PHY 5524: Statistical Mechanics, Spring February 11 th, 2013 Midterm Exam # 1

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1 PHY 554: Statistical Mechanics, Spring 013 February 11 th, 013 Midterm Exam # 1 Always remember to write full work for what you do. This will help your grade in case of incomplete or wrong answers. Also, no credit will be given for an answer, even if correct, if you give no justification for it. Write your final answers on the sheets provided. You may separate them as long as you put your name on each of them. We will staple them when you hand them in. Ask if you need extra sheets, they will be provided. Remember to put your name on each of them and add them after the problem they refer to.

2 Part 1 Answer in one or two sentences the following two questions. (1.a) For an isolated system with fixed total energy E, fixed volume, and fixed number of particles N, what is the probability of finding the system in a particular microstate? The statistical mechanics of a system with fixed (E,, N) is built using the microcanonical ensemble. The probability of finding a system in a given microstate α is constant and given by 1 P α = Ω(E,, N), where Ω is the total number of allowed microstates. (1.b) For a system in thermal contact with a heat reservoir at a fixed temperature T, fixed volume, and fixed number of particles N, what is the probability of finding the system in a particular microstate? The statistical mechanics of a system which exchange energy with a heat reservoir at fixed (T,, N) is built using the canonical ensemble. The probability of finding a system in a given microstate α (with degeneracy g α ) is given by P α = g αe βea Z where we have used a discrete spectrum notation and we have denoted by, Z = α g α e βeα the partition function of the system. The generalization to the case of a continuum spectrum of states is obvious.

3 Part Consider a system of N atoms in contact with a thermal reservoir at a temperature T. You may neglect the translational motion of the atoms and consider them as localized (i.e. distinguishable) objects. Each hypotethical atom displays the following spectrum (see figure): a non-degenerate ground state with energy E 0 = 0 and a charge Q 0 = 0; a first (single-ionized) state with energy E 1 = > 0 and a charge Q 1 = 1 (in units of the proton charge); a second (doubly-ionized) two-fold degenerate state with energy E = and a charge Q = (also in units of the proton charge). E = Q = E = Q = 1 E = 0 Q = 0 (.a) Obtain the partition function and Helmholtz free energy of the system. Since the constituents of the system are distinguishable, the partiction function can be written as Z(T, N) Z N (T ) = [Z 1 (T )] N, where Z 1 (T ) = α g α e βeα = 1 + e β + e β, is the partiction function of each single constituent of the system, and E α and g α denote respectively the energy and the degeneracy of each energy level. Then, the Helmholtz free energy is obtained as, F (T, N) = k B T ln[z N (T )] = Nk B T ln[z 1 (T )]. (.b) Obtain the entropy of the system. The Third Law of Thermodynamics can be stated as follows: The entropy of a system at absolute zero is a universal constant, which may be taken to be zero, i.e. S = 0 at T = 0 regardless of the values of any other parameter of which S can be a function. Is your result consistent with the Third Law of Thermodynamics?

4 The entropy of the system is given by, S(T, N) = ( ) F T N = Nk B ln[z 1 (T )] + Nk B T T ln[z 1(T )] or equivalently, = Nk B ln(1 + e β + e β ) Nk B T 1 k B T = Nk B ln ( 1 + e β + e β ) + N T β ln[z 1(T )] (e β + 4e β ) (1 + e β + e β ), S(T, N) Nk B = ln ( 1 + e β + e β ) + β (e β + 4e β ) (1 + e β + e β ). We can see that in the limit of T 0 or β S(T, N) 0, in agreement with the Third Law of Thermodynamics. (.c) Obtain the average energy of the system? The average energy of the system is calculated as, E = N ( ) ln Z1 (T ) β = N (e β + 4e β ) (1 + e β + e β ) = N 1 Z 1 (T ) ( ) Z1 (T ) (.d) Obtain the leading temperature-dependent term of the specific heat of the system in the low-temperature limit (k B T ). β The specific heat of the system is obtained as, C = = ( ) E T ( N k B T ( = 1 ) ( ) E k B T β ) [ (e β + 8e β ) (1 + e β + e β ) + (e β + 4e β ) (1 + e β + e β ) ]. In the limit T 0 (or β ), which is equivalent to considering physically k B T, the leading temperature-dependent term in the specific heat is, C T 0 1 k B T N e β = Nk B ( β) e β.

5 (.e) Compute the temperature of the thermal reservoir required for the average charge of the system to be equal to Q = 1. The average charge of the system is, Q = α Q α P α = (0 + e β + 4e β ) (1 + e β + e β ), and imposing that Q = 1 we get, 1 = (0 + e β + 4e β ) (1 + e β + e β ) 1 = x + 4x 1 + x + x, where we have introduced the notation x = e β. Solving the equation we get, x + 4x = 1 + x + x x = 1 x = ± 1, but of course the negative solution has to be discarded since our x is always x > 0. Therefore the solution of the above equation is, e β = 1. Taking the logarithm of both sides of the equation we obtain that the corresponding temperature is, β = 1 ln T = k B ln.

(# = %(& )(* +,(- Closed system, well-defined energy (or e.g. E± E/2): Microcanonical ensemble

Recall from before: Internal energy (or Entropy): &, *, - (# = %(& )(* +,(- Closed system, well-defined energy (or e.g. E± E/2): Microcanonical ensemble & = /01Ω maximized Ω: fundamental statistical quantity