Conditional homogeneous multi-field FLRW equations

Vary the scalar action in a target chart by ${\mathrm{\Phi }}^i\mapsto {\mathrm{\Phi }}^i+s{\eta }^i$, where each smooth ${\eta }^i$ has compact support in an oriented spacetime coordinate patch. Differentiating the scalar density at $s=0$ gives $\delta {\mathcal{L}}_{\varphi }=-\frac{1}{2}({\mathrm{\partial }}_k{G}_{ij})(\mathrm{\Phi }){\eta }^k{\mathrm{\partial }}_{\mu }{\mathrm{\Phi }}^i{\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j-G_{ij}(\mathrm{\Phi }){\mathrm{\partial }}_{\mu }{\eta }^i{\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j-({\mathrm{\partial }}_{k}V)(\mathrm{\Phi }){\eta }^k\mathrm{.}$ In an oriented chart, the Levi-Civita formula and symmetry of $g^{\mu \nu }$ give ${\mathrm{\Gamma }}^{\mu }{}_{\mu \lambda }=\frac{1}{2}{g}^{\alpha \beta }{\mathrm{\partial }}_{\lambda }{g}_{\alpha \beta }$. For an invertible matrix $A$ and a matrix $B$, the permutation expansion gives $\det (I+sB)=1+s\mathrm{tr}B+O(s^2)$, since only diagonal entries contribute at first order. Applying this with $B=A^{-1}\dot{A}$ gives $\frac{d}{ds}\det A=(\det A)\mathrm{tr}(A^{-1}\dot{A})$ for every differentiable invertible matrix family. With $A=(g_{\alpha \beta })$ and $s=x^{\lambda }$, this identifies the trace above as ${\mathrm{\partial }}_{\lambda }\log \sqrt{-g}$. Hence $\sqrt{-g}{\mathrm{\nabla }}_{\mu }{X}_k^{\mu }={\mathrm{\partial }}_{\mu }(\sqrt{-g}{X}_k^{\mu })$ for $X_k^{\mu }:=G_{kj}(\mathrm{\Phi }){\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j$. Ordinary integration by parts and compact support therefore give $-\int \sqrt{-g}{G}_{ij}(\mathrm{\Phi }){\mathrm{\partial }}_{\mu }{\eta }^i{\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j{d}^{4}x=\int \sqrt{-g}{\eta }^i{\mathrm{\nabla }}_{\mu }(G_{ij}(\mathrm{\Phi }){\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j)d^{4}x$. Combining the terms yields the coefficient $E_k={\mathrm{\nabla }}_{\mu }(G_{kj}(\mathrm{\Phi }){\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j)-\frac{1}{2}({\mathrm{\partial }}_k{G}_{ij})(\mathrm{\Phi }){\mathrm{\partial }}_{\mu }{\mathrm{\Phi }}^i{\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j-({\mathrm{\partial }}_{k}V)(\mathrm{\Phi })$ of ${\eta }^k$; the boundary term vanishes because $\eta$ is compactly supported.

At fixed $\mathrm{\Phi }$, varying the inverse spacetime metric in the scalar action and using [F6] and $\delta \sqrt{-g}=-\frac{1}{2}\sqrt{-g}{g}_{\mu \nu }\delta g^{\mu \nu }$ gives $\delta S_{\varphi }=-\frac{1}{2}\int \sqrt{-g}[G_{ij}(\mathrm{\Phi }){\mathrm{\partial }}_{\mu }{\mathrm{\Phi }}^i{\mathrm{\partial }}_{\nu }{\mathrm{\Phi }}^j-g_{\mu \nu }(\frac{1}{2}{G}_{ij}(\mathrm{\Phi }){\mathrm{\partial }}_{\alpha }{\mathrm{\Phi }}^i{\mathrm{\partial }}^{\alpha }{\mathrm{\Phi }}^j+V(\mathrm{\Phi }))]\delta g^{\mu \nu }{d}^{4}x\mathrm{.}$ By the Hilbert-tensor convention this bracket is the displayed $T_{\mu \nu }$.

Stationarity [A1] makes $\int \sqrt{-g}{E}_k{\eta }^k{d}^{4}x=0$ for every compactly supported test variation. If continuous $E_k$ were nonzero at a point, it would have a fixed strict sign on a coordinate ball $B_r(x_0)$ whose closure lies in the variation chart. The function $\eta (x)=\exp \text{\hspace{0.17em}⁣}(-1\mathrm{/}(1-\mathrm{\mid }x-x_0{\mathrm{\mid }}^2\mathrm{/}{r}^2))$ for $\mathrm{\mid }x-x_0\mathrm{\mid }<r$, extended by zero outside that ball, is smooth, nonnegative and compactly supported; its derivatives tend to zero at the boundary because $e^{-1\mathrm{/}s}$ times every power of $s^{-1}$ tends to zero as $s\downarrow 0$. Taking $\eta$ in the $k$th component and zero in the other components makes the integral nonzero, since $\sqrt{-g}>0$, a contradiction. Thus $E_k=0$ pointwise. The chain rule gives $E_k=G_{kj}(\mathrm{\Phi }){\mathrm{\square }}_g{\mathrm{\Phi }}^j+({\mathrm{\partial }}_i{G}_{kj}-\frac{1}{2}{\mathrm{\partial }}_k{G}_{ij})(\mathrm{\Phi }){\mathrm{\partial }}_{\mu }{\mathrm{\Phi }}^i{\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j-({\mathrm{\partial }}_{k}V)(\mathrm{\Phi })$. Since ${\mathrm{\partial }}_{\mu }{\mathrm{\Phi }}^i{\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j$ is symmetric in $i,j$, multiplying by $G^{\mathrm{\ell }k}(\mathrm{\Phi })$ and symmetrizing its coefficient gives $\frac{1}{2}{G}^{\mathrm{\ell }k}(\mathrm{\Phi })({\mathrm{\partial }}_i{G}_{kj}+{\mathrm{\partial }}_j{G}_{ki}-{\mathrm{\partial }}_k{G}_{ij})(\mathrm{\Phi })={\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{ij}(G)(\mathrm{\Phi })$. Hence ${\mathrm{\square }}_g{\mathrm{\Phi }}^{\mathrm{\ell }}+{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{ij}(G)(\mathrm{\Phi }){\mathrm{\partial }}_{\mu }{\mathrm{\Phi }}^i{\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^j-G^{\mathrm{\ell }j}(\mathrm{\Phi })({\mathrm{\partial }}_{j}V)(\mathrm{\Phi })=0,{\mathrm{\square }}_{g}f:={\mathrm{\nabla }}_{\mu }{\mathrm{\partial }}^{\mu }f\mathrm{.}$

By the explicit factorization ${\mathrm{\Phi }}^i(t,\mathbf{x})={\varphi }^i(t)$, ${\mathrm{\partial }}_j{\mathrm{\Phi }}^i=0$, while [F3] gives $g_{00}=-1$. Hence $T_{00}=G_{ij}(\varphi ){\dot{\varphi }}^i{\dot{\varphi }}^j-g_{00}(-\frac{1}{2}{G}_{ij}(\varphi ){\dot{\varphi }}^i{\dot{\varphi }}^j+V(\varphi ))=\frac{1}{2}{G}_{ij}(\varphi ){\dot{\varphi }}^i{\dot{\varphi }}^j+V(\varphi )=\rho$, $T_{0i}=0$, and $T_{ij}=a^2{\delta }_{ij}(\frac{1}{2}{G}_{mn}(\varphi ){\dot{\varphi }}^m{\dot{\varphi }}^n-V(\varphi ))=p{g}_{ij}$. This is an isotropic stress with the displayed comoving density and pressure.

In the flat metric of [F3], the Levi-Civita formula gives ${\mathrm{\Gamma }}^0{}_{00}=0$ and ${\mathrm{\Gamma }}^i{}_{0j}=\frac{1}{2}{g}^{ik}{\dot{g}}_{kj}=H{\delta }^i{}_j$. For $X^{\mu }:={\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^i$, the factorization ${\mathrm{\Phi }}^i(t,\mathbf{x})={\varphi }^i(t)$ gives $X^0=-{\dot{\varphi }}^i$ and $X^j=0$; therefore ${\mathrm{\square }}_g{\mathrm{\Phi }}^i={\mathrm{\nabla }}_{\mu }{X}^{\mu }={\mathrm{\partial }}_0{X}^0+{\mathrm{\Gamma }}^0{}_{00}{X}^0+{\sum }_{j=1}^3{\mathrm{\Gamma }}^j{}_{j0}{X}^0=-{\ddot{\varphi }}^i-3H{\dot{\varphi }}^i$. Also ${\mathrm{\partial }}_{\mu }{\mathrm{\Phi }}^j{\mathrm{\partial }}^{\mu }{\mathrm{\Phi }}^k=-{\dot{\varphi }}^j{\dot{\varphi }}^k$. Substituting these identities in step 2.2 and multiplying by $-1$ yields $D_t{\dot{\varphi }}^i+3H{\dot{\varphi }}^i+G^{ij}(\varphi )({\mathrm{\partial }}_{j}V)(\varphi )=0$.

By [A2], $G_{\mu \nu }=M_P^{-2}{T}_{\mu \nu }$. Substitution of [F4] with $k=0,c=1$ and step 3.1 gives $3H^2=\rho \mathrm{/}{M}_P^2$ and $(2a\ddot{a}+{\dot{a}}^2)\mathrm{/}{a}^2=-p\mathrm{/}{M}_P^2$ from the temporal and spatial components. Since $(2a\ddot{a}+{\dot{a}}^2)\mathrm{/}{a}^2=2\ddot{a}\mathrm{/}a+H^2=2\dot{H}+3H^2$, these are the two displayed Einstein equations.

Subtracting the temporal equation of step 4.1 from its spatial equation yields $-2M_P^2\dot{H}=\rho +p=G_{ij}{\dot{\varphi }}^i{\dot{\varphi }}^j$. This uses no division by $H$, so it includes static instants and turning points. With $k=\mathrm{\Lambda }=0$ and $c=1$, the first two equations are the Friedmann and acceleration equations of [F5]; their density and pressure are those computed in step 3.1. No barotropic relation was introduced.