PH291 Stellar Astrophysics Problem Sheet 3

 

Question 1

1)      Describe the series of nuclear reactions which make up the p-p chain. State the form of the dependence of the rate of this reaction as a function of temperature and compare with that for the CNO cycle.

 

p-p II

 

p-p III

 
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


For the p-p chain, the temperature dependence is T4 whereas for the CNO cycle it is T16.

 

The figure below shows the theoretical energy spectrum of the neutrinos emitted by the sun. The spectra of the neutrinos produced in the reactions above can be identified in the figure. In most cases, the neutrino energy distribution is a continuum, because the energy available  for the reaction, obtained from the conversion of rest mass into energy, is shared between three particles. This is the case of the  reaction, for example. In other cases, the neutrino spectrum corresponds to a sharp peak.

 

Solar neutrino spectrum

 

 

 

 

2)      Explain how the phenomenon of quantum tunnelling allows the stars to shine

 

As shown in the last problem sheet (see also below), the probability for two protons to overcome the Coulomb barrier and be close enough to produce a fusion reaction is, according to classical mechanics, vanishingly small (probably not even calculable using a pocket calculator). The only way for the fusion reactions to occur, and thus for stars to shine, is through the non-vanishing probability for p-p interaction obtained from the tunnel effect.


Classically, considering a pair of protons or nuclei with thermal velocity, given by:

 

The shape of this distribution is:

 

Looking at the probability distribution, the

maximum probability can be found by

differentiating the expression above and setting

the derivative to zero. For a temperature of 107 K,

and making mI=mp, this gives:

p=4.8x10-22 kg m/s2 i.e. classically: v=2.9x105 m/s  

 


The distance between two classical charged particles with charges Zi and Zj can be found by making the kinetic energy equal to the electrical potential energy. It is is given by:

where mg should be either the reduced mass of the two particles or, if one is much heavier than the other, the mass of the lightest particle. Assuming we’re discussing p-p collisions, i.e. mg=mpmp/(mp+mp)=mp/2, and assuming the most probable thermal velocity at 107 K, v=2.9x105 m/s, this means about: d=6x10-12 m or 6x103 fm, which is around 103 larger than the size of the proton, i.e. 103 larger than a distance at which a nuclear reaction would occur if tunnelling didn’t exist. Tunnelling is illustrated by the figures below. The figure on the right schematically shows the wavefunction for a “projectile” proton (in red) overlapped with the potential energy of a “target” proton (in black).

 

Text Box: energy                                                             

Note the horizontal scale is distance and there are two vertical scales (one is the potential energy and the other is the magnitude of the wavefunction).

 

Question 2

1)      What is the Eddington limit and what are the consequences for a star which exceeds this limit?

2)      In a star, thermal equilibrium means that the rate of nuclear energy production is equal to the star’s luminosity:

Assume now that a small effect induces a perturbation on the star’s state of equilibrium, so that the rate of energy produced in nuclear burning exceeds the rate of energy dissipated in the form of luminosity by a small amount:

Discuss how the fact that the star has a negative heat capacity implies that the star is in a state of thermal secular stability.

 

The rate of change of the star’s energy, E, is given by the difference between the rate of nuclear energy production and the rate of emission of radiation:

If Lnuc-L>0, the total energy will increase and, since E is negative, its absolute value will become smaller.

From the virial theorem, for an ideal gas with negligible radiation pressure, we have:

 

The negative heat capacity means that the average temperature of the star will actually decrease