2
$\begingroup$

The other day I decided to quickly make sure I can derive the supergravity equations of motion for the NS/NS sector using the following action: $$ S=\int_{M_{10}}d^{10}x \space \sqrt{-g} e^{-2\phi}\left(R+4(\partial \phi)^2-\frac{1}{12}H_{abc}H^{abc}\right) $$

Varying the metric, the Ricci-scalar term (R) will just yield an Einstein's equation term $R_{ab}-\frac{1}{2}Rg_{ab}$. The scalar field term should be simple enough since it is really just 4$g^{ab}\partial_a\phi \partial_b \phi$. This would just give a term in the field equations that looks like $4\partial_a\phi \partial_b \phi$. The flux term, noting that $H_{abc}=\partial_a b_{bc}-\partial_b b_{ac}+\partial_c b_{ab}$, will give a term $-\frac{1}{4}(H^2)_{ab}$.

According to various papers (specifically https://arxiv.org/abs/1205.2274) one should get the following equation of motion for the metric variation: $$ R_{ab}+2D_a D_b \phi-\frac{1}{4}(H^2)_{ab}=0 $$

As you can see, I have no problem getting the last term, but I'm failing to see how the Einstein and scalar field term combine to get $R_{ab}+2D_a D_b \phi$.

Am I just missing some simplification here or is does it look like I actually made an error in my calculation?

$\endgroup$
  • $\begingroup$ the problem here is the word ‘quickly’. if you go through the calculation carefully you will have no trouble finding the correct equations of motion. the handwritten derivation can take up to 100 handwritten pages (depending on how much of it you want to derive from scratch. (and don’t forget the $e^{-2\phi}$ factors ..) $\endgroup$ – Wakabaloola Mar 3 '18 at 19:20
  • $\begingroup$ 100 pages? How could that be possible for varying the metric on an action that has three terms? Below the action, I wrote what I calculated for the contributions to the metric variation. This didn't take more than a page. Do these look wrong in some specific way? $\endgroup$ – Kenny H Mar 3 '18 at 19:24
  • $\begingroup$ you aren’t finding what you’re expecting, right? my suggestion is to do it carefully and give it the respect it deserves. one page won’t do it. yes, when i went through it it took up to 100 pages (i derived everything and took nothing for granted), since that’s the approach that works for me. $\endgroup$ – Wakabaloola Mar 3 '18 at 19:29
  • $\begingroup$ I mean no disrespect, but the suggestion to do it carefully isn't really that helpful. I was looking for something more specific like "the variation involving your kinetic term looks wrong." Upon looking back, it looks like my mistake was forgetting the $\sqrt{-g}$ when I varied the second and third terms in the action. I'll recalculate and see. $\endgroup$ – Kenny H Mar 3 '18 at 19:35
  • $\begingroup$ 1. First, write the action in Einstein frame which is related to your string frame metric like $G_{mn}=e^{\Phi/2} g_{mn}$. Here $G_{mn}$ are Einstein frame metric and other is your usual string frame metric. 2. Now write the Einstein equation which is just Einstein tensor on the left-hand side and stress tensor on the right of the equation. All field contribute to stress tensor( Scalars and 3-form). $\endgroup$ – Hare Mar 5 '18 at 10:04
1
$\begingroup$

I think this is not a big deal. Taking it into consideration the variation of your action with respect to $\phi$, these two terms $g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi$ and $g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\phi$ are actually the same.

Let us figure out $$ \delta\int_{\Omega}e^{-2\phi}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\,{\rm d}V=0, $$ where I kept ${\rm d}V=\sqrt{-g}\,{\rm d}^{10}x$ as an invariant volume element. In this way, apply Euler-Lagrange equation (but in its covariant form for simplicity, i.e., $$ \frac{\partial\mathcal{L}}{\partial\phi}=\nabla_{\sigma}\left(\frac{\partial\mathcal{L}}{\partial\left(\nabla_{\sigma}\phi\right)}\right), $$ where $\nabla_{\sigma}$ denotes the covariant derivative; and of course, for the scalar $\phi$, $\nabla_{\sigma}\phi=\partial_{\sigma}\phi$, while for the metric $g^{\mu\nu}$, $\nabla_{\sigma}g^{\mu\nu}=0$) to $$ \mathcal{L}=e^{-2\phi}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi, $$ and we have $$ \frac{\partial\mathcal{L}}{\partial\phi}=-2e^{-2\phi}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi $$ as well as $$ \frac{\partial\mathcal{L}}{\partial\left(\nabla_{\sigma}\phi\right)}=2e^{-2\phi}g^{\mu\sigma}\partial_{\mu}\phi. $$ Therefore $$ \nabla_{\sigma}\left(\frac{\partial\mathcal{L}}{\partial\left(\nabla_{\sigma}\phi\right)}\right)=2e^{-2\phi}\left(-2g^{\mu\sigma}\partial_{\mu}\phi\partial_{\sigma}\phi+g^{\mu\sigma}\nabla_{\mu}\nabla_{\sigma}\phi\right). $$ As a consequence, $\phi$ must satisfy $$ g^{\mu\sigma}\partial_{\mu}\phi\partial_{\sigma}\phi=g^{\mu\sigma}\nabla_{\mu}\nabla_{\sigma}\phi. $$

So you can see, if you replace the $g^{\mu\sigma}\partial_{\mu}\phi\partial_{\sigma}\phi$ term in the original action by $g^{\mu\sigma}\nabla_{\mu}\nabla_{\sigma}\phi$, and take the variational derivative of $g^{\mu\sigma}$, you will obtain the $\nabla_{\mu}\nabla_{\sigma}\phi$ term as is expected.

Hope this could be helpful for you :-)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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