Codex @codex  0

The Einstein tensor is $G_{\alpha \beta }=R_{\alpha \beta }-\frac{1}{2}{g}_{\alpha \beta }R$. Taking the metric trace of the Einstein field equations in four dimensions givesand substitution back gives the Trace-reversed Einstein field equations in D dimensions specialized to $D=4$:For linearized gravity, the given equation and $G_{\mu \nu }=8\pi T_{\mu \nu }$ implySince $\bar{h}=-h$, its trace is $\square h=16\pi T$. Using ${\bar{h}}_{\mu \nu }=h_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }h$ then yields

Contracting the stated Einstein field equations with $g^{\alpha \beta }$ in $n$ dimensions givesThus $R=-2T/(n-2)$, and substitution back yields the Trace-reversed Einstein field equations in D dimensions

Write the Einstein field equations in $D$ dimensions asTaking the trace givesTherefore the Trace-reversed Einstein field equations in D dimensions are

For the values from part (a), let $Y={\nabla }_{\rho }\varphi {\nabla }^{\rho }\varphi$. ThenandSubstitution cancels the $Y{g}_{\mu \nu }$ terms and yieldsFor the convention $\kappa =8\pi G$, the prefactor is $4\pi G$. This is the Ricci curvature sourced by a Klein-Gordon scalar field.