The total number of microstates
is given by
N
!
N
!
N
!
So we can calculate the entropy via
:
S
k
ln
S
k
ln
N
!
ln
N
!
ln
N
!
Now we use Stirling’s approximation (Guenault Appendix 2) which says
ln
x
!
x
ln
x
x
;
this is important you should remember this.
PH261 –
BPC/JS
–
1997
Page 2.7

Hence
S
k
N
ln
N
N
ln
N
N
ln
N
where we have used
. Into this we substitute for
and
since we need
S
in terms of
E
for differentiation to get temperature.
N
N
N
N
N
S
k N
ln
N
1
2
N
E
/
ln
1
2
N
E
/
1
2
N
E
/
ln
1
2
N
E
/
Now we use the definition of statistical temperature:
to obtain the temperature:
1 /
T
S
/
E
N V
1
T
k
2
ln
N
E
/
N
E
/
(You can check this differentiation with
Maple
or
Mathematica
if you wish!)
Recalling the expressions for
and
, the temperature expression can be written:
N
N
1
T
k
2
ln
N
N
which can be inverted to give the ratio of the populations as
N
N
exp
2 /
kT
Or, since
we find
N
N
N
N
N
exp
/
kT
exp
/
kT
exp
/
kT
N
N
exp
/
kT
exp
/
kT
exp
/
kT
16
14
12
10
8
6
4
2
0
0.0
0.2
0.4
0.6
0.8
1.0
N
/
N
N
/
N
kT
/
Temperature variation of up and sown populations
This is our final result. It tells us the fraction of particles in each of the two energy states as a function
of temperature. This is the distribution function. On inspection you see that it can be written rather
concisely as
n
N
exp
/
kT
z
PH261 –
BPC/JS
–
1997
Page 2.8

where the quantity
z
z
exp
/
kT
exp
/
kT
is called the (single particle)
partition function
.
It is the sum over the possible states of the factor
. This distribution among energy levels is an example of the Boltzmann distribution.
exp
/
kT
This distribution applies generally(to distinguishable particles) where there is a whole set of possible
energies available and not merely two as we have considered thus far. This problem we treat in the
next section. In general
z
states
exp
/
kT
where the sum is taken over all the possible energies of a single particle. In the present example there
are only two energy levels.
In the general case
is the
average
number of particles in the state of energy
(In the present
special case of only two levels
and
are uniquely determined).
To remind you, an example
of counting the microstates and evaluating the average distribution for a system of a few particles with
a large number of available energies is given in Guenault chapter 1. The fluctuations in
about this
average value are small for a sufficiently large number of particles.
(The 1/
factor)
n
n
1
n
2
n
N
2.4.2
Magnetisation of
magnet
—
Curie’s law
S
1 / 2
We can now obtain an expression for the magnetisation of the spin 1/2 paramagnet in terms of
temperature and applied magnetic field.
The magnetisation (total magnetic moment) is given by:
M
N
N
We have expressions for
and
, giving the expression for
M
as
N
N
M
N
exp
/
kT
exp
/
kT
exp
/
kT
exp
/
kT
N
tanh
/
kT
or, since
, the magnetisation in terms of the magnetic field is
B
M
N
tanh
B
kT
3.0
2.0
1.0
0.0
0.0
0.2
0.4
0.6
0.8
1.0
B
/
kT
M
/
N
saturation
M
N
2
B
kT
linear region
magnetisation of paramagnet
PH261 –
BPC/JS
–
1997
Page 2.9

The general behaviour of magnetisation on magnetic field is nonlinear.
At low fields the
magnetisation starts linearly but at higher fields it saturates as all the moments become aligned.

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