# The equation of state

by parsifal
Tags: equation, state
 P: 15 I've got some difficulties trying to understand the equation of state derived from Friedmann equations. I'd greatly appreciate it if someone walked me through this. Now if the equation of state is stated as: $$\Large \dot{\rho}+(3\rho +p)\frac{\dot{R}}{R}=0 \ \ |p=\omega \rho$$ Then (in the case of pressure being zero): $$\Rightarrow \rho \propto R^{-3} \Rightarrow \rho = \rho _0 (\frac{R_0}{R})^3$$ I suspect the latter to be correct as it's not a result of my own logic Now what I do not understand is the proportionality. If the equation of state is integrated I get something like this (set p=0): $$\Large \dot{\rho}+(3\rho)\frac{\dot{R}}{R}=0 \Rightarrow \dot{\rho}=-3\rho\frac{\dot{R}}{R} \Rightarrow \frac{1}{\rho}\dot{\rho}=-3\frac{\dot{R}}{R} \Rightarrow \frac{1}{\rho}\frac{d\rho}{dt}=-3\frac{1}{R}\frac{dR}{dt}\ \|\cdot dt$$ $$\Rightarrow \int _{\rho _0}^\rho \frac{1}{\rho}d\rho}=-3\int _{R_0}^R \frac{1}{R}dR \Rightarrow ln(\rho)-ln(\rho _0)=-3(ln(R)-ln(R_0)) \Rightarrow ln \frac{\rho}{\rho _0}=-3ln\frac{R}{R_0}\Rightarrow \frac{\rho}{\rho _0}=e^-3\frac{R}{R_0} \Rightarrow \rho=e^-3\frac{R}{R_0}\rho _0$$ Now is there some part to the theory that causes the equation to flip so that $$\rho =e^-3\frac{R}{R_0}\rho _0 \Rightarrow \rho = \rho _0 (\frac{R_0}{R})^3$$ or don't I just get the mathematics right? Or have I done something wrong right in the beginning deriving the equation of state? Edit: Of course, how could I not see it... forgetting that a log x = log xa. Kurdt already told me that here, but then his message disappeared. Well thanks to Kurdt anyway!
 Emeritus Sci Advisor PF Gold P: 4,980 Simply the laws of logarithms: $$a\log(b) = \log(b^a)$$ and noting $$(\frac{a}{b})^{-3} = (\frac{b}{a})^3$$ There is also no exponential function involved since: $$e^{(\ln x)}=x$$ EDIT: I'd deleted the original post because I thought I'd misread your original post but I had not.
 Quote by parsifal Now if the equation of state is stated as: $$\Large \dot{\rho}+(3\rho +p)\frac{\dot{R}}{R}=0 \ \ |p=\omega \rho$$
$$\Large \dot{\rho}+3\frac{\dot{R}}{R}(\rho +p)=0$$
It follows from $\nabla_{\mu}T^{\mu}_0 = 0$.