3

Yes, your action is of the form \begin{equation} S=\int\text{d}^{4}x\sqrt{-g}(\mathcal{L}_{\text{EH}}+\mathcal{L}_{\text{M}}), \end{equation} where \begin{equation} \mathcal{L}_{\text{EH}}=\frac{R}{2k^{2}} \end{equation} is the part whose variation with respect to $g_{\mu\nu}$ gives you the Einstein tensor in the equations of motion, and \begin{equation} \...


2

Yes. You have to vary $g^{ab}$ everywhere it appears.


2

You forgot to lower the $\mu$ index in ${(\mathcal{J}^{\rho\sigma})^\mu}_\nu$ before using its definition. Doing it makes $g$ appear as you need.


2

This expression: $$(g^{\mu\nu} \nabla_\mu \nabla_\nu f) g_{\mu\nu} = \square f g_{\mu\nu} = (\nabla_\mu \nabla_\nu f)g^{\mu\nu}g_{\mu\nu} =\nabla_\mu \nabla_\nu f,$$ is wrong. This does not hold. The Box operator is a scalar quantity. It is defined as: $$\Box = g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}$$ and for a four dimensional diagonal metric this is: $$\Box = ...


2

There are two possibilities to look at this: Either the exact form of $V(B^2)$ is known and you can simply take the variation with respect to the $B^2$-dependency in that potential. More general, you can use use the chain rule as follows \begin{equation} \frac{\delta V(B^2)}{\delta B^2} \frac{\delta B^2}{\delta g^{\mu \nu}} \end{equation} The same ...


1

You are trying to prove eq (18) in 2009.11827. That has nothing to do with the identity you are asking about in the question (which is completely incorrect, btw). To derive (18) in the ref. we simply need to vary the action w.r.t. the metric ($G_4$ is a function only of $\varphi$) $$ \delta \int \sqrt{-g} G_4 R = \int \left[ \delta \sqrt{-g} G_4R + \sqrt{-g}...


1

Yes. The field strength tensor is $F_{\mu \nu}$. The only two fundamental fields in your action are $g_{\mu \nu}$ and $A_\mu$ (unless you use Palatini variation and treat the connection as independent of the metric). So before preforming the variation, you should write the action in terms of these fundamental fields, and vary w.r.t each of them. The only ...


1

The corresponding action $S=\int_{\sigma_i}^{\sigma_f} \mathrm{d}\sigma~L$ is the arc length between 2 spacetime events. The principle of stationary action (with Dirichlet boundary conditions) therefore leads to geodesics. See also e.g. this related Phys.SE post.


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