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# Problems in Statistical Mechanics

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3. Consider again the lattice of N spin-1/2 particles in an external homogeneous magnetic field, where each particle has two possible states: spin "down" with energy e = 0 and spin "up" with energy e = 1/2". The microstate of the system is specified by the energy states of all the particles, i.e. the list (e1, e2...

https://brainmass.com/physics/beta/problems-statistical-mechanics-428804

## SOLUTION This solution is FREE courtesy of BrainMass!

(a) We have S = k log W(n) = k log NCn = k log (N!/n!(N-n)!). Using Sterling's approximation to the factorial, we have

S is approximately equal to k log (N^N e^-N/((n^n e^-n)((N-n)^(N-n) e^-(N-n)))
= k (N log N - n log n - (N-n) log(N-n))

Since E(n) = n/2, we have

S(E) = k (N log N - 2E log 2E - (N-2E) log(N-2E))

where the above equality is approximate in the limit of large N and n.

(b) We have

1/T = dS/dE = k(-2E/E - 2 log 2E + 2 log(N-2E) - 2(N-2E)/(N-2E))
= 2k log((N-2E)/2E)
whence

T(E) = 1/(2k log(N/2E - 1)).

(c) Inveting the above formula, we find

E(T) = N/(2(1 + e^(1/(2kT))).

We have

C(T) = dE/dT

= N/2 (-1/(2kT^2)) (-1/(1 + e^(1/(2kT)))^2
= N/(4kT^2) (1 + e^(1/(2kT)))^-2.

(d) We have

Z = sum_{n=0}^infty(NCn e^(-beta E(n)))
= sum_{n=0}^infty(NCn e^(-beta n/2))

where beta = 1/(kT).

(e) We have

<E> = -1/Z dZ/dbeta
= 1/(2Z) sum_{n=0}^infty(n NCn e^(-beta n/2)).

(f) We have

C = d<E>/dT
= d<E>/dbeta dbeta/dT
= -1/(kT^2) d<E>/dbeta
= 1/(kT^2) [(1/(2Z^2) dZ/dbeta) sum_{n=0}^infty(n NCn e^(-beta n/2))
+ 1/(4Z) sum_{n=0}^infty(n^2 NCn e^(-beta n/2))]

= 1/(kT^2) [1/(4Z) sum_{n=0}^infty(n^2 NCn e^(-beta n/2)) - E^2/2].

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