Deriving the Cosmic Fluid Equation from Energy Conservation

Local energy conservation in general relativity (GR), expressed as ${T^{0b}}_{;b}=0$, combines with the Friedmann-Lemaître-Robertson-Walker (FLRW) metric to yield the relation $\frac{d}{dt}(\rho R^3)=-p\frac{d}{dt}(R^3)$, where $\rho$ is energy density, $p$ is pressure, and $R$ is the scale factor. This equation, together with the Einstein field equation, is used to derive the Friedmann equations governing the dynamics of a homogeneous, isotropic universe.

• The derivation starts from the local energy conservation law ${T^{0b}}_{;b}=0$ applied in FLRW coordinates, where the metric tensor components are $g_{00}=-1$, $g_{11}=\frac{R^2}{1-kr^2}$, $g_{22}=R^2r^2$, and $g_{33}=R^2r^2\sin^2\theta$.
• The stress-energy tensor for the cosmic fluid is written as $T=(\rho+p)\vec{v}\otimes\vec{v}+p\,g^{-1}$, with the average four-velocity $\vec{v}=(1,0,0,0)$ following from the isotropy assumption.
• Working through the Christoffel symbol terms in ${T^{0b}}_{;b}=0$ produces the continuity equation $\dot{\rho}+3(\rho+p)\frac{\dot{R}}{R}=0$, which rearranges into $\frac{d}{dt}(\rho R^3)=-p\frac{d}{dt}(R^3)$.
• In the matter-dominated phase of the universe, pressure $p$ is low compared to density $\rho$ and can be ignored, giving the approximation $\frac{d}{dt}(\rho R^3)=0$.
• In the early, radiation-dominated universe, the equation of state for a photon gas or a highly relativistic gas, $p=frac{1}{3}rho$, leads to $frac{d}{dt}(rho R^4)=0$.

How does local energy conservation lead to the cosmic fluid equation?

The energy conservation rule is ${T^{0b}}_{;b}=0$. This relation is used to extract information about the equation of state connecting pressure and energy density in a homogeneous, isotropic universe.

FLRW coordinates are used for this derivation, in which the metric tensor components are all zero except for:

$$\begin{align*} g_{00}&=-1, \quad g_{11}=\frac{R^2}{1-kr^2}, \quad g_{22}=R^2r^2, \quad g_{33}=R^2r^2\sin^2\theta\\ \end{align*}$$

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:

$T^{00}=\rho$, $T^{11}=p \frac{1-kr^2}{R^2}$, $T^{22}=\frac{p}{R^2r^2}$, $T^{33}= \frac{p}{R^2r^2\sin^2\theta}$

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:

$$\begin{align*} T^{ab}\Gamma^0_{ab}&=T^{ab}\frac{1}{2}\delta^a_b g_{aa,0}\\ &=\frac{1}{2}T^{aa} g_{aa,0}\\ &=\frac{1}{2}\Big(\rho(g_{00,0})+p \frac{1-kr^2}{R^2}g_{11,0}+\frac{p}{R^2r^2}g_{22,0}+\frac{p}{R^2r^2\sin^2\theta}g_{33,0}\Big)\\ &=\frac{1}{2}\Big(0+p \frac{1-kr^2}{R^2}\frac{\partial}{\partial t}\left(\frac{R^2}{1-kr^2}\right) +\frac{p}{R^2r^2}\frac{\partial}{\partial t}(R^2r^2)+\frac{p}{R^2r^2\sin^2\theta}\frac{\partial}{\partial t}(R^2r^2\sin^2\theta)\Big)\\ &=3p\frac{\dot{R}}{R} \end{align*}$$

The third term is

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

Putting the three terms together, we have:

$$\begin{align*} \dot{\rho}+3(\rho+p)\frac{\dot{R}}{R}=0\\ \end{align*}$$

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.

What does this equation give in the matter-dominated universe?

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 matter-dominated phase of the universe.

What does this equation give in the radiation-dominated 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:

$$\begin{align*} &R\frac{d}{dt}(\rho R^3)+R\frac{1}{3}\rho\frac{d}{dt}(R^3)\\ &=R\frac{d}{dt}(\rho R^3)+R\frac{1}{3}\rho(3R^2)\dot{R}\\ &=R\frac{d}{dt}(\rho R^3)+(\rho R^3)\dot{R}=\frac{d}{dt}(\rho R^4)\\ \end{align*}$$

So, in the early, energy-dominated era, we have:

$$\begin{align*} \frac{d}{dt}(\rho R^4)&=0\\ \end{align*}$$

Frequently Asked Questions

What is the starting point for deriving the cosmic fluid equation?

The derivation starts from the local energy conservation law ${T^{0b}}_{;b}=0$ applied using FLRW coordinates, where the metric components are $g_{00}=-1$, $g_{11}=frac{R^2}{1-kr^2}$, $g_{22}=R^2r^2$, and $g_{33}=R^2r^2sin^2theta$, with $R$ the scale factor depending only on $t$.

What form does the stress-energy tensor take in this derivation?

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 fluid particles, which equals $(1,0,0,0)$ in FLRW coordinates because of the isotropy assumption.

What is the resulting continuity equation?

Combining the three terms of ${T^{0b}}_{;b}=0$ gives $\dot{\rho}+3(\rho+p)\frac{\dot{R}}{R}=0$, which can be rewritten as $\frac{d}{dt}(\rho R^3)=-p\frac{d}{dt}(R^3)$, where the left side is the increase in energy in a co-moving volume and the right side is the negative of the work done in the expansion.

What happens to this equation in the matter-dominated phase of the universe?

In the current universe, matter dominates and is sparsely distributed, so average pressure $p$ is low compared to density $\rho$ and can be ignored, giving the approximation $\frac{d}{dt}(\rho R^3)=0$.

What happens to this equation in the early, radiation-dominated universe?

In the early universe, particles were highly relativistic, so the equation of state for a photon gas or a highly relativistic gas, $p=\frac{1}{3}\rho$, applies. Substituting this into the fluid equation and multiplying by $R$ leads to $\frac{d}{dt}(\rho R^4)=0$.

Glossary

• FLRW metric (Friedmann-Lemaître-Robertson-Walker metric): the metric tensor describing a homogeneous, isotropic universe, with components $g_{00}=-1$, $g_{11}=frac{R^2}{1-kr^2}$, $g_{22}=R^2r^2$, $g_{33}=R^2r^2sin^2theta$.
• Scale factor ($R$): a function of time $t$ only, describing the expansion of the universe in FLRW coordinates.
• Stress-energy tensor (T): the tensor $T=(\rho+p)\vec{v}\otimes\vec{v}+p\,g^{-1}$ describing the energy density $\rho$, pressure $p$, and average four-velocity $\vec{v}$ of the cosmic fluid.
• Matter-dominated phase: the era of the universe in which average pressure $p$ is low compared to density $\rho$ and can be ignored.
• Radiation-dominated phase: the early era of the universe in which the equation of state $p=\frac{1}{3}\rho$, appropriate to a photon gas or a highly relativistic gas, applies.