Prove that negative absolute temperatures are actually hotter than positive absolute temperatures 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).
 A: 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.
A: Ah, but who says that negative absolute temperatures exist at all? This is not without its controversies. There's a nature paper here which challenges the very existence of negative absolute temperatures, arguing that negative temperatures come about due to a poor method of defining the entropy, which in turn is used to calculate the temperature.
Other people insist that these negative temperatures are "real".
So, depending on which side of this debate you align yourself with, these systems can be described with positive temperatures (and behave accordingly), or negative temperatures which have very exotic properties.
A: 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.
A: 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. 
Link
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.
A: For the visually inclined, this article explains it simply. The maximum hotness definition is the middle image instead of the expected right image:

Due to the unintuitive definition of heat, a sample that only includes hot particles is negative kelvin / beyond infinite hot, and as clear from the image would give energy to colder particles.
A: None of the answers above are correct. Matt Thompson's answer is close. 
The OP asks for a mathematical proof 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

There is no proof for this statement because it is incorrect
In statistical mechanics temperature is defined as 
\begin{equation}
\frac{1}{T} = \frac{\partial S}{\partial E}
\end{equation}
i.e. a derivative of $S$. For $\it normal$ systems, like ideal gases, etc. 
$S(E)$ is a highly convex function of $E$ and there is a 1-to-1 relation between the system's macrostate and its temperature. 
However, in cases where $S$ is not a convex function of $E$, $\frac{\partial S}{\partial E}$ can take the same numerical value at different energies $E$ and therefore the same temperature. In other words,  $T$, unlike $E$ does --in general-- not uniquely describe a system's macrostate. This situation occurs in systems that have a negative Boltzmann temperature (detail: for a negative Boltzmann temperature $S$ needs to be non-monotonous in $E$).
An isolated system 1 with a negative Boltzmann temperature $T_B<0$ can have either higher or lower internal energy $E_1/N$ than another isolated system, system 2, that it gets coupled to. 
Depending on which system has a higher $E_i/N, i=1,2$ heat flows either from system 1 to system 2 or vice versa, regardless of the temperatures of the two systems before coupling. 
For details, see 


*

*Thermodynamics in isolated systems
Below I have attached Fig. 1, taken from the arxiv version of this work to illustrate this fact.  

PS


*

*I am not an author of any of the cited papers. 

*Thermodynamics is compatible with the use of the Gibbs entropy, but not with the Boltzmann entropy. Showing this is a four line proof, see this Nature Physics paper Consistent thermostatistics forbids negative absolute temperatures. The Gibbs temperature (unlike Boltzmann temperature) is always positive, $T>0$. 

*The attempt above by @Nathaniel at a purely thermodynamic proof of the OP's statement relies on the premise that $T<0$ is compatible with thermodynamics. This is not the case, see point 2. The proof given is invalid.

*For normal systems the distinction between Gibbs and Boltzmann temperature is practically irrelevant. The difference becomes drastic though, when edge cases are considered, e.g. truncated Hamiltonians or systems with non-monotonous densities of states. In fact, in most calculations in statistical mechanics textbooks the Gibbs entropy is used instead of the Boltzmann entropy. Remember calculating "all states up to energy $E$" instead of "all states in an $\epsilon$ shell at energy $E$"? That's all the difference.

*There is a whole series of attempts to publish comments on the Nature Physics article by Dunkel and Hilbert, but all got rejected. These all follow the pattern of trying to create a contradiction, but none were able to punch a hole into Dunkel and Hilbert's short mathematical argument.
A: Negative temperature - yes I encountered that once: I seem to recall that it's the state that arises when, say, you have a system of magnetic dipoles in a magnetic field, and they have arrived at an equilibrium distribution of orientations ... and then the magnetic field is suddenly reversed and the distribution is momentariy backwards  - basically the distribition given by substituting a negative value of T. Other scenarios can probably be thought of or actually brought into being that would similarly occasion this notion. I think possibly the answer is that the system is utterly out of thermodynamic equilibrium, whence the 'temperature' is just the variable that formerly was truly a temperature, and is now merely an artifact that gives this non-equilibrium distribution when rudely plugged into the distribution formula. So heat is transferred because you now have a highly excited system utterly out of equilibrium impinging upon a system that approximates a heat reservoir. I think there's no question really of accounting for the heat transfer by the usual method, ie when both temperatures are positive, of introducing the temperature difference as that which drives the transfer.
And would it even be heat transfer atall if the energy is proceeding from a source utterly out of thermodynamic equilibrium? It's more that the transferred energy is becoming heat, I would say.
A: Here is a simple argument from first principles.
First, please review the following points from statistical-thermodynamics (please excuse me for glossing over some minor details):

*

*Definition of entropy: Entropy is the number of microstates (i.e microscopic configurations) a system in equilibrium can be in, for a given total energy.
(up to exponentiation).


*Definition of temperature: When the energy of the system is increased, the entropy changes. In its most fundamental form, temperature is defined by how much the entropy increases when the energy is increased (assuming no other transformation is done on the system). Typically temperature is positive, meaning that the entropy increases when the energy increases. But negative temperature is also allowed, whereby the entropy decreases when the energy increases.
(For simplicity, I'm actually thinking about the inverse of the temperature, usually denoted by $\beta$, but this isn't important here).
(Very often people describe temperature as a measure of the average motion energy of the ingredients that the system is composed of, but this is a non-fundamental description, and it doesn't apply well to negative temperature situations)


*Second law of thermodynamics: When two subsystems are brought into thermal contact, energy will flow between them. This is heat. As the energy of each subsystem changes, the number of accessible microstates of the subsystem changes as well. Heat will keep flowing, until the the number of microstates of the combined system will be the maximum possible with the current total energy. This is equilibrium. At this point, there will be no more net exchange of energy (at long time scales). This is statistically very reasonable, because now the number of accessible microstates vastly outnumbers the number of microstates for any other distribution of energy between the two subsystems, and so the probability that the combined system will move to a microstate whose energies correspond to a different entropy, and keep doing so, is ridiculously tiny.
In other words, the system will equilibrate at a configuration that maximizes the entropy.
Now to the argument.
Let's assume that system $P$ and system $N$ are brought into thermal contact. System $P$ has positive temperature, and system $N$ has negative temperature. Let's try to figure out the direction of the heat flow. Let's first guess that heat flows from system $P$ to system $N$. In this case, the energy of system $P$ decreases, and therefore its entropy decreases as well (since it has positive temperature). The energy of system $N$ increases, and so its entropy decreases as well (since it has a negative temperature). In total, we get a decrease in entropy. This contradicts the second law of thermodynamics.
The conclusion is that heat must flow from system $N$ to system $P$. i.e heat will flow from a negative temperature system to a positive temperature system.
