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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4$\begingroup$ Obligatory movie title $\endgroup$– Qmechanic ♦Oct 12, 2017 at 19:37
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8 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.
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1$\begingroup$ 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. $\endgroup$ Oct 30, 2018 at 10:01
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$\begingroup$ 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 $\endgroup$– jkdsOct 30, 2018 at 14:48
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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.
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.
answered Mar 4, 2012 at 15:35
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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 $S$. $\endgroup$– jkdsOct 15, 2018 at 11:26
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2$\begingroup$ @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. $\endgroup$ Oct 15, 2018 at 12:18
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$\begingroup$ 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. $\endgroup$– jkdsOct 22, 2018 at 9:36
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1$\begingroup$ @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. $\endgroup$ Oct 22, 2018 at 13:29
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1$\begingroup$ @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). $\endgroup$ Oct 23, 2018 at 16:40
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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.
answered Mar 4, 2012 at 15:14
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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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3$\begingroup$ 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, 2015 at 10:26
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1$\begingroup$ 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. $\endgroup$ Jul 1, 2015 at 3:04
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$\begingroup$ @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. $\endgroup$– JoshuaMay 22, 2016 at 16:20
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$\begingroup$ @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 $\endgroup$– jkdsOct 15, 2018 at 6:36
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1$\begingroup$ 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.) $\endgroup$ Oct 30, 2018 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:
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. $\endgroup$ Oct 30, 2018 at 10:20
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$\begingroup$ Really!? It's metastable is it? That's really quite remarkable! I feel a need to look at that more closely. Thankyou. $\endgroup$ Oct 30, 2018 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.
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.
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2$\begingroup$ 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. $\endgroup$– N. VirgoOct 30, 2018 at 9:51
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1$\begingroup$ 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. $\endgroup$– N. VirgoOct 30, 2018 at 10:13
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1$\begingroup$ 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. $\endgroup$– N. VirgoOct 30, 2018 at 14:13
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1$\begingroup$ Look, if $T$ is defined via $1/T = \frac{\partial S}{\partial E}$ then for two coupled systems we have $\frac{\partial S}{\partial E_1} = \frac{\partial (S_1+S_2)}{\partial E_1} =\frac{\partial S_1}{\partial E_1} - \frac{\partial S_2}{\partial E_2} = 1/T_1-1/T_2$, and the entropy increases if and only if heat flows from the system with lower $1/T$ to the system with higher $1/T$. This is a really simple, completely incontrovertible consequence of the definition. If your statistical definition of entropy contradicts this then it contradicts the second law, even if you have a Nature paper. $\endgroup$– N. VirgoNov 2, 2018 at 9:21
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2$\begingroup$ Regarding the Nature Physics paper by Dunkel and Hilbert, it's baffling to me that they fail to mention the Gibbs-Shannon or the von Neumann entropy, those being the statistical definitions from which the Boltzmann distribution is derived in the first place. However, it's not surprising to me that the thing they call the Gibbs entropy (which is actually Boltzmann's definition of the entropy) is a better approximation than the thing they call the Boltzmann entropy. So I don't disagree with them on that point. $\endgroup$– N. VirgoNov 2, 2018 at 9:44
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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.
