We are given an action of the form:

$$S=\int d^4x\sqrt{-g}\left(-\frac14F_{\mu\nu}F^{\mu\nu}+V(B_{\sigma}B^{\sigma}) +R\lambda B_{\mu}B^{\mu}\right)$$

where $R$ is the curvature scalar, $\lambda$ is a constant, $F^{\mu\nu}=\partial_\nu B_\mu -\partial_\mu B_\nu$ is the field strength tensor and $V(B_{\sigma}B^{\sigma})$ is the field potential. Writing:

$F_{\alpha\beta}F^{\alpha\beta}=g^{\alpha\lambda}g^{\beta\rho}F_{\alpha\beta}F_{\lambda\rho}\space\space\space\space\space\space$ and $\space\space\space \space\space\space\space\space\space B_{\mu}B^{\mu}=B_{\mu}B_{\nu}g^{\mu\nu}$

we vary all the terms that have the metric in them in order to obtain the metric field equations, but how do I extract $\delta g^{\mu\nu}$ from the $V(B_{\sigma}B^{\sigma})$ term? On the same token, if I want to find the field equations for the vector field $B_{\mu}$ how can I extract a $\delta B_{\mu}$ out of the $V(B_{\sigma}B^{\sigma})$ term?


The equations I obtained were these:

$$ -\frac{1}{2} g_{\mu\nu} \mathfrak{L} + \lambda[B_\mu B_\nu R+B_\sigma B^\sigma R_{\mu\nu}+g_{\mu\nu} \square(B_\sigma B^\sigma) -\nabla_\mu \nabla_\nu(B_\sigma B^\sigma)]-\frac{1}{2}F_{\mu\alpha}F^{\alpha}_{\nu} +\frac{\partial V(B^2)}{\partial B^2} B_\mu B_\nu =0 \space\space\space\space\space\space\space(1)$$


$$ \mathfrak{L} =-\frac14F_{\mu\nu}F^{\mu\nu}+V(B_{\sigma}B^{\sigma}) +R\lambda B_{\mu}B^{\mu} $$


$$ \frac{\delta V(B^2)}{\delta(B^2)} \frac{\delta (B^2)}{\delta g^{\mu\nu}} = \frac{\delta V(B^2)}{\delta(B^2)} {B_\nu B_\mu} \delta g^{\mu\nu}$$

This term from the integral gives us the final term in equation (1) which I converted the variational derivative into a partial derivative. Not sure if this is the right approach

  • 1
    $\begingroup$ Could you please give some reference? What is the final result? $\endgroup$
    – Noone
    Commented Oct 17, 2020 at 15:15
  • $\begingroup$ Unfortunately I dont know I dont have any references, this is all I´ve been given $\endgroup$ Commented Oct 17, 2020 at 15:40
  • $\begingroup$ The closest thing I found, for the Lagrangian density at least, was this: arxiv.org/abs/0803.3041 $\endgroup$ Commented Oct 17, 2020 at 15:58
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/587638/2451 $\endgroup$
    – Qmechanic
    Commented Oct 17, 2020 at 16:40

1 Answer 1


There are two possibilities to look at this:

  1. 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.

  2. 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 strategy can be used to vary to $\delta B^\mu$
The chain rule for functional derivatives actually implies an integral, but I am not sure whether this is needed for the variation or not: \begin{equation} \frac{\delta F[X]}{\delta Y(r)} = \int dz \frac{\delta F[X]}{\delta X(z)} \frac{\delta X(z)}{\delta Y(r)} \end{equation}

  • $\begingroup$ Not sure if I can convert the variational derivative into a partial derivative like I did above (I edited the post) but I think it would be like that since for the other actions I´ve seen that have explicit expressions for the potential the only thing missing would be the constant factors in front of the $B_\sigma B^\sigma$ $\endgroup$ Commented Oct 31, 2020 at 19:24

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