13! (52 13)! 52! =

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1 Thermo and Intro to Stat Mech 018 Homework assignment 1, Problem 1: What is the probability that all 13 cards on a hand (in bridge for example) are of the same kind, for example all spades? There are 5 cards in a deck, 13 of each kind, in particular 13 spades. The probability of drawing a spade in the first card is 13. The probability 5 that the second card is also a spade is 1, etc. The probability that all cards drawn are spade is 13! (5 13)! 5! = Problem : Consider an ideal gas in a box consisting of molecules,. What is the probability that all the molecules will be found in one half of the box? Evaluate the answer for = 100. What if is Avogadros number? The probability that a gas molecule is in one half of the box is 1/. The probability that two non-interacting gas molecules are in one half of the box is 1/. The probability that non-interacting gas molecules are in one half of the box is 1/. For = 100 the probability is If is Avogadro s number = , the probability is 1/ = log = which is practically zero. Problem 3: Compare the exact value of! and the Stirling approximation for = 10, 0, and 100. The Stirling approximation for! is! π exp ( ) ! Stirling difference % % % 1

2 Problem 4: Consider the binary model with four sites, = 4. This could for example represent a small cluster of atoms with four spins. (a) Write down symbolically with arrows up and down all the possible arrangements of the four spins.,,,,,,,,,,,,,,, (b) What is the total number of different arrangements of the four spins? 4 = 16 (c) When a magnetic field is applied, the energy of the system depends on the total spin, U = smb where s is the excess spin (s = ), m is the magnetic moment of one spin and B is the magnetic field. Sketch an energy diagram showing the number of energy levels and the arrangements of spins that correspond to each level. Energy/mB = -s spin configuration degeneracy (d) What is the multiplicity (degeneracy) of each of the energy levels? See the table in the answer to part (c). (e) Evaluate the appropriate binomial coefficients, (! ( t)!t! )

3 for each of the energy levels and compare with your answer in part (d). (f) Sketch a histogram of the multiplicity vs. spin excess. (g) Use the approximate formula for the multiplicity function g(, s) = /π exp ( s /) which was obtained by applying Stirling approximation to estimate the multiplicity of each energy level and compare with your results in part (d). You will find significant deviations. Why is that? s Ω g(4, s) difference 0 6 4π 4 e 0 1 = 16 = % π 1 4 4π 4 e = e = % π 1 4π 4 e 1 4 = π e = % The approximate multiplicity function relies on the accuracy of the Stirling approximation which is accurate if is large. Here, = 4 which is quite small, so the approximation is not so good, but then not so bad either. See plot in the answer to part (f) which shows both the histogram of the exact function, Ω, and the smooth curve for the approximate function g(4, s). Problem 5: Derivation of the multiplicity function, g(, s). (a) Start with the expression for the number of ways that r spins out of a total of n can be arranged to point up Ω(, r), eqn. (1.14). Apply the loga- 3

4 rithm and use Stirling approximation, eqn. C.0, to obtain an approximate expression for ln Ω(, r). Ω(, r) =! ( r)!r! where n is the total number of spins and r is the number of spins pointing up. Use Stirling approximation, ln! ln to simplify the expression for Ω. This gives ln Ω ln ( r) ln ( r) + ( r) r ln r + r = ln ( r) ln ( r) r ln r = ( r) ln (1 r/) r ln (/r) where the relation ln a/b = ln a ln b has been used. (b) Introduce the variable s where s is defined as the number of up spins in excess of /, s r and use a second order Taylor expansion ln(1 + s/) s/ (s/) to simplify the resulting expression for ln Ω(, s). The variable s denotes how many spins have been flipped from down to up starting from the state where equal number of spins point in the two directions (the most likely state) in order to get r spins up, so s r. Replace r in the expression above with s + n/ and factor out 1/ to get ( ) ( ( ln Ω(, s) s ln 1 s ) ) ( ln s + ) ( ( ln 1 + s ) ) ln ( ) ( = s ln 1 s ) ( ) ( + s ln 1 + s ) ( )+ln s + ) + s The last term involving ln is independent of s since the s-dependence cancels out. We are at this point only interested in the s-dependence of Ω so we do 4

5 not need to keep this term. The full expression for Ω including factors that do not involve s will be found by normalization in part (d). Since s <<, a Taylor expansion of the logarithm to second order can be used here to a good approximation (i.e. ln (1 ± x) ±x x /) to give ( ) ( ln Ω(, s) s s 1 +s = ( ( s 1 s 1 ( ) ) ( ) ( s + s s 1 ( ) s + s 1 ( ) s s + 1 ( ) ) s ( ) ) s + constant ( ) ) s = ( s ) 4s + constant = s + constant (c) Exponentiate the result from (b) to obtain the exponential factor in the equation given in problem 4(g) above. is After exponentiating both sides, the result of the calculation in part (b) Ω(, s) g(, s) e s / The proportionality factor is the value of g when s = 0, so this result can be written as g(, s) = g(, 0)e s / (d) Use the fact that g(, s) should be normalized to obtain the preexponential factor in the equation given in problem 4(g) above. The total number of microstates is so the integral over the function g(, s) over all possible values of s should equal. This can be used to 5

6 determine the pre-exponential factor, g(, 0). Since g is small except in the region where s 0 the integration over s can be extended to all real values This gives g(, s)ds =. g(, 0)e s / ds = g(, 0) e s / ds = Using the formula for the Gaussian integral, exp ( αx )dx = π/α, gives g(, 0) = π. This completes the derivation of the expression given for g(,s) in problem 4(g). 6