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Reminds me of a thermodynamics/statistical mechanics class homework problem from way back. Two systems with negative temperatures A and B, where |A| > |B|, are

Reminds me of a thermodynamics/statistical mechanics class homework problem from way back. Two systems with negative temperatures A and B, where |A| > |B|, are put in contact. Which way does the heat flow? As the article alludes, it can be easier to think about inverse temperature (which they called coldness), so heat will flow from less cold system B to more cold system A. In this context, negative epsilon would be the ‘hottest’ temperature, and a negative infinite temperature the same as a positive infinite temperature.

carlob 5y ago
One of my favorite stat mech facts is that you can get to inverse temperature as a Lagrange multiplier using only probability and information theory: Given a system that has many states each with energy E_i, what is the highest Shannon entropy distribution that has average energy E? If you define a Lagrange multiplier \beta to enforce the constraint \sum_i p_i E_i = E and maximize the entropy you get the Boltzmann distribution!
- contravariant 5y ago
  One minor nitpick is that in this derivation your choice of coordinates matters. Luckily (classical) physics provides a canonical set of coordinates which even comes with a measure that is preserved under the equations of motion. This magical property of the phase space can't be emphasized enough if you ask me.
  - happytron 5y ago
    Can you expand on this? Is there a nicer coordinate-free way to define entropy?
    - contravariant 5y ago
      The definition of entropy is not coordinate independent. Very briefly the reason for this is that the uniform distribution isn't coordinate independent. There is a related coordinate independent way to measure distances from one distribution to another, which is the Kullback leibler divergence, and looks like: D_KL(P || Q) = \int dP/dQ log(dP/dQ) dQ = E_P[log(dP/dQ)] (This is by far the most general definition and uses the Radon-Nikodym derivative to be applicable to discrete, continuous and even weirder distributions. For reasons that I'll try to explain I also view this as a generalization of the notion of entropy) If you let Q be a uniform distribution this is equivalent to the entropy (up to some sign changes and a constant). However uniform distributions aren't coordinate independent and therefore the notion of entropy isn't. My personal conclusion is that there's no way of doing statistics without picking a(n improper) prior, since you inevitably need to pick some coordinates which ends up doing the same thing. You can then equivalently talk about the KL-divergence of a distribution from this prior or the entropy of a distribution, both end up being more or less the same thing.
contravariant 5y ago
I'm not sure this is right, when you make energy flow to the colder system the entropy decreases (by definition).
- radford-neal 5y ago
  I'm also not sure it's right. Consider two systems of spins in an external magnetic field. Suppose system A has all the spins in their high energy state (wrong way given the field) - so it has the most negative temperature possible. Suppose system B is similar, except a few of the spins are in their low energy state - so it has a slightly less extreme negative temperature. If the two systems interact, the few low-energy spins ought to even out between the two systems, which means energy has flowed from A to B (lower temperature to higher temperature).
  - jeffwass 5y ago
    Here’s the misunderstanding. Your example above, where all spins are in their highest possible energy state, really has temperature T=-0, an infinitesimal temperature just below absolute zero. This would not be the negative temperature possible. You are correct that energy would flow out of this maximal-energy system. But it corresponds to System B in my example - the system at the less negative temperature. The animation near the top of the article shows this nicely. The particles approach their max possible state as T flips to negative and approaches zero from below. Temperatures of positive and negative infinity are statistically equivalent in this example, where states of any energy are equally likely. It’s only at smaller less-negative temperatures where the population inversion occurs.
    - radford-neal 5y ago
      Thanks. I see now that you're right.
- jeffwass 5y ago
  Two systems at different temperatures, when put into thermal contact, will come to thermal equilibrium at an intermediary temperature. Total entropy (over both systems) will have increased.

carlob 5y ago

One of my favorite stat mech facts is that you can get to inverse temperature as a Lagrange multiplier using only probability and information theory: Given a system that has many states each with energy E_i, what is the highest Shannon entropy distribution that has average energy E? If you define a Lagrange multiplier \beta to enforce the constraint \sum_i p_i E_i = E and maximize the entropy you get the Boltzmann distribution!

contravariant 5y ago
One minor nitpick is that in this derivation your choice of coordinates matters. Luckily (classical) physics provides a canonical set of coordinates which even comes with a measure that is preserved under the equations of motion. This magical property of the phase space can't be emphasized enough if you ask me.
- happytron 5y ago
  Can you expand on this? Is there a nicer coordinate-free way to define entropy?
  - contravariant 5y ago
    The definition of entropy is not coordinate independent. Very briefly the reason for this is that the uniform distribution isn't coordinate independent. There is a related coordinate independent way to measure distances from one distribution to another, which is the Kullback leibler divergence, and looks like: D_KL(P || Q) = \int dP/dQ log(dP/dQ) dQ = E_P[log(dP/dQ)] (This is by far the most general definition and uses the Radon-Nikodym derivative to be applicable to discrete, continuous and even weirder distributions. For reasons that I'll try to explain I also view this as a generalization of the notion of entropy) If you let Q be a uniform distribution this is equivalent to the entropy (up to some sign changes and a constant). However uniform distributions aren't coordinate independent and therefore the notion of entropy isn't. My personal conclusion is that there's no way of doing statistics without picking a(n improper) prior, since you inevitably need to pick some coordinates which ends up doing the same thing. You can then equivalently talk about the KL-divergence of a distribution from this prior or the entropy of a distribution, both end up being more or less the same thing.

contravariant 5y ago

I'm not sure this is right, when you make energy flow to the colder system the entropy decreases (by definition).

radford-neal 5y ago
I'm also not sure it's right. Consider two systems of spins in an external magnetic field. Suppose system A has all the spins in their high energy state (wrong way given the field) - so it has the most negative temperature possible. Suppose system B is similar, except a few of the spins are in their low energy state - so it has a slightly less extreme negative temperature. If the two systems interact, the few low-energy spins ought to even out between the two systems, which means energy has flowed from A to B (lower temperature to higher temperature).
- jeffwass 5y ago
  Here’s the misunderstanding. Your example above, where all spins are in their highest possible energy state, really has temperature T=-0, an infinitesimal temperature just below absolute zero. This would not be the negative temperature possible. You are correct that energy would flow out of this maximal-energy system. But it corresponds to System B in my example - the system at the less negative temperature. The animation near the top of the article shows this nicely. The particles approach their max possible state as T flips to negative and approaches zero from below. Temperatures of positive and negative infinity are statistically equivalent in this example, where states of any energy are equally likely. It’s only at smaller less-negative temperatures where the population inversion occurs.
  - radford-neal 5y ago
    Thanks. I see now that you're right.
jeffwass 5y ago
Two systems at different temperatures, when put into thermal contact, will come to thermal equilibrium at an intermediary temperature. Total entropy (over both systems) will have increased.