My question is: is the following thinking pretty much correct?
I'm interested hearing about anything here that is fundamentally wrong. Of course there is a bit of hand waving and approximation.
If a balloon of ideal gas in equilibrium at a finite temperature T suddenly bursts while in space, the gas cloud will immediately begin to expand.
If the initial average kinetic energy per particle was $\frac{3}{2}{k_b}T$, the initial total kinetic energy would be just $\frac{3}{2}N{k_b}T$. There shouldn't be much loss of energy anywhere, so I'm assuming the final total kinetic energy would be about the same.
The much-larger gas cloud is now cold "due to expansion", but what's really happened is that the current position of each atom is determined by it's velocity at the last scatter event, when the mean free path became larger than the diameter of the cloud. Now, if a group of atoms are still near each other, it's because their velocities have been similar for a while.
That means the relative velocity of a group of nearby atoms with respect to each other is much lower than the average velocity they are moving, and therefore the local temperature is low.
my question: The average kinetic energy per atom hasn't changed. However their velocities are now highly correlated locally, and so we say the temperature has decreased - it is "cold".
Answer
After writing this answer I recalled asking this question two years ago. I'll answer my own question just to wrap this up.
My question is: is the following thinking pretty much correct?
Yes. For a given small region of the expanding gas, the temperature would be defined by the distribution of velocities around the collective center of mass velocity of that region.
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