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Take the tensor contraction of the Einstein field equations with the inverse metric $g^{\mu \nu }$. In four-dimensional spacetime, $g^{\mu \nu }{g}_{\mu \nu }=4$, henceHere $T$ is the tensor contraction of the stress-energy tensor. Substituting this result back into the Einstein field equations, while keeping both free indices lowered, gives the trace-reversed stress-energy tensor:The factor $1/2$ uses the dimension four: in dimension $n>2$, the Trace-reversed Einstein field equations in D dimensions have $1/(n-2)$ in place of $1/2$.

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.

For a real Klein-Gordon field in $D>2$ dimensions, fix metric signature $(-+\cdots +)$, zero cosmological constant and $G_{ab}=\kappa T_{ab}$. Write the stress-energy tensor with an explicit normalization $A$:Its trace is $T=A[(1-D/2)(\nabla \varphi {)}^2-D{m}^2{\varphi }^2/2]$. Substitution into the Trace-reversed Einstein field equations in D dimensions cancels the metric-proportional gradient term and givesThe convention $A=1/2$ gives the coefficient $\kappa /2$; the usual canonically normalized scalar field has $A=1$. This factor is a normalization choice for the scalar field, not a consequence of trace reversal. The Klein-Gordon equation $(\square -m^2)\varphi =0$ gives stress-energy conservation, since ${\nabla }^a{T}_{ab}=A(\square \varphi -m^2\varphi ){\nabla }_b\varphi$.^[ref]