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HOMEWORK #4 & SOLUTIONS
Problem 1 (Pathria 3.5) Solution
F ( F (T , V , N ) = N f (v, T ) T ) N
v = V /N
∂F ∂f V = N ( N (f + N ) ∂N ∂v N 2 ∂F ∂f 1 V = V N ∂V ∂v N
−
Adding these two gives the desired equation.
1
Problem 2 The Einstein solid picture assumes that atom i of a solid sits at lattice position r0i and can oscillate harmonically around this position. It is described by the Hamiltonian N
H = i=1
2 pi
1 + mω2 2m 2
N
(
ri
i=1
− r0 )2
(1)
i
(a) What shape does the phase space volume with energy less than E , i.e., the volume H (r, p)
≤ E have?
(b) Calculate the entropy S of the Einstein solid at a given energy E using classical theory. Note, the oscillators have different origins and thus the atoms are distinguishable! Solution
(a) H (r, p) (b) Use xi
≤ E defines a 6N dimensional ellipsoid. Axis lengths are = mω(r − r 0). For distinguishable particles: i
√
2mE and
i
1 Σ(E ) = 3N h
d3N pd3N q
2E/mω . 2
(2)
P3N
( p2i +x2i )≤E i=1
This is the volume of a 6N dimensional sphere of radius
By differentiation with respect the energy g(E , V , N ) = By using Ω = gE and ln(Γ(3N )) = ln((3N
1 E 3N −1 3N Γ(3N ) ω
(4)
− 1)!) ≈ N ln(N ) − N
E S (E , V , N ) = 3Nk 1 + ln( ) N ω Classical oscillators with quantum result!
2
(5)
Problem 3 A simple model for a polymer in two dimensions is that of a path on a square lattice. At every lattice point the polymer can either go straight or choose between the two directions in a right angle with respect to its current direction. Each time it bends in a right angle, it pays a bending energy . Thus, for a given shape of the polymer the total bending energy of the polymer is times the number of right angle turns. We assume that the starting segment of the polymer is fixed somewhere on the lattice and that the polymer consists of N + 1 segments. Each possible shape of the polymer is a state of this statistical mechanics system. (a) How many polymer shapes have a total bending energy E where we assume E = m with some integer 0
≤ m ≤ N ?
(Hint: First count how many ways there are to position
the m right angles on the polymer of length N +1 segments and then take into account that there are 2 possible choices for each right angle, namely left and right.) (b) What is the entropy S (E, N ) of this system? Approximate all factorials with the help of Stirling’s formula. (c) Calculate the temperature of this system as a function of the total bending energy E and the length N of the polymer. (d) Calculate the energy E of the polymer as a function of the temperature T and of the length N of the polymer. (e) Calculate the heat capacity at constant length C N as a function of the temperature T and the length N of the polymer. Solution
(a) There are m right angle turns and N
− m straight steps. There are
N ! (N m)!m!
(6)
−
ways to arrange the turns. Each right angle turn can go into two directions, so the number of polymer shapes are N (E ) =
N ! 2m (N m)!m!
(7)
−
(b) Using Stirling:
m m m m m S = kln(N (E )) = kN ln(2) − ln( ) − (1 − )ln(1 − ) N
N
3
N
N
N
(8)
(c) Remembering that m = E/: 1 = T
∂S ∂E
N
m m = k ln(2) − ln( ) + ln(1 − ) N
N
(9)
(d) From the result derived in (c), using β = 1/kT β
