The Hilbert stress-energy tensor of the scalar-field action on flat spacetime

Fix a smooth symmetric compactly supported $(0,2)$ field $h$ and set $g_{ab}(s,x):={\eta }_{ab}+s{h}_{ab}(x)$ for $s\in \mathbb{R}$, $x\in \mathrm{\Omega }$. Since $h_{ab}=0$ off $\mathrm{supp}h$, one has $g(s,x)=\eta$ off $\mathrm{supp}h$. The function $(s,x)\mapsto \det g(s,x)$ is, by [F3], a polynomial in $s$ with coefficients continuous in $x$ and equals $-1$ at $s=0$; for each $x$ continuity in $(s,x^{\mathrm{\prime }})$ at $(0,x)$ yields $\epsilon (x)>0$ and a neighbourhood $U(x)$ of $x$ with $\det g(s,x^{\mathrm{\prime }})<0$ for $\mathrm{\mid }s\mathrm{\mid }<\epsilon (x)$, $x^{\mathrm{\prime }}\in U(x)$, and compactness of $\mathrm{supp}h$ gives finitely many such sets and a common $s_0>0$ with $\det g(s,x)<0$ for all $\mathrm{\mid }s\mathrm{\mid }<s_0$, $x\in \mathrm{\Omega }$. Hence for $\mathrm{\mid }s\mathrm{\mid }<s_0$ the matrix $g(s,x)$ is invertible, its inverse entries are $g^{ab}(s,x)=\det (g(s,x){)}^{-1}\mathrm{adj}(g(s,x){)}^{ab}$ by [F4], and $D_h(s,x):=[\frac{1}{2}{g}^{ab}(s,x){\mathrm{\partial }}_a\phi {\mathrm{\partial }}_b\phi +V(\phi )]\sqrt{-\det g(s,x)}$ is defined; the maps $(s,x)\mapsto g^{ab}(s,x)$ and $(s,x)\mapsto \det g(s,x)$, being polynomials in the continuous entries and quotients with nowhere-vanishing denominator $\det g$, are continuous on $(-s_0,s_0)\times \mathrm{\Omega }$ by [F7], and so are $\sqrt{-\det g(s,x)}$ by [F5] and $D_h(s,x)$ by [F7], and $D_h(s,x)=D_h(0,x)$ whenever $h_{ab}(x)=0$, in particular off $\mathrm{supp}h$.

Derivative identities at $s=0$. (i) Differentiating the components of the identity $g(s,x)g(s,x{)}^{-1}=I$ with respect to $s$ at $s=0$ by [F5] and [F8], and using $g_{ab}(0)={\eta }_{ab}$, $g^{ab}(0)={\eta }^{ab}$ and ${\eta }_{ac}{\eta }^{cb}={\delta }_a^b$, gives ${\mathrm{\partial }}_s{g}^{ab}(0,x)=-h^{ab}(x)$, where $h^{ab}:={\eta }^{ac}{\eta }^{bd}{h}_{cd}$. (ii) Column multilinearity [F3] expands $\det (\eta +sh)=\det \eta +s\mathrm{\Sigma }+s^2(\cdots )$ with $\mathrm{\Sigma }:={\sum }_j\det (\eta \text{ with its }j\text{-th column replaced by the }j\text{-th column of }h)$; expanding each such determinant along the replaced column shows $\mathrm{\Sigma }={\sum }_{i,j}{h}_{ij}{C}_{ij}(\eta )=\mathrm{tr}(\mathrm{adj}(\eta )h)$, and by [F4] $\mathrm{adj}(\eta )=\det (\eta ){\eta }^{-1}=-{\eta }^{-1}$, so $\mathrm{\Sigma }=-\mathrm{tr}({\eta }^{-1}h)=-{\eta }^{ab}{h}_{ab}$, because $\mathrm{tr}({\eta }^{-1}h)={\eta }^{ac}{h}_{ca}={\eta }^{ab}{h}_{ab}$ by symmetry. (iii) Hence $\frac{d}{ds}\det g(0,x)=-{\eta }^{ab}{h}_{ab}(x)$, and since $-\det g(0,x)=1$ the chain rule [F6] with the power rule [F5] gives $\frac{d}{ds}\sqrt{-\det g(s,x)}{\mid }_{s=0}=\frac{1}{2}{\eta }^{ab}{h}_{ab}(x)$.

The field $T^{ab}:=K[{\mathrm{\partial }}^a\phi {\mathrm{\partial }}^b\phi -{\eta }^{ab}(\frac{1}{2}{\mathrm{\partial }}_d\phi {\mathrm{\partial }}^d\phi +V(\phi ))]$ is smooth on $\mathrm{\Omega }$ by iterating the product and chain rules [F5], [F6], because $\phi$ and $V$ are smooth, $V(\phi )$ is a smooth composition, and ${\eta }^{ab}$ is constant; it is symmetric, because ${\eta }^{ab}={\eta }^{ba}$ and ${\mathrm{\partial }}^a\phi {\mathrm{\partial }}^b\phi ={\mathrm{\partial }}^b\phi {\mathrm{\partial }}^a\phi$.

Apply [F1] to $F(s,x):=-\frac{K}{c}[D_h(s,x)-D_h(0,x)]$ on $(-s_0,s_0)\times \mathrm{\Omega }$: step 1.1 shows that $F$ is continuous with the stated compact support in $x$ for every $s$, and ${\mathrm{\partial }}_{s}F$ is continuous by the same arguments applied to the derivative of $D_h$; hence ${\mathcal{A}}_h$ is $C^1$ and ${\mathcal{A}}_h^{\mathrm{\prime }}(0)=-\frac{K}{c}{\int }_{\mathrm{\Omega }}{\mathrm{\partial }}_s{D}_h(0,x)d^{4}x$. By the product and chain rules [F5], [F6] and step 1.2, ${\mathrm{\partial }}_s{D}_h(0,x)=-\frac{1}{2}{h}^{ab}{\mathrm{\partial }}_a\phi {\mathrm{\partial }}_b\phi +\frac{1}{2}{\eta }^{ab}{h}_{ab}(\frac{1}{2}{\mathrm{\partial }}_d\phi {\mathrm{\partial }}^d\phi +V(\phi ))\mathrm{.}$ Therefore ${\mathcal{A}}_h^{\mathrm{\prime }}(0)=\frac{K}{2c}{\int }_{\mathrm{\Omega }}[h^{ab}{\mathrm{\partial }}_a\phi {\mathrm{\partial }}_b\phi -{\eta }^{ab}{h}_{ab}(\frac{1}{2}{\mathrm{\partial }}_d\phi {\mathrm{\partial }}^d\phi +V(\phi ))]d^{4}x=\frac{1}{2c}{\int }_{\mathrm{\Omega }}{T}^{ab}{h}_{ab}{d}^{4}x$, since $h^{ab}{\mathrm{\partial }}_a\phi {\mathrm{\partial }}_b\phi ={\mathrm{\partial }}^a\phi {\mathrm{\partial }}^b\phi h_{ab}$.

Uniqueness in (b). Let $Q$ be a continuous symmetric $(2,0)$ field with ${\int }_{\mathrm{\Omega }}{Q}^{ab}{h}_{ab}{d}^{4}x=0$ for every smooth symmetric compactly supported $h$. Fix index pairs and a smooth compactly supported $\psi$: for $a=b$ take $h_{aa}:=\psi$; for $a\mathrm{\ne }b$ take $h_{ab}:=h_{ba}:=\psi$ and all other components zero. In the first case ${\int }_{\mathrm{\Omega }}{Q}^{aa}\psi d^{4}x=0$, in the second ${\int }_{\mathrm{\Omega }}2Q^{ab}\psi d^{4}x=0$ by symmetry of $Q$; in both cases the continuous function $Q^{ab}$ pairs to zero against every smooth compactly supported $\psi$, so $Q^{ab}=0$ on $\mathrm{\Omega }$ by [F2]. All sixteen components vanish, hence $Q=0$. Applying this to the difference of two continuous symmetric fields with the same pairing against every $h$ shows that the field $T$ of steps 1.3 and 2.1 is the unique such field satisfying ${\int }_{\mathrm{\Omega }}{T}^{ab}{h}_{ab}{d}^{4}x=2c{\mathcal{A}}_h^{\mathrm{\prime }}(0)$.

Claim (a) is step 2.1, claim (b) is steps 1.3, 2.1 and 3.1, and claim (c) is a statement about the hypotheses used: the argument allows any nonempty open $\mathrm{\Omega }$, integrates only the compactly supported density difference $D_h(s,\cdot )-D_h(0,\cdot )$ (identically zero off $\mathrm{supp}h$), requires no convergence of $D_h(0,\cdot )$ on $\mathrm{\Omega }$ and invokes no boundary condition; if the background action ${\int }_{\mathrm{\Omega }}{D}_h(0,x)d^{4}x$ converges, ${\mathcal{A}}_h$ is the difference of the two actions and yields the same derivative. Every choice made is finite; no choice principle is used.