1 Ensemble of three-level systems

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1 PHYS 607: Statistical Physics. Fall Home Assignment 2 Entropy, Energy, Heat Capacity Katrina Colletti September 23, Ensemble of three-level systems We have an ensemble of N atoms, with N 1, each atom with 3 energy levels 0, ± a Find energy and entropy using statistical means For N + particles in energy state, N particles in energy state, and N 0 particles in energy state 0, the energy is trivial to write: E = N + N = N + N ) Here we will also pause to note some relations that will be useful later on. N + N = E These conditions lead to: N + + N = N N 0 N + = 1 2 N N 0 + E ) N = 1 2 N N 0 E ) Proceeding, we can calculate the number of states by realizing that we will have to get the number of ways you can pick N + from N and then also pick N from the remaining free N N + since we have already chosen N + of the available particles). In math, this is written: N ΩE) = N+ ) ) N N+ = N N! N +!N N + )! But we want to write the number of states in terms of E, N, and N 0 : ΩE) = N N + )! N!N N + N )! = N! N +!N!N 0! N! N 0! 1 2 N N 0 + E ))! 1 2 N N 0 E ))! Now we plug this in to our formula for entropy, and then use Sterling s approximation log N! N log N N): S = k logωe)) = k logn!) logn 0!) log 1 2 N N 0 + E ))!) log1 2 N N 0 E ) ))!) S k = N log N N N 0 log N 0 + N N N 0 + E )) log1 2 N N 0 + E )) N N 0 + E )) 1 2 N N 0 E )) log1 2 N N 0 E ))) N N 0 E )) = N log N N 0 log N N N 0 + E )) log1 2 N N 0 + E )) 1 2 N N 0 E )) log1 2 N N 0 E ))) + N + + N + N 0 ) N = N log N N 0 log N N N 0 + E )) log1 2 N N 0 + E )) 1 2 N N 0 E )) log1 2 N N 0 E ))) 1

2 We have entropy as a function of energy E, total number of particles N and number of particles in the 0 energy state N 0. Since we have fixed E and N, we want to find the value of N 0 that maximizes entropy. So to find what value of N 0 will maximize the entropy, we take the derivative and set equal to 0: 1 ds 1 = N 0 log N 0 1 k dn 0 N 0 2 N N 0 + E )) N N 0 + E )) log1 2 N N 0 + E ))) 1 2 N N 0 E )) N N 0 E )) log1 2 N N 0 E ))) 0 = 1 log N log1 2 N N 0 + E ))) log1 2 N N 0 E ))) = log N log1 2 N N 0 + E ))) log1 2 N N 0 E )) log N 0 = 1 2 log1 2 N N 0 + E )) 1 2 N N 0 E ))) N 0 2 = 1 2 N N 0 + E )) 1 2 N N 0 E )) 4N 2 0 = N 0 N E )N 0 N + E ) = N 0 N) 2 E2 2 = N N 2 2N 0 N E2 2 0 = 3N NN 0 + E2 2 N 2 ) N 0 = 2N ± 4N 2 12 E2 N 2 ) 2 6 = N 3 ± 4N 2 3 E2 2 3 Number of particles must be positive, so we find that N 0 = 1 3 N + 4N 2 3 E2 2 ) with the additional constraint that N > E. Plugging back into entropy, we get S = N log N 1 3 N + 2N) 2 E/) 2 ) log 1 3 N + 2N) 2 E/) 2 )) 1 6 4N 2N) 2 E/) E )) log1 6 4N 2N) 2 E/) E )) 1 6 4N 2N) 2 E/) 2 3 E )) log1 6 4N 2N) 2 E/) 2 E )) 2

3 b Find energy and entropy using the partition function We first write the partition function for one atom, say the i th atom: Z i = Σ n e βn = 1 + e β + e β = coshβ) Now since all particles are independent of each other, we get the total partition function simply by multiplying all N identical partition functions together, to give us: We find energy: Z = coshβ)) N log Z = N log coshβ) ) < E > = β log Z E = 2N sinhβ) 2 coshβ) + 1, and the entropy: S = k T S k ) T log Z = log Z + T log Z T = N log 2 cosh/ ) 2N sinh/ )) + T 2 2 cosh/ [ 2/ sinh/ ] S = kn log2 cosh/). 2 cosh/) c d Find the dependence of entropy on temperature [ S = kn log2 cosh/) 2/ sinh/ ] 2 cosh/) Find the dependence of energy on temperature, and find the temperature at which E = 0. Find the temperature range that corresponds to E > 0 E = 2N sinh Solving for T when E = 0 note that since cosh is always greater than 0, the denominator is always greater than 0 as well, thus we will have no divide by zero problems): 2N sinh 0 = ) = sinh = T k This makes sense because at high temperature, all states for an atom should be equally likely, and so the number of atoms in each state should be N/3, and the total energy sums to 0. 3

4 For E > 0 again keep in mind that the denominator is always positive, thus we just require that the numerator be positive; and also note that both and k are fixed positive numbers): sinh < 0 < 0 T < 0 4

5 e Find the specific heat and plot specific heat vs. temperature To get the specific heat, I m defining m to be the mass of one of the atoms. C = 1 E Nm T = 2N 2/ 2 ) Nm = = C = 2 2 m 2 )2 2 2 m 2 ) cosh ) m 2 )2 sinh / 2 ) 1)2 2sinh cosh ) cosh2 cosh )) cosh ) This is the general behavior up to constants) of the heat capacity varying with temperature. This is expected: it starts at 0, and then rises to a maximum and then drops off again as temperature increases. It first rises exponentially, has a maximum around the characteristic energy scale /k, then drops off slower than exponential as T. ) Figure 1: heat capacity general behavior as a function of temperature - courtesy of Mathematica - ignore the actual numerical values that are plotted 5

6 f Find the average values of probabilities p +, p, and p 0 for an atom to be in each of the three states We use the general formula: e En/ pn) = Σ m e Em/ The denominator for our three state system is going to be We can therefore find e / + e / + 1 = p + = e / p = e / p 0 = 1 6

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