# Temperature negative thermodynamics

## Homework Statement

Hi!
I have this question "how is the curve of U(internal energy) with volume if the temperature is negative"?

## The Attempt at a Solution

i cnsidered T=(partial U/partial S) at constant V and N (number of particules, but now i cant say the change of U with V if T<0...can somebody help me?
i also thought that T = slope of the tangent line to the function U(S) but i cant say how is the curve U(V) when T<0 as required

(note: i know the temperature cant be negative but we want to know what theoricaly happens)

thanks!

Well, let's talk about Boltzmann factor a little bit. You know that the number of particles in an energy level is proportional to $e^{-\frac{E_i}{kT}}$, where $E_i$ is the energy of that level. So If I compare the number of particles in two levels i and j where $E_j>E_i$, I have $\frac{N_j}{N_i}=\frac{ e^{-\frac{E_j}{kT}} }{ e^{-\frac{E_i}{kT}} }$. But its obvious that for any finite positive temperature, $N_j<N_i$ and even when we have $T\to \infty$, its only that $N_j=N_i$. But let's just assume(as you did), that its possible to have $T<0$. We then see that for negative temperatures, we have $N_j>N_i$. But wait, there is something weird happening here. Of course when particles go to upper energy levels, it means they are getting energy from somewhere. But when the system has an infinite number of energy levels, as particles get more energy, they just keep moving up. So when we have infinite number of levels(as it is with the usual systems we think about), negative energy can't be accommodated into the theory and so if we want to think about negative temperatures, we should think about systems with finite number of levels. Its hard to explain but physicists actually found such systems(in fact not actually found them, but built them with some tricks) and achieved negative temperatures. So only in this restricted class of systems, we can talk about negative temperatures. But honestly I don't know how to get volume into play here. So I hope other people will contribute to this thread.