Problem involving Negative Temperature

[tanaygupta2000]

Homework Statement

Consider an isolated system of N>>1 distinct particles, each with possible energy states 0 and ε.
(1.) Give the range of energies for which the system has negative temperature.
(2.) If A and B are two subsystems of the given system, each having N/2 particles, and A having energy E(A) = Nε/8 and B having energy E(B) = 5Nε/8, which of these has positive/negative temperature?
(3.) The two systems are kept in thermal contact, what is the equilibrium temperature of two subsystems?

Relevant Equations

Population inversion, n(2)/n(1) = exp[-(ε2 - ε1)/kT]
n(1`) < N/2

The partition function of the given system is given by, Z = 1 + e^(-ε/kT)
So in the '0' energy state, number of particles, n₁ = [1/(1 + e^(-ε/kT))]N
and in the 'ε' energy state, number of particles, n₂ = [e^(-ε/kT)/(1 + e^(-ε/kT))]N

Now according to condition of population inversion, n₁ < N/2
Upon substituting the values, I am simply getting e^(-ε/kT) > 1
How am I supposed to find the 'range of energies for negative temperature'?

 

[DrClaude]

Now according to condition of population inversion, n₁ < N/2
Upon substituting the values, I am simply getting e^(-ε/kT) > 1
How am I supposed to find the 'range of energies for negative temperature'?

What are the possible values for the total energy if $n_1 < N/2$?

 

[tanaygupta2000]

What are the possible values for the total energy if $n_1 < N/2$?

If n₁ < N/2
=> 1/(1 + e^-ε/kT) < 1/2
=> e^-ε/kT > 1
=> ε > 0

 

[DrClaude]

If n₁ < N/2
=> 1/(1 + e^-ε/kT) < 1/2
=> e^-ε/kT > 1
=> ε > 0

No.

You have to calculate the total energy of the $N$ particles.

 

[tanaygupta2000]

No.

You have to calculate the total energy of the $N$ particles.

Total energy, E = n₁ε₁ + n₂ε₂
= e^-ε/kt/(1 + e^-ε/kT) × Nε

which is Nε/(1 + e^ε/kT)

 

[DrClaude]

Total energy, E = n₁ε₁ + n₂ε₂
= e^-ε/kt/(1 + e^-ε/kT) × Nε

which is Nε/(1 + e^ε/kT)

There is no need to introduce temperature explicitly. You need to consider only that it is negative, and therefore that $n_1 < N/2$.

 

[tanaygupta2000]

There is no need to introduce temperature explicitly. You need to consider only that it is negative, and therefore that $n_1 < N/2$.

Using ε > 0, I am getting total energy, E < Nε/2
It means that the energy of the upper level must be greater than (2/N) times the total energy E of the system.
Is this the answer of first part ?

 

[haruspex]

It means that the energy of the upper level must be greater than (2/N) times the total energy E of the system.

Small edit: deleted "yes, but"

that is not the form of answer requested. Your answer should be a range of values of E in terms of N and ε

 

[tanaygupta2000]

Yes, but that is not the form of answer requested. Your answer should be a range of values of E in terms of N and ε

It means 0 < E < Nε/2

 

[haruspex]

It means 0 < E < Nε/2

How do you get that?
You have n₁ < N/2 < n₂. What is E in terms of n₁, n₂ and ε?

 

[tanaygupta2000]

How do you get that?

 

[haruspex]

View attachment 260277

You are forgetting that T<0.
Anyway, it is much simpler: you know n₂>N/2 and you have E in terms of n₂ and ε.

 

[tanaygupta2000]

You are forgetting that T<0.
Anyway, it is much simpler: you know n₂>N/2 and you have E in terms of n₂ and ε.

On applying n₁ < N/2 < n₂ and using E = n₂ε = Nε/(1 + e^ε/kT)
I am getting 2ε/(1 + e^-ε/kT) < ε < 2εe^-ε/kT/(1 + e^-ε/kT)

which becomes e^ε/kT < ε/2 < 1

 

[haruspex]

On applying n₁ < N/2 < n₂ and using E = n₂ε = Nε/(1 + e^ε/kT)
I am getting 2ε/(1 + e^-ε/kT) < ε < 2εe^-ε/kT/(1 + e^-ε/kT)

which becomes e^ε/kT < ε/2 < 1

You are making it far more complicated than it is. You need an inequality involving E, N and ε. Only. Don't involve T.
You were fine this far:
n₁ < N/2 < n₂ and using E = n₂ε

 

[tanaygupta2000]

You are making it far more complicated than it is. You need an inequality involving E, N and ε. Only. Don't involve T.
You were fine this far:
n₁ < N/2 < n₂ and using E = n₂ε

Sir I cannot obtain an inequality involving E, just as I cannot obtain for T, since both of them are not defined in the question.
The best I am getting from n₁ < N/2 < n₂ is -ε < 0 < ε.

 

[haruspex]

Sir I cannot obtain an inequality involving E, just as I cannot obtain for T, since both of them are not defined in the question.
The best I am getting from n₁ < N/2 < n₂ is 0 < ε < 0

N and ε are given. You are asked to find the range for the total energy. You were not given a label for that, but you have chosen to call it E. Therefore the answer should be a range for E in terms of N and ε.

0 < ε is uninteresting. It is necessarily true.

 

[tanaygupta2000]

N and ε are given. You are asked to find the range for the total energy. You were not given a label for that, but you have chosen to call it E. Therefore the answer should be a range for E in terms of N and ε.

0 < ε is uninteresting. It is necessarily true.

 

[haruspex]

You are still making it far too complicated and you should not have T in your answer.

Just use n₁ < N/2 < n₂ and E = n₂ε, no other equations, to obtain an inequality relating only E, N and ε.

 

[tanaygupta2000]

You are still making it far too complicated and you should not have T in your answer.

Just use n₁ < N/2 < n₂ and E = n₂ε, no other equations, to obtain an inequality relating only E, N and ε.

n₁ < N/2 < n₂
=> (N-n₂) < N/2 < n₂

Using E = n₂ε => n₂ = E/ε
=> N - E/ε < N/2 < E/ε

Adding (E/ε - N/2) all sides,
=> N/2 < E/ε < 2E/ε - N/2
=> Nε/2 < E < 2E - Nε/2

Is this correct?

 

[DrClaude]

=> Nε/2 < E < 2E - Nε/2

Is this correct?

No. You can't have E on both sides of an inequality for E.

 

[tanaygupta2000]

n1 < N/2 < n2
=> (N-n2) < N/2 < n2

Sir was this approach correct for obtaining an inequality relating only E, N and ε ?

 

[DrClaude]

Sir was this approach correct for obtaining an inequality relating only E, N and ε ?

Yes, although it is not how I would set up the problem. Since $E = n_2 \varepsilon$, I would write an inequality for $n_2$ and convert it for one for $E$.

 

[tanaygupta2000]

Yes, although it is not how I would set up the problem. Since $E = n_2 \varepsilon$, I would write an inequality for $n_2$ and convert it for one for $E$.

Sir how do I write an inequality for n₂ starting from n₁ < N/2 < n₂ ?

 

[DrClaude]

Sir how do I write an inequality for n₂ starting from n₁ < N/2 < n₂ ?

Don't use $n_1$, just set bounds on $n_2$.

 

[tanaygupta2000]

Don't use $n_1$, just set bounds on $n_2$.

If I have to use just N/2 < n₂
then N/2 < E/ε
=> Nε/2 < E < ∞

 

[DrClaude]

If I have to use just N/2 < n₂
then N/2 < E/ε
=> Nε/2 < E < ∞

Ins't there an upper bound on $n_2$?

 

[tanaygupta2000]

Ins't there an upper bound on $n_2$?

N/2 < n₂ < N
=> N/2 < E/ε < N
=> Nε/2 < E < Nε

 

[tanaygupta2000]

N/2 < n₂ < N
=> N/2 < E/ε < N
=> Nε/2 < E < Nε

Can I proceed with this solution ?

 

[haruspex]

Can I proceed with this solution ?

Looks right to me.

 

[tanaygupta2000]

How do I solve the third part?
I know that for A, N/2 = [e^{(-5Nε/8kT_A)}/ {e^{(-5Nε/8kT_A)} + e^(-Nε/8kT_B)} ]N

and for B, N/2 = [e^(-Nε/8kT_B)/ {e^{(-5Nε/8kT_A)} + e^(-Nε/8kT_B)} ]N

how do I find the equilibrium temperature of these two ?
Do I have to first add T_A and T_B and then equate the derivative to zero?