What makes a star shine? Energy Sources in Stars (§10.3)

Sun’s Energy Output L sun = 4 x W (J s -1 ) t sun 4.5 x 10 9 yr (1.4 x s) - oldest rocks - radioactive dating L sun ~ constant over geological timescale (fossil evidence) Therefore total energy output E tot = L sun x t sun = 6 x J How might the Sun have generated this energy? What makes a star shine?

Energy Sources in Stars Energy Source #1:Gravitational contraction We believe Sun collapsed from a large gas cloud (R ~ ∞) to its present size No change in Mass, M  R cloud RR Potential energy released is:  E grav = -(E now - E initial ) P.E. is obtained from integrating grav. potential over all points in the sphere

Energy Sources in Stars However, from Virial Theorem (E th = -1/2 E grav ), only 1/2 of the gravitational energy is available for energy release, the other half heats the star. So,  E grav = -(E now -E initial ) = 3/10(GM sun 2 )(1/R sun - 1/  ) = 3/10(GM sun 2 )(1/R sun - 1/  ) ~ J ~ J ~ 1/600 E tot needed over Sun’s lifetime ~ 1/600 E tot needed over Sun’s lifetime By this process, Sun could only radiate for t =  E grav /L sun ~ 10 7 years t =  E grav /L sun is Kelvin-Helmholtz timescale Conclusion: Gravitational Contraction not the main energy source in Sun Energy Source #1:Gravitational contraction

Energy Sources in Stars e.g. How much energy would be released if Sun was completely ionized and then all gas recombined into neutral atoms? (binding energy of the H atom 13.6 ev: 1 ev = 1.6 x J  13.6 ev ~ 2 x J) Assume Sun is 100% H (X = 1) Total # H atoms N H = M sun /m H ~  E tot ionization ~ x (2 x ) J ~ 2 x J  E tot ionization ~ x (2 x ) J ~ 2 x J This is only ~ 1/30,000 E total of Sun’s energy output Sun’s energy output ~ burning 7000 kg coal each hour on every sq. metre Sun’s surface! Conclusion: Chemical Reactions cannot be source Sun’s energy Energy Source #2:Chemical Reactions

Energy Sources in Stars Fission? Z = 0.02 (2% Sun consists of heavy elements). Fissionable isotopes like 235 U are rare (isotopes are nucleii with the same # protons but different # neutrons) Fusion? e.g. 4 protons get converted into 2p + 2n i.e. 4H  1 4 He H is very abundant in the Sun (X = 0.73) 4 H nuclei: mass = 4 x m p = x kg 1 4 He nucleus: mass = x kg (binding energy included) Difference is x kg = 0.7% of original mass Where does this mass go? Energy Source #3:Nuclear Energy

Energy Sources in Stars Example: Suppose that Sun was originally 100% H and only 10% of that was available for fusion. Thus… M H available = 0.1 M sun  M H destroyed through fusion = 0.1 M sun x total nuclear energy available E tot = (7 x M sun )c 2 ~ 1.3 x J Sun’s total lifetime energy output = 6 x J E tot ~ 2 x the total energy Sun has emitted!! E tot ~ 2 x the total energy Sun has emitted!!  Sun could shine for at least another 5 x 10 9 yr at its current luminosity Hence t nuclear for Sun ~ yrs Energy Source #3:Nuclear Energy The “lost” mass is converted into energy according to Einstein’s equation for the rest energy of matter: E = mc 2 (E energy, m mass, c speed light)

Energy Sources in Stars Fission? Z = 0.02 (2% Sun consists of heavy elements). Fissionable isotopes like 235 U are rare (isotopes are nucleii with the same # protons but different # neutrons) Fusion? e.g. 4 protons get converted into 2p + 2n i.e. 4H  1 4 He H is very abundant in the Sun (X = 0.73) 4 H nuclei: mass = 4 x m p = x kg 1 4 He nucleus: mass = x kg (binding energy included) Difference is x kg = 0.7% of original mass Where does this mass go? Energy Source #3:Nuclear Energy

Can Fusion Occur in Sun’s Core? Can Fusion Occur in Sun’s Core? Classical Approach: ä E kin proton > E coulomb ä 1/2 m p v 2 > e 2 /r (cgs form) ä LHS is thermal gas energy, thus, 3/2 kT >e 2 /r ä At T central ~ 1.6 x 10 7 K  E kin ~ 3.3 x J ä For successful fusion, r ~ m (1 fm)  m (1 fm)  Protons must overcome mutual electrostatic repulsion: Coulomb Barrier ä E coulomb ~ 2.3 x J ä E kin ~ E coulomb Classically, would need T~10 10 to overcome Coulomb barrier So, no Fusion??

Can Fusion Occur in Sun’s Core? Can Fusion Occur in Sun’s Core? Can we be helped by considering the distribution of energies? - not all protons will have just E th = 3/2 kT Proton velocities are distributed according to the Maxwellian equation: P(v) = 4  (m/2  kT) 3/2 v 2 e -mv 2 /2kT At the high energy tail of the Maxwellian distribution, the relative number of protons, with E > 10 3 x E th is: N(E coulomb ~ 10 3 )/N( ) ~ e -  E/kt ~ e = !! Number protons in Sun = M sun /M H ~ so 1 in of these will have enough energy to overcome Coulomb barrier. So again, no nuclear reactions??

Can Fusion Occur in Sun’s Core? Can Fusion Occur in Sun’s Core? Called Quantum Tunnelling ä Particles have wavelength, ( = h/p ), associated with them (like photons) called de Broglie wavelength ä Proved in many experiments - for example diffraction electrons (wave phenomenon) ä Eg. free electron (3 x 10 6 m/s) = nm (size atom) ä person (70 k gm) jogging at 3 m/s = 3 x m (negligible - person won’t diffract!) ä Increased wavelength reflects loss momentum. Quantum Approach: ä Heisenberg Uncertainty Principle -  p  x > h/2  ä If  p small, then  x may be large enough that protons have non-negligible probability of being located within 1 fm of another proton inside Coulomb barrier Classically, the proton is reflected at R 0. Quantum mechanically, there exists a finite probability for a proton to reach interaction radius, R.

Can Fusion Occur in Sun’s Core? Can Fusion Occur in Sun’s Core? Even with tunnelling, are stars hot enough for nuclear reactions to proceed? = h/p is the wavelength associated with a massive particle (p = momentum) = h/p is the wavelength associated with a massive particle (p = momentum) In terms momentum, the kinetic energy of a proton is: 1/2 m p v 2 = p 2 /2m p Assuming proton must be within 1 de Broglie of its target to tunnel set distance, r, of closest approach to, (where barrier height = original K.E. set distance, r, of closest approach to, (where barrier height = original K.E. incoming particle) gives: e 2 /r = e 2 / = p 2 /2m p = (h/ ) 2 /2m p Solving for (=h 2 /2m p e 2 ) and substituting r = into 3/2 kT = e 2 /r, we get the QM estimate of the temperature required for a nuclear reaction to occur: T QM = 4/3(e 4 m p /kh 2 ) - putting in numbers T QM = 10 7 K which is comparable to T core. Therefore, fusion is feasible at centre of Sun

Proton-Proton Chain Proton-Proton Chain Conservation Laws in nuclear reactions: (1) mass-energy (2) charge (3) difference between number particles and anti-particles conserved - ie particle cannot be created from anti-particle or vice-versa but a pair can be formed or destroyed without violating this rule ParticleSymbol Rest mass (kg) Charge (e-) Spin photon  0 01 neutrino >0 01/2 anti-neutrino ->0 01/2 electrone- 9 x /2 positrone+ +11/2 proton ( 1 1 H) p+ 1.6 x /2 neutron n 01/2 deuteron 21H 21H 21H 21H 3.2 x Basic particles involved in nuclear reactions

Proton-Proton Chain Proton-Proton Chain Types of nuclear reactions Beta Decay: n p + + e proceeds spontaneously - also for neutron inside nucleus Beta Decay: n  p + + e proceeds spontaneously - also for neutron inside nucleus eg [Z-1, A] - (Z = # p +, N = # n, A = Z+N) eg [Z-1, A]  [Z, A] + e (Z = # p +, N = # n, A = Z+N) Inverse Beta Decay: p + + e - Inverse Beta Decay: p + + e -  n + [ 13 N eg [ 13 N  13 C + e + + ] ) process: A Z + p + (p +,  ) process: A Z + p +  A+1 (Z+1) +  eg [ 12 C + p +  13 N +  ] ( ,  ) process:  particle ( 4 He) added to nucleus to make heavier particle [ A Z + 4 He  A+4 (Z+2) +  ] eg. [ 8 Be + 4 He  12 C +  ]

Proton-Proton Chain Proton-Proton Chain Since the Sun’s mass consists mostly of H and He, we anticipate nuclear reactions involving these two elements eg p + + p +  2 He p He  5 Li p He  5 Li 4 He + 4 He  8 Be 4 He + 4 He  8 Be But there is a problem here - what is it? All these reactions produce unstable particles eg p + + p +  2 He  p + + p + p He  5 Li  p He p He  5 Li  p He 4 He + 4 He  8 Be  4 He + 4 He 4 He + 4 He  8 Be  4 He + 4 He So among light elements there are no two-particle exothermic reactions producing stable particles. We have to look to more peculiar reactions or those involving rarer particles

The proton-proton chain PPI Branch NET RESULT Total of 6 protons 1 He nucleus + 2 protons + 2e + + neutrinos + energy Deuterium (p + + n) positronneutrino Slowest step Weak nuclear force Low-mass isotope of He (2p+ + n) Gamma ray photon m e+ ~ 5 x m p+ Very low mass m e  m p+ Proton-Proton Chain Using considerations of: ä minimizing Coulomb barriers, ä crossections, ä and making sure conservation laws are obeyed the nuclear reaction chain at right produces the energy observed in the Sun. 4

Proton-Proton Chain Summary of proton-proton chain: 6p +  4 He + 2p + + 2e  or 4p +  4 He + 2e 6p +  4 He + 2p + + 2e  or 4p +  4 He + 2e  Mass difference between 4p + and 1 4 He = 26.7 Mev x x ev/J = 4.2 x JMass difference between 4p + and 1 4 He = 26.7 Mev x x ev/J = 4.2 x J ~3% of this energy (0.8 Mev) is carried off by neutrinos and does not contribute to the Sun’s luminosity~3% of this energy (0.8 Mev) is carried off by neutrinos and does not contribute to the Sun’s luminosity 2 e + immediately annihilate with 2 e - and add 2 x Mev (= 1.02 Mev)2 e + immediately annihilate with 2 e - and add 2 x Mev (= 1.02 Mev) So, the total energy available for the Sun’s luminosity per 4 He formed is: ( ) Mev = 26.9 Mev = 4.2 x JSo, the total energy available for the Sun’s luminosity per 4 He formed is: ( ) Mev = 26.9 Mev = 4.2 x J # 4 He formed/sec = L sun /4.2 x J = 3.9 x10 26 J/s / 4.2x J = 9.3 x He/s# 4 He formed/sec = L sun /4.2 x J = 3.9 x10 26 J/s / 4.2x J = 9.3 x He/s Increase of 4 He mass/time = dm He /dt = 9.3 x He/s x 6.68 x kg/ 4 He = 6.2 x kg 4 He /sIncrease of 4 He mass/time = dm He /dt = 9.3 x He/s x 6.68 x kg/ 4 He = 6.2 x kg 4 He /s After years (3 x sec), M( 4 He) = 1.9 x kg = 10% mass Sun

Proton-Proton Chain The previous nuclear reaction chain (PPI) is not the only way to convert H into He. eg last step: 3 He + 3 He  4 He + 2p + can proceed differently if there is appreciable 4 He present A second branch (PPII) PPI + PPII + PPIII operate simultaneously A third branch (PPIII) Very unstable

(PPI) (PPII) (PPIII) 69% 31% 99.7% 0.3% PPIIPPIPPIII Fraction of He 4 produced T (10 6 )K) Proton-Proton Chain: Summary

Alternate Fusion Reactions H  4 He First step in PP chain has very low reaction rate (weak interaction). First step in PP chain has very low reaction rate (weak interaction). 12 C can act as a catalyst in fusion reaction. 12 C can act as a catalyst in fusion reaction. Net result: 12 C + 4p +  12 C + 4 He + 2e  Net result: 12 C + 4p +  12 C + 4 He + 2e  Note: 12 C neither created nor destroyed. Also isotopes of N & O are temporarily produced. Note: 12 C neither created nor destroyed. Also isotopes of N & O are temporarily produced. CNO cycle dominates over PP chain if T core > 1.8 x 10 7 (slightly hotter than Sun) CNO cycle dominates over PP chain if T core > 1.8 x 10 7 (slightly hotter than Sun)  10 6 yr 14 min 3 x 10 5 yr 3 x 10 8 yr 82 s 10 4 yr X as frequently Once for every 2500 instances of Cycle 1 THE CNO BI-CYCLE (T < 10 8 K) CNO Cycle (or bi-cycle)

Other Fusion Reactions p. 313 ~10 9 K Mg, Si, S, etc. 16 O p x 10 8 K O, Ne, Na, Mg 12 C p. 312 ~10 8 K Be, C 4 He triple  process Ref. at T central into Fusion of At higher T centre, heavier nuclei can fuse to produce energy We believe Universe began with only H, He, (Li, Be) All other elements created in core of stars (stellar nucleosynthesis) We are all made of “STARSTUFF”