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Prove that negative absolute temperatures are actually hotter than positive absolute temperatures
Proof of existence of lowest temperature $0 K$

On the Kelvin temperature scale, zero is the lowest possible temperature. Why there is no defined negative temperature?

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marked as duplicate by Qmechanic, David Z Oct 3 '12 at 17:39

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Essentially a duplicate of e.g. physics.stackexchange.com/q/28601/2451 and physics.stackexchange.com/q/21851/2451 –  Qmechanic Oct 3 '12 at 15:44
    
@Qmechanic: Hi there Qmechanic, Good point spotting it out.. :) –  Waffle's Crazy Peanut Oct 3 '12 at 15:59
    
Not a duplicate, because there is negative temperature. –  Ron Maimon Oct 3 '12 at 16:26
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The thermodynamic definition of temperature is the reciprocal derivative of the entropy with respect to the energy. This definition, if you know that the entropy is the volume of phase space, defines the thing only using microscopic quantities.

The derivative of entropy with respect to E (at constant volume and constant every other conserved quantity and external field) is:

$$ {\partial S \over \partial E} = \beta = {1\over T}$$

When you set the Boltzmann constant to 1 (so that entropy is dimensionless, and temperature is measured in units of energy). This definition shows you that $\beta$ is the fundamental quantity.

The reason temperature is not exactly zero in the thermodynamic limit is because the entropy derivative is large, the smallest increment of energy will increase the entropy enormously.

But the temperature can be negative. When talking about $\beta$, a negative $\beta$ is just smaller than 0 $\beta$, which in terms of $T$ means that negative temperatures are hotter than infinite temperature. At infinite temperature, all states are equally likely, so you need a system with only finitely many states for negative temperature to make sense.

An example of such a system is nuclear spins. Some nuclei have multiple spin states, and these can have a separate temperature from the electrons because the spins equilibrate with each other much faster than they equilibrate with the electrons. In such a system, the temperature can be negative for a while, and this means that higher energy states of the nuclei are more likely than lower energy states.

The negative zero temperature is also unattainable, as it is maximum hot. The temperature line only makes sense when you turn it inside out to make a $\beta$ line, since this is the correct fundamental quantity.

This is covered in standard sources on thermodynamics.

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really dumb question here, wouldn't it be easier to inverse that equation, and say that temperature is the derivative of energy with respect to entropy? Or is there a subtlety I'm neglecting? –  AlanSE Oct 3 '12 at 16:48
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@AlanSE: This is the way it was originally defined, and thermodynamics is permanently stuck with bad conventions because of this. but the energy is a microscopic fundamental quantity, and the entropy is a thermodynamic information quantity, and it is just philosophically wrong to write it this way. The very idea assumes that E(S) makes sense as a single valued function--- that there is a unique energy at every entropy, and this is only true locally. For negative temperature systems, the entropy rises to a maximum with energy, then falls, so that there are two values of E for every value of S. –  Ron Maimon Oct 3 '12 at 17:00
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