KALMAN KNIZHNIK - KITTEL AND KROEMER SOLUTIONS

K&K 2.1Part aSuppose g(U) = CU3N/2, where C is a constant and N is the number of particles. Showthat U = 32N .We use the definition of temperature as

U

N

=1

(1)

So, lets calculate .

= ln(g) =3N

2ln(U) + ln(C) (2)

Therefore,

U=

3N

2U=

1

(3)

So finally,

U =3

2N 2 (4)

Part bShow that 2/U2|N is negative. This form of g(U) actually applies to an ideal gas.Referring to the definition of U in equation 3 and taking another derivative:

2

U2= 3N

2U2< 0 2 (5)

This suggests that the temperature of the system will increase with increasing U.K&K 2.2Find the equilibrium value at temperature of the fractional magnetization M/Nm =2s/N of the system of N spins each of magnetic moment m in a magnetic field B.The spin excess is 2s. Take the entropy as the logarithm of the multiplicity as givenin 1.35: (s) log{g(N, 0)} 2s2/N .The energy of a single magnetic moment m in a magnetic field B is U = m B. A spin excess of2s means that there are 2s spins with this magnetic moment. In other words, there could be, say,3s particles of magnetic moment m and 1s particles of magnetic moment m. The energy wouldcancel for all but 2s particles. Thus, the total energy of the system is U = 2smB. Therefore,s = U/2mB. So the entropy is

= 0 2s2

N= 0

U2

2m2B2N(6)

Since 1/ = /U , we find

U= U

m2B2N=

1

U = m

2B2N

(7)

Finally, we know that U = MB since the energy is the total magnetization times the magneticfield. Therefore,

m2B2N

= MB M

Nm=mB

2 (8)

Note that the magnetic susceptibility obeys Curies law M/B 1/ .K&K 2.3Part aFind the entropy of a set of N oscillators of frequency as a function of the total

quantum number n. Use the multiplicity function 1.55 and make the Stirling approx-imation. Replace N 1 by N .The general expression for the possible ways to obtain the energy nh amongst a system of Nharmonic oscillators is (equation 1.55):

g(N,n) =(N + n 1)!n!(N 1)!

(9)

Making the approximation that N is large, we get:

g(N,n) =(N + n)!

n!N !(10)

from which the entropy = ln(g) is:

= ln(g) = ln{(N + n)!} ln(n!) ln(N !) (11)

Invoking Stirlings approximation:

(N + n)ln(N + n)Nln(N) nln(n) 2 (12)

where I have canceled out some terms.

Part bLet U denote the total energy nh of the oscillators. Express the entropy as (U,N).Show that the total energy at temperature is

U =Nh

exp(h/) 1. (13)

This is the Planck result for the energy of a system of N harmonic oscillators.The total energy of the system is U = nh. Thus, we replace n in expression 12 with U/h:

= (N +U

h)ln(N +

U

h)Nln(N) U

hln(

U

h) (14)

Now we can do a little rearranging:

= Nln(N + UhN

) +U

hln(

N + UhUh

) = Nln(1 +U

Nh) +

U

hln(1 +

Nh

U) (15)

Next, we use the thermodynamic definition of temperature /U = 1/ , to calculate 1/ :

U=

N

1 + UNh

1

Nh+

1

hln(1 +

Nh

U) U

h

Nh

U21

1 + NhU=

1

(16)

Cleaning up the expression a bit, we see that the first and last terms in the derivative actuallycancel:

U=

N

Nh + U+

1

hln(1 +

Nh

U) N

U +Nh=

1

hln(1 +

Nh

U) =

1

(17)

which we can easily solve for U:

1 +Nh

U= exp(

h

) U = Nh

exp( h ) 12 (18)

K&K 2.4Suppose that 1010 monkeys have been seated at typewriters throughout that age of theUniverse, 1018 seconds. This number of monkeys is about three times greater than the

2

present human population of the earth. We suppose that a monkey can hit 10 type-writer keys per second. A typewriter may have 44 keys; we accept lowercase letters inplace of capital letters. Assuming that Shakespeares Hamlet has 105 characters, willthe monkeys hit upon Hamlet?Part a: Show that the probability that any given sequence of 105 characters typed atrandom will come out in the correct sequence is of the order of 44100000 = 10164345.There are 44 keys on a typewriter, so the probability that the first letter is correct is 1/44. Theprobability that the second letter is correct is 1/44, etc. Since we need all the letters to be correct,we need to raise 1/44 to the power of the number of letters in Hamlet:

1

44

100000

= 44100000 = 10log10(44)100000 = 10164345 2 (19)

where I have used the fact that log10(4) = 1.64345.Part b: Show that the probability that a monkey-Hamlet will be typed in the age ofthe Universe is approximately 10164316.Since a monkey types 10 letters a second, and the age of the Universe is 1018 seconds, the totalnumber of letters that can be typed by one monkey in that time is 1019. The monkey is able tostart with any of them except for the last 1019 (105 1) 1019 of them. Now, the probabilitythat a sequence of 105 letters will yield Hamlet is given in equation 19. Since there are 1019characters that it can start typing with in order to create Hamlet, the probability that one monkeywill type Hamlet in the age of the Universe is

P (Hamlet, 1Monkey) = 101910164345 (20)

so the probability that 1010 monkeys will do it is (just multiplying by 1010):

P (Hamlet) = 1010101910164345 = 10164316 2 (21)

K&K 2.5Given two systems of N1 = N2 = 10

22 spins with multiplicity functions g1(N1, s1) andg2(N2, s s1), the product g1g2 as a function of s1 is relatively sharply peaked at s1 = s1.For s1 = s1 + 10

12, the product g1g2 is reduced by 10174 from its peak value. Use the

Gaussian approximation the multiplicity function.Part aCompute g1g2/(g1g2)max for s1 = s1 + 10

11 and s = 0.We use equation 17 in K&K, which describes the number of states in a configuration of deviation from equilibrium:

g1(N1, s1 + )g2(N2, s2 ) = (g1g2)maxexp(22

N1 2

2

N2) (22)

with = 1011 and N = 1022, we get

g1g2(g1g2)max

= exp(2 (1011)2

1022 2 (10

11)2

1022) 0.0183 2 (23)

If the deviation is even an order of magnitude different, the multiplicity is incredibly different.0.0183 corresponds to a deviation of 1011 from equilibrium, while for a deviation of 1012 the resultwould be 10174. Plotting equation 22 on a log-log plot with N1 = N2 = 10

22 reveals basically aline that drops almost precisely at = 1011.K&K 3.1Part aFind an expression for the free energy as a function of of a system with two states,one at energy 0 and one at energy .The partition function is:

Z = exp(0/) + exp(/) = 1 + exp(/) (24)

3

and the free energy is related to the partition function by F = ln(Z), so:

F = ln(Z) = ln(1 + exp(/)) 2 (25)

Part bFrom the free energy, find expressions for the energy and entropy of the system.The energy is calculated from the derivative of the partition function, but I am just going to writethe normalized energy explicitly:

U =0 exp(0/) + exp(/)

exp(0/) + exp(/)=

exp(/)1 + exp(/)

2 (26)

and the entropy can be calculated by = F/ at constant volume:

=

ln(1 + exp(/)) = ln(1 + exp(/)) + exp(/)

1 + exp(/)(

) 2 (27)

As a sanity check, note that U reproduces F . Furthermore, note that 0 as 0,consistent with the third law of thermodynamics. As , ln(2).K&K 3.2Part aUse the partition function to find an exact expression for the magnetization M and thesusceptibility = dM/dB as a function of temperature and magnetic field for the modelsystem of magnetic moments in a magnetic field. The result for the magnetization isM = nmtanh(mB/), where n is the particle concentration.The energy of a single magnetic moment is = mB, depending on the orientation of the moment.Therefore the partition function is:

Z = exp(mB/) + exp(mB/) = 2cosh(mB/) (28)

and the average magnetic moment at temperature is

m = m exp(mB/)m exp(mB/)exp(mB/) + exp(mB/)

= mtanh(mB/) (29)

So to find the total magnetization, we need to multiply by the particle concentration n:

M = mntanh(mB/) 2 (30)

Thereofre, the susceptibility is

=dM

dB=nm2

sech2(mB/) 2 (31)

Part bFind the free energy and express the result as a function only of and the parameterx M/nm.The free energy is

F = ln(Z) = ln(2cosh(mB/)) (32)

To get this in the desired form, we used the trig identity sech2x = 1 tanh2x, so coshx =1/1 tanh2x. Therefore, the free energy can be written as:

F = ln(2

1

1 tanh2(mB/)) =

2ln(

4

1 ( Mnm)2) =

2ln(

1 x2

4) 2 (33)

Part cShow that the susceptibility is = nm2/ in the limit mB .

4

This follows directly from equation 31 and the fact that sech2x 1 for x 1. 2K&K 3.3Part aA one dimensional harmonic oscillator has an infinite series of equally spaced energystates, with s = sh, where s is a positive integer or zero, and is the classicalfrequency of the oscillator. We have chased the zero of energy at the state s = 0.Show that for a harmonic oscillator, the free energy is

F = ln(1 exp(h/)) (34)

Note that at high temperatures such that h we may expand the argument of thelogarithm to obtain F ln(h/).The partition function for this system is:

Z =s=0

exp(sh/) = 11 exp( h )

(35)

The free energy is related to the partition function by F = ln(Z), so:

F = ln{ 11 exp( h )

} = ln{1 exp(h/)} 2 (36)

Part bFrom equation 34 show that the entropy is

=h/

exp(h/) 1 ln(1 exp(h/)) (37)

The entropy is rel