17 Stars notes 2017/10/4 - Wed - Nuclear cross sections,
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1 17 STARS NOTES 2017/10/4 - WED - NUCLEAR CROSS SECTIONS, S-FACTOR 1 17 Stars notes 2017/10/4 - Wed - Nuclear cross sections, S-factor 17.1 Coulomb Tunneling continued Last time we obtained the probability of the proton being inside the barrier 2mr c 2 1/2 P 0/P = exp παz 1 Z 2 E Called the Gamov tunnelling factor. There is an energy scale that we can write as the Gamov energy E G = παz 1 Z 2 2 2m r c 2 so that P 0 exp E G /E 1/2 Looking at some numbers: for p+p we have E G = 494 kev and for p+ 12 C have E G = 33 MeV. This gives for p + p, 1/2 494 P 0 = exp = this is how likely you are to see the nuclear force, but even if you do because we want a weak interaction it is suppresed by another large factor for E = 1 2 m rv 2 v = relative velocity can rewrite P 0 = exp 2π e2 Z 1 Z 2 v There is in some sense a maximum coulomb energy E coul = Z 1Z 2 e 2 λ = Z 1Z 2 e 2 m r v h then P 0 scales like exp E coulλ 1 m 2 rv 2 Now want to convolute this probability with the thermal distribution to see what energy particles are important.
2 17 STARS NOTES 2017/10/4 - WED - NUCLEAR CROSS SECTIONS, S-FACTOR Nuclear cross-sections We ve only calculated the overlap. There s reason to expect a resonance which helps the cross section. For now consider non-resonant reactions. The only length scale in the problem is λ = h p cm at kev this is much greater than R nuc = 1.3 fm A 1/3. The reaction physics depends only on the COM energy E COM = 1 2 m r v 1 v 2 2 For s-wave interactions no angular momentum in COM the cross section, which is what one measures with an accelerator, looks like σe = 4πλ 2 dimensionless stuff exp E G /E 1/2 the way this is defined consistent with textbook 4πλ 2 = 4π 1 k = 2π 2 kev = 2000 barns 2 m r E c where 1 barn = cm 2 and E com = 2 k 2 /2m. Then the conventional way to write the cross section is E com σe = 1 E SEe E G/E 1/2 Where SE is called the S-factor, and is what is usually obtained in the lab and extrapolated more on this next time. If the dimensionless stuff was = 1 then SE = 2000 barns kev This is a VERY large cross section, rarely are they this big. Now the experimentalists must measure the dimensionless stuff for the reaction we want, that is get SE. To do this experimentalists must measure σe and then extrapolate to low energies. The cross section σe is what is appropriate for an experiment. i.e. a beam of welldetermined energy pointed at a target. By contrast, stars will have a broad thermal energy distribution. An example of a measured cross-section for p+ 12 C from Clayton chapter 4:
3 17 STARS NOTES 2017/10/4 - WED - NUCLEAR CROSS SECTIONS, S-FACTOR 3 Another example from 12 C+ 16 O, which is very important for thermonuclear supernovae Martinez-Rodriguez et al. 2017ApJ M is the following cross section from Jiang et al
4 17 STARS NOTES 2017/10/4 - WED - NUCLEAR CROSS SECTIONS, S-FACTOR 4
5 17 STARS NOTES 2017/10/4 - WED - NUCLEAR CROSS SECTIONS, S-FACTOR 5 And from pre-release data: We need to figure out what energy range is important for stars thermal Nuclear reactions for star Need to go from cross section σe, which depends on incoming particle energy, to rate ɛt, which depends on gas temperature. Now we want to go about convolving with the thermal distribution of the protons. Usually we write down time between collisions t 1 σnv or a rate r = σnv We have σe, dependent on energy of incoming particle, but we really want the thermal average σv = 0 v r σv r P v r dv r where P v r dv r = probability that two particles have a relative velocity between v r and v r + dv r. The energy generation rate will then be ɛ nuc = n 1n 2 σv E nuc ρ where E nuc is the energy released per reaction. For an ideal gas obeying maxwell-boltzmann: P vd 3 v = [ mr ] 3/2 exp m rvr 2 2πkT 2kT 0 d 3 v now just need to put this in and do the integral. The integral becomes [ mr ] 3/2 σv = exp m rvr 2 4πv 2 2πkT 2kT rdv r v r σv r
6 17 STARS NOTES 2017/10/4 - WED - NUCLEAR CROSS SECTIONS, S-FACTOR 6 convert integral to E c, center of mass energy, σv = 1/2 1 8 SE exp kt 3/2 πm r 0 E kt EG E so what does this look like? at low energyies the Gamov term gets you 1/2 de exp E/kT exp E G /E 1/2 product Gamov Peak Peaked at characteristic energy E 0. Will compute next time.
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