Why is there no absolute maximum temperature?

If temperature makes particles vibrate faster, and movement is limited by the speed of light, then temperature must be limited as well I would assume. Why there is no limits?

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Forget SR considerations and lets focus on low velocity/KE particles. If I am standing, with my thermometer, in a stream of unidirectional particles going on average 1 mile per second I measure a certain temperature. If I am accelerated by an outside force to 1 mps then the particles appear stationary except for some wiggle. Has the temperature measured by my thermometer dropped?? Related to the above does the measured temperature depend on the random spread of energies about the mean or is it solely related to the mean regardless of spread – user2607 Mar 17 '11 at 18:18

6 Answers

I think the problem here is that you're being vague about the limits Special Relativity impose. Let's get this clarified by being a bit more precise.

The velocity of any particle is of course limited by the speed of light c. However, the theory of Special Relativity does not imply any limit on energy. In fact, as energy of a massive particle tends towards infinity, its velocity tends toward the speed of light. Specifically,

$$E = \text{rest mass energy} + \text{kinetic energy} = \gamma mc^2$$

where $\gamma = 1/\sqrt{1-(u/c)^2}$. Clearly, for any energy and thus any gamma, $u$ is still bounded from above by $c$.

We know that microscopic (internal) energy relates to macroscopic temperature by a constant factor (on the order of the Boltzmann constant), hence temperature of particles, like energy, has no real limit.

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Yeah. So it should be explicitly noted (and I am perhaps blind but I don't see this anywhere in your answer) that the appearance that temperature relates to the velocity (as opposed to the energy) is just a low-energy approximation. In SR concepts of energy and velocity depart greatly whereas in classical mechanics they are connected by simple kinetic energy law. – Marek Dec 9 '10 at 23:54
@Marek: Well I think it's noted quite clearly in the SR equation for `E`. Saying that, it might not be immediately apparent that $\gamma$ (that appears in the equation for E) depends on velocity u. – Noldorin Dec 10 '10 at 0:17
@Noldorin: I was thinking more along the lines that $E$ doesn't depend on velocity at all for photons so the two concepts totally depart in SR (and your $\gamma$ formula falls apart). And the reason why I am talking about stating this explicitly is that apparently OP asked his question precisely because he thought temperature has to do with velocity. – Marek Dec 10 '10 at 0:33
@Marek: Massless particles don't come into the question here. I don't want to get broader than need be... – Noldorin Dec 10 '10 at 0:34
@Noldorin: well sure, they don't come in if you don't mention them. But I have a feeling that something is missing. On the other hand, this answer isn't intended for me so be it. Just one last remark: if I were to answer OP's question (which I probably won't anymore), I'd point out the black body which makes it obvious that velocity has nothing to do with temperature. – Marek Dec 10 '10 at 0:41
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The speed of light is an upper limit for the speed of a massive object, but there is no upper bound on the kinetic energy of an object. In fact, that's why the speed of light is an upper limit (one of many reasons, anyway)-- an object moving at the speed of light would have infinite kinetic energy.

The temperature is a measure of the average kinetic energy of particles in a sample. Since kinetic energy does not have an upper limit, temperature does not have an absolute maximum.

(In equations, the kinetic energy is: $K=(\gamma - 1)mc^2 = (\frac{1}{\sqrt{1-v^2/c^2}}-1)mc^2$ which becomes infinitely large as v gets very close to the speed of light c.)

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There is an absolute maximum temperature, and it is $0^{-}$. :)

Okay, that sounds silly, but look it up in L&L: Statistical Physics I.

Think about an Ising paramagnet in an external field: At "zero" temperature (or actually $0^{+}$) the free energy of a system will be minimized by a unique minimum energy configuration. As we raise the temperature, the number of microstates with slightly higher energy grows rapidly, so we have a lower free energy in these entropically favorable configurations. Now we continue all the way to infinite temperature, at which point the system becomes completely disordered.

But wait, what if we drive the system to even higher energy? In that case there are fewer microstates and so the derivative that defines temperature goes negative, and the temperature that corresponds to these configurations is $-\infty$. This actually corresponds to the principle of "population inversion" in lasers. Anyway, higher and higher energy configurations (with their continually decreasing entropy) correspond to decreasing negative temperatures, until all of the spins point against the external field at $T=0^-$.

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That's a fantastic answer. There might be some skepticism about the "negative temperature" part however you do mention its relation with "population inversion" a routine occurrence in laser physics. Question: Has anyone setup an experiment which can 'measure' these negative temperatures? – user346 Jan 23 '11 at 0:05
None that I know of, and if such a set-up does exist it must be exceptionally clever. I guess the point is that, to do thermometry in the usual sense, the system that you're measuring has to act as a reservoir with respect to your probe - population inversion is easy enough, but to maintain such an unstable state? and with enough degrees of freedom to behave as a thermal reservoir? It seems unreasonably difficult. – wsc Jan 23 '11 at 0:15
you need to be careful that a system can have different temperatures; for instance, one could say that galactic halos have pretty uniform movement relative to the disk, so one would say that the $\Delta E$ is small and hence small temperature, but the halo might be composed of stars will high temperatures themselves! so a temperature may be adequate only to a specific scale of the system – lurscher Mar 17 '11 at 19:52

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@Deepak : An experimental negative temperature of -350 K was demonstrated in 1951 paper "A Nuclear Spin System at Negative Temperature" link.aps.org/abstract/PR/v81/p279 , found via en.wikipedia.org/wiki/Negative_temperature . – Frédéric Grosshans Mar 18 '11 at 18:01
Note that negative temperatures are possible in systems like this because there is a limit on the energy available per particle in the context of the system. In the context of kinetic temperatures the necessary condition does not obtain. – dmckee♦ Jan 26 '12 at 2:52

If there is a maximum possible physical temperature it is well above anything we can reach experimentally and would require a complete theory of quantum gravity to understand it fully.

Neutron stars are some of the hottest objects in the universe today with temperatures up to around 10 trillion degrees Kelvin ($10^{12} K$). Similar temperatures have been reached in heavy ion collisions recently at the Large Hadron Collider for very small volumes and times. At these temperatures even the protons and neutrons in nuclear matter are torn apart leaving just a plasma of quarks and glouns.

But these temperatures are cool compared to the earliest moments of the big bang. According to our incomplete theories something really odd happens when you get to the Planck temperature which is around $10^{32} K$, so a good 20 orders of magnitude higher than anything we can produce.

At such temperature spacetime itself must be highly energised by gravitational interactions with hot matter. Some people think that spacetime passes through some kind of phase transition at this point, but if it does we have very little understanding of what kind of phase state lies beyond or whether temperatures can be raised further. Such understanding is in the realm of quantum gravity which is not yet fully developed. Such physics may describe the very earliest moments of the big bang and perhaps nowhere else in the universe.

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While special relativity does not, a priori, place any constraints on the maximum temperature a system can attain, the situation changes when we consider the quark-gluon plasma - a stage you will eventually reach if you heat up any hadronic matter sufficiently. Rolf Hagedorn realized that for hadronic matter there exists a maximum temperature above which the partition function of the system is not well-defined. In other words you can only heat up hadronic matter to a maximum given by the Hagedorn temperature $T_H$.

Since hadronic-matter constitutes the vast majority of the matter we interact with (excluding dark matter and dark energy), in some sense $T_H$ is the maximum temperature that ordinary matter can attain, though this is by no means the end of the story ...

Of course, even with special relativity alone, one can see that when the temperature of a gas of particles becomes comparable to the rest energy of the particles in question, any attempt to increase the temperature beyond that point will only lead to pair creation. This was, vaguely speaking, the reasoning behind Hagedorn's work.

You might also find this Nova column on the Hagedorn phase enlightening.

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But this is like saying that the maximum temperature of liquid water is 100 degrees C; it's, strictly speaking, correct, but sort of misses the point, which is that a phase transition happens and you can have the same matter at higher temperatures in a different phase. For hadronic matter, heating it to higher temperatures produces a deconfined plasma. – Matt Reece Dec 10 '10 at 16:28
This is incorrect, Hagedorn later realized that the "maximum temperature" is a sign of a phase transition. There is no maximum hadronic temperature, because the exponentially growing number of states are spatially more and more extended. – Ron Maimon Aug 22 '11 at 2:50

We have two reasons for there not being a limit. As everyother commenter here has said, SR does not limit the energy per particle. Actually energy per degree of freedom would be a more precise statement. In any case temperature does not equate directly to the energy per particle DOF, but rather to the staistical probabilities, namely that the relative probabilities of a particular state being occupied is proportional to e(- deltaE/kT ). (Even that only applies to the low density limit, fermions are limited to one particle per allowable state, so in some high density low temperature limits (solid state, and degenerate state (some stellar interiors, white dwarfs etc.)) the lowest energy states are almost fully occupied. But, in any case, temperature applies to the probability distribution of the occupation of states with different energies, average energy per particle is just the normalized integral of this density time energy.

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