# Energy conservation in a FLRW spacetime

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1. Aug 23, 2014

### andrewkirk

Consequences of Energy conservation in a FLRW spacetime

This entry uses the local energy conservation law to derive an equation that can be used, together with the Einstein field equation, to derive Friedman's equations for the dynamics of a homogeneous, isotropic universe.

The energy conservation rule is ${T^{0b}}_{;b}=0$.

We want to use this to get information about the equation of state

We use FLRW coordinates for this, in which the metric tensor components are all zero except for:
\begin{align*}
g_{00}&=-1, g_{11}=\frac{R^2}{1-kr^2}, g_{22}=R^2r^2, g_{33}=R^2r^2sin^2\theta\\
\end{align*}
Here $R$ is the scale factor, which depends only on $t$.

The stress-energy tensor is T$=(\rho+p)\vec{v}\otimes\vec{v}+p$g-1 where $\vec{v}$ is the average four-velocity of the particles of the fluid which, in the FLRW coordinates, is $(1,0,0,0)$, because of the isotropy assumption.
The representation of T in FLRW coordinates is a diagonal matrix with elements:
$T^{00}=\rho$, $T^{11}=p \frac{1-kr^2}{R^2}$, $T^{22}=\frac{p}{R^2r^2}$, $T^{33}= \frac{p}{R^2r^2sin^2\theta}$

Now we write $0={T^{0b}}_{;b}={T^{0b}}_{,b}+T^{ab}\Gamma^0_{ab}+T^{a0}\Gamma^b_{ab}$
We calculate the three terms in turn.

First, we have ${T^{0b}}_{,b}={T^{00}}_{,0}=\frac{\partial\rho}{\partial t} = \dot{\rho}$ because T inherits a diagonal nature in the FLRW coordinates from $\vec{v}\otimes\vec{v}$ and g.

For the next step we use
\begin{align*}
\Gamma^0{ab}&=\frac{1}{2}g^{0s}(g_{as,b}+g_{bs,a}-g_{ab,s})\\
&=\frac{1}{2}g^{00}(g_{a0,b}+g_{b0,a}-g_{ab,0})\\
&=\frac{1}{2}g^{00}(g_{00,b}+g_{00,a}-\delta^a_b g_{aa,0})\\
&=\frac{1}{2}\delta^a_b g_{aa,0}\\
\end{align*}
since $g^{00}$ is constant at $-1$. Hence the second term is:
\begin{align*}
T^{ab}\Gamma^0_{ab}&=T^{ab}\frac{1}{2}\delta^a_b g_{aa,0}\\
&=\frac{1}{2}T^{aa} g_{aa,0}\\
&=\frac{1}{2}\Big(\rho(g_{00,0})+p \frac{1-kr^2}{R^2}g_{11,0}+\frac{p}{R^2r^2}g_{22,0}+\frac{p}{R^2r^2sin^2\theta}g_{33,0}\Big)\\
&=\frac{1}{2}\Big(0+p \frac{1-kr^2}{R^2}\frac{\partial}{\partial t}(\frac{R^2}{1-kr^2}) +\frac{p}{R^2r^2}\frac{\partial}{\partial t}(R^2r^2)+\frac{p}{R^2r^2sin^2\theta}\frac{\partial}{\partial t}(R^2r^2sin^2\theta)\Big)\\
&=3p\frac{\dot{R}}{R}
\end{align*}

The third term is
\begin{align*}
T^{a0}\Gamma^b_{ab}&=T^{00}\Gamma^b_{0b}\\
&=\frac{\rho}{2}g^{bs}(g_{0s,b}+g_{bs,0}-g_{0b,s})\\
&=\frac{\rho}{2}g^{bb}(g_{0b,b}+g_{bb,0}-g_{0b,b})\\
&=\frac{\rho}{2}g^{bb}g_{bb,0}\\
&=\frac{\rho}{2}\Big(0+\frac{1-kr^2}{R^2}\frac{\partial}{\partial t}(\frac{R^2}{1-kr^2})+\frac{1}{R^2r^2}\frac{\partial}{\partial t}(R^2r^2)+\frac{1}{R^2r^2sin^2\theta}\frac{\partial}{\partial t}(R^2r^2sin^2\theta)\Big)\\
&=3\rho\frac{\dot{R}}{R}
\end{align*}

Putting the three terms together, we have:
\begin{align*}
{\rho}+3(\rho+p)\frac{\dot{R}}{R}=0\\
\end{align*}

We can express this in a manner that better reflects the relationship of work and energy by considering the rate of change of energy in a co-moving volume:
\begin{align*}
\frac{d}{dt}(\rho R^3)&= 3\rho R^2\dot{R}+\dot{\rho}R^3 =R^3\Big(\dot{\rho}+3\rho\frac{\dot{R}}{R}\Big) = -3p R^3\frac{\dot{R}}{R}
=-3p\dot{R}R^2=-p\frac{d}{dt}(R^3)\\
&\textrm{In summary:}\\
\frac{d}{dt}(\rho R^3)&=-p\frac{d}{dt}(R^3)\\
\end{align*}

The left-hand side is the increase in energy in the co-moving volume as it expands and the rightmost formula is the negative of the work done in the expansion.

In the current universe, in which most mass-energy is matter, and it is very sparsely distributed on average, average pressure $p$ is low compared to $\rho$ and can be ignored, so we get the approximation $\frac{d}{dt}(\rho R^3)=0$ in the \textbf{matter-dominated }phase of the universe.

In the early universe, there was not much matter and such particles as there were would be travelling very fast. Hence the equation of state for a photon gas or a highly relativistic gas applies, which is $p=\frac{1}{3}\rho$. Inserting that in the above equation we get:
\begin{align*}
\frac{d}{dt}(\rho R^3)= -\frac{1}{3}\rho\frac{d}{dt}(R^3)\\
\end{align*}
Multiply by $R$ and collect on one side to get:
\begin{align*}
0&=R\frac{d}{dt}(\rho R^3)+R\frac{1}{3}\rho\frac{d}{dt}(R^3)\\
&=R\frac{d}{dt}(\rho R^3)+R\frac{1}{3}\rho(3R^2)\dot{R}\\
&=R\frac{d}{dt}(\rho R^3)+(\rho R^3)\dot{R}=\frac{d}{dt}(\rho R^4)\\
\end{align*}
So, in the early, energy-dominated era, we have:
\begin{align*}
\frac{d}{dt}(\rho R^4)&=0\\
\end{align*}

V.361.

Last edited: Aug 23, 2014