Divergence identity for the scalar-field stress-energy tensor

Differentiating the Hilbert tensor. By [F1] and [F2], using that the components ${\eta }^{ab}$ are constant, ${\mathrm{\partial }}_a{T}^{ab}=K\left[{\mathrm{\partial }}_a\right({\mathrm{\partial }}^a\phi {\mathrm{\partial }}^b\phi )-{\mathrm{\partial }}^b(\frac{1}{2}{\mathrm{\partial }}_d\phi {\mathrm{\partial }}^d\phi +V(\phi )\left)\right]=K[(\mathrm{\square }\phi ){\mathrm{\partial }}^b\phi +{\mathrm{\partial }}^a\phi {\mathrm{\partial }}_a{\mathrm{\partial }}^b\phi -{\mathrm{\partial }}^b(\frac{1}{2}{\mathrm{\partial }}_d\phi {\mathrm{\partial }}^d\phi +V(\phi )\left)\right]\mathrm{.}$

Cancellation of the second-order terms. For a smooth $\phi$ the two second-order terms of step 1.1 coincide: the first is $X^b:={\mathrm{\partial }}^a\phi {\mathrm{\partial }}_a{\mathrm{\partial }}^b\phi ={\eta }^{ad}{\eta }^{bc}{\mathrm{\partial }}_d\phi {\mathrm{\partial }}_a{\mathrm{\partial }}_c\phi$, while expanding the derivative of the kinetic term by [F2] gives ${\mathrm{\partial }}^b\left(\frac{1}{2}{\mathrm{\partial }}_d\phi {\mathrm{\partial }}^d\phi \right)={\eta }^{de}{\mathrm{\partial }}_d\phi {\mathrm{\partial }}^b{\mathrm{\partial }}_e\phi =:Y^b$ and adding $V^{\mathrm{\prime }}(\phi ){\mathrm{\partial }}^b\phi$ for the potential term; after substituting ${\mathrm{\partial }}^b{\mathrm{\partial }}_e\phi ={\eta }^{bc}{\mathrm{\partial }}_c{\mathrm{\partial }}_e\phi$ and relabelling the dummy indices, both $X^b$ and $Y^b$ equal the same contraction ${\eta }^{ad}{\eta }^{bc}{\mathrm{\partial }}_a\phi {\mathrm{\partial }}_c{\mathrm{\partial }}_d\phi$, the identification of the mixed second partials ${\mathrm{\partial }}_a{\mathrm{\partial }}_c\phi ={\mathrm{\partial }}_c{\mathrm{\partial }}_a\phi$ being [F3]. Hence the second-order terms cancel and ${\mathrm{\partial }}_a{T}^{ab}=K(\mathrm{\square }\phi -V^{\mathrm{\prime }}(\phi )){\mathrm{\partial }}^b\phi$ on $\mathrm{\Omega }$.

Consequences. If $\mathrm{\square }\phi =V^{\mathrm{\prime }}(\phi )$ then the right-hand side vanishes, so $T$ is conserved; conversely if ${\mathrm{\partial }}_a{T}^{ab}=0$ for all $b$ then $(\mathrm{\square }\phi -V^{\mathrm{\prime }}(\phi )){\mathrm{\partial }}^b\phi =0$ for all $b$, and since ${\mathrm{\partial }}^b\phi ={\eta }^{bc}{\mathrm{\partial }}_c\phi$ with ${\eta }^{bc}$ invertible this is equivalent to $(\mathrm{\square }\phi -V^{\mathrm{\prime }}(\phi )){\mathrm{\partial }}_c\phi =0$ for every $c$, i.e. to the vanishing of the pointwise product of the scalar $\mathrm{\square }\phi -V^{\mathrm{\prime }}(\phi )$ with the covector $d\phi$; at a point where some ${\mathrm{\partial }}_c\phi \mathrm{\ne }0$ this forces $\mathrm{\square }\phi =V^{\mathrm{\prime }}(\phi )$ there, hence on any open set on which $d\phi$ is nowhere zero. Off shell the identity is a balance with the four-force density $f^b=K(\mathrm{\square }\phi -V^{\mathrm{\prime }}(\phi )){\mathrm{\partial }}^b\phi$.

The divergence identity and all its consequences are proved in steps 2.1 and 3.1; the identity holds for every smooth $\phi$, the on-shell statement is the specialization of [F4], the converse is stated with its exact $d\phi =0$ caveat, and no choice principle is used.