives

δ S = \int_{M} (\nabla_{a} \nabla^{a} ψ - μ^{2} ψ) δ ψ d vol_{g},

\nabla_{a} \nabla^{a} ψ - μ^{2} ψ = 0 .

ii

Solution

T^{ab} = \nabla^{a} ψ \nabla^{b} ψ - \frac{1}{2} g^{ab} (\nabla_{c} ψ \nabla^{c} ψ + μ^{2} ψ^{2}) .

\nabla_{a} T^{ab} = (\nabla_{a} \nabla^{a} ψ - μ^{2} ψ) \nabla^{b} ψ,

because the two Hessian terms cancel by symmetry. It vanishes on every solution of the Klein-Gordon equation.

\nabla_{(a} K_{b)} = 0

J^{a} = T^{a}_{b} K^{b}

\nabla_{a} J^{a} = (\nabla_{a} T^{ab}) K_{b} + T^{ab} \nabla_{a} K_{b} .

The first term vanishes on shell, while the second contracts the symmetric tensor

T^{ab}

with the antisymmetric derivative selected by Killing's equation. Thus

\nabla_{a} J^{a} = 0

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