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Variation of action S in GR

  1. Apr 26, 2017 #1

    Dan

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    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. jcsd
  3. May 1, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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