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 $R_{\mu\nu}$?

  • $\begingroup$ Please edit in your calculations. $\endgroup$
    – J.G.
    Jul 31, 2021 at 11:58
  • $\begingroup$ I have included in my calculations, as requested. $\endgroup$
    – user161157
    Jul 31, 2021 at 12:17
  • $\begingroup$ Have you tried with $\sigma R$ instead of only $R$ in the action? $\endgroup$
    – KP99
    Jul 31, 2021 at 14:11
  • $\begingroup$ @Thomas i deleted the previous answer, which although described the procudure had some misconceptions in the comments. $\endgroup$
    – ApolloRa
    Aug 1, 2021 at 17:55

2 Answers 2


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.


ApolloRa's answer is spot on. I just wanted to give a slightly different approach that illustrates a trick that can be very powerful for debugging calculations. (Also there is a tricky factor of 2 I want to point out).

The trick is to try and decouple the part you are having trouble with, from the full calculation. In this case, we can notice that since your question has to do entirely with the gravity-scalar sector, without the gauge field, it is a useful simplification to take the limit $b \rightarrow -\infty$. That gets rid of the vector field completely. Then the action is GR coupled to a massless scalar field (I am going to keep the explicit factors of $\pi$ and Newton's constant $G_N$ that were dropped by Rasheed for the moment) \begin{equation} S = \int {\rm d}^4 x \sqrt{-g} \left(\frac{R}{16\pi G_N} - 2 (\partial \sigma)^2\right) \end{equation} By standard results we can immediately write down the equations of motion \begin{eqnarray} R_{\mu\nu}-\frac{1}{2} g_{\mu\nu} R &=& 8\pi G_N T^{(\sigma)}_{\mu\nu}\\ \square \sigma &=& 0 \end{eqnarray} where the stress-energy tensor of a massless scalar field is given by \begin{equation} T^{(\sigma)}_{\mu\nu} = 4 \partial_\mu \sigma \partial_\nu \sigma - 2 (\partial \sigma)^2 g_{\mu\nu} \end{equation} where $(\partial \sigma)^2=\partial_\mu \sigma \partial^\mu \sigma$ (to compare with the wikipedia conventions, note that we can replace $\hbar^2/m$ in their notation with $2$). Note the trace is given by \begin{equation} T^{(\sigma)} = - 4 (\partial \sigma)^2 \end{equation}

By following the same steps in Apollo Ra's answer (rewriting the Einstein equations as $R_{\mu\nu}=8\pi G_N \left(T_{\mu\nu}-\frac{1}{2} T g_{\mu\nu} \right)$ and plugging in the expressions for $T_{\mu\nu}$ and $T$), we arrive at \begin{equation} R_{\mu\nu} = 32 \pi G_N \partial_\mu \sigma \partial_\nu \sigma \end{equation} Now the final step is to note that Rasheed has chosen units where $16\pi G_N=1$ (this is a bit of a tricky factor of 2 since it is more standard to set $8\pi G_N=1$, which gives the Einstein equations $G_{\mu\nu}=T_{\mu\nu}$ instead of Rasheed's $G_{\mu\nu}=\frac{1}{2} T_{\mu\nu}$). Using this fact, we have the desired result \begin{equation} R_{\mu\nu} = 2 \partial_\mu \sigma \partial_\nu \sigma \end{equation} Now that the scalar sector is working, it's easy to go back and make $b$ finite with your results from the vector sector.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy