H and He Burning Prof John Lattanzio Centre for Stellar and Planetary Astrophysics School of Mathematical Sciences

Reaction Rates Lets look at reactions involving two different kinds of particles, i and j (or and ) Suppose there are two of each in a box There are thus four possible reaction pairs

Rate formula In general each i can react with all the other j particles so the number of reaction pairs is n i n j For identical particles this becomes n i n j-1 except that it counts each pair twice; so the number of pairs is really n i (n i -1)/2 in this case Now in general n i is ~ N A ~ –So n i -1 = n i to high accuracy!

New rate formula So for identical particles the number of pairs is n i 2 /2 to very high accuracy But for different particles its n i n j We can combine these into one formula by using the Kronecker delta  ij –  ij = 1 if i = j –  ij = 0 otherwise r ij = n i n j  ij 1 +  ij

H Burning There are two main ways of burning H –Proton-Proton chains (pp) –CNO cycle(s) PP chains are the simplest We deal with those first

The Proton-Proton Reaction The first step is adding two protons 1 H +  2 D + + + + e Deuterium = heavy H Z=1 N=1 Nucleus = d = deuteron Hydrogen Z=1 N=0 Nucleus = p = proton Positron or anti-electronNeutrino (electron) One proton becomes a neutron with a positron for charge conservation and a neutrino for energy conservation The key is really: p  n +  + + e

Next Step: H + D The next step is adding H and D 1 H + 2 D  3 He +  Deuterium = heavy H Z=1 N=1 Nucleus = d = deuteron Hydrogen Z=1 N=0 Nucleus = p = proton Helium-3 Z=2 N=1 Gamma ray (photon) No changes of particles here…2p + 1n on each side

Last Step: 3 He + H? You may think that the next step is to ad a proton to 3 He to get 4 He, the most common isotope of helium But it turns out that the nuclear structure is such that this is highly unlikley! You cannot tell this from pure thought!

Last Step: 3 He + 3 He The last step is adding two 3 He together 3 He +  4 He + pp Helium-3 Z=2 N=1 Returning 2 protons No changes of particles here…4p + 2n on each side Helium-4 Z=2 N=2

Overall PPI Chain 1 H + 1 H  2 D +  + + e 1 H + 2 D  3 He +  3 He + 3 He  4 He +  p

What are the rates? 1 H + 1 H  2 D +  + + e 1 H + 2 D  3 He +  3 He + 3 He  4 He +  p r pp = n p 2  pp /2 r pd =n p n d  pd r 33 = n 3 2  33 /2 r ij = n i n j  ij 1 +  ij

What are the DEs for the abundances? 1 H + 1 H  2 D +  + + e 1 H + 2 D  3 He +  3 He + 3 He  4 He +  p r pp = n p 2  pp /2 r pd =n p n d  pd r 33 = n 3 2  33 /2 d n p dt = production rate – destruction rate = 2 r 33 – 2 r pp – r pd

For Deuterium 1 H + 1 H  2 D +  + + e 1 H + 2 D  3 He +  3 He + 3 He  4 He +  p r pp = n p 2  pp /2 r pd =n p n d  pd r 33 = n 3 2  33 /2 d n d dt = production rate – destruction rate = r pp – r pd

For 3 He 1 H + 1 H  2 D +  + + e 1 H + 2 D  3 He +  3 He + 3 He  4 He +  p r pp = n p 2  pp /2 r pd =n p n d  pd r 33 = n 3 2  33 /2 d n 3 dt = production rate – destruction rate = r pd – 2 r 33

For 4 He 1 H + 1 H  2 D +  + + e 1 H + 2 D  3 He +  3 He + 3 He  4 He +  p r pp = n p 2  pp /2 r pd =n p n d  pd r 33 = n 3 2  33 /2 d n 4 dt = production rate – destruction rate = r 33

Summary of rates and DEs 1 H + 1 H  2 D +  + + e 1 H + 2 D  3 He +  3 He + 3 He  4 He +  p r pp = n p 2  pp /2 r pd =n p n d  pd r 33 = n 3 2  33 /2 d n p dt = 2 r 33 – 2 r pp – r pd d n d dt = r pp – r pd d n 3 dt = r pd – 2 r 33 d n 4 dt = r 33

Deuterium in equilibrium Note that the 2 D equation is self-correcting If n d is low, the first term dominates –n d increases If n d is high the second term dominates –n d decreases d n d dt = r pp – r pd = n p 2  pp /2 – n p n d  pd

Deuterium in equilibrium d n d dt = r pp – r pd = 0 n p 2  pp /2 = n p n d  pd npnp ndnd ( ) eqm =  pp  pd ( )

3 He in equilibrium Note that the 3 He equation is also self-correcting If n 3 is low, the first term dominates –n 3 increases If n 3 is high the second term dominates –n 3 decreases d n 3 dt = r pd – 2r 33 = n p n d  pd – n 3 2  33

PPI in equilibrium Suppose the deuterium and 3 He are in eqm d n p dt = 2 r 33 – 2 r pp – r pd d n d dt = r pp – r pd =0 d n 3 dt = r pd – 2 r 33 =0 d n 4 dt = r 33 = r pd /2 = r pp /2 = - ¼ d n p dt

PPI Energy Generation One can show (but not quickly!) that the energy generation rate for PPI is     X 2 T 6 4  where T 6 = T / 10 6 K

More on PPI: 2 D DE for deuterium

2 D equation Now define 

2 D equation  p (D) ~ secs or mins! Can almost always assume the 2 D is in equilibrium Can remove 2 D from the equations

2 D equilibrium: T = 10 million K H burns into He 4 in ~10 11 years D 2 destroyed in y Or 300 secs Or 5 mins

2 D equilibrium: T = 20 million K D 2 destroyed in y Or 10 secs H burns into He 4 in ~ years

3 He equation Assume 2 D in equilibrium: r pp = r pd ( 3 He/H) e ~ to low T high T

3 He equilibrium: T = 10 million K He 3 increases due to D 2 destruction Reaches eqm later H burns into He 4 in ~10 11 years 3 He eqm ~

3 He equilibrium: T = 20 million K He 3 increases due to D 2 destruction Reaches eqm later H burns into He 4 in ~10 10 years 3 He eqm ~

Exercise

PP Chains at various T As T rises: H burns more quickly D 2 eqm happens earlier He 3 eqm happens earlier He 3 eqm value decreases

PP Chains (I, II and III) PPI Chain But once there is some 4 He… And 7 Be can Capture a proton Or capture an electron Makes PPII and PPIII chains

PP Chains Fate of 3 He determines PPI or PPII and III Fate of 7 Be determines PPII and III

DEs for the PP chains

Equations for PP Chains Some simplifications! – 2 D in equilibrium (seconds) – 7 Be and 7 Li in equilibrium in a few years – few thousand years Further details –This enables some simplifications At low T we get PPI At higher T we get PPII and PPIII

7 Li and 7 Be: T = 10 million K H burns into He 4 in ~10 11 years Li and Be reach eqm values in ~10 6 years

7 Li and 7 Be: T = 20 million K H burns into He 4 in ~10 10 years Li and Be reach eqm values in ~10 4 years

PP Chains PPI PPII PPIII

PP Chains

Chart of the Nuclides: The “big boys/girls” periodic table Z = number of protons N = number of neutrons

PP Chains PPI PPII PPIII

Notation Time to become real nuclear physicists! We have been writing a+X  Y+b From now on we will write X( a,b )Y the heavy species X( a,b )Y lighter particles

Examples 2 D +p  3 He +  becomes 2 D(p,  ) 3 He 3 He + 3 He  4 He +    becomes 3 He( 3 He,  p) 4 He 1 H + 1 H  2 D +  becomes 1 H(p,  ) 2 D 3 He + 4 He  7 Be +  becomes 3 He( ,  ) 7 Be 12 C +p  13 N +  becomes 12 C(p,  ) 13 N 12 C + 4 He  16 O +  becomes 12 C( ,  ) 16 O

Decays Decays are different 8 B  8 Be +   + becomes 8 B(   ) 8 Be NB No comma! 13 N  13 C +   + becomes 13 N(   ) 13 C

CN Cycle At slightly higher temperatures the dominant H burning reactions are the CN cycle 12 C + p  13 N +  13 N  13 C +     13 C + p  14 N +  14 N + p  15 O +  15 O  15 N +     15 N + p  12 C + 

CN Cycle Add up both sides: 12 C + p  13 N +  13 N  13 C +     13 C + p  14 N +  14 N + p  15 O +  15 O  15 N +     15 N + p  12 C +  4p  + energy

CN Cycle Temperature sensitivity? –It starts at about million K –PP starts at 5-10 million K Approximate rate formula     X H X CN T 6 20     X 2 T 6 4 for PP

The CN cycle But sometimes we get: 15 N(p,  ) 16 O

ON cycle Once there is some 16 O present… (or 16 O may be present initially in any case…)

The CNO bi-cycle

At high T, say > 100 million we must add more reactions! The elements that suffer  decay might capture a proton at high T… The CNO bi-cycle

CNO cycles  + decays are very quick t 1/2 ( 13 N)=10 min t 1/2 ( 15 O)=122 secs t 1/2 ( 17 F)= 65 secs Assume instantaneous for 13 N, 15 O and 17 F Accurate after a few minutes! Simplifications…?

Separation: CN and ON cycles Lets ignore branching –Take branching ratio r = 1 –Its really about 0.999, depending on T Thus we get a CN cycle (no O) So go around CN 999 times per O produced Clearly the CN cycle part is ~10 4 times faster! We’ll later deal with the ON part

CN Cycle This is basically the original cycle with the  decays assumed instantaneous

CN Cycle This can be written as an eigenvalue problem 0+ +0

CN Cycle  = 1/( N) is a function of T and  Assume T and  constant: solution is When cycle in equilibrium there is no time-dependence So one of the eigenvalues must be zero! Its eigenvector must contain equilibrium abundances! The other eigenvalues must be negative… … t   these terms must  0 so we are left with the eq m value

CN cycle solution Equilibrium abundances Terms sum to zero…

An example Take T = 25 million K and  X=25 14 N+p is slowest… T > 10 4 y gives eqm Conservation of nuclei

CN Cycle at 20 million K  decay species negligible 12 C down 13 C up 14 N produced

Executive Summary of CN Cycle CN cycle burns H into He After ~ 10 4 years we have equilibrium As a by-product – 12 C/ 13 C ~ 4 –~98% of C and N ends up as 14 N

CN Cycle at various T

ON Cycle We saw that the CN cycle goes around ~10000 times for each nucleus that “leaks” into 16 O So before there is much cycling in the second cycle we may expect that the first has already come to equilibrium Assume CN cycle in equilibrium Look at remaining reactions

ON Cycle Assume CN cycle in equilibrium Assume all C has become 14 N Assume 15 N decays instantly Look at remaining reactions 3’ 5’ and 6’

CN and ON symmetry

CN ON cycle matrices

ON cycle at 20 million K 16 O down 14 N up CN cycle in eqm

ON cycle at 20 million K 17 O produced 18 O destroyed Constant ratio of 17 O/ 16 O ~ O/ 16 O ~ 10 -4

ON Cycle results Again, everything ends up as 14 N – 14 N + p is slow reaction… ON cycle requires higher T than CN ON cycle takes longer to reach equilibrium – years… Approx for CNO cycles:

ON cycle at 20 million K CN eqm: 10 4 y ON eqm: 10 8 y

Summary of H burning Low T: pp Higher T: CN and then ON cycles

CN cycle

CNO cycles

CNO cycles – including high T

More H burning: Ne-Na & Mg-Al chains CNO CyclesNe-Na ChainMg-Al Chain Leakage

More H burning: Ne-Na cycle

Relatively complete network

Helium Burning After H burning the next step is He burning You may expect But in fact the Be decays into two 4He almost instantly… But if a third He collides with the Be before it decays…. 4 He + 4 He  8 Be

Helium Burning: Triple Alpha 4 He + 4 He + 4 He  12 C +        Y 2 T 8 41 NB Its T/10 8 K Helium burning takes about 100 million K to start…

Complete Helium Burning But there is another important reaction… 3 4 He  12 C +  12 C + 4 He  16 O +  So once some 12 C exists it can burn into 16 O XiXi time Helium Carbon Oxygen

Solar abundances The aim of nucleosynthesis is to explain all of this graph!

X, Y and Z Astronomers are pretty lazy people –X = mass fraction of H –Y = mass fraction of He –Z = 1=X-Y = mass fraction of “metals”

Periodic Table

Enough for today!