 Solution

For a variation of the covariant metric tensor, $\delta g^{\alpha \beta }=-g^{\alpha \mu }{g}^{\beta \nu }\delta g_{\mu \nu }$ and $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{\mu \nu }\delta g_{\mu \nu }$. Writing $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$, one obtainsThe specified metric-variation convention therefore givesFor $\varphi =\varphi (t)$, its invariant gradient square is $-{\dot{\varphi }}^2$. Hence $T^{00}={\dot{\varphi }}^2-({\dot{\varphi }}^2/2-V)={\dot{\varphi }}^2/2+V$, while $T^{ij}=a^{-2}{\delta }^{ij}({\dot{\varphi }}^2/2-V)$ and $T^{0i}=0$. Lowering one index makes $T^{\mu }{}_{\nu }=\mathrm{diag}(-\rho ,P,P,P)$, withPotential energy has negative pressure, whereas scalar kinetic energy contributes positive pressure. This sign distinction is what permits accelerated expansion.