Compact potential and metric variation of the Maxwell action

Interpret a compactly supported variation as a smooth family equal to its background off one fixed compact set $K$ for all sufficiently small parameters. Connection variations $A+s\alpha$ have $K=\mathrm{supp}\alpha$ and $F(s)=F+sd\alpha$ in the fixed sector. For a prescribed metric direction $q$, put $h_{ab}=-g_{ac}{g}_{bd}{q}^{cd}$ and $g(s)=g+sh$. If $K=\mathrm{\varnothing }$, then $h=0$ and the metric is unchanged. Otherwise, to choose one parameter interval valid on all of $K$, fix $p\in K$ and make a constant linear coordinate change so that $g(p)$ has matrix $\eta =\mathrm{diag}(-1,1,1,1)$. By continuity choose a relatively compact chart neighborhood on which $\mathrm{\parallel }g(x)-\eta {\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<1\mathrm{/}2$ and $\mathrm{\parallel }h(x){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ is bounded. Choose ${\delta }_p>0$ so $\mathrm{\mid }s\mathrm{\mid }\mathrm{\parallel }h(x){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<1\mathrm{/}2$ there for $\mathrm{\mid }s\mathrm{\mid }<{\delta }_p$. Then $A(x,s):=g(x)+sh(x)$ is positive on the fixed three-dimensional spatial coordinate subspace and negative on the time coordinate line. It is nondegenerate: if $A(x,s)v=0$, then $v=-{\eta }^{-1}(A(x,s)-\eta )v$, contradicting $\mathrm{\parallel }{\eta }^{-1}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=1$ and $\mathrm{\parallel }A(x,s)-\eta {\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<1$. Thus its inertia has three positive and one negative directions, exactly $(-,+,+,+)$. Such neighborhoods cover $K$; compactness gives a finite subcover and the minimum of their parameter bounds works on all of $K$. Outside $K$, $h=0$ and the metric is unchanged. Thus the action-density difference $\omega (s)-\omega (0)$ is a smooth top form supported in a common $K$, even when neither background action is finite. Its integral in [F12] is the relative action; additivity of integrals makes successive relative differences consistent. No subtraction of infinite action values is involved.

Fix the chart cover and partition $({\rho }_i)$ of [F12] independently of $s$. Local finiteness gives a neighborhood of each point meeting only finitely many supports; a finite such neighborhood cover of $K$ shows that only finitely many ${\rho }_i$ meet $K$. Each ${\rho }_i(\omega (s)-\omega (0))$ has compact support inside its chart. Extend its coefficient by zero and enclose that support in a coordinate rectangle. On a closed parameter interval about zero the coefficient derivative is uniformly continuous on the product rectangle; the mean-value argument in [F3] makes its difference quotients converge uniformly. The estimate $\mathrm{\mid }{\int }_{Q}u\mathrm{\mid }\le \mathrm{vol}(Q){\sup }_Q\mathrm{\mid }u\mathrm{\mid }$ from [F7] permits differentiation inside this chart integral. Summing finitely many terms proves $\delta {\int }_M(\omega (s)-\omega (0))={\int }_M\delta \omega$. Choice independence is exactly [F12].

For any smooth compactly supported vector field $W$, the same partition has only finitely many terms meeting its support. Each ${\rho }_{i}W$ is compactly supported inside one chart. In that chart its density divergence is ${\mathrm{\partial }}_a(w{\rho }_i{W}^a)$ by [F4]. Zero extension is smooth, and FTC/Fubini on a containing rectangle gives zero integral, because its coefficients vanish near every rectangle face. The global integral agrees with this chart integral by [F12]. On a neighborhood of $\mathrm{supp}W$, ${\sum }_i{\rho }_i=1$ and ${\sum }_{i}d{\rho }_i=0$, so ${\sum }_i{\mathrm{\nabla }}_a({\rho }_i{W}^a)={\mathrm{\nabla }}_a{W}^a$; off the support both sides vanish. Finite additivity therefore gives ${\int }_M{\mathrm{\nabla }}_a{W}^a{\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$. This proves the compact divergence rule without requiring the support to fit one chart.

For a connection variation in [A1], $\delta F=d\alpha$; in components $\delta F_{ab}=2{\mathrm{\nabla }}_{[a}{\alpha }_{b]}$ because the Levi-Civita Christoffel symbols are symmetric in their lower indices. Holding $g$ fixed gives $\delta (F_{ab}{F}^{ab})=2F^{ab}\delta F_{ab}=4F^{ab}{\mathrm{\nabla }}_a{\alpha }_b$. Differentiating the coefficient in the action therefore yields ${\delta }_A{S}_{\mathrm{E}\mathrm{M}}=-({\mu }_{0}c{)}^{-1}{\int }_M{F}^{ab}{\mathrm{\nabla }}_a{\alpha }_b{\mathrm{v}\mathrm{o}\mathrm{l}}_g$.

Let $q^{ab}=\delta g^{ab}$ and hold $F_{ab}$ fixed. From $F_{cd}{F}^{cd}=g^{ca}{g}^{db}{F}_{cd}{F}_{ab}$, symmetry of $q$ and antisymmetry of $F$ give $\delta (F_{cd}{F}^{cd})=2F_{ac}{F}_b{}^c{q}^{ab}$. The volume variation in [F3] then gives ${\delta }_g(F_{cd}{F}^{cd}{\mathrm{v}\mathrm{o}\mathrm{l}}_g)=(2F_{ac}{F}_b{}^c-\frac{1}{2}{g}_{ab}{F}_{cd}{F}^{cd})q^{ab}{\mathrm{v}\mathrm{o}\mathrm{l}}_g\mathrm{.}$ By [F2], this is the variation of the same action density as its wedge-form representative. Multiplying by $-1\mathrm{/}(4{\mu }_{0}c)$ and comparing with the definition in part (b) gives the displayed $T_{\mathrm{E}\mathrm{M},ab}$. The variation is local and compactly supported, so no boundary term is discarded.

Set $W^a:=F^{ab}{\alpha }_b$. The product rule gives ${\mathrm{\nabla }}_a{W}^a=({\mathrm{\nabla }}_a{F}^{ab}){\alpha }_b+F^{ab}{\mathrm{\nabla }}_a{\alpha }_b$. By step 3.1, ${\int }_M{\mathrm{\nabla }}_a{W}^a{\mathrm{v}\mathrm{o}\mathrm{l}}_g=0$, for arbitrary compact support of $\alpha$. Substitution into step 3.2 gives ${\delta }_A{S}_{\mathrm{E}\mathrm{M}}=({\mu }_{0}c{)}^{-1}{\int }_M({\mathrm{\nabla }}_a{F}^{ab}){\alpha }_b{\mathrm{v}\mathrm{o}\mathrm{l}}_g$, proving the connection-variation formula.

Adding [A2] to step 4.1 gives ${\delta }_A(S_{\mathrm{E}\mathrm{M}}+S_{\mathrm{m}})[\alpha ]=({\mu }_{0}c{)}^{-1}{\int }_M({\mathrm{\nabla }}_a{F}^{ab}+{\mu }_0{j}^b){\alpha }_b{\mathrm{v}\mathrm{o}\mathrm{l}}_g$. If ${\mathrm{\nabla }}_a{F}^{ab}+{\mu }_0{j}^b=0$, this integral vanishes for every allowed $\alpha$. Conversely, if it vanishes for every smooth compactly supported one-form supported in one chart rectangle, fix a point and suppose a component $K^{b_0}:={\mathrm{\nabla }}_a{F}^{a{b}_0}+{\mu }_0{j}^{b_0}$ is nonzero there. By continuity, shrink to a coordinate ball with compact closure where $K^{b_0}$ has one strict sign and $w>0$. In coordinates centered at that point define $\varphi (x)=\exp (-1\mathrm{/}(r^2-\mathrm{\mid }x{\mathrm{\mid }}^2))$ for $\mathrm{\mid }x\mathrm{\mid }<r$ and $\varphi (x)=0$ for $\mathrm{\mid }x\mathrm{\mid }\ge r$. Every derivative approaching $\mathrm{\mid }x\mathrm{\mid }=r$ is a finite sum of a polynomial in $1\mathrm{/}(r^2-\mathrm{\mid }x{\mathrm{\mid }}^2)$ times the same exponential, and $s^{-m}{e}^{-1\mathrm{/}s}\to 0$ as $s\downarrow 0$ for every $m\ge 0$; hence $\varphi$ is a smooth nonnegative compactly supported function, positive inside the ball. Set $\alpha =\varphi d{x}^{b_0}$ in the chart and extend it by zero outside the compact support; this gives an admissible global one-form in [A1]. The bump and rectangle are explicitly constructed, so the ambient convention [F5] introduces no arbitrary choice. The product $K^{b_0}\varphi w$ is continuous and supported in a compact rectangle containing the bump support, so [F8] makes its full integral well-defined. At the center, $\mathrm{\mid }{K}^{b_0}\varphi w\mathrm{\mid }>0$; continuity gives a smaller closed nondegenerate rectangle $R$ on which this coefficient has one strict sign and absolute value at least some $\eta >0$. By [F8] it is Riemann integrable there. Applying [F11] to its sign-adjusted coefficient shows that the absolute value of the integral over $R$ is at least $\eta \mathrm{vol}(R)>0$. On the entire bump support the product has the same sign, since $K^{b_0}$ has one sign on the ball and $\varphi ,w$ are nonnegative; therefore its full integral over the support cannot vanish. This contradicts the assumed stationarity integral. Thus every component $K^b$ vanishes at every point. Conversely its pointwise vanishing makes the variation zero for every compactly supported global one-form by step 4.1. This proves the stated global stationarity equivalence under [A2]–[A3].

When the Euler–Lagrange equation in step 5.1 holds, apply [F6] to obtain $d({\star }_{g}F)={\mu }_0{\iota }_j{\mathrm{v}\mathrm{o}\mathrm{l}}_g$. This uses the conditional current interface of [A1] and the exact current-sign identity proved in the earlier item; it adds no global potential or matter-action claim.