1. Compact Objects:White Dwarf stars
Newtonian stars.
Equation of state.
Toy model of white dwarf stars
Numerical methods.
Mass, radius, density profiles.
Chandasekhar Mass.
European Journal of Physics 26 (2005)
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2. White dwarfs are the final evolutionary state of all stars whose mass is not high enough to become supernovae, over 97 % of the stars in our galaxy. It is generated when a star lower than 9 or 10 solar masses finishes its nuclear fuel.
They are very dense; a white dawrf’s mass is comparable to that of the Sun and its volume is comparable to that of the earth.
Typical mass  1.4 MS (MS2 1030 Kg)
Typical radius  104 Km.
No nuclear reactions take place, and therefore there is not thermal energy
to stop gravitational collapse. It is the pressure produced by the ionized
electrons that supports the star.
99 % of the white dwarfs is made by C and O which are the products of
Helium fusion.
There are few made by He.
At the begining they have a high temperature, but as there is not energy
production, they slowly cool. T  several thousands of Kelvins.

3. Comparison of a typical white dwarf with the earth.
The scape velocity in a white dwarf is larger than in the earth,
For a white dwarf with the same radius than the earth and 1 solar mass
no relativistic treament is required. Newtonian star.

4. At 1926, Fowler gave the first explanation of the structure
Sirio B, it has a superficial temperature
T K.
At 1926, Fowler gave the first
explanation of the structure
of white dwarfs, using the electron
degenarate pressure, only a few months later than the formulation of Dirac statistics.
A litle later Chandrasekhar introduced special relativity to describe the electrons and find out a limiting mass for white dwarfs.
Image taken from the Hubble Space
telescope. Discovered in 1915 at
the Mount Wilson observatory.

5. Radial force acting on a small mass element
Δm=ρ(r) ΔV located at a distance r from the center
M(r) is the enclosed mass within a radius r.
P(r) is the pressure at that point.
Newton’s law:

6. Assuming hidrostatic equilibrium
One arrives at the fundamental equations describing the structure of
Newtonian stars.
ρ(r) is the density mass
not the particle density
In hydrostatic equilibrium, the pressure in the star is a decreasing function of r.
We need some gradient of pressure . Otherwise the star collapses.
The radius of the star, R, is defined as the value of r at which the pressure goes
to zero, P(r)=0.
Mass, is the enclosed mass at r=R.
The equations are two coupled, first order differential equations. They
look simple. However, ρ(r) is unkown and no solution is possible until the equation
of state (EoS) , i.e. a relation between the density and the pressure P=P(ρ) ! ,
is known,

7. For white dwarf stars, the system is approximated as a plasma containing
positively charged nuclei and electrons.
The nuclei provide almost all the mass and no pressure. (He, C, Si)
The electrons provide all the pressure and almost no mass.
The system on a global scale is electrically neutral.
The electrons have been stripped from the atoms because the binding energy
of the atomic electrons is significantly lower than T106 K which is the
the temperature inside the star ( 106 K 100 eV).
As EoS we use the EoS of a free Fermi gas of electrons at zero temperature.
Can we consider T106 K to be zero temperature? Yes! The density is high 
Fermi energy large, of the order of 1 MeV which corresponds to a Fermi
temperature of TF 10 10K. The system is very degenerate. This means that the
momentum distribution function is very similar to the step function (kF-k).
Thermal transitions are Pauli blocked except for a small tiny fraction of
electrons at the Fermi surface. For the purpose of computing the pressure
of the system, this assumption is extremely accurate!
However, the density is high enough that the Fermi momentum is relativistic,
and the kinetic energy is large compared with the rest mass of the electron.
Quantum mechanics (indistinguisibility and Pauli principle) and special relativity
play a crucial role in preventing white dwarf stars from collapsing!!

8. At T=0 And the Fermi momentum
All the levels (momenta) are filled up to the Fermi level.
Relation between the number of particles ( macroscopic quantity) and the
Fermi momentum (microscopic quantity)
and
And the Fermi momentum

9. The single particle energy in the non-relativistic case:
Evaluation of the energy. Non relativistic case.
The single particle energy in the non-relativistic case:
Total energy
Which can be written in terms of the energy per particle

10. Notice that a gas of particles at zero temperature can generate a pressure
due to the Pauli principle.
And the chemical potential coincide with the energy of the Fermi level.
Ultra-relativistics case
Total energy

11. For ultra relativistic , P ρ4/3 For non relativistic, P ρ5/3
And the pressure
The equation of state is less stiff
For ultra relativistic , P ρ4/3
For non relativistic, P ρ5/3
An equation of state of the form P ρГ is said to be polytropic.

12. Intermediate case
Where the dimensionless paremeter x,
The energy density
with
And the pressure,

13. Finally an explicit expresion for the pressure is

14. Toy model of white dwarf stars
We assume a uniform spherically symmetric mass distribution.
M and R are the mass and the radius of the star.
For such spherically symmetric star, the gravitational energy released
during the process of building the star:

15. Finally , the gravitational energy,
Therefore, without a source of gravitational support, a star with a fixed mass
M would minimize its energy by collapasing into an object of zero radius.
We should consider the energy of the electrons that are fully ionized from the
atoms and provide the pressure and no mass. In fact the electronic
contribution to the mass is very small, as the ratio of electron to nucleon
mass is approximately 1:2000.
The total energy of a degenerate electron gas:

16. The total energy is the sum of the gravitational energy and the energy
The total energy density, E/V :
where
We also need the pressure,
and the derivative of the pressure
notice that
The total energy is the sum of the gravitational energy and the energy
of the degerenate relativistic electron gas

17. For a given M we can minimize the energy respect to R
For a given M we can minimize the energy respect to R. To do that, we need
to know which atoms (nuclei) are present in order to calcule Ne and xF
Given M and knowing the type of atoms  Ne  xF minimization of E(M,R)
To perform this process we must address an enormous range of scales! in the white dwarf stars, the pressure generated by the degenerate electron gas
must support stars with masses comparable to the sun
60 orders of magnitud  convenient to SCALE the equations
HOW TO SCALE THE EQUATIONS?

18. quantifies the importance of the relativistic effects , if xF 1 
We measure the energies in terms of the total rest mass of the electrons
Dimensionless and intensive, i.e., independent of the size system.
In addition , the scaled Fermi momentum ,
quantifies the importance of the relativistic effects , if xF 1 
corrections are negligible. For white dwarf stars the most interesting
physics occurs at xF 1. We also scale the gravitational energy,
Where we have assumed that the mass of the star, M= A mn , may be written In terms of its baryon number A and the nucleon mass mn . Ye  Z/A represents
the electron per baryon fraction of the star , Ye=1/2 for 4He and 12C.

19. The final step for the scaling is to introduce M0 and R0 that are unkown and
will be chosen as the natural mass and length scales that will simplify the
equations
In terms of these natural mass and length scales, the gravitational energy
of the system takes the following form:
In the Fermi gas contribution to the energy, the dependence on
is through xF, .

20. Now we choose M0 and R0 by imposing
and finally we get
Dimensionless strength of the gravitational
coupling between two nucleons.

21. By scaling the equations we have learn that white dwarf stars have masses comparable to that of the Sun but typical radii of only km, recall that the radius of the Sun is around km. Further we observe that while R0 scales with the inverse electron mass, the mass M0 is independent of it. This suggests that neutron stars, where the neutrons provide all the pressure and all the mass, will also have masses comparable to that of the Sun but radii of only 10 km.
Now that the expresions have been simplified we go back to the
minimization of the energy
The graviational energy looks very simple:
= 1

22. At the end we want to minimize respect to xF therefore we should express
the energy in terms of xF 
Therefore,
The minization condition will define the mass-radius relation of the star:

23. It is instructive to analyze the non-relativistic and the ultra-relativistic limits
of the function fF(xF), which are given by
Let’s work first the non relativistic limit 
Has always solution!!! For any mass, we can always find the radius that
Minimizes the energy!

24. What happens in the ultrarelativistic case?
In this case, the derivative of the Fermi energy is a constant equal
to 3/4, independent of the mass of the star.
Defines the maximum value of
M for which we can find the
minimization radius when we use
the exact relation for the energy
of the Fermi sea.
Therefore, the value
Prediction of a maximum mass!
This upper limit, known as the Chandrasekhar mass, is predicted in this simple model to be equal to:

25. M= 1 MS
Relativistic case
RR=4968 Km.
In the NR case, there
Is always solution,
(R/R0)=3/5 (M/M0)1/3,
The correct dispersion
Relation predicts the
Existence of an upper
limit for the mass.
The toy model predicts
MCh=(3/4)3/2M0= 1.72 MS
Remember that for Ye=1/2, M0=2.650 MS, therefore for M=MS, (M/M0)2/3=0.522,
the solution for the equilibrium density of the star, that will define the radius
of the start for a given mass, is obtained from the intersection of the red (
relativistic) and black (non-relativistic) with the blue line. In the non relativistic
case, the solution may be computed analytically: xF0= (M/M0)2/35/3. However, the
non-relativistic prediction overestimates the Fermi pressure and consequently
the radius of the star, RNR=7162 Km.

26. Relation R(104Km) and Mass (MS). Smaller mass  Larger radius
The maximum mass corresponds to a R=0 (collapse!). The non-relativistic
description never collapses, has not a maximum mass.

27.  Which is directly related to the zero temperature incompressibility.
Numerical analysis of the exact treatment
We assume that the EoS is that of a simple degenerate Fermi gas at zero
temperature. In this case the EoS is known analytically and it is convenient to
Incorporate it directly into the differential equation. 
For the relativistic free Fermi gas,
Which is directly related to the zero temperature incompressibility.

28. The mass density ρ(r) is also easily expressed in terms of xF,
Now we introduce the scaled variables
We get
Using the previous definitions of M0 and R 0 the two long expressions
in brackets reduce to the simple numerical values of 5/3 and 1/3
respectively.

29. Where the two functions on the right-hand side of the equations
Finally, the equations of hydrostatic equilibrium describing the structure
of white dwarf stars are given by the following expressions:
Where the two functions on the right-hand side of the equations
(f and g) are given by:
These coupled set of first-order differential equations may be solved
using standard numerical techniques.

30. For this low-order approximation of the derivative, we should use
Using the definition of the derivative of a funcion F(x), one can approximate
the value of the funcion at a point x+h close to x,
For this low-order approximation of the derivative, we should use
very small value of h in order to ensure numerical accuracy.
Doing this procedure for the system of equations we get:
We obtain recursive relations that allow to build the solution from point
To point from the center to the surface. Notice also that the right hand side
of the two recursive relations are completly known at the origin

31.  And continue the iterative procedure until xF(r) becomes negative!
Then the values of the functions f and g at
are:
Then we can calculate the values of
And continue the iterative procedure until xF(r) becomes negative!
A safe Δr that produces reasonable results is Δr=0.0001

32. MCh=1.44 MS
Radius (km) mass (M/MS) relation. Notice the existence of maximum mass
that can be supported by the pressure of the degenerate electron gas.

33. MCh=1.44 MS
Mass (M/MS) contained in a sphere of radius R (km), for three different total
masses, smaller than the limit mass. Larger the mass  smaller the radius

34. Density in kg/m3 as a function of the radial coordinate, for three total
masses , (M= MS, R= 9637 km), (M=1.009 MS, R= 5510 km) and
(M=1.421 MS, R= 1145 km). When the total mass increases, the radius
decreases and the central density increases.