In many presentations of statistical mechanics where we have a system of particles having mass, such as the molecules of an ideal gas, the temperature is often equated to the average relative velocity of the particles. However no such velocity can exceed that of light. Is there therefore an absolute upper bound to temperature as well as an absolute zero?
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1$\begingroup$ If that were really how temperature worked, then all photon gases would have the same temperature... $\endgroup$– Brian BiCommented Apr 23, 2014 at 20:46
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1$\begingroup$ Possible duplicates: physics.stackexchange.com/q/1775/2451 , physics.stackexchange.com/q/46397/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Apr 23, 2014 at 21:41
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3 Answers
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The definition of temperature from kinetic theory gives temperature as a function of the average energy of the massive particles, not necessarily their speed (this is how you can have photon gasses with different temperatures). If the particles are point particles with no internal structure, that would imply that there is no upper bound on the temperature for a given mass because the kinetic energy of a massive particle increases without bound as $v \rightarrow c$.
The more formal thermodynamic definition of temperature is given by the derivative of the entropy of the system with respect to its energy: $\frac{1}{T} = \frac{dS}{dE}$. To understand this, imagine that you fix the energy of the system. Given that energy, there will be some number of states the system could find itself in. That number of states allows you to calculate the entropy of the system. Now increase the energy of the system by just a bit, and recalculate the entropy. Divide the change in the entropy by the change in the energy, invert that result, and you have the temperature.
Under this thermodynamic definition, there is neither a maximum attainable temperature, nor a minimum attainable temperature. Negative temperatures are even allowed. An easy to understand example system is the one-dimensional Ising model. Take some number $N$ particles, which can each have one of two orientations. For the $i$-th particle, one orientation has energy $E_i = 0$, while the other orientation has energy $E_i = \epsilon$. Now examine the three situations of total energy $E = 0$, $E = N\epsilon/2$, and $E = N\epsilon$. The first and the last have only one possible state each, so they both have zero entropy. The second is the maximal entropy state possible. After working things out, you get that $T = +0, \infty, -0$, respectively for those three cases. Paradoxically, $T = -\infty$ is "hotter" than $T = +\infty$, in the sense that energy would flow from a system at $T = -\infty$ to a system at $T = +\infty$. The details are worked out in Ramsey, Phys. Rev. 103, 20 (1956)
answered Apr 23, 2014 at 21:48
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Temperature is proportional to the average kinetic energy, not velocity, of the particles. Kinetic energy is unbounded; it goes to infinity as velocity approaches light-speed, proportional to $(1 - v^2/c^2)^{-1/2}$.
answered Apr 23, 2014 at 21:42
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$\begingroup$ Many thanks for your answer which really clarifies the situation for me. $\endgroup$ Commented Apr 24, 2014 at 20:14
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Simplest answer: no, there's no upper limit. Temperature is actually proportional to the average energy per particle in the system, and energy is unlimited in special relativity. (The velocities, of course, will all be below $c$.)
Here's a cool example I just found: https://en.wikipedia.org/wiki/Relativistic_plasma
