0
$\begingroup$

I'm trying to find the component equation of motion for the action in a paper. The action for the system is, $$S=\frac{m_P^2}{8\pi}\int d^4x\sqrt{-g}\bigg(\frac{R}{2}-\frac{1}{2}\partial_\mu\phi\partial^\mu\phi+\alpha\phi\mathcal{G}\bigg),$$ where $\mathcal{G}=R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}-4R^{\mu\nu}R_{\mu\nu}+R^2$ is the Gauss-Bonnet invariant. The modified Einstein equation of motion for the action is, $$G_{\mu\nu}=T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-\frac{1}{2}g_{\mu\nu}(\partial\phi)^2-\alpha(g_{\rho\mu}g_{\delta\nu}+g_{\rho\nu}g_{\delta\mu})\nabla_{\sigma}(\partial_{\gamma}\phi\epsilon^{\gamma\delta\alpha\beta}\epsilon^{\rho\sigma\lambda\eta}R_{\lambda\eta\alpha\beta}),$$ where the final term in the brackets is the Riemann Dual Tensor (it's divergenceless I think). The scalar field is a function of $r$ only and hence $\gamma=r$ for the third term to be non-zero. In the first part of the paper I have linked, the metric, $$ds^2=-e^{A(r)}dt^2+e^{B(r)}dr^2+r^2(d\theta^2+sin^2(\theta)d\varphi^2),$$ is used as an ansatz as a solution. This is then subbed into the Einstein equation and the $tt, rr$ and $\theta\theta$ equations are found.

I have tried doing the contractions in MAPLE for the $tt$ component, ensuring that the indices are correct (eg. $\rho=t$ since $\mu=t$ etc.). However, I keep getting terms of the form, $$-8\alpha e^A\frac{(1-e^B)(\phi''-\frac{B'}{2}\phi')}{e^{2B}r^2}+\frac {1}{2}e^{A}\phi'^2,$$ which is close to the answers they produce in the appendix except the $B'\phi'$ term in the appendix carries a ($e^{B}-3$) and I don't know where this 3 comes from. In finding my answer I use the divergenceless nature of the Riemann dual ($*R*$), in order to write the last term on the RHS of the Einstein equation as

$$\nabla_{\sigma}(\partial_{\gamma}\phi\epsilon^{\gamma\delta\alpha\beta}\epsilon^{\rho\sigma\lambda\eta}R_{\lambda\eta\alpha\beta})=\epsilon^{rt\varphi\theta}\epsilon^{tr\theta\varphi}R_{\theta\varphi\varphi\theta}\nabla_{r}\partial_r\phi=(*R*)_{\theta\varphi\varphi\theta}\bigg(\phi''-\frac{B'}{2}\phi'\bigg),$$ where in the last equality I have expanded out the covariant derivative and use the $\Gamma^{r}_{rr}$ Christoffel symbol.

Further problems develop when I look at the $rr$ term as I missing at least 4 terms from the appendix.

I'm unsure if there is a problem in my understanding, or if there is something I should know about the Riemann Dual that I don't have here, or whether my use of only the $\Gamma^r_{rr}$ symbol is correct. If anyone could give me a helping hand seeing where my calculations are going awry I would really appreciate it.

$\endgroup$
1
$\begingroup$

The extra terms come from

  1. the fact that the Levi-Civita is completely anti-symmetric and so there are multiples of the same term, i.e. we can swap two LC symbols and two Riemann symbols, and they will add together;

  2. $\gamma\neq r$ all of the time due to the covariant derivative.

$\endgroup$

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.