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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).

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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.

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  • $\begingroup$ "From a fundamental (i.e., statistical mechanics) point of view, the physically relevant parameter is coldness". I am afraid, that is not correct. It is energy, as shown in this paper. For instance, (inverse) temperature does generally not allow determining the direction of heat flow, because it is only a derivative of $S$. $\endgroup$ – jkds Oct 15 '18 at 11:26
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    $\begingroup$ @jkds: Of course, internal energy, temperature, pressure, etc. are all physically relevant. What I had meant is that coldness (inverse) temperature is more relevant than temperature itself. $\endgroup$ – Arnold Neumaier Oct 15 '18 at 12:18
  • $\begingroup$ Sure, but what the authors showed was that temperature is not in one-to one correspondence to a system's macrostate. The same system can have the same temperature at completely different internal energies. So temperature, unlike $E/N$, can be a misleading descriptor of the system. $\endgroup$ – jkds Oct 22 '18 at 9:36
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    $\begingroup$ @jkds: In the canonical ensemble, the macrostate is determined by the temperature; in other ensembles (such as the grand canonical one), one needs of course additional parameters. Then temperature and internal energy are no longer in 1-1 correspondence but related by an equation of state involving the other parameters. But my answer is anyway independent of heat flow. $\endgroup$ – Arnold Neumaier Oct 22 '18 at 13:29
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    $\begingroup$ @jkds: Temperature is a property of the thermodynamic limit where the microcanonical ensemble is equivalent to the canonical ensemble. In the canonical ensemble the 1-1 correspondence is self-evident. Moreover one can prove convexity. Thus if you assume a non-convex entropy functional you are in the thermodynamic situation only after performing the Maxwell construction (corresponding here to taking the convex envelope). $\endgroup$ – Arnold Neumaier Oct 23 '18 at 16:40
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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.

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    $\begingroup$ This is right, and central point can be stated like this: when heat energy leaves a body at negative temperature, the entropy of that body increases. $\endgroup$ – Andrew Steane Oct 30 '18 at 10:01
  • $\begingroup$ Your thermodynamic proof is wrong, because in thermodynamics $T<0$ breaks the consistency of thermodynamics, see this Nature Physics paper Consistent thermostatistics forbids negative absolute temperatures $\endgroup$ – jkds Oct 30 '18 at 14:48
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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.

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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.

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    $\begingroup$ This does not answer the question (the proof that is asked for does not rely on whether such systems actually exist or not). $\endgroup$ – ACuriousMind Jun 30 '15 at 10:26
  • $\begingroup$ The one thing that everyone agrees on is that their behavior is a bit surprising, and that is to be expected as we don't encounter systems with temperature ceilings in day-to-day life. In any case, that paper is cited in the comments on most of our "negative absolute temperature" questions. I can assure you that most of the answer authors are aware of it. But the question presupposes the definition of temperature which generates 'negative' values and this post doesn't really address it. $\endgroup$ – dmckee Jul 1 '15 at 3:04
  • $\begingroup$ @ACuriousMind: What of E=-mcc? Matt Thompson's answer is to claim the negative temperatures are the similar beast of spurious mathematical solutions and have no meaning whatsoever. $\endgroup$ – Joshua May 22 '16 at 16:20
  • $\begingroup$ @matt-thompson: you are spot on. In fact, "temperature" as opposed to energy is only a derived quantity (a derivative of $S$) and nowhere near as fundamental. By looking at non-monotonously growing densities of states it is easy to construct paradoxa, like systems in which heat is flowing from the colder to the hotter bath, regardless of which entropy definition is used, see the authors' follow-up paper $\endgroup$ – jkds Oct 15 '18 at 6:36
  • $\begingroup$ For negative temperature, you require a thermal equilibrium in which dS/dU < 0. This can happen, but only in a metastable sense. However, much of equilibrium thermal physics can apply to long-lived metastable equilibria. The concept of negative temperature is consistent with this. (And by the way, if it were true that someone had found a way for heat to flow from a colder to a hotter bath (correctly defined) without entropy increasing elsewhere, then we would all know about it because they would be rich and our energy problems would be over.) $\endgroup$ – Andrew Steane Oct 30 '18 at 10:13
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For the visually inclined, this article explains it simply. The maximum hotness definition is the middle image instead of the expected right image:

absolute zero, infinite hot, and beyond infinite hot

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.

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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.

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  • $\begingroup$ Just to say, in the spin example the system is not "utterly out of equilibrium". Surprising as it may seem, the situation with spins more "up" than "down" is a metastable equilibrium, because the second derivative of the entropy is negative. This means that after a small fluctuation the system will move back or 'relax' to the negative temperature state, and this is the sense in which we can speak of thermal equilibrium here. $\endgroup$ – Andrew Steane Oct 30 '18 at 10:20
  • $\begingroup$ Really!? It's metastable is it? That's really quite remarkable! I feel a need to look at that more closely. Thankyou. $\endgroup$ – AmbretteOrrisey Oct 30 '18 at 10:23
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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

Below I have attached Fig. 1, taken from the arxiv version of this work to illustrate this fact.

enter image description here

PS

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

  2. 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, $T>0$.

  3. 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.

  4. 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.

  5. 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.

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    $\begingroup$ It is not necessary for $S$ to be nonconvex in order to have a negative temperature. The canonical ensemble for a simple 2-state system has a negative temperature regime, but $S(E)$ is convex in that case. It is surely the case that if you move to the microcanonical ensemble then nonconvexity can make things more complicated, but that's tangential to this question. $\endgroup$ – Nathaniel Oct 30 '18 at 9:51
  • $\begingroup$ I had a quick look at the paper just in case, but I didn't change my mind. The proof in my answer really is a mathematical proof - it says that (i) if temperature is defined as $1/T=\frac{\partial S}{\partial E}$, and (ii) if the first and second laws hold, then (iii) heat must always flow from lower $1/T$ to higher $1/T$. If it doesn't then you're using the wrong ensemble or have made some other mistake - there is no other possibility. Neither non-convexity of the entropy nor non-uniqueness of $E(T)$ can change this. $\endgroup$ – Nathaniel Oct 30 '18 at 10:13
  • $\begingroup$ @Nathaniel the research result I quoted here, including an specific example, is precisely that temperature (regardless which entropy is used) does not allow to deduce the direction of heat flow. My answer is specific to the OPs question and short, because I did not want to go into all the details. Please see the linked paper and others by the same authors for answers to your questions. $\endgroup$ – jkds Oct 30 '18 at 11:10
  • $\begingroup$ Yes, I read the paper, albeit briefly, as I said. They review multiple statistical definitions of the entropy and temperature, and claim that for some of them the temperature doesn't predict the direction of heat flow. But that implies a violation of the second law, so it just means those definitions are not the correct ones for the system in question. I do agree with them that the temperature doesn't uniquely determine the thermodynamic state if the entropy isn't convex, but they seem to say this implies it can't predict the direction of heat flow, which doesn't actually follow at all. $\endgroup$ – Nathaniel Oct 30 '18 at 14:13
  • $\begingroup$ @Nathaniel "But that implies a violation of the second law ". Not correct. The second law is discussed in the paper in Sec. V. Of the reviewed entropy definitions only one --the Gibbs entropy-- satisfies the second law strictly. $\endgroup$ – jkds Nov 2 '18 at 8:42

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