Use Stirling’s approximation to show that the multiplicity of an Einstein solid, for any large values of N and q, is approximately

The square root in the denominator is merely large, and can often be neglected. However, it is needed in below Problem (Hint: First show that Do not neglect the in Stirling’s approximation.)

Problem:

This problem gives an alternative approach to estimating the width of the peak of the multiplicity function for a system of two large Einstein solids.

a) Consider two identical Einstein solids, each with N oscillators, in thermal contact with each other. Suppose that the total number of energy units in the combined system is exactly 2N. How many different macrostates (that is, possible values for the total energy in the first, solid) are there for this combined system?

b) Use the result of above Problem to find an approximate expression, for the total number of microstates for the combined system. (Hint:Treat the combined system as a single Einstein solid. Do not throw away factors of “large” numbers, since you will eventually be dividing two “very large” numbers that are nearly equal.

c) The most likely macrostate for this system is (of course) the one in which the energy is shared equally between the two solids. Use the result of above Problem to find an approximate expression for the multiplicity of this macrostate.

d) You can get a rough idea of the “sharpness” of the multiplicity function by comparing your answers to parts (b) and (c). Part (c) tells you the Height of the peak, while part (b) tells yon the total area under the entire graph. As a very crude approximation, pretend that the peak’s shape is rectangular. In this case, how wide would it be? Out of all the macrostates, what fraction have reasonably large probabilities? Evaluate this fraction numerically for the case N = 1023.

Step 1:

To show that the multiplicity of an Einstein solid, for any large values of N and q, is approximately

=

Step 2:

We know that

= -----(1)

The basic formula related to n! = n(n-1)!

This gives (n-1)! = -----(2)

Applying the (2) in the appropriate terms in (1)

= -----(3)

(N-1)! = -----(4)

Step 3:

Substituting (3) and (4) in (1)

We get

= -----(4)

Step 4:

We know that from Stirling’s approximation

n! = -----(5)

Therefore

= ----(6)

= -----(7)

= -----(8)