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Prove that negative absolute temperatures are actually hotter than positive absolute temperatures

+ 8 like - 0 dislike
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Could someone provide me with a mathematical proof of why, a system with an absolute negative Kelvin temperature (such that of a spin system) is hotter than any system with a positive temperature (in the sense that if a negative-temperature system and a positive-temperature system come in contact, heat will flow from the negative- to the positive-temperature system).

This post imported from StackExchange Physics at 2014-04-04 16:14 (UCT), posted by SE-user ramanujan_dirac
Closed as per community consensus as the post is not graduate-level.
asked Mar 4, 2012 in Closed Questions by ramanujan_dirac (235 points) [ no revision ]
recategorized Apr 19, 2014 by dimension10

3 Answers

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Arnold Neumaier's comment about statistical mechanics is correct, but here's how you can prove it using just thermodynamics. Let's imagine two bodies at different temperatures in contact with one another. Let's say that body 1 transfers a small amount of heat $Q$ to body 2. Body 1's entropy changes by $-Q/T_1$, and body 2's entropy changes by $Q/T_2$, so the total entropy change is $$ Q\left(\frac{1}{T_2}-\frac{1}{T_1}\right). $$ This total entropy change must be positive (according to the second law), so if $1/T_1>1/T_2$ then $Q$ has to be negative, meaning that body 2 can transfer heat to body 1 rather than the other way around. It's the sign of $\frac{1}{T_2}-\frac{1}{T_1}$ that determines the direction that heat can flow.

Now let's say that $T_1<0$ and $T_2>0$. Now it's clear that $\frac{1}{T_2}-\frac{1}{T_1}>0$ since both $1/T_2$ and $-1/T_1$ are positive. This means that body 1 (with a negative temperature) can transfer heat to body 2 (with a positive temperature), but not the other way around. In this sense body 1 is "hotter" than body 2.

This post imported from StackExchange Physics at 2014-04-04 16:14 (UCT), posted by SE-user Nathaniel
answered Mar 4, 2012 by Nathaniel (495 points) [ no revision ]
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From a fundamental (i.e., statistical mechanics) point of view, the physically relevant parameter is coldness = inverse temperature $\beta=1/k_BT$. This changes continuously. If it passes from a positive value through zero to a negative value, the temperature changes from very large positive to infinite (with indefinite sign) to very large negative. Therefore systems with negative temperature have a smaller coldness and hence are hotter than systems with positive temperature.

Some references:

D. Montgomery and G. Joyce. Statistical mechanics of “negative temperature” states. Phys. Fluids, 17:1139–1145, 1974.
http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19730013937_1973013937.pdf

E.M. Purcell and R.V. Pound. A nuclear spin system at negative temperature. Phys. Rev., 81:279–280, 1951.
http://prola.aps.org/abstract/PR/v81/i2/p279_1

Section 73 of Landau and E.M. Lifshits. Statistical Physics: Part 1,

Example 9.2.5 in my online book Classical and Quantum Mechanics via Lie algebras.

This post imported from StackExchange Physics at 2014-04-04 16:14 (UCT), posted by SE-user Arnold Neumaier
answered Mar 4, 2012 by Arnold Neumaier (12,385 points) [ no revision ]
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Take a hydrogen gas in a magnetic field. The nuclei can be aligned with the field, low energy, or against it, high energy. At low temperature most of the nuclei are aligned with the field and no matter how much I heat the gas I can never make the population of the higher energy state exceed the lower energy state. All I can do is make them almost equal, as described by the Boltzmann distribution.

Now I take another sample of hydrogen where I have created a population inversion, maybe by some method akin to that used in a laser, so there are more nuclei aligned against the field than with it. This is my negative temperature material.

What happens when I mix the samples. Well I would expect the population inverted gas to "cool" and the normal gas to "heat" so that my mixture ends up with the Boltzmann distribution of aligned and opposite nuclei.

This post imported from StackExchange Physics at 2014-04-04 16:14 (UCT), posted by SE-user John Rennie
answered Mar 4, 2012 by John Rennie (470 points) [ no revision ]




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