Trivial-potential classical charged scalar system

Since $A$ and $\phi$ are global smooth fields, $F=dA$, $\pi =D\phi$, and $D\pi$ are global smooth fields of the stated real or complex tensor types. The composition $V(\mathrm{\mid }\phi {\mathrm{\mid }}^2)$ is smooth and real because $\mathrm{\mid }\phi {\mathrm{\mid }}^2(M)\subset \mathcal{U}$ and $V$ is smooth and real-valued. The kinetic contraction $g^{ab}\overline{{\pi }_a}{\pi }_b$ is real: complex conjugation followed by exchanging the dummy indices $a,b$ leaves it unchanged because $g^{ab}=g^{ba}$. Thus $L_{\phi }$ is a global smooth real scalar, and its product with ${\mathrm{v}\mathrm{o}\mathrm{l}}_g$ is a real smooth four-form. On a compact set its integral is finite; on noncompact $M$ the Definition uses only finite integrals or compactly supported differences, so it assigns no divergent action value.

The tensor $\mathrm{Re}(\overline{{\pi }_a}{\pi }_b)$ is symmetric because interchanging $a,b$ conjugates the scalar and real part is unchanged. The Maxwell bilinear $F_{ac}{F}_b{}^c$ is also symmetric: after lowering the contracted index it is $g^{cd}{F}_{ac}{F}_{bd}$, which is unchanged under $a\leftrightarrow b$ by symmetry of $g^{cd}$ and commutativity of scalar multiplication. Hence $T^{\phi }$, $T^{\mathrm{E}\mathrm{M}}$, and $T$ are smooth symmetric covariant two-tensors, while $j$ is a smooth vector field. The equations therefore compare tensors of the displayed types, with no variation or empirical assertion added to the definition.

In length coordinates, $[D_a\phi ]=[{\pi }_a]={\mathrm{m}}^{-1}$, so both terms of $L_{\phi }$ have units ${\mathrm{m}}^{-2}$ and $[S_{\phi }]=(\mathrm{J}\mathrm{s}{\mathrm{m}}^{-2})({\mathrm{m}}^{-2})({\mathrm{m}}^4)=\mathrm{J}\mathrm{s}$. The scalar stress has units $[K_{\phi }][\pi {]}^2=(\mathrm{J}{\mathrm{m}}^{-1})({\mathrm{m}}^{-2})=\mathrm{J}{\mathrm{m}}^{-3}$. The electromagnetic stress has units $[F{]}^2\mathrm{/}[{\mu }_0]=(\mathrm{W}{\mathrm{b}}^2{\mathrm{m}}^{-4})\mathrm{/}(\mathrm{H}{\mathrm{m}}^{-1})=\mathrm{J}{\mathrm{m}}^{-3}$, using $\mathrm{H}=\mathrm{W}\mathrm{b}{\mathrm{A}}^{-1}$ and $\mathrm{W}\mathrm{b}=\mathrm{J}{\mathrm{A}}^{-1}$. Also $[j^a]=(\mathrm{J}{\mathrm{m}}^{-1})(\mathrm{W}{\mathrm{b}}^{-1})({\mathrm{m}}^{-1})=\mathrm{A}{\mathrm{m}}^{-2}$. Finally $[\kappa ]=8\pi [G_N]\mathrm{/}[c{]}^4={\mathrm{s}}^2\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{m}}^{-1}$, hence $[\kappa T_{ab}]={\mathrm{m}}^{-2}$, matching curvature and $\mathrm{\Lambda }$. These are dimensional identities under the assigned SI bookkeeping.

In a smooth chart, inversion of a nondegenerate $C^3$ matrix is $C^3$; the Christoffel formula therefore gives a $C^2$ connection and the curvature formula a $C^1$ tensor. Smooth coordinate changes preserve these regularities. The ordinary chain and product rules give $F,P\in C^1$, the stated continuous Euler–Lagrange expressions, and $C^1$ current and stress. Symmetry of $g^{-1}$ makes $g^{ab}\overline{P_a}{P}_b$ real and $\mathrm{Re}(\overline{P_a}{P}_b)$ symmetric; the Maxwell quadratic is also symmetric. The coordinate transformation laws follow by the chain rule, so all displayed equations compare global tensors. Commuting the continuous second partial derivatives of $A$ gives $dF=0$.

Let $K=\mathrm{supp}\omega$. Apply [F2] and multiply $\omega$ by its finitely many cutoffs. Each continuous coefficient has compact support strictly inside its chart, so zero extension is continuous and has a bounding rectangle; [F4] defines its finite integral. On a chart overlap the top-form coefficient transforms by the determinant of the smooth coordinate change. Its orientation sign converts that determinant to the absolute determinant, so [F3] gives the same signed integral in either chart. For two finite localizations $({\chi }_i)$ and $({\tau }_j)$, insert ${\chi }_i\omega ={\sum }_j{\chi }_i{\tau }_j\omega$ and the corresponding identity with $i,j$ exchanged. Every product has compact support in the chart intersection. Finite linearity from [F5] and the overlap identity make both sums equal to the same double sum. Thus ${\mathcal{I}}_M$ is well defined, linear and local. Empty support gives the empty sum zero.

Each action density has continuous coefficients. Configurations agreeing off a compact set have density difference supported in that set, including the curvature term, because their derivatives agree on its open complement. Step 1.2 therefore defines every stated relative action. For three mutually compactly differing configurations the density differences add on the union of their compact supports, so their relative actions add. On smooth fields the construction is precisely the same finite chart integral and the formulas in [F1]; hence it agrees with the original smooth relative action.

On the union $K$ of the three direction supports, nondegeneracy and the chosen time cone persist under a uniformly small perturbation: use finitely many chart neighborhoods of the compact set and take the minimum of their positive parameter bounds. The attained set $\mathrm{\mid }\phi {\mathrm{\mid }}^2(K)$ is a compact subset of the open set $\mathcal{U}$, so uniform convergence of $\mathrm{\mid }\phi +\epsilon \eta {\mathrm{\mid }}^2$ keeps it inside $\mathcal{U}$ for small $\epsilon$; outside $K$ nothing changes. The resulting metric is $C^3$ and the matter fields are $C^2$, with $\delta g_{ab}=-g_{ac}{g}_{bd}{q}^{cd}$. The action densities and their parameter derivatives are continuous on compact parameter-chart rectangles. Consequently these directions are actual variations of the declared class, not only formal symbols.

A real $C^3$ function $\chi$ gives $A^{\mathrm{\prime }}=A+d\chi \in C^2$ and ${\phi }^{\mathrm{\prime }}=e^{i\lambda \chi }\phi \in C^2$. The product rule gives $P^{\mathrm{\prime }}=e^{i\lambda \chi }P$ and $F^{\mathrm{\prime }}=F$, so the action density is unchanged. Pullback by a smooth orientation-preserving diffeomorphism also preserves the classical class and the density formulas. For compactly supported transformations, step 1.2 and [F3] give invariance of the relative action. These are mathematical invariances, not a physical adoption or a solution-existence theorem.

Fix the compact support $K$ of the directions. Use [F2]'s finite smooth localization near $K$, independently of the variation parameter. In each chart the localized density difference and its parameter derivative are continuous on a fixed compact parameter-rectangle product: every density is a smooth algebraic expression in the nondegenerate metric, its derivatives through order two, and the matter fields through order one. The families in [F1] have the required joint derivatives. The one-variable fundamental theorem gives the difference quotient as the average of those parameter derivatives along the parameter segment. Uniform continuity on this compact product makes that quotient converge uniformly to the derivative at zero. Monotonicity and the absolute-value estimate [F6], compared with the constant ${\sup }_Q\mathrm{\mid }f\mathrm{\mid }$, give $\mathrm{\mid }{\int }_{Q}f\mathrm{\mid }\le \mathrm{vol}(Q){\sup }_Q\mathrm{\mid }f\mathrm{\mid }$; this passes the limit through each finite Riemann integral. Thus the relative actions can be differentiated by differentiating their densities, even when the background global actions diverge.

Let $W$ be any compactly supported $C^1$ vector field. Write $w=\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$. The Christoffel formula and determinant derivative give ${\mathrm{\Gamma }}^a{}_{ab}={\mathrm{\partial }}_b\log w$, hence ${\mathrm{\nabla }}_a{W}^a{\mathrm{v}\mathrm{o}\mathrm{l}}_g={\mathrm{\partial }}_a(w{W}^a)d^{4}x$ in a positive chart. For each finite cutoff ${\chi }_i$ in [F2], $w{\chi }_i{W}^a$ extends by zero to a $C^1$ compactly supported coefficient on a bounding rectangle. Its partial derivatives and all sections are continuous, so [F3]–[F5] integrate its coordinate divergence to zero: the opposite-face values vanish. Summing over $i$ gives the same result for $W$, because ${\sum }_i{\chi }_i=1$ and ${\sum }_{i}d{\chi }_i=0$ near its support. Therefore ${\mathcal{I}}_M(({\mathrm{\nabla }}_a{W}^a){\mathrm{v}\mathrm{o}\mathrm{l}}_g)=0$ for this $C^1$ field. No smoothness approximation or unproved rough Stokes assertion is used.

Put $h_{ab}=\delta g_{ab}=-g_{ac}{g}_{bd}{q}^{cd}$. Differentiation of $g^{ac}{g}_{cb}={\delta }_b^a$ and of the determinant gives $\delta w=-\frac{1}{2}w{g}_{ab}{q}^{ab}$. Differentiating the coordinate formula for the Christoffel symbols and collecting the terms containing $h$ gives $\delta {\mathrm{\Gamma }}^a{}_{bc}=\frac{1}{2}{g}^{ad}({\mathrm{\nabla }}_b{h}_{cd}+{\mathrm{\nabla }}_c{h}_{bd}-{\mathrm{\nabla }}_d{h}_{bc})\mathrm{.}$ To differentiate Ricci, use $R_{ab}={\mathrm{\partial }}_c{\mathrm{\Gamma }}^c{}_{ab}-{\mathrm{\partial }}_b{\mathrm{\Gamma }}^c{}_{ac}+{\mathrm{\Gamma }}^c{}_{cd}{\mathrm{\Gamma }}^d{}_{ab}-{\mathrm{\Gamma }}^c{}_{bd}{\mathrm{\Gamma }}^d{}_{ac}$. The differentiated quadratic terms are exactly the connection terms that complete the two derivatives to $\delta R_{ab}={\mathrm{\nabla }}_c\delta {\mathrm{\Gamma }}^c{}_{ab}-{\mathrm{\nabla }}_b\delta {\mathrm{\Gamma }}^c{}_{ac}$. Metric compatibility in this coordinate connection then gives $\delta ((R-2\mathrm{\Lambda })w)=w(G_{ab}+\mathrm{\Lambda }{g}_{ab})q^{ab}+{\mathrm{\partial }}_a(w{V}^a),V^a=g^{bc}\delta {\mathrm{\Gamma }}^a{}_{bc}-g^{ab}\delta {\mathrm{\Gamma }}^c{}_{bc}\mathrm{.}$ All expressions exist classically: $g,h$ are $C^3$, $\delta \mathrm{\Gamma }$ is $C^2$, and $V$ is a compactly supported $C^2$ vector field. Steps 1.1–1.2 therefore yield the Einstein–Hilbert first variation $c^3(16\pi G_N{)}^{-1}{\mathcal{I}}_M((G+\mathrm{\Lambda }g{)}_{ab}{q}^{ab}{\mathrm{v}\mathrm{o}\mathrm{l}}_g)$.

At fixed metric, $\delta F_{ab}={\mathrm{\partial }}_a{\alpha }_b-{\mathrm{\partial }}_b{\alpha }_a=2{\mathrm{\nabla }}_{[a}{\alpha }_{b]}$; torsion-freeness follows directly from symmetry of the Christoffel formula. Thus $\delta (F_{ab}{F}^{ab})=4F^{ab}{\mathrm{\nabla }}_a{\alpha }_b$. The $C^1$ compact vector $W^a=F^{ab}{\alpha }_b$ satisfies ${\mathrm{\nabla }}_a{W}^a=({\mathrm{\nabla }}_a{F}^{ab}){\alpha }_b+F^{ab}{\mathrm{\nabla }}_a{\alpha }_b$. Steps 1.1–1.2 give ${\delta }_A{S}_{\mathrm{E}\mathrm{M}}=({\mu }_{0}c{)}^{-1}{\mathcal{I}}_M(({\mathrm{\nabla }}_a{F}^{ab}){\alpha }_b{\mathrm{v}\mathrm{o}\mathrm{l}}_g)$. At fixed lower $F$, direct inverse-metric differentiation gives $\delta (F_{cd}{F}^{cd})=2F_{ac}{F}_b{}^c{q}^{ab}$. Combining it with $\delta w$ from step 2.1 gives ${\delta }_g{S}_{\mathrm{E}\mathrm{M}}=-(2c{)}^{-1}{\mathcal{I}}_M(T_{ab}^{\mathrm{E}\mathrm{M}}{q}^{ab}{\mathrm{v}\mathrm{o}\mathrm{l}}_g)$. This uses only $F\in C^1$ and the declared metric class.

For $\eta =\delta \phi$, direct differentiation gives ${\delta }_{\phi }{L}_{\phi }=2\mathrm{Re}(\overline{{\mathcal{D}}_a\eta }{P}^a+V^{\mathrm{\prime }}\bar{\eta }\phi )$. The product rule is ${\mathrm{\nabla }}_a(\bar{\eta }{P}^a)=\overline{{\mathcal{D}}_a\eta }{P}^a+\bar{\eta }{\mathcal{D}}_a{P}^a$: the opposite gauge signs cancel. Its real and imaginary parts are $C^1$ compact vector fields. Integrating their divergence by step 1.2 yields ${\delta }_{\phi }{S}_{\phi }=(K_{\phi }\mathrm{/}c)\mathrm{Re}{\mathcal{I}}_M(\bar{\eta }{E}_{\phi }{\mathrm{v}\mathrm{o}\mathrm{l}}_g)$. At fixed $g,\phi$, ${\delta }_A{P}_a=-i\lambda \phi {\alpha }_a$, so ${\delta }_A{L}_{\phi }=-2\lambda \mathrm{Im}(\bar{\phi }{P}^a){\alpha }_a$, giving the claimed $j$. At fixed lower $A$ and fixed $\phi$, ${\delta }_g{L}_{\phi }=\mathrm{Re}(\overline{P_a}{P}_b)q^{ab}$; the volume term then gives the displayed $T^{\phi }$. These calculations require no derivative beyond $\phi \in C^2$ and $A\in C^2$.

Add steps 2.1, 3.1 and 3.2. The equality $(2c{)}^{-1}=[c^3\mathrm{/}(16\pi G_N)]{\kappa }_E$ gives exactly the total first variation in the Statement, with its three independent directions. Its coefficients are continuous; the metric coefficient is symmetric. To see that testing detects a nonzero continuous component, take a positive chart where that component has one strict sign. A nonnegative smooth bump supported inside that chart, positive on a smaller closed rectangle, has a strictly positive signed pairing with the component times $w>0$. Such a bump can be taken as $e^{-1\mathrm{/}(r^2-\mathrm{\mid }x{\mathrm{\mid }}^2)}$ inside a coordinate ball and zero outside. The resulting integral is nonzero: on the smaller rectangle the continuous sign-adjusted coefficient has a positive lower bound $a>0$, so its integral is at least $a$ times that rectangle volume by [F6]. Cutting a containing rectangle along the smaller rectangle faces and using coordinate-slice additivity in [F6] shows the full integral is at least that positive value, since the integrand is nonnegative on the remainder. A one-form test isolates any connection component; diagonal and symmetrized off-diagonal metric tests isolate every symmetric component. Scalar tests $\eta =f$ and $\eta =if$ isolate the real and imaginary parts of $E_{\phi }$. Therefore stationarity for all these tests forces all three coefficients to vanish pointwise. Conversely their pointwise vanishing makes every first variation zero. The identity $dF=0$ in [F1] supplies the remaining homogeneous equation.

The charged product rule gives ${\mathrm{\nabla }}_a{j}^a=K_{\phi }\lambda \mathrm{Im}(\overline{P_a}{P}^a+\bar{\phi }{\mathcal{D}}_a{P}^a)$. The first contraction is real and $V^{\mathrm{\prime }}$ is real, giving the stated current identity. Differentiating $T^{\phi }$ and using ${\mathrm{\partial }}_b{L}_{\phi }=2\mathrm{Re}(\overline{P^a}{\mathcal{D}}_b{P}_a)+2V^{\mathrm{\prime }}\mathrm{Re}(\bar{\phi }{P}_b)$ gives ${\mathrm{\nabla }}_a(T^{\phi }{)}^a{}_b=K_{\phi }\mathrm{Re}(\overline{E_{\phi }}{P}_b+\overline{P^a}({\mathcal{D}}_a{P}_b-{\mathcal{D}}_b{P}_a))\mathrm{.}$ The symmetric Christoffel terms cancel in the difference. Expanding $P=\mathrm{\partial }\phi -i\lambda A\phi$ and commuting the continuous second scalar partial derivatives gives ${\mathcal{D}}_a{P}_b-{\mathcal{D}}_b{P}_a=-i\lambda F_{ab}\phi$. Its contribution to the real part is $-K_{\phi }\lambda F_{ab}\mathrm{Im}(\bar{\phi }{P}^a)=F_{ba}{j}^a$. This proves the two scalar off-shell identities using only first derivatives of $P$ and $C^2$ scalar partials.

Since $F=dA$ with $A\in C^2$, its homogeneous identity is ${\mathrm{\nabla }}_a{F}_{bc}+{\mathrm{\nabla }}_b{F}_{ca}+{\mathrm{\nabla }}_c{F}_{ab}=0$. Contraction with $F^{ac}$ gives $2F^{ac}{\mathrm{\nabla }}_a{F}_{bc}=\frac{1}{2}{\mathrm{\partial }}_b(F_{cd}{F}^{cd})$. The product rule in the explicit Maxwell tensor therefore yields ${\mathrm{\nabla }}_a(T^{\mathrm{E}\mathrm{M}}{)}^a{}_b={\mu }_0^{-1}[({\mathrm{\nabla }}_a{F}^{ac})F_{bc}+F^{ac}{\mathrm{\nabla }}_a{F}_{bc}-\frac{1}{4}{\mathrm{\partial }}_b(F_{cd}{F}^{cd})]={\mu }_0^{-1}{F}_{ba}{\mathrm{\nabla }}_c{F}^{ca}\mathrm{.}$ These are continuous identities for $F\in C^1$. If the sourced Maxwell equation holds, the last expression is $-F_{ba}{j}^a$, which cancels the scalar force term in step 4.2 when $E_{\phi }=0$. This proves total-stress conservation and current conservation before any Einstein equation. No smooth-only stress or Noether theorem is invoked at the classical regularity.

Expanding the divergence equation and retaining only derivatives of the independent $\pi$ variables gives $g^{00}{\mathrm{\partial }}_s{\pi }_0+g^{0i}{\mathrm{\partial }}_s{\pi }_i+g^{i0}{\mathrm{\partial }}_i{\pi }_0+g^{ij}{\mathrm{\partial }}_i{\pi }_j$. Substitute ${\mathrm{\partial }}_s{\pi }_i={\mathrm{\partial }}_i{\pi }_0-i\lambda A_i{\pi }_0-i\lambda F_{0i}\phi$ from the compatibility equations. Since $g^{i0}=g^{0i}$, the derivative terms become $g^{00}{\mathrm{\partial }}_s{\pi }_0+2g^{0i}{\mathrm{\partial }}_i{\pi }_0+g^{ij}{\mathrm{\partial }}_i{\pi }_j$; the substituted source terms contain no derivatives of $\pi$. Multiplying this row by $-1$ gives the top row of the stated principal system.

For each output row $r$, summing the $i$th compatibility equation against $g^{ri}$ gives time term $g^{ri}{\mathrm{\partial }}_s{\pi }_i$ and spatial term $-g^{ri}{\mathrm{\partial }}_i{\pi }_0$. Thus the coefficient of ${\mathrm{\partial }}_k{\pi }_0$ in row $r$ is $-g^{rk}$. The complete $\pi$-block has exactly the displayed $A^0,A^k$. The time block is positive because $-g^{00}>0$ and $(g^{ij})$ is positive definite. Each spatial block is symmetric because $g^{ij}=g^{ji}$, including the off-diagonal entries $-g^{kj}$ and $-g^{jk}$. The same calculation applies separately to the real and imaginary parts.

The equations ${\mathrm{\partial }}_s\phi ={\pi }_0$ and ${\mathrm{\partial }}_s{A}_i=F_{0i}$ append identity time blocks and zero spatial blocks. Write $v\in {\mathbb{R}}^{13}$ for, in order, the real parts of the four ${\pi }_a$, their imaginary parts, $\mathrm{Re}\phi ,\mathrm{Im}\phi$, and the three real $A_i$. Denote these scalar-potential matrices by $S^{\mu }$: $S^0=\mathrm{diag}(A^0,A^0,I_5)$ and $S^k=\mathrm{diag}(A^k,A^k,0_5)$ in this component order. They are symmetric, with $S^0$ positive definite.

To verify the full coefficient domain in [L1], introduce independent real coordinates $z=(G_{ab},Q_{cab},E_i)\in {\mathbb{R}}^{53}$, with $G_{ab}=G_{ba}$ and $Q_{cab}=Q_{cba}$, for $0\le a,b,c\le 3$ and $1\le i\le 3$. Here $Q$ represents the first metric derivatives and $E_i$ represents $F_{0i}$. Let $O\subseteq {\mathbb{R}}^{66}$ consist of $u=(v,z)$ such that $G$ has Lorentzian signature $(-,+,+,+)$, $G^{00}<0$, $(G^{ij})$ is positive definite, and $\mathrm{\mid }\phi {\mathrm{\mid }}^2$ lies in the open domain of $V$. This is open: inversion is smooth by the adjugate-over-determinant formula, and invertibility, fixed signature, and strict definiteness persist under sufficiently small changes of a symmetric matrix. Set ${\mathrm{\Gamma }}^c{}_{ab}=\frac{1}{2}{G}^{cd}(Q_{abd}+Q_{bad}-Q_{dab})$, $A_0=0$, $c_i=-i\lambda A_i{\pi }_0-i\lambda E_i\phi$, and $R=G^{ab}{\mathrm{\Gamma }}^c{}_{ab}{\pi }_c+i\lambda G^{ab}{A}_a{\pi }_b+V^{\mathrm{\prime }}(\mathrm{\mid }\phi {\mathrm{\mid }}^2)\phi$. The transformed scalar source vector $b(v,z)$ has complex components $R-G^{0i}{c}_i$ in the ${\pi }_0$ row and $-G^{ri}{c}_i$ in the ${\pi }_r$ rows, components $-{\pi }_0$ in the $\phi$ rows, and $-E_i$ in the $A_i$ rows, all split into real coordinates. Indeed the original divergence equation is $G^{ab}{\mathrm{\partial }}_a{\pi }_b=R$, so these signs follow by the substitution in step 1.1. All these maps are smooth on $O$, using only finite sums, products, inversion and composition with the smooth function $V^{\mathrm{\prime }}$.

On $\mathbb{R}\times {\mathbb{T}}_{2\pi }^3\times O$ define ${\widehat{A}}^0=\mathrm{diag}(S^0,I_{53})$, ${\widehat{A}}^k=\mathrm{diag}(S^k,0_{53})$, and $\widehat{B}=(b,0_{53})$, with every formula independent of the base point $(s,x)$. The lifted maps are therefore smooth and periodic on the entire base, and the matrices are symmetric with positive-definite time matrix. Restricting the time base to any interval $J$ of [L1] gives a coefficient family exactly of its stated type. Evaluating its first thirteen rows along $(G,Q,E)=(g,\mathrm{\partial }g,F_{0i})$ on $U$ gives precisely the claimed scalar-potential reduction. The extra identity and zero blocks are a coefficient completion only: no auxiliary evolution equation, global extension of the given fields, or existence of a coupled solution is asserted. Thus the local block has the required coefficient realization, while the actual metric and Maxwell evolution blocks must be supplied separately in a coupled system.