Subido por Marc Asenjo


Master Universitario en Física Nuclear UAM, UCM, UB, UGR,
USAL y la US
Marc Asenjo Papiol
UB Juny 2020
White Dwarfs
Describe the main features of a white dwarf and, using the program discussed in class,
determine numerically the relationship between the mass and the radius of the star. To this
end, you have to solve the hydrostatic equilibrium equations by varying the central density
and extract the total mass and radius for each central density.
Also called a degenerate dwarf, a white dwarf is a stellar core remnant composed mostly of
electron in degenerated state. Its average mass is comparable with that of the Sun but its volume
is similar to that of the Earth, reaching extremely high densities and becoming one of the most
compact objects yet discovered in the Universe.
The white dwarf luminosity comes entirely from the emission of thermal energy that is stored
inside the core so no nuclear fusion is given inside and the only supporting force against the
gravitational collapse is that corresponding to the electrons degeneracy (Fermi gas degeneracy
pressure). Related to these extremely dense environment, there is the so called Chandrasekhar’
mass limit that predicts a sufficient mass condition to withstand the gravitational collapse. Once
this limit is surpassed and the mass exceed that of the Chandrasekhar’s, which consists of 1.40
Solar masses approximately, a white dwarf can explode like a supernovae type-Ia or become
even denser and turn itself into a neutron star.
White dwarf is the final evolutionary state of stars whose mass is not high enough to become a
neutron star. After the burning of hydrogen through the Main Sequence, a typical low or
medium mass star will expand becoming a red giant. In the aftermath of that red giant state if
the mass is insufficient to start the burning of Carbon, which requires temperatures as high as
10^9 K, a planetary nebula is going to be formed and an inner inert mass of carbon and oxygen
will start to be formed in its center.
A) Make the calculations for white dwarfs with cores composed of 4-He(Ye=1/2). Plot the
results for the mass-radius relation in a figure and extract the corresponding
Chandrasekhar mass. In the figure express the mass of the star in solar masses and the
radius in units of 10^4 km. Note that the electrons are to be described using relativistic
Using the program white_dwarf.f provided in the campus, it was demanded to compute the
relation between Radius and Mass for different white dwarfs cores. The program needs an input
file (.dat) that provides information about the total number of different stellar cores in this case
2000, and the different central densities from 0 to a final central density up to 200.
The program solves the different hydrostatical equations corresponding to the differential mass
and differential radius that are characteristic of the equilibrium between gravitational collapse
and the radiative pressure.
The Chandrasekhar mass limit resulted from our calculus (showed in the plot) is approximately
1.43 M/Mo where Mo is the mass of the Sun.
B) Determine the density profile p(r) of the star in Kg/m^3 and make a comparative graph
of the density profiles for the three representative central densities. Express the radial
distance in Km. Compare these density profiles with those you would get using nonrelativistic kinematics for the electrons.
Now we are interested in getting the density profiles of three different stars for two different
cases: relativistic and non-relativistic kinematics. Three new central densities have been set. We
are considering a star with an Xf = 3, another with an Xf = 9 and the last one with an Xf = 30.
The contribution of the electrons is a determinant factor for the pressure to compensate the
gravitational force.
It can easily be seen that the more lesser a density profile is, the bigger the average radius of
the star is.
Stablishing differences between the relativistic and non-relativistic kinematics, we can see that
for the same central density, in the non-relativistic case the radius of the same white dwarf is
bigger in magnitude than that of the relativistic case. This is due to the energetic difference when
adding the contribution of the electrons. The higher it becomes the energy, the higher the
pressure to compensate the gravitational force is.
C) Would the Chandrasekhar mass change if the white dwarf were composed of 12C core?
And if the star were composed of 56Fe core? Justify your answer.
As shown in the classes, the Chandrasekhar mass limit depends principally on de Ye (electronbaryon density fraction). In the case of Helium and Carbon, the Ye parameter is the same so the
Chandrasekhar mass does not change. So for our He-4 white dwarf the mass limit is the same.
Conversely in the case of Fe-56, the Ye parameter is different so it is the Chandrasekhar mass.
Skyrme Energy Functional
Consider the following nuclear energy density:
We first want to determine the coefficients t0, x0, t3, в and W0. For this purpose, we will start by
using data on symmetric nuclear matter.
A) Let us set ϒ=1/3. In symmetric nuclear matter we have ρn= ρp =p/2 and the kinetic
energy density and the kinetic energy density is given by (non-relativistic Fermi gas
model), with q=n,p for neutrons and protons. Recall that in nuclear matter
𝛁 ·ρn= 𝛁 ·ρp=0.
Determine the t0 and t3 parameters using the saturation condition
and the following value for the energy per nucleon in symmetric nuclear matter.
With ρ0 =0.16 fm-3 for the saturation density. Use h2/2m = 20.7335 MeVfm2 in the
numerical calculations.
First we are going to consider that in nuclear matter:
∇ ·ρn= ∇ ·ρp=0.
We set the right term of the functional to 0, obtaining the following:
Substituting for the initial conditions given in the enunciate, saturation and the symmetric
nuclear matter, and taking into account the form of the kinetic energy density for the neutrons
and protons, which are equal, we get the following expression:
Once we have that expression, we divide it by ρ and substitute ϒ=1/3 getting (1)
Then we derivate with respect to ρ and for ρ0, or saturation density, we have (2):
After obtaining expressions (1) and (2), we stablish the following equalities:
(1) = -16.0 MeV
(2) = 0 (corresponding to the minimum of the energy density Skyrme functional
for the value)
From these equations we are able to compute the values of the parameters t3 and t0
t3 = 12914.7 MeVfm4
t0 = -1803.7 MeVfm3
B) Obtain the expression for the incompressibility of symmetric nuclear matter as a
function of t3 and g, knowing that is defined as
Compute the value of K(ρ0)
From expression (1), we derivate it two times with respect to ρ
Finally, after multiplying by 9ρ2:
After the substitution of the pertinent values of the parameters:
K(ρ0) = 236.24 MeV
To calibrate the remaining parameters x0, в and W0, we may adopt the following
strategy. In symmetric and spin-saturated light nuclei, such as 16O and 40Ca, the terms
with x0, в and W0 contribute negligibly to the total energy, because x0 is related to
symmetry energy and W0 is related to the spin-orbit contribution, which is small in
these nuclei. In the following, you have to use the code for finite nuclei discussed in
C) Assuming provisionally x0 and W0 = 0, vary the в parameter in order to find a value
that reasonably reproduces de experimental binding energies of 16O (-127,62 MeV)
and 40Ca (-342,05 MeV). A possible starting value for в is given in the data file of the
program. In your written report provide в with units (recall that the units of the energy
density H are Mevfm-3)
Given the binding energy of the 16O, we must compute the energy per nucleon, dividing by the
mass number of the oxygen. In that case E/A = (-127.62 / 16) MeV. So the binding energy per
nucleon E/A = -7.976 MeV
The в parameter we fix for the model with 16O is 55.18
In the case of the 40Ca the energy per nucleon E/A = (-342,05/40) MeV. E/A= -8.551 MeV
Our final value for the в parameter in the case of 40Ca is 56.1
We are going to compute the average of the two parameters to have the same measuring
associated error. Our final в parameter value is 55.64
D) With the previous в value, fix the W0 parameter by reproducing the experimental
splitting of approximately 6.15 MeV for the 1p levels of 16O, see the paper by
Vautherin and Brink. A possible starting value for Wo is given in the data file of the
program. In your written report provide W0 with units.
Our final W0 parameter value is 123
E) Finally, fix x0 as to reproduce the experimental binding energy of 208Pb (-1636.43 MeV).
In the case of the 208Pb the energy per nucleon E/A = (-1636.43,05/208) MeV. E/A= -7.867 MeV
Our final x0 parameter value is 0.160
F) Next, iterate steps (c) – (e) in order to improve your fit, by starting from the в, x0 and
W0 values determined in the first stage.
With the final parametrization, compute the binding energies of the 48-Ca, 90-Zr and
Sn nuclei. Compare your results with the experimental data (-415.99 MeV, -783.89
MeV and -1102.85 MeV, respectively).
The results are really satisfying. The error associated to some of these measures is really low
B-stabble matter and neutron stars
A) Using the Skyrme energy density described in the previous exercise, compute the
equation of state of asymmetric nuclear matter, i.e, the pressure P(ρ,δ) as a
function of the density p and asymmetry δ of a uniform infinite system of neutrons
and protons. It should be taken into account that in asymmetric nuclear matter
ρn=ρ(1+δ)/2 and ρp =ρ(1-δ)/2 where ρ= ρn + ρp and δ= ( ρn - ρp)/ ρ. Plot your
equation of state in a density range between ρ =0.1 fm-3 and ρ =0.5 fm-3 for
asymmetries δ=0, 0.3 and 0.6.
From the previous exercise, we use the Skyrme functional and make the corresponding
substitutions provided in the enunciate A)
We divide the former expression by ρ:
And then we introduce the energy density in the equation of state when computing the pressure
associated to that system:
The EoS obtained need to be evaluated for different asymmetries δ=0, 0.3 and 0.6.
The corresponding plot in the density range stated in A)
P(ρ,δ) vs ρ
B) The outer region of the core of a neutron star is composed of a uniform fluid of
neutrons, protons and electrons in a single phase. Employing the energy density
for asymmetric nuclear matter of the previous point and adding a contribution
from the electrons (they can be assumed to be ultra-relativistic), derive the
equation of state P(ρ,δ) of beta stable matter. Note that for each value of the
nuclear density ρ you need to compute the asymmetry δ such that matter satisfies
the condition of beta equilibrium. Include P(ρ,δ) both the nuclear contribution and
the electron contribution. Plot P(ρ,δ) in a density range between ρ =0.1 fm-3 and ρ
=0.5 fm-3. Compare with the results obtained in the previous point.
Assuming a ultra-relativistic electrons contribution to the total energy of the system we use the
energy density of these electrons to compute the chemical potential related to the beta stability
equilibrium (electron capture and beta decay), Mainly the system is composed by neutrons,
protons and electrons because neutrinos’ mean free path is on average larger than the radius of
the neutron star, so they escape out of the system.
Let us compute the pressure associated to these ultra-relativistic electrons. From the equation
of energy purely related to the Fermi momentum KFe:
We also used the relation between ρe and ρp, and then we substitute the expression in terms of
We add the contribution of the electrons to the total pressure of the system
Substituting for h/2m, π and the different parameters of the Skyrme functional (t0, t3)
Now we have to stablish the b-stable matter condition in the EoS. Try to find a relation between
ρ and δ. We must use the chemical potentials so first we will need the different hamiltonians
(ultra-relativistic electrons and the Skyrme functional one)
Given the energy density of the electrons, we must derivate with respect to ρe.
We compute the difference between chemical potentials of the neutrons and protons in order
to equate it to the chemical potential of the electrons.
And given the H:
We derivate with respect to the asymmetry parameter δ
And all this results in the в-stable matter condition
The plot corresponding to the pressure vs density with and without the electron contribution is
placed at the end of these sheets after section D) of the problem 3.
C) With the code tov_master.f discussed in class and the equation of state of betastable matter we can compute the neutron star mass-radius relation predicted by
the nuclear interaction that we have fitted in the previous exercise using
terrestrial data. The code tov_master.f reads the equation of state of the crust of
the star from the table crust.dat, which reaches up to a density close to ρ0/2,
where ρ0=0.16 fm-3. You do not need to modify this table. The first line of
EOS_npe.dat must be copied to crust.dat (it is the column line with the title of the
columns). Next, the first column of data in EOS_npe.dat will contain the massenergy density e(ρ) of the beta-stable matter: e(ρ)= HSkyrme + Helectron + mp* ρ,
the second column in the EOS_npe.dat will contain the pressure P(p) of the betastable matter, and the third column will contain ρ/ ρ0.
Run the code tov_master.f and plot the calculated neutron star ;-R relation. What
is the predicted maximum mass? Is it compatible with the astrophysical
observations of the heaviest neutron stars? What is the radius of the star at
maximum mass?
Mass vs Radius for Neutron Stars
As It can easily be observed in the figure above, the maximum mass value of a neutron star
corresponds to a mass of approximately 1.92 solar masses and a radius of 10.21 Km (this value
of the radius is coherent with the spectrographic X-ray observations which determine the
biggest theoretical mass value for a radius comprehended between 9.9 and 11,2 Km.
Our results are satisfying provided that they are compatible with the most massive already found
neutron stars.
D) The compactness parameter of a star is defined as ε=GM/Rc2. It is a measure of
the ratio between gravitational energy (GM2/R) and the mass energy (Mc2) of the
stellar body. Compute the compactness parameter ε for the neutron star of
maximum mass of the previous point, for a white dwarf of M=Mo mass,and for
the Sun. Compare the results.
Taking into account that the Gravitational Universal Constant G = 6.67 x10-11 m3/Kg·s2 , that the
solar mass Mo = 1.989 x1030 Kg and that c = 3 x108 m/s
White Dwarf
Neutron Star
The results are congruent with the theory. The highest value of the three computed belongs to
the Neutron Star. The second object with highest density in the entire Universe.