# Problem involving Negative Temperature

Homework Helper
Gold Member
Sir but how do I solve the third part of the question? What can I say about the equilibrium temperature?
See your post #5. You can get T from E.

tanaygupta2000
See your post #5. You can get T from E.
We deduced, Nε/2 < E < Nε
Since E = Nε/(1+e(ε/kT))
=> Nε/2 < Nε/(1+e(ε/kT)) < Nε
=> 1/2 < 1/(1+e(ε/kT)) < 1
=> 2 > 1+e(ε/kT) > 1
=> 1 > e(ε/kT) > 0
=> 0 > ε/kT > -∞
=> 0 > 1/T > -∞
Is this correct ?

Homework Helper
Gold Member
Once the two subsystems A and B are in equilibrium with each other, you can treat the whole system as one system in a state of thermal equilibrium. So, you can apply the relation E = Nε/(1+e(ε/kT)) to the whole system. You know the value of E for the whole system. Therefore, you can solve the relation for T to get the equilibrium temperature of the system.

tanaygupta2000
Once the two subsystems A and B are in equilibrium with each other, you can treat the whole system as one system in a state of thermal equilibrium. So, you can apply the relation E = Nε/(1+e(ε/kT)) to the whole system. You know the value of E for the whole system. Therefore, you can solve the relation for T to get the equilibrium temperature of the system.
Using E = Nε/(1+e(ε/kT)), I am getting T = (ε/k) /ln(Nε/E - 1)

Substituting E = Nε/8 + 5Nε/8 = 3Nε/4
=> T = (ε/k) /ln(1/3) = (ε/k) /(-1.098) = -0.91ε/k

Hence the system as a whole is having NEGATIVE TEMPERATURE.

Homework Helper
Gold Member
Using E = Nε/(1+e(ε/kT)), I am getting T = (ε/k) /ln(Nε/E - 1)

Substituting E = Nε/8 + 5Nε/8 = 3Nε/4
=> T = (ε/k) /ln(1/3) = (ε/k) /(-1.098) = -0.91ε/k

Hence the system as a whole is having NEGATIVE TEMPERATURE.
Yes. That looks right if the total energy is 3Nε/4.

However, as I mentioned earlier, I don't believe it's possible for system B to have an initial energy of 5Nε/8. The maximum possible energy of B (or A) is (N/2)ε, since B has only N/2 particles.