3.20 Materials at Equilibrium (SMA 5111): Solutions to Problem Set 5 (2002)

Problem Set 5 Additional Problems Solutions 3.20 MIT Dr. Anton Van Der Ven Fall 2001 Fall 2002 Problem 1 (a) A system of N non-interacting particles with two possible states either 0 or ε. A good rule is too assume ’’particles’’ (e.g. atoms, electrons, etc) are indistinguishable unless they are localized in a crystal or on a surface. The number of atoms in the excited state can be determined using Boltzmann statistics under the assumption that we are working at high temperature and/or low density:: Nε = N  ε where  ε is the probability an atom will be in state ε. This probability is determined using the single particle partition function and can be written as 
ε  exp kT   ε =  exp
ε; kT (McQuarrie 4-14) i But our system can be in only two states, so the sum in the denominator can be found explicitly: 
ε  exp kT 
ε  ε = exp [0] + exp kT So Nε can be written as 
ε  exp kT N 
ε  =  ε  Nε = N  ε = N 1 + exp kT 1 + exp kT Nε = (b) The total energy is simply U = N ε = N  N ε 1+exp[ kT ]  i εi (McQuarrie 4-12 and 4-13) i U =N   i εi = N ( 1 ε1 +  2 ε2 ) = N [(1
 2 )  0 +  2  ε] i 
ε   exp kT   ε 1 + exp
ε kT U = N ( 2 ε) = N U= Nε ε 1+exp[ kT ] Note: Since many particles will occupy the same state (either 0 or ε) these particles must be Bosons. At lower temperatures we would have to use Bose-Einstein statistics (McQuarrie 4-26) which would lead to a much more complicated problem since we would have to determine the chemical potential . 1 Problem 2 The magnetization is given as M = N  ni o i=1 This is basically saying that we have atoms localized in a crystal and the magnetic moment at each site can be either up or down. The problem asks us to determine the thermodynamic properties as a function of T, N, H. Let us also assume for simplicity that we can work at constant volume. Therefore, our controlling variables are T, N, V, H. We need to make the appropriate Legendre transform to the entropy. Remember the entropy can be written starting from E: E = T S
pV + HM + N rearranging to get things in terms of S and  we get S = E + pV
HM
N k Legendre transform such that our controlling variables are V, N, T, H S
E + HM =  (T S
E + HM ) =
 = ln  k where is the characteristic potential for this ensemble with V, N, T, H constant and  is the partition function.  can be written as  = exp [
 (Estate
Mstate H)] states where we sum over all possible energy states and magnetizations Mstate. Since the particles are non-interacting, the energy at N,  = constant and H = 0 is constant. E is independent of the number or arrangement of up versus down spins. Since the absolute scale of energy is not important for thermodynamics, we can arbitrarily set the constant energy equal to zero giving us  N  N      ni o H = = exp [Mstate H] = exp  exp [nio H ] n1 ,n2,...n N states =  N  n1 ,n2 ,...n N i=1 i=1 exp [ni o H ] = N  +1  exp [ni o H ] i=1 ni =
1 n1 ,n2 ,...n N i=1 We can evaluate the sum since ni = 1 so,  = (exp [o H ] + exp [
o H ]) N We know the characteristic potential of an ensemble is related to the partition function for that ensemble according to
 = ln  → =
kT ln  Furthermore, we know from thermo that d =
SdT
pdV + dN
M dH which gives us the following relationships for the properties of the system:
S =

M =
2 ∂ ∂T ∂ ∂H  V,N,H  V,N,T
 ∂ ∂N V,T ,H
 ∂
∂V T ,N,H  = p = and from stat mech we have from above =
kT N ln [exp [o H ] + exp [
o H ]] Lets get the entropy, S
S=
∂ ∂T  V,N,H  S = kN ln (exp [o H ] + exp [
o H ]) + kT N  o H (exp [o H ]
exp [
o H ])
kT 2 exp [o H ] + exp [
o H ] S = kN {ln (exp [o H ] + exp [
o H ])
o tanh (o H )} Now for the magnetization
   ∂ exp [o H ]
exp [
o H ] M =
= kT N o = N o tanh (o H ) ∂H V,N,T exp [o H ] + exp [
o H ] M = N o tanh (o H ) The energy E = E
T S
HM E= + T S + HM = 0 However if you define a quantity called the internal magnetic energy (which is a quantity analogous to the enthalpy in the T, p, N ensemble) EH = E
HM you can get EH =
N o H tanh (o H ) The last part of this problem asks you to determine the behavior of the energy and entropy as T → 0. EH (T → 0) =
N o H lim S = lim kN {ln (exp [o H ] + exp [
o H ])
o tanh (o H )} = 0 →∞ →∞ S (T → 0) = 0 is in accordance with the third law of thermodynamics. Problem 3 (a) M = N  o ni i=1 (see solution to Problem 2 - Method 1)
   exp [o H ]
exp [
o H ] ∂ M =
= kT N o = N o tanh (o H ) ∂H V,N,T exp [o H ] + exp [
o H ] M = N o tanh (o H ) 3 (b) The partition function for the N, V, H, T fixed ensemble is  = exp [
 (Es
Ms H )] s where s = states We want to determine the fluctuations in the extensive quantity M. Use the 3-step procedure developed in class. Step 1: Multiply both sides by the partition function  M = Ms exp [
 (Es
Ms H )] states Because Es is always constant (see comment in Problem 2) we can therefore arbitrarily set it to zero and write N  N  n1 ,n2,...n N i=1 M =   N  exp o H o ni  ni i=1 Step 2: Get derivative with respect to mechanical variable’s conjugate. ∂M +M ∂H  N N   ∂M ∂ +M = ni  o ∂H ∂H n ,n ,...n i=1 N  n1 ,n2 ,...n N  N   1 N 2  exp o H o ni i=1 N  2  exp o H N  = ni  ni i=1  n1 ,n2 ,...n N i=1 N   o N  i=1 2 ni  exp o H N  i=1 Step 3: Divide through by the partition function ∂M 2 + M = M 2 ∂H or we can write it like this & 2 ( M) ' 2 = M2
M = 1 ∂M  ∂H with:
∂M ∂H  ) ( = N 2 1
tanh2 (H ) ,N & ' 2 = 2 {1
1} = 0. In other (c) As  → ∞, tanh (H ) → 1. Therefore M T →0 = N  and ( M ) T →0 words, the ground state with all the spins aligned has no fluctuations. 4  ni Problem 4 From the first and second law we have dE = T dS
pdV + HdM + dN (Euler Form) E = T S
pV
HM + N We are working with fixed N, T, V, M in this problem. We need to make a Legendre transform because T is a controlling variable instead of E. S
E =
F = ln Q k  Q= exp [
Estate ] states F =
kT ln Q From the differential form of F, namely dF =
SdT
pdV + HdM + dN we get that
H= ∂F ∂M  T ,V,N We are working under constant magnetization so M = o N  ni = o (n+
n
) i=1 where n+ = number of up spins and n
= number of down spins. That means that the sum in the expression of Q must be performed over only those states with fixed M (i.e. fixed n+ and n
) Also the atoms in the system do not interact, meaning that the energy is independent of the number and arrangement of up/down spins and is therefore constant = Eo.   Q= exp [
Estates] = exp [
Eo ] =  exp [
Eo ] states with magnetization M states with magnetization M where  is the number of states that are consistent with a magnetization M = o (n+
n
) . N! n+ ! (N
n+ )! N! Q= exp [
Eo ] n+ ! (N
n+ )!

 N! N! F =
kT ln exp [
Eo ] = Eo
kT ln n+ ! (N
n+ )! n+ ! (N
n+ )! = using Stirling’s approximation, F = Eo
kT {(N ln N
N )
n+ ln n+ + n+
(N
n+ ) ln (N
n+ ) + (N
n+ )} but we know from above that M = o (n+
n
) = o (2n+
N ) =⇒ n+ = M + o N 2o we can substitute this back into our expression for F to get F as a function of M 


 M + o N M + o N o N
M o N
M F = Eo
kT N ln N
ln ln
2o 2o 2o 2o 5  ∂F  Remembering that H = ∂M T ,V,N
 

  1 1 1 1 ∂F M + o N o N
M H= =
kT
ln
+ + ln 2o 2o 2o 2o 2o 2o ∂M T ,V,N 

 kT M + o N o N
M ln H=
ln 2o 2o 2o
 M + o N 2o H = ln o N
M and it is easy to show that M = o N tanh (o H ) which is the same result as in the other ensemble of Problems 2 and 3. This is due to the equivalence of ensembles in the thermodynamic limit. Problem 5 (a) Grand canonical ensemble:
= ∞   N =0 j exp [
Ej ] ∞  exp [N ] = Q (N, V, T ) exp [N ] N =0 This sum extends over allstates with total # of particles = N  exp [
Ej ] = Q(N, V, T ) = j qN N! for a one component gas of non-interacting particles. We can write q in closed form as we have done before as
3 2mkT 2 q= V h2 So now
can be written as:
= ∞ ∞   N   qe qN exp [N ] = N! N! N=0 N=0 But if we let a = qe then we can write this sum in a form for which the solution is know, namely ∞  aN = ea N! N=0 If we now substitute in for q we can write
as
= ezV z= with e h2
2m  (b) The characteristic potential for the grand canonical ensemble is
pV =  =
kT ln
 = E
T S
N d =
SdT
pdV
N d 6  32

 ∂ ∂ (zV ) = kT = kT z p=
∂V T , ∂V

 ∂ ∂z = kT N =
V = zV ∂ T ,V ∂ putting those two parts together and eliminating z gives us the familiar ideal gas equation p = kT N . V (c) * ( ) 2 2 * 2 ( N) N2 = (because V fixed) 2 2 We need to get ( N ) = N 2
N . Follow the 3-step procedure. Step 1: Multiply both sides by the partition function N
=
= ∞  N N=0 q N N e N! ∞ ∞   q N N q N N = N N e e N! N! N=0 N=0 Step 2: Get derivative with respect to mechanical variable’s conjugate.
∂N ∂ 
+N
∞  N=0 ∂N ∂  q N N N e N!  = ∞  N 2 N=0
+ N  N
= N 2
Step 3: Divide through by the partition function
 ∂N 2 + N = N 2 ∂
 ∂N 2 2 N
N = kT ∂ 3  2  from (b) we have that N = zV with z = eh2 2m . 

 ∂N ∂z = V = zV ∂ ∂ 2 N 2
N = zV divide by V to get the densities z N = 2 V V 2
2 N = 2 2 N + * 2
2 1 kT = √ = pV N 2
2 = 7 q N N e N! (d) To second order in N  ∂ 2 ln P (N ) ∂N 2 N =N  2 assuming a Gaussian distribution around N with variance N we get   7 P (N )
10
12  N = exp = e
10 which is very unlikely. 2 P (N ) ln P (N ) = ln P (N ) + 1 2 ( N) 2 8