# Reif, problem 2.4

1. Feb 20, 2008

### gc2004

[SOLVED] reif, problem 2.4

1. Statement of the Problem

Consider an isolated system consisting of a large number N of very weakly interacting localized particles of spin 1/2. Each particle has a magnetic moment 'm' which can point either parallel or antiparallel to an applied field H. The energy E of the system is then E= - (n1 - n2)mH, where n1 is the number of spins aligned parallel to H and n2 the number of spins aligned antiparallel to H.

(a) consider the energy range between E and E+del E where del E is very small compared to E but is microscopically large so that del E >> mH. What is the total number of states W lying in this range?

(b) Write down an expression for ln W(E) as a function of E. Simplify this expression by applying Stirling's formula in its simplest form.

(c) Assume that the energy E is in a region where W(E) is appreciable, i.e., that it is not close to the extreme possible values NmH or - NmH which it can assume. In this case apply gaussian approximation to part (a) to obtain a simple expression for W(E) as a function of E.

2. Relevant equations

Stirling formula : lnx! = xlnx - x

3. Attempt to a solution

(a) naturally, when the spin points antiparallel, its energy is mH and when it points parallel, its energy is - mH. Thus, the ground state has energy -NmH, since in this state all the N number of spins are parallel. The state next to the ground state has (N-1) number of parallel spins and 1 antiparallel spin and hence has energy -(N-1)mH+mH=-(N-2)mH. Thus, the difference in energy between ground state and the next state is 2mH, which is the difference in energy between any two macrostates. Therefore, in energy range del E, there are (del E)/2mH number of energy gaps or (del E)/2mH + 1 number of energy levels, which is approximately equal to del E/2mH since since del E >> mH and hence we can neglect 1. Now, total energy is E = (-n1 + n2)mH and total number of spins is n1 + n2 = N. Solving, we get n1=(1/2)(N-(E/mH)) and n2=(1/2)(N+(E/mH)). Now, for a given state, when n1 spins are parallel and n2 spins are antiparallel, the state can occur in N!/n1!n2! ways, which is the degeneracy of each state. Thus, the total number of available states is number of states in range del E multiplied by the degeneracy of each state. Therefore W(E) = (N!/n1!n2!)(del E/2mH). Substituting the expressions for n1 and n2, the result matches with the answer given in the book.

Question: When i consider the total number of states = number of macrostates in range del E multiplied by degeneracy of each state, i assume that the degeneracy of each state is the same for each of the macrostates? Is it justified? i have got the answer rather by force, but i'm not too sure about the logic behind it.

(b) Using Stirling's formula directly, i get NlnN - (n1)ln(n1) - (n2)ln(n2) + ln(del E/2mH).

Question: Can I further simplify this? Answer is not given in the book.

(c) I am rather stumped out by this part!! I first tried to find W(E) near E=0. Strictly speaking, when E = 0, n1 = n2 = N/2 so that the total degeneracy (and hence the total number of states) should be N!/(N/2)!(N/2)! but substituting E = 0 in part (a) gives me an addition term of (del E/2mH).

Question: why is part (a) giving an answer different from what appears physically?

i anyhow continued with the physical answer and found, after applying stirling's formula that W(E=0) = 2^N.

Question: Am i right? How am i supposed to find out W near E = 0 (which is where i think the gaussian approximation holds true)?

by the way, i'm new in this forum... so if there's anything wrong with this post, feel free to tell me about it...

Last edited: Feb 21, 2008
2. Feb 23, 2008

### gc2004

ah, i solved it myself... i just followed the way one arrives at the gaussian distribution from the binomial distribution and the answer is W=(2^N)exp[-(2(m^2)(H^2)N)]... though this answer is not given in the book, there are problems based on this answer in chapter 3, the answers of which have matched... anyway, thanks to those who have viewed the page...

3. Jan 19, 2011

### Nonovero

reif, problem 7.19

This problem deals with mean values, quadratic mean value, etc. Can someone help me to solve the exercise. Should I to start with v = 3kT^1/2?
Ref.: Reif, Thermal Physics, 7.19.
Thank you!