PS2Sol_f10 - Physics 112 Fall 2007 Professor William...

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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 ....
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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.

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PS2Sol_f10 - Physics 112 Fall 2007 Professor William...

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