# Multiplicity of a Einstein Solid, Low Temperature Limit

1. Feb 4, 2009

### 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):

$$\Omega (N,q) = (\frac{(N - 1 + q)!}{(n - 1)!q!})$$

N large:

(N - 1)! approx = N!

$$\Omega (N,q) = (\frac{(N + q)!}{N! q!})$$

Take logs

$$ln(\Omega (N,q)) = ln(N + q)! - ln(N!) - ln(q!)$$

Use Stirling aprrox:

$$ln N! \approx N ln(N) - N$$

$$ln(\Omega (N,q)) = (N + q)ln(N + q) - (N + q) - (Nln(N) - N) - (qln(q) - q)$$

Cancels down to:

$$ln(\Omega (N,q)) = (N + q)ln(N + q) - Nln(N) - qln(q)$$

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?

TFM

2. Feb 4, 2009

### 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

$$ln(N+q)=ln(N(1+q/N))$$

3. Feb 5, 2009

### TFM

Okay, so now:

$$ln(\Omega (N,q)) = (N + q)ln(N + q) - Nln(N) - qln(q)$$

Use Taylor Expansion:

$$ln(N+q)=ln(N(1+q/N))$$

$$ln(N+q)=ln(N) + ln(1+q/N))$$

now the notes say:

$$ln(N+q) = ln(N) + q/N$$

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

Now plug into omega to give:

$$ln(\Omega (N,q)) = (N + q)(ln(N) + q/N)- Nln(N) - qln(q)$$

Multiply out:

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

Cancel down:

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

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:

$$ln(\Omega (N,q)) = q + qlnN - qln(q)$$

take exponentials

$$\Omega (N,q) = e^q + (N/q)^q$$

Thus

$$\Omega (N,q) = (eN/q)^q$$

Does this look okay?

4. Feb 5, 2009

### Vuldoraq

This is just rearranging (not using a Taylor series)

You should say, $$ln(N+q)=ln(N(1+q/N))=ln(N) + ln(1+q/N)$$

You can then take the Taylor expansion of the last term in the above, namely
$$ln(1+q/N)$$, 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 $$e^x+e^y \neq e^{x+y}$$

5. Feb 5, 2009

### TFM

Okay so:

$$ln(\Omega (N,q)) = q + qlnN - qln(q)$$

so I have to take exponentials,

$$\Omega (N,q) = e^q + e^qlnN - e^qln(q)$$

$$\Omega (N,q) = e^q + Ne^q - qe^q$$

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

$$\Omega (N,q) = (eN/q)^q$$

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?