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Solar Interior (continued) Role of convection energy transport in the solar interior 1.

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Presentation on theme: "Solar Interior (continued) Role of convection energy transport in the solar interior 1."— Presentation transcript:

1 Solar Interior (continued) Role of convection energy transport in the solar interior 1

2 The onset of convection Convection occurs in liquids and gases when the temperature gradient > some critical value, i.e. it begins because the state of the fluid is unstable. In a gas, convection occurs if a rising element is less dense than its surroundings. This depends on the rate at which the element expands due to decreasing pressure, and the rate at which the surrounding density decreases with height. In the Sun and other stars the temperature falls going from the core outwards. Thus both dT/dr and dP/dr are negative. For r < 0.71 R, dT/dr is determined by the Rosseland absorption coefficient (approximating free-free and free-bound absorption). When r > 0.71 R, as T falls still further, the free electrons start to recombine with nuclei to form ions which are effective absorbers of radiation. As a result the opacity increases and so dT/dr becomes steeper (i.e. T falls off with r more quickly). 2

3 3 ONSET OF CONVECTION Element of gas after rising Element of gas before rising Values of pressure (P), density (ρ) outside gas element Illustrating an element of gas rising in the solar interior and principle of convection. Temperature = T 1, T 1 initially, then T 2, T 2 after element has risen.

4 Lets assume that the element rises adiabatically and does not exchange heat with its surroundings. So, in the figure, for 2 T 2, the element keeps rising towards the photosphere. Here it radiates, cools, becomes denser, and falls. Hence a convective cycle is established. Evidence for convection can be seen in the patterns of granulation and supergranulation in the surface layers of the Sun. 4 Onset of convection (contd.)

5 Figure shows the gas element starting out at P 1 = P 1 and 1 = 1. It moves to a new position with P 2 = P 2 but 2 2 and T 2 T 2. The transfer is adiabatic if the element rises quickly compared with the time to absorb or emit radiation. Hence, or where γ is the ratio of specific heats at constant pressure & volume: (as we had before) for a fully ionized gas. 5

6 In terms of density (ρ 1 /ρ 2 = V 2 /V 1 ): Hence the condition for instability is (13a) i.e. density inside the element < density of surroundings. Before the element starts rising, it is indistinguishable from its surroundings, so: Also, after the element rises, the pressures inside & outside the element must be equal: Now let 1 =, 2 = ρ + d and P 1 = P, P 2 = P + dP, then inequality (13a) is: (13b) using the binomial expansion, (1 + α) n ~ 1 + nα if α is small. 6

7 Now if we consider radial gradients (subscript r) as in the Suns interior, the condition for instability becomes: Eliminate ρ r using ideal gas equation ( μ = molecular weight, m H = H atom mass): or First recall differential calculus rule for quotients: (13) 7

8 So inequality (13) becomes: and expanding: or But note that both dT/dr and dP/dr are negative and hence the criterion for instability is: (14) 8

9 The RHS of inequality (14) is called the adiabatic temperature gradient. So there is convection when the magnitude of dT r /dr is greater than the adiabatic value. Proceeding outwards in the solar interior, the absorption (opacity) increases because of the increased recombination of free electrons with ions (particularly Fe ions). This causes the radial temperature gradient dT r /dr to steepen. Onset of convection occurs at a level in the solar interior where the temperature ~ 10 6 K. The condition for convective instability is called the Schwarzschild Criterion. It implies that when the actual T gradient, | dT r /dr | is greater than the | dT r /dr | adiabatic convection will begin. (See Schwarzschilds book pp. 44 et seq. for further description of convective energy transport.) 9

10 We have now derived the basic stellar structure equations: Equation of hydrostatic equilibrium (3) Equation of state (8) Mass as a function of radius (4) Radiation flux L r as a function of radius (9) Radiative temperature gradient (12) Condition for convective transport (14) The boundary conditions for the solution of these equations are: M(r=0) = 0; L(r=0) = 0; M(r=R ) = M ; radius of Sun at age 4.6 Gyr is R. These can be integrated numerically to give T, P, M, and L as a function of radius r – a model of the solar interior. A standard solar model has been developed by J. N. Bahcall (see his book Neutrino Astrophysics, chap. 4). 10

11 Solar evolution Progressive gravitational contraction of an interstellar gas cloud – some gravitational energy is converted to thermal energy Temperature (T ) and particle density (n) increase at the centre of the collapsing cloud: when T and n become large enough, nuclear fusion of H begins (nucleosynthesis) and contraction stops. Sun is on main sequence, where it spends nearly all its lifetime (next slide). After years = 10 Gyr, the H is exhausted, contraction will resume. Fusion continues in a thin H shell whose radius increases with time. The layers outside the H shell are heated so that the Sun expands to become a red giant (see next but 1 slide). As core contraction continues the core temperature will increase until He fusion begins (T ~ 10 8 K) and contraction halts again. When He is exhausted, contraction continues until electron degeneracy provides sufficient pressure to halt the collapse – the white dwarf phase. A tenuous envelope of gas is shed to form a planetary nebula. 11

12 Hertzsprung-Russell Diagram 12 4M ʘ 1M ʘ 0.25 M ʘ 0.5 M ʘ Stellar masses indicated in units of 1 solar mass = 1 M ʘ

13 Evolution of the Sun: (a)Contraction (b)Main Sequence (c)Red giant (d)He-burning stage (e)White dwarf 13

14 NGC 3132 nebula Cats Eye nebula 14 Examples of Planetary nebulae

15 Core temperature Core density Solar evolution: core temperature and density Shaded area = convectively unstable 15 MAIN SEQUENCE PRESENT

16 Nuclear fusion reactions For nuclear reactions to occur, the K.E. of the interacting particles must be large enough to overcome the Coulomb (charge) barrier and bring them within the range of the (residual) strong nuclear force: 1.7 × m ( = 1.7 femtometers = 1.7 fm). Four factors determine the most likely nuclear reactions: 1.Abundance of nuclear species 2.Reaction probability at the core temperature 3.Nuclear charge, Z 4.Production of stable nuclei as reaction products Since the Coulomb force scales with Z 2, reactions occur preferentially between low-Z nuclei. The most abundant are: H, He, C, N, O, Ne, Mg & Si. At the temperature of the Suns core, some reaction rates are very slow – only one reaction per particle occurs in 1.4×10 10 years. 16

17 Particle energies in the core (temperature = 15.6 MK) are at the few keV level, but typical Coulomb barriers for light elements are ~ MeV. Fusion in main sequence stars only begins as the result of quantum mechanical tunnelling (i.e. wave functions of interacting particles penetrate each other). Two nuclear reaction sequences occur in the Sun: 1.Proton-proton (p-p) chain – gives 99% of the energy 2.Carbon-Nitrogen (CN) cycle – gives 1% of the energy. Nuclear reaction sequences in Sun **17

18 Protonproton chain There is a series of reactions. First, deuterium ( 2 H) is formed from the collision of two protons. 1a) p + p 2 H + e + + e with Q = MeV, E < MeV Alternatively, the pep reaction can start the chain: 1b) p + e - + p 2 H + e with Q = 0.4 MeV, E =1.442 MeV where Q is the amount of energy produced (e.g. in -ray photons, positron annihilation, and particle kinetic energy) which is available for heating the Sun (i.e. to produce its luminosity); E ν is the neutrino energy (neutrinos escape from the Sun and so carry away this energy). Both reactions proceed very slowly – first, 2 protons must penetrate each other to within 1.7 x m (only ~10 -8 of them do), secondly, one must undergo an inverse β decay during the collision process: p n + e + + ν e 18

19 The next step happens within ~10s: 2) 2 H + p 3 He + Q = MeV followed by: 3) 3 He + 3 He 4 He + 2p Q = MeV Note that (1a) and (2) or (1b) and (2) must occur twice for each time (3) occurs. The net result is: 4p 4 He + 2e e plus energy available for heating the Sun.. 19 Protonproton chain (contd.)

20 Important side reactions (producing neutrinos – observable if in blue): 4) 3 He + 4 He 7 Be + with Q = MeV followed by either of these two reactions: 5) 7 Be + e - 7 Li + e with Q = MeV and E = MeV (89.7%) or 6) 7 Li + p 2 4 He with Q = MeV and E = MeV (10.3%) Or these reactions: 7) 7 Be + p 8 B + Q = MeV 8) 8 B 8 Be * + e + + e Q = MeVE < MeV 9) 8 Be * 2 4 He Q = MeV In addition there is a very rare reaction (hep reaction): 10) 3 He + p 4 He + e + + e E < MeV Protonproton chain (contd.) 20

21 This set of reactions contributes ~1% of the solar luminosity and is only important in the interiors of higher temperature stars. The summary reaction is: 4p 4 He + 2 e e Q = MeV with C and N nuclei involved, acting as catalysts. Carbon-Nitrogen cycle 21

22 Reaction Rates To get energy generation rates per unit mass per second, we evaluate the reaction rates (units: m -3 s -1 ): where N 1 and N 2 are the number densities of nuclei with atomic number Z 1 and Z 2 and atomic mass A 1 and A 2. The factor 12 is a rate coefficient or temperature-averaged cross section, and is an integral of the product of the cross section (v), the particle velocity v, and the Maxwell-Boltzmann velocity distribution (velocities v are in the centre-of-mass system of the reacting nuclei) – 22

23 whereis the reduced mass, E is the centre-of-mass energy, T is the temperature, S(E) is a function which defines the cross sections and f 0 is a factor that describes the screening effect of electrons. The additional exp (-2 ) term expresses the probability of penetrating the Coulomb barrier and is called the Gamow penetration term: where A is related to the atomic masses A 1, A 2 by S(E) can be measured, but only at E 200 keV. S(E) for lower energies are obtained by extrapolation to lower energies. 23

24 The energy generation rate,, is set by the reaction rate per unit volume, the energy per reaction, and the density. Hence, ε = f 0 X 1 X 2 ρ T α where X 1 and X 2 are reactant fractions (X 1 = X 2 for p-p chain) and α = 4.5 for the p-p chain (it is α = 20 for the C-N cycle). So ε is extremely dependent on the core temperature. Fusion suddenly turns on during the solar contraction phase when T reaches a particular value. 24

25 Standard solar model The points discussed so far allow us to specify the standard solar model describing the physical conditions in the solar interior as follows. 1)Abundances The model begins with a homogeneous chemical composition (H or X = 0.71, He or Y = 0.27, metals or Z = 0.02) – changes that occur during the 4.6 × 10 9 years to the present day are assumed only to occur due to fusion reactions. 2)Hydrostatic equilibrium The Sun is assumed to be in hydrostatic equilibrium. 3)Energy transport In the deep interior (out to 0.71 R ), energy transport is by radiation. Outside this region, energy transport is by convection. 4)Energy generation Nuclear fusion reactions (mostly p-p chain) are the primary source of energy. 25

26 Conditions inside Sun: Bahcalls models 26 r/R 0 Luminosity Hydrogen fraction Temperature Density

27 Solar neutrinos There are uncertainties in the solar standard model because of the Suns internal composition and age, and from errors in opacities, nuclear reaction rates and the equation of state (relating pressure and density) in the dense central regions. In the p-p chain, neutrinos of the following energies are produced: < MeV MeV MeV MeV < MeV < MeV … and for the C-N cycle: < MeV < MeV < MeV 27

28 Calculated Neutrino Fluxes Spectra of solar neutrinos: p-e-p and 7 Be reactions produce neutrinos with single energies. p-p chain reactions produce neutrinos with continuous energies. Energy range of Neutrino experiments: 28 hep neutrinos are those produced by 3 He + p reactions.

29 Predicted neutrino fluxes at Earth (standard solar model) SourceReaction no.Neutrino energy (MeV) Flux (10 14 m -2 s -1 ) Proton-proton p-p1a< p-e-p1b< hep10< x Be50.862, B8B8< x10 -4 CN cycle 13 N< O< F< x

30 Neutrino production as a function of solar radius 30

31 The first solar neutrino experiment was set up in 1968 by Raymond Davis and collaborators. It operated continuously from 1970 to The experiment was 1.5 km deep in the Homestake gold mine in South Dakota (to shield it from cosmic rays). Its operation was based on the reaction: 37 Cl + e 37 Ar + e - The neutrino energy threshold is MeV. From the standard model most (77%) detectable neutrinos for this reaction are from reaction (8) in the pp chain, i.e. decay of 8 B to 8 Be. ~13% are from reaction (5). The detector consisted of a large (400 m 3 ) tank of the cleaning fluid perchloroethylene (C 2 Cl 4 ) containing 2.2 × Cl atoms. Just under 2 solar neutrino-induced reactions were expected per day. The Homestake Neutrino Detector 31

32 The Homestake Mine Neutrino Detector in South Dakota. The tank contained 400 m 3 of perchloroethylene. Raymond Davis Jr ( ), Nobel Prize winner in

33 A typical run lasted 80 days, after which He gas was bubbled through the tank to pick up 37 Ar atoms. A measured volume of 37 Ar gas was then placed in a proportional counter (similar to a Geiger counter but with pulse height proportional to photon energy). 37 Ar decays by electron capture and emits 2.82 keV electrons that are detected by the proportional counter. Counting was carried out over 8 months to estimate the background - 37 Ar decays in ~ 3 months. New unit introduced by Bahcall - the solar neutrino unit or SNU: 1 SNU 1 neutrino capture per sec in target atoms 5.35 SNU 1 37 Ar atom per day Observed 37 Ar production rate = 2.55 ± 0.25 SNU Prediction rate from Bahcalls standard solar model = 9.3 ± 1.3 SNU Observed rate = 3.6 times less than predicted by theory. This is (or was) the solar neutrino problem. 33

34 Other experiments include: Chemical detection in large Ga targets, either aqueous (GALLEX) or solid (SAGE). Ge decays by electron capture with a half-life of 11 days – the resulting electrons are detected as for the 37 Cl experiment. Neutrino scattering by electrons orbiting atoms in water molecules in highly purified water (KAMIOKANDE and SUPER KAMIOKANDE) or heavy water (Sudbury Neutrino Observatory, SNO). The electron generates Čerenkov radiation which is detected by arrays of photomultipliers. SUPER KAMIOKANDE and SNO also detect muon neutrinos. Čerenkov radiation is generated by electrons with v > c / n where n is the refractive index. Radiation is emitted in a cone with half-angle cos = 1 / n, where = v / c. KAMIOKANDE thus determines the direction of neutrinos. 34

35 SUPER KAMIOKANDE neutrino detector (Japan) The detector, which measures 40m (tall) × 40m (wide), is filled with purified water, and has 13,000 photomultiplier tubes that detect Cerenkov radiation from recoiling electrons struck by incoming neutrinos, in the form of a cone of light. The detector is 1km underground. Unlike other neutrino detectors, it can determine the neutrino direction. 35

36 Super Kamiokande under construction: filling up with water 36

37 Suns image in neutrinos: 500 days of Super Kamiokande data **37 ~90 °

38 TargetExperimentThreshold energy (MeV) Measured neutrino flux (SNU) Predicted neutrino flux (SNU) Ratio: measured /predicted Chlorine 37 Homestake WaterKamiokande Gallium 71 Gallex Gallium 71 SAGE Measured and predicted neutrino fluxes 38

39 Possible explanations for the discrepancy: 1.The standard solar model is wrong 2.The experiments are wrong 3.Standard model of particle physics is wrong The neutrinos detected in the 37 Cl + 37 Ar + e - reaction (in the Homestake expt.) are mostly those due to 8 B decays. The 8 B neutrino production rate T 15, so only a slightly wrong T in the standard solar model could account for the discrepancy. Observed neutrino rate is times less than predicted by theory. 39

40 1.Low Z model: (dT/dr) rad hence a lower central temperature can be achieved by decreasing. This can be done with a lower abundance of heavy elements in the core. Need to assume Z/X ~ 0.1 (Z/X) surface to obtain 1/3 of standard model neutrino flux. But! This is ruled out by helioseismology (p-mode oscillations). 2.Rapid core rotation: If the core were to rotate 1000 times faster than the surface, the thermal pressure required to support the Sun against gravity would be reduced and the neutrino flux would decrease. But! analysis of p-mode oscillations shows no increase of rotation rate between r = 0.2 R and r = R. This is outside the region of the core where 8 B neutrinos are produced. 3. Mixing in the Sun: the core progressively depletes H by fusion (note diffusion of nuclear species is negligible). Now the energy generation rate X 2 T 4.5, so somehow returning H to the core would allow a reduced value of T. To obtain a factor 3 reduction in neutrino flux requires 60% of the Suns mass to have been mixed for the Suns lifetime. But! There is no evidence of mixing at this level. 40

41 Neutrino mixing – the Neutrino Problem solved There are 3 different neutrino states or flavours: e,, associated with the electron, muon, and tau particle in weak interaction decays. Most solar neutrino experiments so far have only detected electron neutrinos. Mikheyev and Smirnov (1985) and Wolfenstein (1978) showed that neutrinos can change state in the presence of other matter if they have a tiny rest mass. This change of state is called the MSW effect. Masses are very small: the e has a mass that is < 2.2 eV/c 2. Hence the emitted electron neutrinos could be mixed into muon or tau particle neutrinos modified as they leave the Sun. So electron neutrinos would only be 1/3 of the total number of neutrinos. The solar neutrino problem has now been resolved by mixing of the neutrino flavours as they emerge from the solar core many detectors only see electron neutrinos which are ~ 1/3 of the total neutrinos arriving at Earth. 41


43 Solar rotation All stars must rotate - random eddy motion in protostellar clouds transfers angular momentum. The Sun rotates rather slowly compared with many similar stars. Pressure differences drive flows between the poles and the equator. The Sun does not rotate rigidly (rotates faster at equator). In the convection zone convective motions are much faster than circulation currents - this leads to a distribution of angular velocity that varies both with depth and latitude differential rotation. 43

44 Measured rates of solar rotation Solar rotation rates observed by Doppler shifts at the solar limb and by sunspots. Latitude N or S ( o )Period from Doppler shift (days) Spot period (days)

45 Flows in the solar interior 45

46 Faster equatorial rotation ascribed to large-scale motions– meridional flows– from equator to pole near surface, return flows from pole to equator well beneath surface. Flow velocities measured by the Michelson Doppler Imager instrument on SOHO are ~10 m s -1 Observations from sunspots are slightly ambiguous – small sunspots rotate slightly faster than large spots. Flows in the solar interior (contd.) 46

47 Solar spectral irradiance ( f ν or f λ ) 47 Theory (HRSA = Harvard Reference Standard Atmosphere) and observed distributions + Black body

48 f ν as a function of wavelength (or frequency) is measured with bolometers, either Earth-based or on spacecraft. The values of f are corrected to the mean Earth-Sun distance (149,600,000 km). Its integral over frequency: is the total solar irradiance (formerly known as the solar constant) = total amount of radiation per unit energy per unit time per unit area reaching the top of the Earths atmosphere. 48 Photospheric radiation

49 49 The total solar irradiance is slightly variable and depends on the 11-year sunspot cycle. Total range of variations is 0.3%, and is at a maximum at sunspot maximum. But when there are large sunspots the total solar irradiance may dip by 0.25%. Its mean value is about × 10 3 W m -2 (1.368 kW m -2 ). Variability of Total Solar Irradiance

50 Irradiance (W m -2 ) 50

51 More recent information D. Pesnell (AGU 2008)

52 The Suns Effective Temperature The Suns mean total irradiance is × 10 3 W m -2 This is incident on a sphere, radius = 1 A.U. = × m. So total energy/second received by sphere is × 10 3 × 4 π (1.496 × ) 2 W. Radiation is uniformly emitted from the entire surface of the Sun. Radius of Sun isR = 6.96 × 10 8 m. If the Sun were a perfect black body, the radiation emitted would be 4 π R 2 × σ T eff 4 W where T eff is the Suns effective temperature and σ = Stefan-Boltzmann constant = 5.67 × W m -2 K -4. This is the Suns luminosity. Therefore i.e. T eff = 5778 K. 52

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