CHE 485: STATISTICAL MECHANICS AND THERMODYNAMICS

ASSIGNMENT # 7 SOLUTIONS

TOPICS: (1) FLUCTUATIONS (2) ISING MODEL

1. (Chandler 3.15) For an open multicomponent system show that ⟨ ⟩

( ⟨ ⟩

)

where ⟨ ⟩ is the fluctuation from the average of number of

particles of type i and is the chemical potential of type i. Similarly relate ⟨ ⟩ to

a thermodynamic quantity. Finally, for a one component system in the grand canonical

ensemble evaluate ⟨ ⟩ and relate this quantity to the constant volume heat capacity

and compressibility.

Solution: Using the given definition of we get:

⟨ ⟩ ⟨ ⟨ ⟩ ( ⟨ ⟩)⟩

Expanding, taking averages and canceling out terms we get

⟨ ⟩ ⟨ ⟩ ⟨ ⟩⟨ ⟩ ∑

(∑

) (∑

)

Here the probability distribution is given by:

where ∑

Now taking derivatives of with respect to and we get

(

)

∑

(

)

∑

Taking the mixed derivative we get

∑

Substituting these values in the above equation we get

⟨ ⟩

(

(

)

)

( ⟨ ⟩

)

Similarly ⟨ ⟩ ⟨ ⟨ ⟩ ( ⟨ ⟩) ⟨ ⟩ ⟩

⟨( ⟨ ⟩ ⟨ ⟩ ⟨ ⟩⟨ ⟩) ⟨ ⟩ ⟩

⟨( ⟨ ⟩ ⟨ ⟩ ⟨ ⟩⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩

⟨ ⟩⟨ ⟩ ⟨ ⟩⟨ ⟩⟨ ⟩)⟩

⟨ ⟩ ⟨ ⟩⟨ ⟩ ⟨ ⟩⟨ ⟩ ⟨ ⟩⟨ ⟩ ⟨ ⟩⟨ ⟩⟨ ⟩

We only have to evaluate the first expression, which is given by:

⟨ ⟩ ∑

So putting the other expressions we get the equation as:

⟨ ⟩

(

)

( ⟨ ⟩

)

For a one component system in the grand canonical ensemble the probability distribution is given

by:

and ∑

The fluctuation in energy is given by:

⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ∑

∑

∑

(

)

(

)

[

]

Now

⟨ ⟩ ⟨ ⟩

Hence *

+

⟨ ⟩ *

⟨ ⟩

+

* ⟨ ⟩

+

⟨ ⟩ * ⟨ ⟩

+

Taking the second derivative we get:

⟨ ⟩

[

]

⟨ ⟩

*

⟨ ⟩

+

*

⟨ ⟩

+

Also *

+

*

+

*

+

*

+

*

+

Here and are the isothermal compressibility and co-efficient of volume expansion with

temperature.

2. (Chandler 3.17a) x is a random variable with distribution p(x) in a<x<b and the average over

the distribution of x is given by:

⟨ ⟩ ∫

To prove that ⟨ ⟩ ⟨ ⟩⟨ ⟩ if and only if where is the Dirac delta

function.

Solution: For for all x in (a,b) this is trivially true, so we consider

From the definition we get for the two sides of the equality:

⟨ ⟩ ∫

⟨ ⟩⟨ ⟩ (∫

) (∫

) ∫ ∫

So the two sides will be equal only when ∫

∫

Rearranging we get ∫ ( )

hence we have (

) now as so we must have

(Chandler 3.17b) For two random variables x and y the joint probability distribution is given by

p(x, y). Prove that ⟨ ⟩ ⟨ ⟩⟨ ⟩ only when where and are the distributions of x and y.

Solution: Again, the equation is trivially true if or for all x or y. The two

sides of the equation are

⟨ ⟩ ∫ ∫

⟨ ⟩⟨ ⟩ (∫

) (∫

) ∫ ∫

Hence subtracting we get ∫ ∫ ( )

which gives

as before.

3. (Chandler 3.18) Consider a system of N distinguishable non-interacting spins in a

magnetic field H. The energy of a particular state is given by

∑

Here is the magnetic moment in the direction of the field.

a) Determine the internal energy of this system as a function of , H and N by employing

an ensemble of these variables

Solution: A particular state of the system is completely characterized by values of the indices ,

so state . Only the energy of the system is fluctuating based on the directions

of the spins so this ensemble is a canonical type ensemble with the probability distribution of

a state given by:

∑

The Boltzmann factor is given by:

∑ ∏

Hence the partition function is ∑ ∑ ∏

Now all possible states are generated by varying the indices of the molecules which can take 2

values -1 or 1. This will give 2N possible states of the system. The partition function Q:

∑∏

∑( )

As we put we get the following terms:

and : One of each

N and : There are N ways of replacing a positive or negative sign from N

of each.

N(N-1)/2 and : There are N(N-1)/2 ways of replacing two positive or

negative sign from N.

…and so on.

So the summation can be written as (using the shorthand ( ) for replacing m signs from n)

( )

Which we recognize as the binomial expansion of ( )

So for the current system we have ( )

hence ( )

Using the relation for internal energy ⟨ ⟩ with the partition function Q given by:

⟨ ⟩ (

)

Also, energy ⟨ ⟩ can be obtained from using just the definition of average:

⟨ ⟩ ∑

∑

∑

Putting the expression for energy we get

⟨ ⟩

∑(∑

)

∏

∑

( )

In this case we can write the sum in the same way as explained above for the partition function

but it is clear that same number of +1 and -1 in the sum will give rise to terms with equal and

opposite coefficients and we can use the identity ( ) (

):

⟨ ⟩

[ (

)

(

)

]

The above series is just what is obtained if we differentiate the series for Q with respect to , so

we can write:

⟨ ⟩

( )

This gives the same result as above.

b) Determine the entropy of this system as a function of , H and N.

Solution: Using the definition of the Helmholtz energy ⟨ ⟩ we get

⟨ ⟩

Now, ( ), for this system and ⟨ ⟩ from the previously derived expression. Hence we get:

( )

Again we can also derive S from the Gibb’s formula:

∑

∑

∑

Separating terms we get:

∑

∑

∑

( )

( )

( ) as before.

c) Determine system behavior as T->0

Solution: As or we get for the energy and entropy the limits as:

⟨ ⟩

( )

( )

( )

The last form is obtained by applying L Hopital rule for indeterminate limits (form 0/0).

Problem 4. (Chandler 3.19 a) For the system mentioned above find the average total

magnetization M

⟨ ⟩ ⟨∑

⟩

as a function of , H and N

Solution: Using the definition of average we get:

⟨ ⟩ ∑

∑

∑

∑(∑

)∏

This can be written in expanded form as:

⟨ ⟩

∑

Again using the combination of +1 and -1 we get the different energy states as below:

⟨ ⟩

( (

) )

( (

)

)

We recognize the series from previous problem as the derivative of Q which gives:

⟨ ⟩

b) Determine ⟨ ⟩ where ⟨ ⟩ and compare results with the susceptibility

( ⟨ ⟩

)

Solution: ⟨ ⟩ ⟨ ⟨ ⟩ ⟩ ⟨ ⟩ ⟨ ⟩ . We only need to find ⟨ ⟩.

⟨ ⟩ ∑

∑(∑

)

∏

∑

Again using the combination of +1 and -1 we get the different energy states as below:

⟨ ⟩

( (

)

)

( ( )

(

) ( )

)

This is recognizable as just the second derivative of the partition function with respect to :

⟨ ⟩

Then the fluctuation is given by:

⟨ ⟩

(

)

(

)

⟨ ⟩

Now susceptibility is given by: ⟨ ⟩

⟨ ⟩

c) Determine system behavior as T->0

⟨ ⟩

This is just using the previously shown limit of .

⟨ ⟩

Now based on the definition of ⟨ ⟩ ⟨∑ ⟩ we see that when for all i,

which is the state when all the spins are aligned with the magnetic field. This shows that there is

not fluctuation for this ground state.

Problem 5 (Chimowitz 7.3) The energy if interacting spins in a mean field model is given

by:

∑

To show that for a single spin ⟨ ⟩ fluctuating in this field the mean field strength is

given by

For a single spin we see that the energy is given by

⟨ ⟩ The fluctuation term is zero as there are no neighboring interactions. Then partition function is

given by:

∑

For a single spin there are only two states with , so the partition function is given by:

Hence the expectation value will be given by:

⟨ ⟩

For we get the equation as

We consider the function

The derivative is given by

The equation has real solution only when the slope of

exceeds the slope of the later is just 1. So this only requires that for which