Stability of a White Dwarf Against Gravitational Collapse

In summary, the conversation discusses the stability of a white dwarf against gravitational collapse. It is assumed that the star is made up of an equal number of electrons and protons, and also an equal number of neutrons and protons. The conversation then goes on to discuss the kinetic and potential energy of the star, and how they relate to its radius. The critical value of N for the collapse of the star is also discussed.
  • #1
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Homework Statement


I have a problem with question (b), in what follows there's a quote of the problem and its solution. (again from Bergersen's and Plischke's textbook).

Stability of a White Dwarf Against Gravitational Collapse.
It is energetically favorable for a body held together by gravita-
tional forces to be as compact as possible. We take the star to be made
up of an approximately equal number ##N## of electrons and protons, since
otherwise the Coulomb repulsion would overcome the gravitational in-
teraction. Somewhat arbitrarily we also assume that there is an equal
number of neutrons and protons. On Earth the gravitational pressure
is not large enough to overcome the repulsive forces between atoms and
molecules at short distance. Inside the sun, matter does not exist in the
form of atoms and molecules, but since it is still burning there is radi-
ation pressure which keeps it from collapsing. Let us consider a burnt
out star such as a white dwarf. Assume that the temperature of the
star is low enough compared to the electron Fermi temperature that the
electrons can be approximated by a ##T = 0## Fermi gas. Because of their
large mass the kinetic energy of the protons and neutrons will be small
compared to that of the electrons.
(a) Show that, if the electron gas is non-relativistic, the electron mass
is ##m_e## , and the radius of the star is ##R##, the electron kinetic energy of
the star can be written:
$$E_{kin} = \frac{3\hbar^2}{10m_e}\bigg( \frac{9\pi}{4} \bigg)^{2/3} \frac{N^{5/3}}{R^2}$$

(b)
The gravitational potential energy is dominated by the neutrons and
protons. Let ##m_N## be the nucleon mass. Assume the mass density
is approximately constant inside the star. Show that, if there is an
equal number of protons and neutrons, the potential energy will be
given by:

$$E_{pot}=-\frac{12}{5}m_N^2G N^2/R$$
where ##G## is the gravitational constant.

(c)
Find the radius for which the potential energy plus kinetic energy
is a minimum for a white dwarf with the same mass as the sun
(##1.99 x 10^{30} kg##), in units of the radius of the sun (##6.96 x 10^8 m##).

(d)
If the density is too large the Fermi velocity of the electrons becomes
comparable to the velocity of light and we should use the relativistic
formula,
$$(2.107)\ \ \ \ \ \epsilon(p) = \sqrt{m_e^2c^4+p^2c^2}-m_e c^2$$
for the relationship between energy and momentum.
It's easy to see, that in the ultra relativistic limit (##\epsilon \approx cp##), the electron kinetic energy will be proportional to ##N^{4/3}/R##, i.e. the ##R## dependence is the same as for the potential energy. Since for large ##N## we have ##N^2\gg N^{4/3}##
we find that if the mass of the star is large enough the
potential energy will dominate. The star will then collapse. Show that the critical value of ##N## for this to happen is:
$$ N_{crit} = \bigg(\frac{5\hbar c}{36\pi m_N^2 G}\bigg)^{3/2} \bigg( 9\pi /4 \bigg)^2$$

Homework Equations

The Attempt at a Solution


[/B]
My problem is with the last expression in (d) I get 54 instead of 36.

To get the ##N_{crit}## I equated to 0 the next expression:

$$\frac{d}{dN} \bigg[ -12/5 m_N^2 GN^2/R+\frac{N^{4/3}\hbar c}{3\pi R}\bigg(9\pi / 4\bigg)^{4/3} \bigg]$$

Is this wrong? if so, then why?

Thanks.
 
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  • #2
Why differentiate with respect to N?

In order for the star to continue to collapse, what must happen to the total energy as R continues to decrease?
 
  • #3
The total energy should be equal to zero in this case.
 
  • #4
Thanks
 
Last edited:

1. What is a white dwarf?

A white dwarf is a type of stellar remnant, formed when a star with a mass between 0.07 and 10 times that of the Sun runs out of nuclear fuel and collapses. It is extremely dense, with a mass comparable to that of the Sun but a size similar to that of Earth.

2. How does a white dwarf remain stable against gravitational collapse?

A white dwarf remains stable against gravitational collapse due to two opposing forces: the inward pull of gravity and the outward pressure from electron degeneracy. Electron degeneracy is a quantum mechanical effect that occurs when electrons are packed too closely together, creating a repulsive force that counteracts gravity.

3. Can a white dwarf collapse into a black hole?

No, a white dwarf does not have enough mass to collapse into a black hole. The maximum mass a white dwarf can have, known as the Chandrasekhar limit, is about 1.4 times the mass of the Sun. Beyond this limit, the electron degeneracy pressure is not strong enough to counteract the force of gravity, and the star will collapse into a neutron star or black hole.

4. How does the stability of a white dwarf change as it cools?

As a white dwarf cools, its internal pressure decreases and it becomes less stable against gravitational collapse. This is because the electrons lose energy and slow down, causing the electron degeneracy pressure to decrease. Eventually, the internal pressure will no longer be strong enough to counteract gravity, and the white dwarf may undergo a thermonuclear explosion or collapse into a neutron star.

5. What is the role of mass in the stability of a white dwarf against gravitational collapse?

The mass of a white dwarf is directly related to its stability against gravitational collapse. As the mass increases, so does the pressure from electron degeneracy, making the star more stable. However, once the Chandrasekhar limit is reached, the star cannot support itself against gravity and will collapse. Therefore, the mass of a white dwarf is a crucial factor in determining its stability against gravitational collapse.

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