Virializing the Night Away #4

By David Vartanyan
Coauthors: Mee and John

4) In this problem we will show that the sun would rapidly collapse under its own weight if there were no pressure countering the collapse. We will introduce degeneracy pressure as a dominant pressure countering collapse.

a. The rate at which the sun generates energy is defined as the sun's luminosity L_s, 3.9*10^33 erg/s.

b. Assume the sun is powered entirely by fusion with a 0.7% efficiency, and that fusion occurs only in the sun's core which is 10% of the solar mass.
Then, defining M_s as the solar mass we have the total energy available to the sun:



=1.2*10^51 ergs.

The sun will burn out at time t = E/L_s, the sun's total energy divided by the luminosity (rate of energy consumption), so:

t= 9.9 billion years

The more massive a star, the more energy it has available to burn through and the longer it typically stays on the main sequence.

We initially tried to find the total energy by summing all the individual reactions that occurred. We considered the total number of fusion reactions in the core multiplied by the energy released in each reaction. For a core with n hydrogen atoms available for fusion, there will be n fusion reactions

4 ¹H + 2 e -->; ⁴He + 2 neutrinos + 6 photons (Fusion Steps)


which each release 26 MeV. The sun's core is .1 M_s. Dividing the core's mass by the mass of a hydrogen atom (assuming it consists of only hydrogen before fusion has begun), we have

n= 1.2*10^56, so total energy is n*26MeV = 5*10^51 ergs

c. We solve this similarly to the problem of an elliptical orbit with eccentricity of one rather than integrating over the sun's surface to calculate collapse. Define the sun so that it is free-falling towards the core. Label the core as one focus of the ellipse and a point on the sun's surface as the other focus. Assuming the sun is a uniform homogenous sphere, all points on the surface should have the same 'period' of collapse.

From Kepler's third law:

where R_s is the sun's core radius (.25* sun's radius), M_c is the mass of the sun's core (M_c=.1*M_s),
The time for collapse for an eccentricity 1 orbit is half the period.
This comes to about 31 minutes (Nathan's group got 24 minutes using a slightly different radius). That is quite fast.

d. After some initial collapse, degeneracy pressure becomes significant in countering core collapse. Degeneracy pressure is an extension of the Pauli exclusion principle: 2 electrons won't occupy same quantum state so after some collapse this degeneracy pressure takes effect. When particle spacing is on the order of the de Broglie wavelength

degeneracy pressure becomes significant. Assuming the particles velocities are roughly the same, electrons have a much smaller mass so they have a larger de Broglie wavenelength.
More rigorously, kinetic energy scales as k*T where T is the temperature at the sun's core. So mv^2~T and v~Sqrt(T/m).
A 2-D image of the spacing of degeneracy pressure:

For an electron in the sun's core:


For proton:
 
 At the same temperature, electron wavelengths near the de Broglie limit scale with the inverse square root of electron mass, and proton wavelengths with the inverse square root of proton mass. Since an electron is much less massive than a proton, electrons have a larger de Broglie wavelength.
Therefore electrons reach their critical density first. For a system (like our sun) dominated by hydrogen such that electrons and protons are roughly equal, electron degeneracy pressure dominates over proton degeneracy pressure.

In the case where the star's density is too high for degeneracy to support, we have a black hole!

e. Degeneracy becomes important at an electron number density ~, with mass density scaling as where m is the mass of a hydrogen atom.
 In terms of fundamental units, mass density becomes important at the de Broglie wavelength, or when mass density ~


Since \lambda ~ h/p ~h/(m_e*v) and 1/2 m_e*v^2 =3/2 k*T for n = 1 (a single electron in a core at critical density), so v~ Sqrt(3k*T/m_e).


mass density ~ m_H/(\lambda^3)~m_H*p^3/h^3~m_h*(m_e)^3*v/h^3. Substituting for velocity gives us the mass density equation above.


According to Wolfram-Alpha, the density of the sun's core is around 150g/cm^3 (approximate by density = mass/volume = .1 * solar mass/(4*pi*.25 solar radius^3).
Find \lambda at the sun's core temperature T_c = 1.5*10^7 K.

This comes out to be around 80 g/ cm^3, half of the actual value. Before we claim content with an OoM estimate, consider sources of inaccuracy. The sun's density isn't uniform, we likely have some other unknown quantum effects at this scale, the sun's temperature isn't uniform, we have some proton degeneracy contribution (but very, very little), and critical density itself is an OoM equation.

Now we can be content.

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Thanks to John and Mee for their cooperation, and to Professor Johnson for pointing out that the reason we were getting critical density to be higher than is possible given all the particles in the universe was that we were using the sun's mass to find mass density rather than the mass of a hydrogen atom.