A relatively quick way to calculate this is to use simple approximation for collisions of N particles in a gas.
In those terms, the collision rate, in terms of particles per second is:
N = 10^{26}
v = average velocity of proton = 3/2kT; for T = 10^{7}, v = 3 x 10^{7} cm/sec.
A = cross sectional area for collisions. Just take this to be radius^{2} for a proton which is
A = 10^{-26} cm^{2}
And so we can recover the estimate of 10^{7} easily.
Okay, no one part of the story we haven't told you yet is the following. Although 3/2kT specifies the average kinetic energy of a proton, there clearly is some distribution of energies around that value.
The distribution of particle speeds in a gas follows what is called a Maxwell Boltzman distribution. That distribution is like a Gaussian but has a longer tail towards high particle speeds. Graphically, for the case of stars, that distribution looks as follows:
The equation behind that distribution is of the form:
e^{-v2/T }
with a whole bunch of constants out front.
So, given this distribution can one avoid quantum mechanics? That is, what is the probability that a proton has a kinetic energy, in this distribution, that is larger than the columb barrier (e.g. 700 keV).
The answer (and you can work it out yourself if you like) is no. So we need quantum tunnelling as well and its really the convolution of the tunnelling probablity and the maxwell distribution that yields the most probable range of proton kinetic energies taht will start the fusion cycle.
This is called the gamow peak and its shown below. Your homework assignment asks you to derive this peak energy.
Reaction | Energy Released | Reaction Timescale |
---|---|---|
H + H D + e^{+} + neutrino e^{+} + e^{-} photon |
0.16 Mev 1.02 Mev | 14 billion yrs a few seconds |
H + D ^{3}He + photon | 5.49 Mev | 6 seconds |
^{3}He + ^{3}He ^{4}He +H +H | 12.86 Mev | 10^{6} years |
You should convince yourself, like we did in class, that the origin of antimatter in the first step (e.g. the positron) must occur to satisfy the rules of conservation of baryon #, lepton #, and charge.
The total energy released is
2(0.16 + 1.02 + 5.49) + 12.86 = 26.20 Mev
The mass difference between 4 protons and 1 ^{4}He nucleus is 0.0287 amu which yields 26.73 MeV.
Where is the missing 0.53 Mev its carried off by the neutrinos (0.265 Mev each time).
When the density of H and He become comparable in the core, the PP reaction changes (as described in the previous notes) to favor the production of Be, Li, and B.
The decay of B(oron) into Be(rrylium) emits a neutrino of energy 7.2 Mev, much higher than the neutrinos emitted by the first step of the PP chain.
In principle, if we had good neutrino telescopes (which we don't, for obvious reasons), measuring the ratio of 7.2 Mev neutrinos to 0.26 Mev neutrinos would yield the age of a stellar core.