There's an absolute zero, is there an "absolute hot"?

This might be more of a physics question, but is there a ceiling on how hot things can get?


What happens at this temperature?




In the actual theories of physics the highest temperature which has a physical meaning is the [Planck's temperature](https://en.wikipedia.org/wiki/Planck_temperature). 


$$T\_\mathrm{P} = \frac{m\_\mathrm{P} c^2}{k} = \sqrt{\frac{\hslash c^5}{G k^2}} \approx \pu{1.4e32 K}$$


For the moment no theory predict higher temperature because of the limit of our theories.




---


There is a Wikipedia article about [absolute hot](https://en.wikipedia.org/wiki/Absolute_hot) with some references. You must have a look.



> 
> Contemporary models of physical cosmology postulate that the highest possible temperature is the Planck temperature, which has the value $\pu{1.416785(71)e32}$ kelvin [...]. Above about $\pu{10^{32} K}$, particle energies become so large that gravitational forces between them would become as strong as other fundamental forces according to current theories. There is no existing scientific theory for the behavior of matter at these energies. A quantum theory of gravity would be required. The models of the origin of the universe based on the Big Bang theory assume that the universe passed through this temperature about $10^{−42}$ seconds after the Big Bang as a result of enormous entropy expansion.
> 
> 
> 


 30

It depends on what you mean by ceiling. Are we talking about a practical or theoretical limit?


At a high enough energy, the stress-energy tensor will be large enough that you're going to make a black hole. I'm not sure we understand the astrophysics well enough to know what this will look like in the limits you refer to.


Also, at some temperature, you're going to mess up the separation of forces, and fundamental forces are going to start to merge again, like in the primordial mix present in the brief moments after the big bang. What does this look like? No one knows for sure.


 14

Depends on what you mean by "temperature".


In **statistical mechanics**, a system of interacting parts is in thermal equilibrium if the probability of finding a given part in a state with energy $E$ is proportional to $e^{aE}$ for some constant $a$ that is the same for all of the parts. Usually, $a$ is negative, and this becomes a *Boltzmann distribution* if we declare the temperature of the system to be the $T$ such that $a=\frac{-1}{kT}$.


If the system has an upper bound of the energy of possible states, there may be equilibria where $a$ is zero or positive, which formally corresponds to "infinite temperature" or "negative temperature". It turns out that $T=\infty$ is hotter than finite $T$, and negative $T$ is *hotter yet* -- in the sense that if you let two systems which are internally in equilibria with $T$s of different signs interact, then energy will flow from the negative-$T$ one to the positive-$T$ one.


My understanding is that this can be observed experimentally with systems of spin states whose interaction with their environment is slow enough that they have time to reach a (good approximation of a) Boltzmann equilibrium internally.


In such a system the "hottest possible temperature" is reached when its energy is maximal, in which case all the parts must be in their most energetic state at the same time. For this macrostate, the best value to assign to $T$ would be "$-0\,\rm K$".




---


From a mathematical point of view, it might perhaps be neater to measure the equilibrium in terms of $a$ rather than $T$. Then the direction of the scale would make more sense: $a=-\infty$ corresponds to $T=+0$, as the system gets hotter we move towards $a=0$ (corresponding to $T=\infty$), and hotter yet sees $a$ go from $0$ towards $+\infty$ while $T$ increases from $-\infty$ to $-0$.


For practical purposes, this is a non-starter, however, because the everyday temperatures an *unbounded* system can have would correspond to negative $a$. Macroscopic laws would look rather strange in terms of such a scale.


 13