We will discuss Bose-Einstein condensation
today. We know from Bose-Einstein statistics
that the BE distribution ni is 1 by B e to
the power beta epsilon i minus 1, where B
is e to the power minus beta mu. This is for
non-degenerate case. When there is degeneracy,
n i is g i 
by e to the power beta epsilon i minus 1,
where g i is the degeneracy of ith quantum
state. Let us now consider 
the behavior of the constant B 
that appears in the BE distribution just mentioned
above.
This quantity 
corresponds to the partition function in the
Boltzmann distribution or the chemical potential
in the Fermi-Dirac distribution. Now, we need
to calculate the value of B. We 
know sum over i n i gives N, suppose this
is our equation 1, where N is the total number
of particles. Now we will evaluate how does
B depend on temperature.
As we discussed before, we can consider the
density of states for spinless particles moving
in a box of volume V.
We can represent g epsilon is V by 4 pi to
the power 2 times twice m by h cross to the
power 2 whole to the power 3 by 2 times root
over epsilon, which is our equation 2.
Let us assume that we can replace the summation
over discrete energy levels 
by an integration over a continuum of energy
levels. Then the constraint of equation 1,
which is sum over i n i equals to capital
N can be approximated to integral 0 to infinity
n epsilon d epsilon and we represent this
equation as equation number 3.
Using Bose-Einstein distribution and the density
of states of equation 2, we obtain
V by 4 pi to the power 2 twice m by h cross
to the power 2 to the power 3 by 2 0 to infinity
root over epsilon by B e to the power beta
epsilon minus 1 d epsilon equals to N. This
is our equation number 4. Now, we consider
y is beta epsilon, which is our equation number
5. When epsilon goes to 0, y goes to 0 and
epsilon goes to infinity then y goes to infinity.
If we substitute y equals to beta epsilon
in equation 4, we get V by 4 pi square times
twice m by h cross to the power 2 whole to
the power 3 by 2 integral 0 to infinity root
over y dy by B e to the power y minus 1 equals
to N. Suppose this is our equation number
6.
We consider another function, F which is function
of B is 2 by root over pi 0 to infinity root
over y dy by B e to the power y minus 1 and
this is our equation 7. By substituting F
B of equation 7 into equation 6, we get, V
m kB T by twice pi h cross to the power 3
by 2 times F B is equal to N, and this our
equation number 8. Next we plot F B Versus
B, which looks like this and we will explain
this in a while.
For large value of ‘B’ we can see, F B
is 1 by B. This is because for B much much
larger than 1, we can write 2 by root over
pi times integral 0 to infinity root over
y dy by B e to the power y minus 1 can be
approximated as 2 by root over pi times 0
to infinity root over y dy by B e to the power
y. Since B is very very large, we can neglect
minus 1 and this gives 1 by B.
Thus, from equation 8, for large ‘B’ value,
we can write B is V by N m kB T by twice pi
h cross to the power 2 whole to the power
3 by 2. Suppose this is our equation number
10. But we find that the integral cannot be
evaluated below B equals to 1 or when B 
is less than 1. This is because there is a
singularity 
at B equals to 1. This is important to note
that the value of B cannot be less than 1
and there is a singularity at B equals to
1.
Thus, the Bose-Einstein distribution is n
i equals to g i by B e to the power epsilon
i by kB T minus 1. If B is less than 1, there
is a possibility 
of an energy level epsilon i such that n i
is negative. Clearly we cannot have a negative
number of particles or occupation number in
any energy level. So, we expect B has to be
greater than or equals to 1 and for this region
the above plot B starts from 1. Next we see
that F B takes a maximum value at B equals
to 1.
The maximum value of 'B' is (when B equals
to 1) F 1 is 2 by root over pi 0 to infinity
root over y dy by e to the power y minus 1
which is equal to zeta 3 by 2, which a value
of 2.612. This is our equation number 11 Zeta
x is the Riemann zeta function defined by
zeta x equal to summation over n equal to
1 to infinity 1 by n to the power x. This
is our equation number 12. The fact that FB
has a maximum value of 2.612 is somewhat disturbing.
Because ‘N’ is constant 
and
the equation number 8 suggests that as the
temperature decreases, F B should increase.
But, if FB has a finite maximum, then below
some temperature TB, which we define as Bose
temperature, the number of particles 
we find by integrating 
the Bose-Einstein distribution over all energies
is 
less than N, the number of particles in the
system. Thus, in that condition that is when
T is less than TB and for all B values, we
can write V by 4 pi square times twice pi
m kB T by h cross to the power 2 whole to
the power 3 by 2 times integral 0 to infinity
root over y dy by B e to the power y minus
1 is less than 0.
The situation becomes clear 
if we plot 
the population density n epsilon as a function
of energy epsilon for different temperatures.
The plot looks like this. This is the case
when T is 1.1 times TB and for this case T
is 1.5 times TB and this is the case for T
equals to 0.3 TB. We consider 3 different
(temperatures) temperature regions one is
much larger than TB, one is much lower than
TB, and one is very close to TB. What we get
from this plot?
For T is greater than TB, we see a smooth
distribution curve, that for high temperatures,
approaches the distribution 
expected 
for a Maxwell-Boltzmann gas. The area under
the curves for T greater than TB are independent
of temperature. We are more interested for
when T is less than TB.
When T is lower than TB, strictly speaking
we 
cannot evaluate the population density, since
we do not have a solution for the parameter
B. 
At 
the best we can do is 
we set B equals to 1 for T less than TB and
we find a distribution that is sharply peaked
at low energy. But the peak is not high enough
for the area under the curve to give the correct
result for the number of particles in the
system. In the above plot we can just replace
this beta epsilon by epsilon by kb times TB
because we cannot evaluate for T less than
TB. The temperature TB is important because
at this temperature the particles start to
disappear, so, V m kb Tb by twice pi h cross
to the power 2 to the power 3 by 2 zeta 3
by 2 gives equals N. This is our equation
number 13.
Thus the value of TB is 
twice pi h cross to the power 2 by m k N by
zeta 3 by 2 times V to the power 2 by 3. This
is our equation number 14. Now, the question
is where are the missing particles 
for T less than TB? The answer is 
in the ground state. The problem, which we
now need to rectify, is that we replaced the
summation over discrete 
energy levels by an integral over a continuum
of energies in equation 3.
(Refer slide time 42:22)
When we replaced the summation by integration,
we integrated over 0 to infinity and by doing
so we basically omitted the particles in the
ground state. Since the lower limit of the
integral is zero, we should assume for consistency
that the ground state has zero energy. But,
the distribution function g epsilon is proportional
to root over epsilon, so at zero energy the
density of states 
is zero and the population in the small energy
range 0 and d epsilon is 0. It means that
the integral in equation 3 only estimates
particles in the excited states omitting those
in the ground state.
Remember we did similar exercise in FD distribution,
but in the FD distribution in the ground state
at the max we can have 2 particles or when
there is no spin of the particle only 1 particle
can be present in the ground state for FD
statistics. Considering the value of N, number
of particles, is very very large, if we neglect
that 1 term or at the max we can have 2 particles
in the ground state when particles will have
plus or minus half spin. So for FD statistics
basically we can safely ignore the number
of particles in the ground state because the
total number of particles N is much much greater
than 1. Whereas in case of the BE statistics,
since there is no restriction in the number
of particles in the ground state we cannot
neglect the number of particles in the ground
state.
(Refer slide time 47:17)
We made the same mistake 
when we considered the Fermi gas. But, in
that case, there could be only two particles
in the ground state, assuming degeneracy 2,
for spin half particles. In other words, we
omitted two particles out of a large number
of particles ‘N’ and this did not affect
our conclusions. But for a Bose gas or gas
molecules which are following Bose-Einstein
statistics, there is no limit in the number
of particles 
that can be present in the ground state. So,
if we ignore 
the ground state 
from the integration we can make a large error
for BE distribution. Thank you.
