 Solution

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