# Why does the temperature of the early Universe decrease as it expands? [duplicate]

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According to the Big Bang theory, the universe cooled down as it expanded. I would like to know why does the temperature decrease? And by "why" I mean how. What causes the temperature to decrease, what's is the physical process?

For simplicity, let's consider the early Universe after protons and neutrons formed but before the molecules appeared, so we have something resembling a monatomic gas. Temperature of such gas is defined in terms of the average kinetic energy $\epsilon_{avg}$ of the particles: $$T = \frac{2}{3}\frac{\epsilon_{avg}}{K_{boltzmann}}$$

If the temperature $T$ decreases, then the $\epsilon_{avg}$ must decrease. By what causes $\epsilon_{avg}$ to decrease? A popular explanation for an ideal gas in a cylinder is that the particles do work on the retracting piston and lose energy. But can we apply this explanation to the expanding Universe? I don't think so. If the particles lose energy, where does this energy go? The particles do not collide with the edge of the universe (the imaginary "piston"), since there is nothing for them to collide with. There is no edge, since the whole space is expanding.

## marked as duplicate by Ben Crowell, Community♦Feb 5 '18 at 21:12

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• damtp.cam.ac.uk/user/db275/Cosmology/Chapter3.pdf – anna v Feb 5 '18 at 13:33
• Maybe the cooling is due to something similar to the Joule-Thomson Effect, where an expansion of e real gas into vacuum produces a cooling due to a conversion of kinetic energy of the molecules to potential energy. – freecharly Feb 5 '18 at 16:01
• In the epoch you're describing, the universe was radiation-dominated, not matter-dominated. In a radiation-dominated universe, there is a pretty appealing semi-hand-wavy explanation for cooling, which is simply that cosmological expansion stretches the wavelengths of photons. If the particles lose energy, where does this energy go? General relativity doesn't have any conserved, scalar measure of energy, except in special cases like asymptotically flat spacetimes. This is why, for example, a photon in the present-day universe can lose energy due to cosmological expansion. – Ben Crowell Feb 5 '18 at 16:22

## 1 Answer

I think you've already identified the concept to think about, it's not that the particles are losing energy, it's that the expanding universe is decreasing the pressure of this theoretical gas, and Temperature is pressure dependant, being more or less a measure of energy density.

• This cannot be the answer! If you expand an ideal gas into a larger vacuum vessel, the pressure decreases without a change in temperature. Only of you adiabatically expand the volume by moving the walls of the gas vessel, you will get a temperature decrease because of the work done by the gas on the walls. – freecharly Feb 5 '18 at 15:51
• I believe this answer is discussing this in more detail than I managed. physics.stackexchange.com/a/57080/93704 – Robert Laverick Feb 5 '18 at 16:02