This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 112 Fall 2007 Professor William Holzapfel Homework 2 Solutions Problem 1: Magnetization a) For the paramagnet, we can use the Gaussian approximation to the multiplicity in terms of the total number of spins N and the spin excess s : g ( N,s ) = g ( N, 0) e 2 s 2 /N The entropy is easily calculated: σ ( s ) = log g ( N,s ) = log( g ( N, 0)) 2 s 2 /N = log( g ( N, 0)) U 2 2 Nm 2 B 2 where we’ve substituted U for s using U = 2 smB . From the definition of temper ature we have: 1 τ = ∂σ ∂U = U Nm 2 B 2→ τ = m 2 B 2 N U . Now we can solve for U ( τ ) and plug it into the definition of fractional magneti zation: M Nm = 2 s N = U NmB = mB τ . b) Entropy and and temperature as functions of energy are plotted in Figures 1 and 2. We can see that temperature is negative when U > 0. The concept of negative temperature is less mysterious when we consider the definition of temperature: 1 τ = ∂σ ∂U V , (1) 1 Figure 1: This is the plot of entropy vs. energy for the case σ ≈ 100 and m 2 B 2 N ≈ 100. Figure 2: This is the plot of temperature vs.energy for the case m 2 B 2 N ≈ 100. Notice that the temperature is negative for U > 0. 2 Negative temperature means that dσ/dU is negative, and so the entropy increases when the energy decreases. If such a system is brought into contact with a positive temperature system, then energy will flow from the negative T system to the positive T system since that will increase the entropy of both! This is in contrast to what happens when two positive T systems are brought together; in that case one will lose entropy and the other will gain, but the energy will flow in such a way that the net entropy change is still positive. In the case of negative T and positive T , both systems increase their entropy. Problem 2: Crystal Disorientations a) First we need to choose which M of the total of N atoms will be removed from their lattice sites; clearly there are ( N M ) ways of doing this. Then we need to choose which interstitial sites to put the M atoms in. For a square lattice of N sites, there are N interstitial sites, so there are also ( N M ) ways to place the displaced atoms in the interstitial sites. So the total multiplicity is: g ( N,M ) = N M N M = N ! M !( N M )! 2 . b) This part of the problem is analogous to problem 1: first we write σ as a function of U and then take the derivative to find τ : σ = log g ( N,M ) = 2(log N ! log M ! log( N M )!) = 2( N log N M log M ( N M ) log( N M )) = 2 N log N N M M N log M N M = 2 N log 1 1 U U U U log U U 1 U U ! = ⇒ 1 τ = ∂σ ∂U U = N = 2 N 1 U U 1 U h log U U log(1 U U ) i 1 U U U ( U U ) = 2 N U log U U U = 2 log M N M where we have used U = N , U = M . We can solve for M/N by taking the exponential of the above expression: 3 M N M = e / 2 k B T = ⇒ M N = 1 e / 2 k B T + 1 ....
View
Full
Document
This note was uploaded on 11/18/2010 for the course PHYSICS 112 taught by Professor Steveng.louie during the Fall '06 term at Berkeley.
 Fall '06
 StevenG.Louie
 Physics, Work

Click to edit the document details