# Equation of motion for cyclic model of the universe

I recently started to study about cyclic universe. I came across this article . My question is about the action that used for describing the cyclic model: $$S = \int d^{4}x\sqrt{-g}(\frac{1}{16\pi G}R-\frac{1}{2}(\partial\phi)^{2}-V(\phi)+\beta ^{4}(\phi)(\rho _{M}+\rho _{R}))$$ where $R$ is Ricci scalar and $g$ is the metric. I solve Euler-Lagrange equation for this action and find the equation of motion for $\phi$: $$\partial _{\mu }\frac{\partial L}{\partial (\partial _{\mu }\phi )} - \frac{\partial L}{\partial _{\mu }\phi } = 0$$ $$\Rightarrow \partial _{\mu }\left [ \frac{1}{2} \sqrt{-g} g^{\alpha \beta }\left ( \delta _{\mu }^{\alpha } \partial _{\beta }\phi + \delta _{\mu }^{\beta }\partial _{\alpha}\phi \right )\right ] = \sqrt{-g}\left ( -V_{,\phi }+4\beta ^{3} \beta _{,\phi }(\rho _{M}+\rho _{R})\right )$$ The radiation term is actually independent of $\phi$ so only $\rho _{M}$ enters the equation of motion. The zero component: $$3 a^{2} \dot{a} \dot{\phi } + a^{3}\ddot{\phi }=a^{3}(-V_{,\phi }+4 \beta ^{3}\beta _{,\phi }\rho _{M})$$ $$\Rightarrow\ddot{\phi }+3H\dot{\phi }= -V_{,\phi }+4 \beta ^{3}\beta _{,\phi }\rho _{M}$$ where $H$ is Hubble parameter. But this result is different from that was written in the article. The difference is the coefficient of the last term. In addition, I have problem to find Friedmann equations for this action (again in finding coefficients). Can anybody elaborate the reason?

Reference:

 P.J. Steinhardt and N. Turok, "A Cyclic Model of the Universe," Science 296 (2002), available at here

• For the problem of the factor $4$, from the equation $(5)$ (fluid equation of motion), you see that there is a dependence $\rho_M \sim \hat a^{-3}\sim (a\beta(\phi))^{-3}$, so the term $\beta^4(\phi)\rho_m$ is (secretely...) proportionnal to $\beta(\phi)$ – Trimok Dec 4 '13 at 12:20
• Thanks very much, @Trimok. That explains the factor 4, but what about the sign? I read this article. In this paper the sign of the coupling term is negative. Which one is correct? – Fatima Dec 4 '13 at 13:26
• It is not completely clear for me what $(\partial \phi)^2$ is exactly. If it is $(\partial \phi)^2 = -(\partial_0 \phi)^2+(\vec \partial \phi)^2$, then the Euler-Lagrange equations give a $\square \phi = \ddot \phi - \triangle\phi$, and not a $\ddot \phi$. So, I think you need to reexpress, if possible, $\triangle\phi$ as a function of $\beta, \beta_{,\phi}$, etc... – Trimok Dec 5 '13 at 11:04
• I think that $\triangle\phi$ is zero and $\phi$ only depends on time @Trimok. – Fatima Dec 5 '13 at 19:13
• Seems strange. You have $2$ papers with same authors (J. Steinhardt and Neil Turok) and a different sign... Sounds like a typo for the equation ($1$) of the first paper. – Trimok Dec 6 '13 at 21:28