Okay, so, I had stopped somewhere here.
So, I just want to refresh your memory.
So, the point is that you know I was forced
to re-derive the expression for the pressure
of a degenerate Fermi gas because till now
I had only studied or discussed non-relativistic
quantum gases or classical gases regardless.
So, in other words, the energy versus momentum
dispersion was quadratic, so E was p squared
by 2m, but in the case of white dwarf, as
I told you that we will have occasion to study
both the limits.
So in other words, we will have occasion to
study the conventional non-relativistic limit,
which I am going to actually skip because
that is also important for reasons that I’ll
mentioned later, but what is more interesting
is the ultra-relativistic limit, where the
energy is close to, so in other words, the
energy versus momentum relation in general
in relativity, as you know, is this.
So, you have 2 different limits.
So, when you have cp much less than mc squared,
you get a certain limit, so you get this limit,
and then you get cp much greater than mc squared.
So, these are the 2 limits.
So, this is called a, this is called b.
So, the energy in the a case is going to be
mc squared + p squared by 2m.
So, in other words, this is the non-relativistic
limit, where the momentum is small compared
to mc.
So that is in a non-relativistic limited,
the energy is mc squared plus the non-relativistic
correction, which is p squared by 2m.
However, in the ultra-relativistic limit where
momentum is much greater than mc, the energy
is going to be roughly cp.
So, actually, we should be studying both,
but I am going to focus mainly on this.
So, this is ultrarelativistic limit, but regardless
you know, I have done it in general, you see
I have taken this general result, and so if
you recall that I had mentioned that this
will be what we had derived earlier.
So, this would be pressure and the point is
that at 0 temperature, beta tends to infinity
and because beta tends to infinity, this is
log of 1 + something.
When beta tends to infinity of this whole
thing, this whole thing either goes to infinity
or goes to 0.
If it goes to zero, this becomes log 1 and
it will not contribute.
So the only time it is going to contribute
is when this thing goes to infinity rather
than goes to 0.
So, when does it go to infinity, it goes to
infinity when mu minus eXilon j is greater
than 0.
So, in other words, that is this condition.
So mu is greater than eXilon j, and then only
this 1 can be ignored in comparison with this,
and when we take the log, the beta cancels
out and you get this expression.
So, now I am going to become ambitious and
do the general case where I write eXilon j
in this fashion and then I can write down
the formula for the degeneracy pressure.
So it turns out that the general expression
is of this nature, which I already mentioned,
and you can go ahead and recast this.
So notice that this is in terms of the Fermi
momentum or the Fermi wave number kF, the
Fermi momentum will be h bar kF.
So, however, it is more useful to rewrite
everything in terms of the density of fermions
and as you know, the density of fermions is
just given by the Fermi distribution.
In this case, the Fermi distribution is a
step function.
So this is important for you to appreciate
why it is a step function.
So recall that this was my Fermi distribution
and then recall that beta is actually large,
in case of 0 temperature, this beta is large,
so when beta is large, you have this situation
that if eXilon j is greater than mu, this
whole thing becomes very large, but it is
in the denominator, so it is going to vanish,
so the whole thing is going to vanish.
So the only situation when it is not going
to vanish is when beta is large and mu is
greater than eXilon j, in which case this
is exponentially suppressed and becomes close
to 1.
So, mu is greater than eXilon j is the only
situation where that nj contributes and in
those situations, nj is actually close to
1 because beta tends to infinity.
So that is the situation that I have in mind.
So now I am going to perform this summation
as you very well know how to do this, because
I told you how to do this already, it involves
replacing the summation over eXilon j by integration
over the quantum numbers, in this case, the
quantum numbers are the case.
So,I have in mind, you know particle in a
box type of situation, where you have the
k’s n pi by L, and then finally L tends
to infinity, so that the k’s become continuous,
and then you can integrate rather than sum
over discrete case, and then when you integrate,
you get this expression.
So, I have proved this already.
So please, if you forgotten, just go towards
the beginning of this statistical mechanics
lecture series, and then you will find it.
So, I have already proved this to you.
So now of course, this integration is such
that I am going to only integrate within this
radius, which is kF.
So when I do that, I get this answer.
So this is going to be the density of fermions.
So, this is how density of fermions is related
to the wave number.
So this is very general.
So, I want to impress upon you that this expression
is very general, in the sense that it is valid
for Fermi gas definitely, but it is valid
for Fermi gas in three dimensions, but there
is no restriction about what type of Fermi
gas, if it is a non-relativistic or relativistic
or something in between, it is always valid.
So, so long as it is a Fermi gas and this
2 is because of the spins, I have assumed
that the fermions have spin half.
So, up spin I have to count the states and
for down spin also have to count the state.
So I get a factor of 2 there.
So, this is actually valid for spin half fermions,
so that means that 2 spin projections up and
down.
So, this is valid for spin half fermions and
in 3 dimensions okay and it does not matter
whether it is relativistic, non-relativistic,
or whatever.
So now, what Chandrasekhar assumed was that
in the star, you see it is possible to kind
of think of, see remember that the star is
a, you know, it is a star, I mean it is a
macroscopic object to put it mildly.
So the point is that we are looking at the
atomic description, that we are looking at
the electronic degrees of freedom in a star.
So obviously, if you are at some point r and
you can of course, you know, demarcate a certain
volume around that point and whatever volume
you demarcate is going to be necessarily macroscopic,
but however, it is going to be miniscule.
You can always have a situation where the
volume that you mark out is incredibly tiny
compared to the size of the star, but incredibly
large compared to the electronic length scales
which are involved.
So in other words, like the thermal wavelength
and that sort of thing.
So in other words, you can suspect that there
will be enormous number of electrons in this
small volume.
So as a result, what Chandrasekhar said was
that it is possible to naïvely put an r dependence
here so that you assume that this as a slowly
varying function of r.
So, the density of electrons vary slowly as
a function of r, so you can suspect that close
to the center of the star, the density is
huge and as you go farther and farther away,
it falls, and then when you reach the boundary,
you get a density which is 0 okay.
So, like I was saying, so Chandrasekhar suggested
that it is possible to and of course grain
the region inside the star, so you kind of,
divide it up into small pieces, then each
piece is very small compared to the size of
the star, but still has enormous number of
electrons.
So as a result, you can kind of assume that
the density of electrons is a slowly varying
function of the distance from the center.
So as a result, what happen is we can imagine
that for a star like this, the density close
to the center is going to be huge and it is
going to fall off and become zero once you
reach the boundary.
So, this was, of course, a very general result
and it would be nice if we could deal with
this, but unfortunately, it is not possible
to deal with this in the sense that, so let
me tell you what I mean by deal with this.
So remember that this is the degeneracy pressure.
So, this is the pressure that is being exerted
by the degenerate Fermi gas as a result of
Pauli exclusion principle.
Now, this pressure is going to balance the
inward pressure caused by the gravity that
is trying to collapse the star.
So as a result, when the two become equal,
the star is going to reach an equilibrium.
So that is what I mean by dealing with it.
So that means, I want to equate this degeneracy
pressure with the pressure of the incoming
the falling star at a given point r.
So, then I will have to so that will give
me implicitly, it will tell me what should
be the distribution of density versus distance
in the star and so on.
So, it will tell me everything about the detailed
distribution of matter inside the star.
So that is very ambitious and it would be
nice if we could do this in general, but it
is not possible.
So, what we have to do is we have to make
further approximations.
So in fact, Chandrasekhar did both the calculations,
namely, he studied the non-relativistic limit
first, so this is relativistic limit.
So ultrarelativistic limit, see the non-relativistic
limit would be the opposite, which is h bar
kF is much less than mc.
So in this case, E is close to p squared by
2m plus a constant and you can go ahead and
do that and you get a result, I am going to
tell you what results we will get for that,
I will just, I will not derive it, I will
just mention it, but I will derive, what I
am going to derive is this ultrarelativistic
limit, which leads to the famous Chandrasekhar’s
mass limit of the white dwarf.
So let us get on with it.
So let us look at the Chandrasekhar’s calculation
where he did the second part where h bar kF
which is the Fermi momentum is much greater
than mc.
So in this limit, the fermions are moving
close to speed of light.
So, instead of dealing with this complicated
equation, we can just go ahead and approximate
this by h bar kC and then just do this integral.
So, that is what I have done here.
So, when I do this integration, I get this
result.
So it is going to be kF to the fourth power
and recall that kF is nothing but, so from
here you can see that kF is proportional to
density of fermions raised to one-third.
So, kF is going to be density of fermions
raised to one-third.
So because pressure is kF to the power 4,
so for an ultrarelativistic gas, please recall
that, remember that this is ultra-relativistic
only.
So for an ultrarelativistic Fermi guess, the
pressure is proportional to the fourth power
of the wave number and the wave number itself
is always given in 3 dimensions by rho, density
raised to one-third.
So putting those together, you get expression,
which is density raised to four-thirds.
So, this is an example of what is known as
a polytrope.
So, the pressure that is exerted by a degenerate
Fermi gas or in general, any type of gas,
so if you express it in terms of the density,
so that equation is referred to as a polytrope.
So in this case, well, specifically if it
is a power law, so that means if it is a power
law, it is called a polytrope.
So, in the extreme limits, in extreme non-relativistic
limit and the extreme relativistic limit,
in both cases, it is going to be a polytrope.
So, I would not do the non-relativistic calculation.
So, you will have to believe the result that
I tell you later, maybe if we have time, we
will leave it for the exercises later on tutorials
and so on okay.
So, now remember that what I am supposed to
do is that this is the pressure exerted by
the gas, which is basically due to Pauli’s
exclusion principle.
So in other words, this is the pressure the
electrons will exert if you try to compress
those electrons.
So, the origin of this pressure is basically
Pauli’s exclusion principle and this is
valid when the electrons are moving close
to speed of light.
So, now this pressure has to balance the pressure,
what is going to push it, so what is going
to push it is basically the gravitational
force.
So what we have to do is we have to calculate
the pressure caused by gravitational force
and we have to equate that with the pressure
that we just calculated, which is the outward
pressure caused by the Pauli exclusion principle
and degenerate Fermi gas where the electrons
are trying to avoid being pushed together
too closely.
So how do we do that?
It is very easy to do that.
So imagine, you have a small area like this.
So this is my dA and this is my dr.
So I have a volume dA to dr and this volume
is experiencing a force.
So, this volume contains this much mass.
So okay I will have to say what is all this?
So, this is clearly the volume dv okay.
So, this is how much volume there is in this
square, I mean this kind of cubicle or whatever
box that I have written.
So, now the point is that the rest of it is
basically the mass density of the star.
So, the mass density of the star times the
small volume is basically the small mass that
is contained in this box.
So what is the mass density of the star?
See, recall that rho is basically the electron
density okay, so this is the number of electrons
per unit volume.
So remember that for every negative charge,
there is a positive charge, because overall,
the star is electrically neutral.
So in other words, there are as many protons
as there are electrons, but then, so let us
assume that we have a situation where the
number of neutrons is equal to the number
of protons.
So notice that so what I am implying therefore
is that see I have to calculate or I have
to find out what is the mass, not the number
of electrons, I want to find how heavy this
box is.
So how heavy it is, so let us see what is
all in the box.
So in the box, there are this many electrons,
there are rho into dV number of electrons,
but then electrons are very light, right.
So electrons are very light, so they do not
contribute to the mass, but then along with
the electrons for every electron, there is
a proton also because otherwise the system
will not be electrically neutral.
There is a neutral atom sitting there after
all, but electrons have been dislodged from
the atoms, they are freely moving about, yes,
but still there are as many protons as there
are electrons.
So in fact, and protons being enormously heavy
compared to the electrons, they are actually
going to contribute to the mass, more or less
fully to the mass.
So, we are going to ignore the mass of the
electrons, we are only going to consider the
mass of the nucleons.
So, nucleons means protons and neutrons, but
one should not forget the neutrons because
they do not contribute to any electrical forces,
but they contribute to gravitational force
definitely because they are as massive as
protons.
So, let us assume a situation where you have
an atom, which has as many neutrons as there
are protons.
So that is kind of typical for a stable atom.
So typically, number of protons matches the
number of neutrons and matches, of course,
the number of electrons, that is always true
because of charge neutrality.
So, let us assume that this is the situation.
So in that case, the mass of the box is actually
the number of nucleons which is like two,
so one proton, one neutron, and I have assumed
that the mass of the proton and mass of the
neutron are roughly the same as two times
the mass of the nucleon times the number of
nucleon per unit volume, which is equal to
the number of electrons per unit volume times
the volume.
So that is the mass that is contained in this
box.
So, I hope that is clear.
So, that is the dm the mass that is contained
in this box, and this box, the mass that is
contained in this box experiences a gravitational
force and of course you know Newton showed
this 300 or more years ago, that the force
on this is actually due to a force contained
inside this region.
So, it does not matter what is outside.
So, if you have a spherical symmetry, then
the gravitational force exerted on this mass
is all because of the mass inside, so this
Mr is the mass that is inside this dotted
line, and what is that mass?
So, it is basically volume integrated from
0 to r times the density, so that is what
it is.
So, it is dmM by r squared, but then this
is the dm there.
So now, this is the force, but you know that
force per unit area is pressure.
So what I have to do, so I have to divide
by dA, so dF by dA is my pressure.
So, I am going to divide by dA.
So, I am going to calculate dFr by dA, which
is my pressure.
So, I am going to actually end up getting
this answer, so I am going to get this okay,
so let me go and get that.
So when I get that, so I will get a pressure
to be this result.
So let me do that GMr by r squared and so
this is still dp by the way because there
is a dr there.
So, it is the pressure due to that really
small dr.
So, the pressure is 2mn rho r dr.
So, I want the pressure at some point r.
So, what I have to do is I have to integrate
from a point where the pressure I know to
be 0, so which is pressure at r equal R is
0.
So, I am going to integrate from a place where
I know the pressure to be, so r equal R to
r, so that is going to be integral dr to r
GM(r) dash over.
So I have just replaced r by r dash because
it is a dummy variable, because I want to
reserve r for the point of interest.
So this is the pressure at r.
So, this is what it is.
So, that is what I have written there, so
I have integrated from.
So the point is of let us fix the sign here.
See, the point is that if I take the derivative
with respect to r, what do I expect?
So from here, you can see it is negative.
So that means the pressure is actually decreasing
as I go away from this, so I should have said
a minus.
So this force is attractive, so there should
be a minus there, so that is the reason why
I am not getting the right answer.
So it is an attractive force, so pressure
is like this, I mean the attractive means
it is in minus r cap.
So this is if you put a vector there, it will
be actually minus r cap, so r cap is out,
minus r cap is in.
So P dash r should be negative because pressure
is decreasing as you go from the center through
the surface.
So now, I can rewrite this in this form.
So I can first differentiate with respect
to r, then I multiply by r squared divided
by rho and then again differentiate by r.
So basically, look, it is a little bit of
algebra.
So what I can do here is that if I take P
dash r, so P dash r is going to be minus GMr
2mn rho r by r squared.
So now, I multiply by r squared and divide
by rho of r, I get this.
So I get minus GMr into 2m.
Now, keep in mind that M itself is nothing
but this integration.
So because of this, M dash r is actually going
to be 2mn rho of r into 4 pi r squared okay
right.
So if I just take the derivative of M of r,
it is going to look like this.
So, then I can take d by dr of this, so that
is going to be M dash r.
So, it is going to be minus G 2mn M dash r.
So what is m M dash r?
It is again to 2 mn.
So it is going to be this to cut a long story
short, okay.
So, I am going to skip the rest of the details,
so you can figure it out yourself.
So finally, I can rewrite this by appropriately
differentiating however many times I want
and I can always, so please convince yourself
that this is what I got from my physics considerations
by looking at this, the gravitational force
exerted by this cube and all that.
Now, this is from here to here is just some
algebra.
So you just make sure that you substitute
this P here and show that it is an identity.
So now, remember that for a polytrope, we
just derived that the degeneracy pressure,
so this is because of gravity.
So this is the pressure that gravity exerts
and this is the pressure that is due to the
Pauli’s exclusion principle.
So that is why I have called it by the same
P, so even though this is due to different
reasons.
This pressure is due to gravity, this pressure
is due to Pauli’s exclusion principle, but
I want the two to be equal because I want
the equilibrium.
So in equilibrium, these 2 pressures are the
same.
So now, I equate these two and so after this,
it is a little bit technical, it is just a
bunch of algebra.
So what you will have to bear with me because
I find that the technical nuances are actually
skipped in many of the books and because Chandrasekhar
limit is such an exciting topic, you know,
we all learn about it in our school days and
we take great pride as being Indians to say
that we know Chandrasekhar limit.
So, we would be fooling ourselves if we did
not make an effort to actually go through
some of the steps that Chandrasekhar himself
went through and recall that he was only 19
years old when he did these calculations.
It is amazing how he managed to do all this
when he was only 19 and sailing a ship.
Okay, so let me continue.
So what I have to do is I am going to define
dimensionless quantity, which depends on r.
So rho of zero is the density of electrons
near the center of the star.
So, I am going to define the ratio rho of
r by rho zero, which is basically the ratio
of which is the density of electrons at point
r measured in units of the density at the
origin okay, so and then there is a raised
to one-third for reasons that are obvious
because the polytrope always involves these
type of ratios.
So, what I am going to do is I am going to
use this correspondence and rewrite P in terms
of this theta rather than in terms of the
rho.
So, then I end up getting these 2 formula.
So the P becomes related to the fourth power
of theta and the rho becomes related to the
third power of theta, so that is the beauty
of this theta definition because the pressure
is some integer power of theta and the density
is some other integer power of theta.
Okay, so now, I am going to go ahead and substitute
this pressure here, rather here, I am going
to substitute it there.
So when I do that, I get this result okay.
Well, I am going to do both, so I am going
to substitute this pressure here and I am
going to substitute this density there and
then a whole bunch of thetas I am going to
cancel out and finally I end up with this
result.
So see that this ugly looking constant has
actually come out of the equation.
So that is why I am going to take this down
and call this whole thing as some alpha.
So this you can easily convince yourself this
as the dimensions of length.
So this is the length, so it has dimensions
of length, some kind of a length.
So, now I am going to define a new function.
So I am going to define what is called alpha
into Xi, so where this is now dimensionless.
So this is distance and this is also distance,
so both these are lengths.
So both these are lengths and this is dimensionless.
So, I am going to start calling, so this was
my r, so theta of r, I am going to call it
as U of Xi.
So if I choose to do that, then this equation
can be written in this very beautiful form
and this is called the Lane-Emden equation
okay.
So, this has to be solved.
So this is a completely dimensionless equation,
but then you see it is a second order nonlinear
ordinary differential equation with non-constant
coefficients, so that is a mouthful, and it
is nonlinear.
So if it was linear at least, if it is non-constant
coefficients, it is already not easy to do
as you very well know probably, those of you
who have studied differential equations.
You know that if your non-constant coefficients,
you have to use something called the Frobenius
method to solve it, where if it is constant
coefficients and it is second order, it is
very easy.
The solutions are always exponential or oscillatory
or whatever, but if it is non-constant coefficients
and linear, you can still if you get one solution,
you can get the second by Wronskian method,
but then none of those methods are going to
work here because this is actually a nonlinear
equation, so the superposition principle does
not work.
So you know the second order equation, you
expect 2 linearly independent solutions, but
then if you know one, there is no way of figuring
out the second one because you cannot use
superposition principle, but fortunately,
we do not need to go through all those, there
is only going to be one physically meaningful
solution.
Alright, so this is the equation we have to
solve.
So now, what are the boundary conditions,
of course, I mean I am not really looking
for a general solution, I want a specific
solution consistent with my physical boundary
condition.
So recall that theta of r it was nothing but
this ratio.
So if I put r equal to 0, I will get clearly
theta of 0 equals 1, just by definition.
So now that is one boundary.
So this is a second order equation, and theta
of 0 means U of 0 okay, so U of 0 and theta
of 0 are the same things.
So U of 0 is 1, but then that is just one
boundary condition, but then I need one more
because this is a second order equation.
So how do I deal with the second boundary
condition?
So in this equation, so if I go back here,
so if I look at the small r limit, so if I
look at the limit as r tends to 0, so when
r tends to 0, you see the dominant term is
going to be so when for r small, we can expect
that this is going to be close to, P dash
r is going to be close to P dash 0.
So in this limit, p dash r is approximately
P dash 0 and rho of r is approximately a rho
of 0.
So the thing is that in this limit, this is
going to be 1 by r, so this whole thing is
going to become something like 2 because it
is d by dr of r squared because all this will
become constant and go out of the derivative.
So it is d by dr of r squared, which is basically
2r, so 2r by r squared is 2 by r, 2 by r P
dash 0 by rho 0, that is what I get, but then
that is equal to some rho of 0.
So in other words, it is actually what is
telling me that the P dash 0 should tend to
0, that is P dash r should tend to 0 in the
limit as r tends to 0 in such a way that P
dash r is proportional to r.
So from here, you can conclude that P dash
r has to be proportional to r as r tends to
0.
So it is a little bit tricky, you should think
about it more deeply.
So as r tends to 0, you should convince yourself
by staring at this equation that P dash r
is proportional to r for when r is very small
okay, just by staring at this you can convince
yourself.
Now so as a result when P dash r is close
to r when r is very small, when r tends to
0, P dash 0 becomes 0 therefore, okay.
So now, what is P dash r.
So P dash r remember that P dash r, P r itself
was theta to the 4 and P dash r is nothing
but theta dash r our times theta cubed and
so on.
So if P dash 0 is 0, so that means that because
theta 0 is 1, we already know that, so that
means theta dash of 0 should be 0 okay.
So, theta dash of 0 is 0 same as saying U
dash of 0 is 0.
So, in other words, the Lane-Emden equation
was this, but then I cannot solve this without
supplying boundary conditions or initial conditions
in this case.
So the initial conditions are going to be
and because it is second order, I have to
supply two initial conditions.
So, the first initial condition has been U
0 is 1.
The second initial condition is U dash 0 is
0.
So it so happens that this equation looks
formidable, but its solution is not that formidable
and one can easily solve it, well it is not
that easy, but it is not that hard either,
and in fact, one can show that there is a
Xi1, value Xi 1 for which U become 0.
So, in other words, one can imagine that,
remember that U dash 0 starts off being 1
and then finally it becomes 0.
So, there will be some Xi1.
So, this Xi axis and this is U axis.
So it starts off when Xi 0 is 1 then finally
becomes.
So the idea is that there is a Xi for which
U become 0 and what is the physical meaning
of that.
So remember that U is nothing but theta, because
U and theta are related.
So theta of alpha Xi 1 equals U of Xi 1 equals
0.
So that means, this is some r1, which I call
basically r.
So this is basically the radius of the star,
why is this the radius of the star?
So the radius of the star is this ugly constant
alpha times this fundamental number Xi 1,
which makes U vanish, and so why is this the
radius of the star because at that value if
you put alpha Xi 1 here, so this theta is
basically going to become 0, so that means
at that value of r, when r equals alpha Xi
1, the pressure is 0, and the density is also
0 because this is 0.
So the theta vanishes, so the pressure and
density of the electrons is basically theta
to some power.
So if theta becomes 0, the pressure becomes
0 and density becomes 0 and we can identify
that that radius to be the radius of the star
because we know that at the radius of the
star, the pressure becomes 0 and there are
no electrons left, that is the boundary of
the star okay.
So, we have found the boundary of the star
and for later use, we will see that we will
require this product.
So Xi 1 square, U dash and Xi 1 can be evaluated
and it is some number and remember that U
dash is negative because the pressure and
so on are basically decreasing functions of
the r.
So our density and pressure are both decreasing
as you go from the center to the surface of
the star.
So, this is the radius of the star okay.
Now, so you can see that the radius of the
star is basically, so it is kind of inversely
related to that density okay.
So larger the density, smaller the radius.
If the density of the star at the center is
very huge, the radius of the star is very
small okay.
So let us see the mass of the star.
So, this is not surprising that the radius
of the star is inversely related to the density
at the center, but what is really amazing
is the mass, suppose you try to calculate
the mass of the star, which is nothing but
4 pi r squared dr times the rho, so it so
happens that, so this of course I have to
put a 2mn there, I forgot the 2mn, so 2 mn
okay.
So, the point is that if I calculate the mass
of the star, okay, you can see that instead
of rho, I can start putting in my this equation.
So I can put my this result, so I will get
this okay and keep in mind that U cubed, okay,
so rho is basically theta cubed, it is rho
zero into theta cubed, but then theta is basically
directly related to U.
So, this whole thing becomes just integral
over U cubed Xi squared d Xi from 0 to Xi
1, which is the radius, I mean the radius
in dimensionless units of the star.
So remember that U cubed actually from this
Lane-Emden equation can be written in terms
of the derivative and the remarkable thing
is this Xi squared is in the denominator in
this Xi squared, they cancel, and then you
end up with.
So if I do this Xi squared, so remember that
if I do this kind of an integral, so it is
going to be Xi squared, and what is U cubed?
I forget the sign here.
So it is 1 by Xi squared d by d Xi Xi squared
okay dU by d Xi.
So, this is what I have to do.
So, now Xi squared Xi squared cancels and
then I will end up having to do this.
So, in other words, this is gone, this is
gone, so I am trying to now integrate the
derivative, so from 0 to Xi 1 and what is
that answer?
So, the lower limit is just, so this is just
going to be Xi squared U dash Xi evaluated
from Xi equal to 0 to Xi equals Xi 1.
So, now, you can see that the lower limit
will not contribute because that makes Xi
0.
So, the upper limit will contribute and upper
limit is going to be U, U of Xi 1 squared,
U of Xi 1, which is why I listed that earlier.
So I told you I will require it and now is
when I require it.
So if I put this number there, you will see
that finally, it will actually cancel out.
So the final answer does not involve rho 0
at all because it has cancelled out okay.
Why it has cancelled out because alpha cubed,
remember that, there is a rho 0 sitting here
with an alpha, which is here.
So you see what is alpha, so alpha is proportional
to rho of 0 raised to minus one-third okay,
so that is what alpha is.
Alpha is rho 0 raised to minus one-third.
Now, whereas here, what is this, this is alpha
cubed.
So alpha is rho 0 raised to minus one-third.
So alpha cubed is this cubed, so times rho
0 okay, so that is what this is.
So it is alpha cubed, which is rho 0 raised
to minus one-third whole cube times rho 0
and that is independent of rho 0.
So that is the amazing thing here.
So now, if you calculate the total mass of
the star, it is completely independent of
the density at the center.
So it only depends on a whole bunch of fundamental
constants like Planck's constant and gravitational,
not a whole bunch, just three of them; speed
of light, Planck's constant and gravitational
constant.
So if you work out the numbers, and this is
the astronomical symbol, the M with a circle
with a dot in the center is a universal ancient
symbol for the sun okay, and this would be
the earth.
I mean, this would be earth and this is the
sun.
So, this is sun and this is the earth, so
that is the ancient astronomical symbol for
the sun.
So now, you know what is the mass of the sun,
you just work out, it becomes 1.44 times the
mass of the sun and this is a kind of a universal
constant independent of anything, I mean this
whole thing is a universal constant, it is
independent of anything else.
So, that was the remarkable result, see the
reason why it is remarkable, contrast this
with this result, which says that the radius
of a white dwarf is actually inversely related
to the density at the center raised to one-third.
So in other words, as the density increases,
the radius keeps shrinking, so that is very
believable and nobody will question that.
So it is this result that made a lot of people
not believed this initially.
So in fact, the famous story goes that the
great astronomer Arthur Eddington who was
very influential at that time, kind of ridiculed
this idea.
So he called this you know, stellar buffoonery.
So, he did not believe that the mass of a
star, so Chandrasekhar himself remarked that,
you know, so being able to write down the
mass of a star in terms of laboratory constants
that you find, you know, in the back cover
of your high school textbook and that is the
mass of a star is somewhat hard to believe,
but it is nevertheless true that what this
is saying is that in the ultrarelativistic
limit, the mass of this star when it is stable.
So if that is a white dwarf that is stable
because it is exerting the degeneracy pressure
due to Pauli principle, which is balancing
the gravity, which is trying to collapse it
and it so happens that the mass of that star
is actually unique, it is a fundamental constant.
So, there is only one mass of that star.
So, only such a star can survive.
So that was the remarkable result of Chandrasekhar,
which people did not believe initially, but
finally, you know, when it was confirmed that
there are no stars heavier than Chandrasekhar’s
limit.
After making detailed observations over a
period of time, then people realized that
there are no white dwarfs which are heavier
than the Chandrasekhar limit, so they were
forced to conclude that this is correct and
as a result, Chandrasekhar won the Nobel Prize
as you very well know.
So now, let me conclude by pointing out why
is this the limit?
So, this calculation is just telling me that
the white dwarf has a mass which is a fundamental
constant, namely this, but then why is this
the upper limit?
The reason is because you see we directly
jumped into the ultrarelativistic calculation
where the energy was cp.
So what I should have strictly done is, I
should have done both the calculations, I
should have done the calculation where energy
is P squared by 2m, and then compared it with
the result when energy is cp.
So you can imagine when that density is very
small, so the smallness of the density is
basically governed by how much this mu, that
means if PFC is much less than mc okay, so
that is when you should be using the non-relativistic
approximation.
So one can in fact define what is called the
critical, it is not really critical but some
kind of a crossover scale where it crosses
over from the non-relativistic to the relativistic
regime.
So you can define what is called the crossover
scale.
So the crossover scale will be exactly when
these two are equal.
So, one can define this crossover density.
So this will implies so this is a Fermi momentum
for the crossover, this implies a certain
density.
In fact, if you use your non-relativistic,
if you use your E equals P squared by 2m calculation
and recalculate the mass, which I am not going
to do, but suppose you redo the whole thing,
all the way up to this point, but using not
equal to cP like I have done till now, but
use E equal to P squared by 2m, I am going
to get a mass, which is not a fundamental
constant, but I am going to get a mass that
depends upon the density at the center raised
to one-half.
So, it is going to be a density at the center
raised to one-half.
So this is of course a fundamental constant
because that is related to all these.
So, there is going to be a situation if the
mass of the star keeps increasing, this rho
is going to keep increasing, but notice that
this result is valid only when the rho is
much less than this crossover okay.
So if the density of the center is much less
than the crossover, then only this is valid.
So if you keep increasing the mass of the
star, there is going to come a situation when
it is going to cross over into the relativistic
regime.
So, the non-relativistic approximation will
fail and then you will gradually start making
it more relativistic.
So, that is the reason why this is called
the limit.
So, what Chandrasekhar did, first he calculated
this way.
So, he did this calculation, where he showed
that the mass of the star in the non-relativistic
limit, so when the mass of the star is small,
you can use non-relativistic limit and you
show that the mass is proportional to the
square root of the density at the center.
So, as you keep increasing the mass, the density
at the center keeps increasing, but then once
it goes much beyond the crossover limit, then
you cannot use non-relativistic limit.
So, you use ultrarelativistic limit and it
will immediately tell you that you should
stop when you reach the mass to be this number,
which is the universal Chandrasekhar limit.
So, you can keep increasing the mass of the
star at least stable, stable, stable, stable,
stable, and it will stop being stable once
you cross this.
So, once you reach this, you are already in
the ultrarelativistic limit.
Once you reach Chandrasekhar limit here, you
are already in the ultrarelativistic limit.
So, what Chandrasekhar limit is telling you
that you cannot cross this.
So if you try to cross this, the star will
not be stable.
So that is the reason why it is the limit
okay.
So it took a long time for people to accept
this because the Chandrasekhar limit is just
a bunch of fundamental constants and it refers
to the mass of a star.
So, I hope you enjoyed this presentation of
Chandrasekhar limit.
So it is a little difficult and you should
show some respect because Chandrasekhar did
this difficult calculation when he was only
19 years old as I keep pointing out.
Okay, so let me close here and move on to
some other topic next time.
Thank you.
