# 15. Pressure and Energy Density Evolution

There is a sense in which energy is conserved in general relativity. We say it is *locally* conserved, which effectively means that in a sufficiently small region of spacetime, the change in energy is equal to the flux of energy across the boundary of the region, including that via any work being done on the region. From this principle, or from the Einstein equations themselves, someone with some skill in general relativity can derive for a homogeneous expanding universe that:

\[dE = -PdV\]

where \(E\) is the energy in some volume \(V\). This *looks* a lot like conservation of energy as we are used to seeing it. Indeed it isthe first law of thermodynamics, at least for the special case of no heat flow across the boundary. If you have a gas in a volume \(V\) and you squeeze it down by \(dV\), the work you do (\(-PdV\)), increases the kinetic energy (and hence total energy) of the gas particles by \(dE=-PdV\). Since \(dV\) is negative, this is an increase in energy, assuming \(P > 0\).

But the simplicity of the result is deceptive. As soon as one starts to ask some obvious questions about it, things can become confusing. In a homogeneous expanding universe, how is the work being done? There are no pressure gradients to push things around. Then is it somehow being done by gravitational potential energy? That is a reasonable guess, since the expansion of space is a gravitational effect.

These questions though are not helpful. They assume energy is conserved, when in fact it is not. We can, however, use our Newtonian intuition to guide us about how the gas will behave given that the region it occupies is expanding. If the volume slowly increases that is containing a gas (with \(P>0\)), then the energy of that gas will decrease no matter if it's because of the expansion of space or the expansion of the walls that contain the gas.

From \(dE=-PdV\) we can derive how energy density evolves as the scale factor evolves. Gas comoving with the expasion and in a region with comoving volume \(V_c\), occupies a physical volume of \(a^3 V_c\). The energy content of this homogeneous gas is \(\rho a^3V_c\). Thus \(dE=-PdV\) leads to \(a^3 d\rho + 3\rho a^2 da = -3Pa^2 da\) or:

\[a\frac{d\rho}{da} = -3(P+\rho)\]

Box \(\PageIndex{1}\)

**Exercise 15.1.1:** Use the above equation to find \(P(\rho)\) for non-relativistic matter, given that \(\rho \propto a^{-3}\).

**Exercise 15.2.1:** Use the above equation to find \(P(\rho)\) for relativistic matter, given that \(\rho \propto a^{-4}\).

For (1) in the above exercise you should find that \(P=0\). This might be surprising since non-relativistic matter can certainly have non-zero temperature. Remember though that our \(\rho \propto a^{-3}\) result came from neglecting all kinetic energy of the gas, because it was so small compared to the kinetic energy. Of course it is not exactly zero so the pressure is also not exactly zero.