I am following this paper here (arXiv here). What I want to do is derive equations ($2.7$) and ($2.8$) given in section $2$. While the authors include the higher order Euler Lagrangian terms in their derivation, I do not.

I understand the procedure the authors have followed - but if I attempt to derive the field equations beginning from a general metric $g_{\mu\nu}$ and put in the weak field form later, I can't recover their equations.

I started with the usual Einstein's and Maxwell's equations, $$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = \frac{\kappa^{2}}{2}T_{\mu\nu} = 8\pi GT_{\mu\nu} \tag{1}$$

$$ \nabla_{\mu}F^{\mu\nu} = 0 \tag{2}$$

With, $$T_{\mu\nu} = F_{\mu\alpha}F_{\nu}^{\alpha} - \frac{1}{4}g_{\mu\nu}F^{\alpha\beta}F_{\alpha\beta} \tag{3}$$

Next, I took the weak field approximation as per equation ($2.4$) in the paper, $$ g_{\mu\nu} = \eta_{\mu\nu} + \kappa h_{\mu\nu} \tag{4}$$ I chose the flat metric with the mostly negative signature (this is different to the paper - but only the EM stress tensor changes sign, if I am not wrong). In this approximation, I have from eqn ($4$), $$ \Gamma^{\alpha}_{\beta\gamma} = \frac{\kappa}{2}\Big(\partial_{\beta}h^{\alpha}_{\gamma} + \partial_{\gamma}h^{\alpha}_{\beta} - \partial_{\alpha}h_{\beta\gamma}\Big) $$ I use the TT gauge mentioned in the paper - this means, $h_{0\mu},\ \partial_{\mu}h^{\mu}_{\nu},\ h^{\mu}_{\mu} = 0$. This gives me from ($1$) (the Wiki page lists the same Ricci tensor if the gauge choice is imposed), $$ \Box h_{\mu\nu} = \frac{\kappa}{2}T_{\mu\nu} \tag{5}$$ $$ \nabla_{\mu}F^{\mu\nu} = \partial_{\mu}F^{\mu\nu} + \Gamma^{\mu}_{\mu\lambda}F^{\lambda\nu} + \Gamma^{\nu}_{\mu\lambda}F^{\mu\lambda} = \partial_{\mu}F^{\mu\nu} = 0 \tag{6}$$

Eqn ($5$) here differs from equation ($2.7$) by a negative sign (that arises out of the metric convention differences) and a factor of $2$ which I do not understand. More importantly, eqn ($6$) here and equation ($2.8$) are completely different - disregarding the EH terms that I did not include, there are additional terms on the left side which include derivatives of both the field tensors and which do not depend on the EH terms.

I don't understand why I don't get the expressions they derived if I do it this way - I varied the fields first and then applied the weak field approximation, while they applied the approximation first and then varied the fields. I went through their procedure as well (though, not in as much detail) and I did end up at equations ($2.7$) and ($2.8$) then.

Any help will be appreciated!


1 Answer 1


From the Einstein equations one has $$ G_{\mu\nu}=-\frac{\kappa^2}{2}T_{\mu\nu} $$ and this is true whatever the energy-momentum tensor is. From Wikipedia we get $$ G_{\mu\nu}=R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = \frac{\kappa}{2}(\partial_\sigma\partial_\mu h^\sigma_\nu + \partial_\sigma\partial_\nu h^\sigma_\mu - \partial_\mu\partial_\nu h - \square h_{\mu\nu} - \eta_{\mu\nu}\partial_\rho\partial_\lambda h^{\rho\lambda} + \eta_{\mu\nu}\square h). $$ Note the $\kappa$ factor on the last side of this equation due to the choice of the authors of the paper, $g_{\mu\nu}=\eta_{\mu\nu}+\kappa h_{\mu\nu}$, that is not so common (one would like to leave the metric dimensionless). Equating these two formulas you will get the right result with your gauge's choice.

  • $\begingroup$ Thanks a lot for the answer - I was confused by that 2 back then, but yes, it's perfectly sensible. Any idea about the rest? $\endgroup$
    – ShKol
    Commented Jan 22 at 17:07

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