Multiplicity of a Einstein Solid, Low Temperature Limit

TFM

1. The problem statement, all variables and given/known data

(a)

The formula for the multiplicity of an Einstein solid in the “high-temperature” limit,
q >> N, was derived in one of the lectures. Use the same methods to show that the multiplicity of an Einstein solid in the “low-temperature” limit, q << N, is

Ω(N,q)=(eN/q)^q (When q≪N)


(b)

Find a formula for the temperature of an Einstein solid in the limit q << N. Solve for the energy as a function of temperature to obtain U=Nϵe^(-ϵ/(k_B T)), where ε is the size of an energy unit.


2. Relevant equations

N/A

3. The attempt at a solution

Okay, I have started (a):

[tex] \Omega (N,q) = (\frac{(N - 1 + q)!}{(n - 1)!q!}) [/tex]

N large:

(N - 1)! approx = N!

[tex] \Omega (N,q) = (\frac{(N + q)!}{N! q!}) [/tex]

Take logs

[tex] ln(\Omega (N,q)) = ln(N + q)! - ln(N!) - ln(q!) [/tex]

Use Stirling aprrox:

[tex] ln N! \approx N ln(N) - N [/tex]

[tex] ln(\Omega (N,q)) = (N + q)ln(N + q) - (N + q) - (Nln(N) - N) - (qln(q) - q) [/tex]

Cancels down to:

[tex] ln(\Omega (N,q)) = (N + q)ln(N + q) - Nln(N) - qln(q) [/tex]

Now I have to use the Taylor Expnasion for q << N, but I got slightly confused here.

Could anyone please offer some assistane what I need to do from here?

Many thanks in advance,

TFM

Vuldoraq

You need to write the (N+q) log like this (below) and then just copy the lecture derivation, I think. You can show how you use the taylor expansion, but I would have said it's not neccessary

[tex]ln(N+q)=ln(N(1+q/N))[/tex]

TFM

Okay, so now:

[tex] ln(\Omega (N,q)) = (N + q)ln(N + q) - Nln(N) - qln(q) [/tex]

Use Taylor Expansion:

[tex] ln(N+q)=ln(N(1+q/N)) [/tex]

[tex] ln(N+q)=ln(N) + ln(1+q/N)) [/tex]

now the notes say:

[tex] ln(N+q) = ln(N) + q/N [/tex]

Althought I am not sure why the last ln has dissapeared?

Now plug into omega to give:

[tex] ln(\Omega (N,q)) = (N + q)(ln(N) + q/N)- Nln(N) - qln(q) [/tex]

Multiply out:

[tex] ln(\Omega (N,q)) = NlnN + q + q^2/N + qlnN - Nln(N) - qln(q) [/tex]

Cancel down:

[tex] ln(\Omega (N,q)) = q + q^2/N + qlnN - qln(q) [/tex]

Now the notes have that:

because q >>N,

N^2/q approx 0

So for

because q <<N,

This should still be

q^2/N approx 0

thus:

[tex] ln(\Omega (N,q)) = q + qlnN - qln(q) [/tex]

take exponentials

[tex] \Omega (N,q) = e^q + (N/q)^q [/tex]

Thus

[tex] \Omega (N,q) = (eN/q)^q [/tex]

Does this look okay?

Vuldoraq

This is just rearranging (not using a Taylor series)

You should say, [tex] ln(N+q)=ln(N(1+q/N))=ln(N) + ln(1+q/N)[/tex]

You can then take the Taylor expansion of the last term in the above, namely
[tex]ln(1+q/N)[/tex], provided you assume q<<N so that q/N<<1. If you look up the Taylor expansion for ln(1+x) you will find that it says this is approximately equal to x in the limit x<<1.
Here there is an error. When you exponentiate an equation the entire right side is raised to the same exponential. Also [tex]e^x+e^y \neq e^{x+y}[/tex]

TFM

Okay so:

[tex] ln(\Omega (N,q)) = q + qlnN - qln(q) [/tex]

so I have to take exponentials,


[tex] \Omega (N,q) = e^q + e^qlnN - e^qln(q) [/tex]

[tex] \Omega (N,q) = e^q + Ne^q - qe^q [/tex]

So now do we use the log/exponential rules to give?

[tex] \Omega (N,q) = (eN/q)^q [/tex]

Also,

Find a formula for the temperature of an Einstein solid in the limit q << N. Solve for the energy as a function of temperature to obtain U=Nϵe^(-ϵ/(k_B T)), where ε is the size of an energy unit.

Where woul be aq good place to start for here?