Recently, I have been researching about modified gravity theories and one of the theories has been the theory of the graviton. If one starts with the metric tensor $g_{\mu\nu}$ and then performs the perturbation $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ one then obtains the perturbation for a linearized gravitational wave, which can also represent a spin-2 particle, the graviton. The Riemann tensor is then simplified in a local inertial frame to be:

$R^\alpha_{\beta\mu\nu}=\partial _\mu \Gamma^a_{\beta\nu}-\partial _\nu \Gamma^a_{\beta\mu}$

The other curvature tensors can then be calculated and the result could construct the Einstein Hilbert Action for the graviton field. Now the question that I have is the following: since the tensor $h_{\mu\nu}$=$kH_{\mu\nu}$, where k is the gravitational constant commonly seen in the Einstein Field Equation, could one instead of the k plug in a scalar field $\phi$ like in Brans-Dicke Theory? Would this yield a result similar to the Brans-Dicke action?

I tried to calculate the Ricci scalar by plugging-in the scalar field into the graviton tensor and got:

$R$=$\partial_{\mu}\partial_{\nu}h^{\mu\nu}-\square h$

$R$=$\partial_{\mu}\partial_{\nu}(\phi H^{\mu\nu})$-$\square (\phi H^\mu_\mu)$=$(\partial_{\mu}\partial_{\nu} \phi)H^{\mu\nu}$+$\partial _\mu H^{\mu\nu}\partial _\nu \phi$+$\partial _\nu H^{\mu\nu}\partial _\mu \phi$+$(\partial _\mu \partial_\nu H^{\mu\nu} )\phi$$-...$

To get:

$\phi R$+$(\partial_{\mu}\partial_{\nu} \phi)H^{\mu\nu}$+$\partial _\mu H^{\mu\nu}\partial _\nu \phi$+$\partial _\nu H^{\mu\nu}\partial _\mu \phi$$-...$

Could one also use the metric tensor instead of the graviton tensor? I am relatively new to this level of physics, so forgive me if I make obvious mistakes.


1 Answer 1


Well, you'd have to linearize $\phi$ as well, so you'd assume that $\phi = \phi_{0} + (c){}^{1}\phi + (c^{2})\,{}^{2}\phi\, + ...$, along with assuming that $g_{ab} = \eta_{ab} + (c) {}^{1}h_{ab} + (c^{2}){}^{2}h_{ab} + ...$, and then you would drop all of the terms in the Brans-Dicke EOM that contain terms where the exponent of $c$ is greater than 1.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.