Temperature negative thermodynamics

rf1
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Homework Statement


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

Homework Equations

The Attempt at a Solution


i cnsidered T=(partial U/partial S) at constant V and N (number of particules, but now i can't 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 can't say how is the curve U(V) when T<0 as required

(note: i know the temperature can't be negative but we want to know what theoricaly happens)

thanks!
 
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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&gt;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&lt;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&lt;0. We then see that for negative temperatures, we have N_j&gt;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.
 
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