Statistical mechanics: Particles with spin


by SoggyBottoms
Tags: mechanics, particles, spin, statistical
SoggyBottoms
SoggyBottoms is offline
#1
Mar5-12, 11:55 PM
P: 61
1. The problem statement, all variables and given/known data

We have N particles, each of which can either be spin-up ([itex]s_i = 1[/itex]) or spin-down ([itex]s_i = -1[/itex]) with [itex]i = 1, 2, 3....N[/itex]. The particles are in fixed position, don't interact and because they are in a magnetic field with strength B, the energy of the system is given by:

[tex]E(s_1, ...., s_n) = -mB \sum_{i=1}^{N} s_i[/tex]

with m > 0 the magnetic moment of the particles. The temperature is T.

a) Calculate the canonic partition function for N = 1 and the chance that this particle is in spin-up state [itex]P_+[/itex].

b) For any N, calculate the number of microstates [itex]\Omega(N)[/itex], the Helmholtz free energy F(N,T) and the average energy per particle U(N, T)/N

3. The attempt at a solution

a) [tex]Z_1 = e^{-\beta m B} + e^{\beta m B} = 2 \cosh{\beta m B}[/tex]
[tex]P_+ = \frac{e^{-\beta m B}}{2 \cosh{\beta m B}}[/tex]

b) The number of possible microstates is [itex]\Omega(N) = 2^N[/itex], correct?

I know that [itex]U = -\frac{\partial \ln Z}{\partial \beta}[/itex], but I'm not sure how to calculate Z here.
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Liquidxlax
Liquidxlax is offline
#2
Mar6-12, 01:31 PM
P: 321
leave Z as the summation Z = Ʃ e-Eiβ where β = 1/KBT


so ∂ln(Z)/dβ = (1/Z)(∂Z/∂β) = [-EiƩe-Eiβ]/Z


i think b) is supposed to be (Z1)N sorry yeah your b) is right


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