I am working through the variation of the Einstein-Maxwell-Dilaton action as stated in The Rotating Dyonic Black Holes Of Kaluza-Klein Theory. Rasheed gives the action as
\begin{equation}
S=\int\mathrm{d}^4x\sqrt{g}\left[R-2(\partial\sigma)^2-2e^{2b\sigma}F^2\right]
\tag{1}
\end{equation}
where $b$ is a constant, $\sigma$ is the dilaton scalar field and $F^2\equiv F^{\mu\nu}F_{\mu\nu}$. By varying $S$, I want to recover the equations of motion as stated in equation (1.2) in the paper. I was able to recover the equations relating to $\delta\sigma$ and $\delta A_{\nu}$. However, I don't seem to be able to recover the equation of motion when varying with respect to $g^{\mu\nu}$ i.e. $\delta g^{\mu\nu}$.
Rasheed gives the equation of motion for $\delta g^{\mu\nu}$ as
\begin{equation}
R_{\mu\nu}=2(\partial_{\mu}\sigma)(\partial_{\nu}\sigma)+2e^{2b\sigma}T_{\mu\nu}
\tag{2}
\end{equation}
where I believe there was a typo in the positioning of the indices for the partial derivative term, in the paper, and $(2)$ should be the correct version.
I was able to find the last term involving $T_{\mu\nu}$. However, for the term with the partial derivatives, I will always have an extra term involving $(\partial\sigma)^2$ which arises from varying $\sqrt{g}$.
My calculations are as follows
\begin{align}
-2\delta\left[\sqrt{-g}(\partial\sigma)^2\right]&=-2\delta\left[\sqrt{-g}g^{\mu\alpha}\partial_{\alpha}\sigma\partial_{\mu}\sigma\right]\\
&=-2\partial_{\alpha}\sigma\partial_{\mu}\sigma\sqrt{-g}\left(\delta g^{\mu\alpha}-\frac{1}{2}g_{\gamma\beta}\delta g^{\gamma\beta}g^{\mu\alpha}\right)\\
&=-2\partial_{\alpha}\sigma\partial_{\mu}\sigma\sqrt{-g}\delta g^{\mu\alpha}+\partial_{\alpha}\sigma\partial^{\alpha}\sigma\sqrt{-g}g_{\gamma\beta}\delta g^{\gamma\beta}
\tag{3}
\end{align}
How do I get rid of the $(\partial\sigma)^2=\partial_{\alpha}\sigma\partial^{\alpha}\sigma$ term? Also, is the left hand side of $(2)$ supposed to be the Einstein tensor $G_{\mu\nu}\equiv R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}$ instead of just $R_{\mu\nu}$?
Best Answer
Einstein equation in general reads: $$R_{\mu\nu} - \cfrac{1}{2}g_{\mu\nu}R = T_{\mu\nu}$$ Tracing ($d=4$) we get $$R=-T$$ and substituting back we have $$R_{\mu\nu} = T_{\mu\nu} -\frac{1}{2}g_{\mu\nu}T$$ where $T$ denotes the trace of $T_{\mu\nu}$. Varying the action (1) with respect to the metric, the first term gives the Einstein tensor the second gives $$2\partial_{\mu}\sigma\partial_{\nu}\sigma - g_{\mu\nu}\partial_{a}\sigma\partial^{a}\sigma$$ while the third term is related to the EM tensor. According to the paper's notation the Einstein equation will read
$$R_{\mu\nu} - \cfrac{1}{2}g_{\mu\nu}R = 2\partial_{\mu}\sigma\partial_{\nu}\sigma - g_{\mu\nu}\partial_{a}\sigma\partial^{a}\sigma + 2\exp(2b\sigma)T_{\mu\nu}$$
$T_{\mu\nu}$ is traceless in $d=4$. Tracing we have
$$ R = 2\partial^{a}\sigma\partial_{a}\sigma$$
Substituting back we have
$$R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}2\partial^{a}\sigma\partial_{a}\sigma = 2\partial_{\mu}\sigma\partial_{\nu}\sigma - g_{\mu\nu}\partial_{a}\sigma\partial^{a}\sigma + 2\exp(2b\sigma)T_{\mu\nu}$$
where the summation terms are cancelled and one obtains the desired equation.