Multiplicity of a Einstein Solid, Low Temperature Limit

[TFM]

Homework Statement

(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.

Homework Equations

N/A

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?

Many thanks in advance,

TFM

 

[Vuldoraq]

Cancels down to:

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

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))

 

[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?

 

[Vuldoraq]

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))

This is just rearranging (not using a Taylor series)

now the notes say:

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

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.

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

take exponentials

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

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}

 

[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?