Variation of action S in GR

1. Apr 26, 2017

Dan

Hello All!
1. The problem statement, all variables and given/known data

Action, where there are Yang-Mills field and Scalar field Lagrangians (as I know, so let me know if it is not), is given as $$S=\int \sqrt -g \left[ \frac {M_p^2} {2} R + F\left( Z \right) + \frac {\bar \kappa} {384} \left( \epsilon^{\alpha \beta \lambda \delta} F_{\alpha \beta}^a F_{\lambda \delta}^a \right)^2 - \Lambda + \frac 1 2 \left( \dot \phi \right)^2 + T \left( \phi \right) \right] \, d^4x$$ where g is the determinant of the metric tensor $g_{\mu\nu}$ , R is the scalar curvature, $\bar \kappa$ is a constant.

I have to find the variation of the action S with restect to $g_{\mu\nu}$.
2. Relevant equations
The SU(2) YM field $A_\mu^b$ has the internal symmetry index a, the field strength tensor being $$F_{\alpha \beta}^a = \partial_\alpha A_\beta^a - \partial_\beta A_\alpha^a + f^{abc}A_\alpha^b A_\beta^c$$
The function $F\left( Z \right)$ is an arbitrary function of $Z = F_{\mu \nu}^a F_{\alpha \mu \nu}$ and $f^{abc} = - \bar g \left[ abc\right],$
Roman indices a, b, c will run over 1, 2, 3, and the Levi-Civita tensor is given by $$\epsilon^{\alpha \beta \lambda \delta} = \sqrt {-g} g^{\rho_1 \alpha} g^{\rho_2 \beta} g^{\rho_3 \lambda} g^{\rho_4 \sigma} \left[ \rho_1 \rho_2 \rho_3 \rho_4 \right], ~~~ \left[ 0123 \right] = 1$$

3. The attempt at a solution

I took $\epsilon^{\alpha \beta \lambda \delta} F_{\alpha \beta}^a F_{\lambda \delta}^a$ as J
$$\frac {M_p^2} {2} \delta R + \delta F \left( Z \right) + \frac {\bar \kappa} {384} \delta \left( J \right)^2 - \delta \Lambda + \frac 1 2 \delta \left( \dot \phi^2 \right) + \delta T \left( \phi \right) = 0$$

where $$\delta R = R_{\mu \nu} - \frac 1 2 g_{\mu \nu} R$$
$$\delta F = F_{\mu \nu} - \frac 1 2 g_{\mu \nu} F = 2F^\prime F_{\mu\rho}^a F_\nu^{a \rho} - \frac 1 2 g_{\mu \nu} F$$
$$\delta \left( J \right)^2 = J^2_{\mu \nu} - \frac 1 2 g_{\mu \nu} J^2 = - \frac 3 2 J^2 g_{\mu \nu} - 8J \sqrt{-g} g^{\rho_2 \beta} g^{\rho_3 \lambda} g^{\rho_4 \sigma} \left[ \mu \rho_2 \rho_3 \rho_4 \right] F_{\nu \beta} ^b F_{\lambda \sigma} ^b$$
$$\delta \Lambda =\Lambda_{\mu \nu} - \frac 1 2 g_{\mu \nu} \Lambda = - \frac 1 2 g_{\mu \nu} \Lambda$$
$$\delta \left( \dot \phi^2 \right) = - \frac 1 2 g_{\mu \nu} \dot \phi^2$$
$$\delta T= T_{\mu \nu} - \frac 1 2 g_{\mu \nu} T$$

So the result is
$$\frac {M_p^2} {2} \left( R_{\mu \nu} - \frac 1 2 g_{\mu \nu} R \right) - \frac 1 2 g_{\mu \nu} F \left( Z \right) + 2F^\prime F_{\mu\rho}^a F_\nu^{a \rho} - \frac {\bar \kappa} {384} \left\{ \frac 3 2 J^2 g_{\mu \nu} - 8J \sqrt{-g} g^{\rho_2 \beta} g^{\rho_3 \lambda} g^{\rho_4 \sigma} \left[ \mu \rho_2 \rho_3 \rho_4 \right] F_{\nu \beta} ^b F_{\lambda \sigma} ^b \right\} + \frac 1 2 g_{\mu \nu} \Lambda - \frac 1 2 g_{\mu \nu} \dot \phi^2 + T_{\mu \nu} - \frac 1 2 g_{\mu \nu} = 0$$

Can you look and let me know if there is a mistake.

2. May 1, 2017

PF_Help_Bot

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