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Solar Physics

Solar Physics. Course PHASM314/SPCEG012 Spring Term 2012 (Semester B). SOLAR PHYSICS PHASM314/SPCEG012 Professor K. Phillips & Dr L. van Driel-Gesztelyi kjhp@mssl.ucl.ac.uk lvdg@mssl.ucl.ac.uk. Aims of the course The place of the Sun in the evolutionary progress of stars;

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Solar Physics

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  1. Solar Physics Course PHASM314/SPCEG012 Spring Term 2012 (Semester B)

  2. SOLAR PHYSICS PHASM314/SPCEG012 Professor K. Phillips & Dr L. van Driel-Gesztelyi kjhp@mssl.ucl.ac.uklvdg@mssl.ucl.ac.uk Aims of the course The place of the Sun in the evolutionary progress of stars; The internal structure of the Sun; Its energy source; Its magnetic fields and activity cycle; Its atmosphere (photosphere, chromosphere, corona); The solar wind; The heliosphere; Sun and its relation to nearby stars Method and Assessment This is a 30-lecture course with 4 problem sheets. Final assessment is derived from written examination.

  3. You will receive handout notes for the 1st set of lectures, NOT ANY OTHERS! (sorry, it’s a UCL rule). Please try to copy the remaining sets of lectures from the PowerPoint files on my web site. Please also take notes during the lectures. • Problem sheets: these form part of the continuous assessment for anybody taking this course. Please hand them in promptly, there is a penalty for late work. • For 4th year MSci / MSc courses, this coursework counts for 10% of the overall course assessment (the remaining 90% being the exam). • If a student does not obtain at least 15% of the overall maximum mark in this coursework he/she will FAIL THE WHOLE COURSE, regardless of his/her performance in the exam. • Feedback. We welcome constructive feedback about the course and will do our best to address any problems.

  4. Solar physics courseInformation on web sites For Dr van Driel-Gesztelyi’s part of course, see http://www.mssl.ucl.ac.uk/~lvdg For my part of course, see http://www.mssl.ucl.ac.uk/~kjhp From there, go to subdirectory: SOLAR_PHYSICS_COURSE_2012 where you will find subdirectories for LECTURES (all 6 batches of lectures in ppt format) PROBLEM_SHEETS (this week’s problem sheet in doc format) SCHWARZSCHILD_BOOK (jpg files) MOVIES (files of movies used in lectures)

  5. Solar Physics Moodle Page Go to ucl.ac.uk/moodle and select MSc Space Science and Engineering Moodle pages: Scroll down to Box 4 Advanced Courses and click on the live link for SS6 Solar Physics (moodle web page).

  6. Solar Physics Moodle pages You should see this: Select item 1 General Admin for lecture schedule and syllabus. Select item 4 for Reading List. Select Problem Sheets to get latest problem sheet handout.

  7. Suggested reading (see Moodle item 4) Solar Astrophysics(2nd revised edition) P. V. Foukal, Wiley-VCH, 2004. ISBN 3-527-40374-4 Neutrino Astrophysics J. Bahcall, Cambridge Univ. Press, 1989. ISBN 0 521 37975X. Stellar AstrophysicsVol. 2 (Stellar Atmospheres), Vol. 3 (Stellar Structure and Evolution). Erika Böhm-Vitense, Cambridge Univ. Press, 1992 (both paperback). ISBN 0 521 34870 6 (vol. 2), 0521 34871 4 (vol. 3). The Stars; their structure and evolution R. J. Tayler, Cambridge Univ. Press, 1994. ISBN 0 521 458854 . The Sun as a Star R. J. Tayler, Cambridge Univ. Press, 1996. ISBN 0 52146837 X Guide to the SunKenneth J. H. Phillips, Cambridge Univ Press, 1992/1995. ISBN 0 521 39483 X The Solar Corona by Leon Golub and Jay M. Pasachoff, Cambridge Univ. Press, 1997. ISBN 0 521 48535 5 The Sun: an Introductionby M. Stix, Astronomy & Astrophysics Libr., Springer, 2004. ISBN 3 540 20741 4

  8. Suggested reading (contd.) Physics of the Solar Corona: An IntroductionMarkus Aschwanden, Springer/Praxis Publ. Ltd., 2004. Hardbound, ISBN 3 540 22321 5. The Cambridge Encyclopedia of the SunKenneth R. Lang, Cambridge Univ. Press, 2001. Hardbound, ISBN 0 521 7803 4. Books on related subjects: Ultraviolet and X-ray Spectroscopy of the Solar Atmosphere by K. J. H. Phillips, U. Feldman, E. Landi, Cambridge Univ. Press, 2008. ISBN 978 0 521 84160 3 Solar & Stellar Magnetic Activity. C. Schrijver and C. Zwaan, Cambridge Univ. Press, 2001. ISBN 0521582865. Structure and Evolution of the StarsMartin Schwarzschild, Dover publications, 1958. Astronomical Spectroscopy J. Tennyson, World Scientific, 2011. Allen’s Astrophysical Quantities A. N. Cox (Ed.), Springer-Verlag

  9. The Sun: some basics

  10. The Sun: some basics (continued) Note: Absolute magnitude is apparent magnitude at distance of 10 parsecs (10 pc) = 3.1 × 1014 km = 2.06265 × 106 AU 1 solar mass = 1 M, 1 solar radius = 1R 1 radian = 206265 arcseconds. 1 year = 31557000 s. For exams, you should learn approximate values of most of these quantities, particularly for those on the previous slide, though the exam paper Rubric will give some of them.

  11. Units

  12. Units – electron volts (energy) For atomic or nuclear reactions, we often use the electron volt for energy units: 1 eV = 1.602 × 10-19 J 1 keV = 1.602 × 10-16 J 1 MeV = 1.602 × 10-13 J 1 electron volt is the energy acquired by an electron when accelerated over a potential of 1 volt. Particle masses are often expressed in terms of electron volts (e.g. electron mass = 511 keV/c2, proton mass = 938 MeV/c2).

  13. Lecture schedules 2012: provisional Lectures are in 3 one-hour groups, Tuesday afternoons, 2pm-5pm, in Foster Court 130 Lecture Theatre Jan. 10 Solar interior - standard solar model, radiative transport (KJHP) Jan. 17 Solar interior – convective transport, nuclear reactions (KJHP) Jan. 24 Helioseismology (LvD-G) Jan. 31 Magnetic fields (LvD-G) Feb. 7 Photosphere, chromosphere, corona (components) (KJHP) Feb. 14 Active region evolution, solar cycle (LvD-G) Feb. 21 Corona – atomic processes, heating (KJHP) Feb. 28 Coronal mass ejections, solar wind (LvD-G) March 6 Solar flares (KJHP) March 13 Solar-stellar connections (KJHP) After the lecture course: Late April (TBC) – Revision lecture (venue TBC) Early May (TBC) – Examination (venue TBC)

  14. THE SOLAR INTERIOR

  15. Solar interior The evolution of the Sun is modelled by following the gravitational collapse of an interstellar cloud and the onset of nuclear fusion of light elements at the core of the Sun.Basic fusion reaction is 4p → He nucleus, with a small amount of mass lost. Is fusion really the source of solar (and stellar) energy? Radiation output from the Sun = 4 × 1026 W Using E=mc2, the implied mass loss = 4 × 109 kg s-1 Age of Sun (from age of oldest meteorites) = 4.6 × 109 years (4.6 Gyr) [from the radioactive decay of 87Rb to 87Sr: half-life = 46 Gyr] Over Sun’s lifetime, total mass loss is 5 × 1026 kg This is only a small fraction (0.025%) of the Sun’s present mass, 2 × 1030 kg. Nuclear fusion could certainly explain the Sun’s radiation output over its lifetime.

  16. Alternatives to nuclear fusion as source of Sun’s energy: Gravitational energy released as Sun contracts Thermal energy of Sun Chemical reactions in the Sun Fission reactions (breaking up of heavy element nuclei)

  17. Gravitational contraction of Sun The Sun was formed from the gravitational collapse of a cloud. Suppose the cloud had radial symmetry, and the starting mass was Mr = M(r) where r = radial distance. The gravitational energy released when an element of mass Δm = ρ x 4πr2 Δr (where ρ = mass density) falls from infinity down to Mr is or integrating: Now integrate over all mass elements up to the Sun’s present radius Rʘ:

  18. Gravitational contraction of Sun Gravitating shell s Δm=ρ4πr2Δr Mr r

  19. Kelvin-Helmholtz time scale Suppose Sun had a constant density: ρr = ρ0. So Therefore But so where Rʘ= present solar radius, Mʘ = present mass. So Half of this energy goes into thermal energy, the other half into radiation. Energy emitted by Sun over its lifetime if shining at its present luminosity Lʘ is Lʘ t . So if gravitation supplies energy,

  20. Kelvin-Helmholtz time scale (contd) t is known as the Kelvin-Helmholtz time, i.e. the time for the Sun to release all its gravitational energy. Put in numerical values: Mʘ = 2.1030 kg , Rʘ = 7.108 m, Lʘ = 4 × 1023 kW , G = 6.7 × 10-11 SI units, we have: Kelvin-Helmholtz timescale So, assuming gravitational potential energy were converted to radiation, the Sun would shine for 10 million years. (This calculation is quite rough.) This time is only 0.004% of the Sun’s actual age (from meteorite ages).

  21. Equation of hydrostatic equilibrium Consider a small cylinder of unit cross section within a shell of the solar interior: Thickness dr Mass dMr Difference in pressure dP on either face of cylinder Difference in density is dρ

  22. Equation of hydrostatic equilibrium (contd) For the elementary cylindrical volume, dPr = - gr dMr where gr is gravitational force at r. Or: dPr = - gr ρr dr (1) According to Newton’s law of gravitation, gr = G Mr / r2 (2) where G = gravitational constant. So dPr /dr = - G Mr ρr / r2(3) [Note: we write MrρrPrfor M(r) ρ(r) P(r) etc.] This is the equation of hydrostatic equilibrium. In words: (gas) pressure is balanced by gravitation.

  23. Now, Mr , r , and r are not independent, since mass contained within r is determined by the density of the material inside r. Consider the mass of a spherical shell between r and r + r. The mass of the shell is given by 4  r2 r r provided r is small. Also the difference between Mr+r and Mr for a thin shell can be written: which gives us: (4) or (5)

  24. Combining equations (3) and (4) and multiplying through by 4 r3 (eliminating ρr) we get: Integrating this over the whole Sun gives: (6) where Vr is the volume contained within radius r (Vr = 4 πr3 / 3), Pc and Ps are core and surface pressures, Mʘ is the total solar mass. Note we use subscripts s=surface, c=centre of Sun. Integrating the LHS by parts; we get:

  25. Term in square brackets vanishes: at the lower limit V(r=0) = Vc = 0, while at the upper limit, if the Sun is surrounded by a vacuum, Ps = 0. The term on the RHS is the Sun’s gravitational PE = + Ω, i.e. the energy released in forming the Sun from its component parts dispersed to ∞. Note also that dMr = r dVr . Thus: which is an expression of the Virial Theorem. (7)

  26. For unit volume of an ideal gas, number density n, the total energy = n × the number of degrees of freedom Nf × kB T /2 (kB is Boltzmann’s constant). Thus the thermal energy per unit volume is: and Nfis related to γ, the ratio of specific heats of the material by: or where  = Cp/Cv. Thus, Nf = 2 / (γ – 1). The pressure of an ideal gas is given by Pr = nr kB Tr . (8) Let the thermal energy per unit mass be ur . Then: or:

  27. Multiply by dMr and integrate: where U is the total thermal energy of the Sun. Substitute this into the Virial Theorem (Eq. 7) to get: The material inside the Sun is totally ionized,so  = 5/3 and thus: i.e. negative gravitational energy = twice thermal energy From earlier, the Kelvin-Helmholtz time (i.e. gravitational collapse) is too small to explain the Sun’s energy. The thermal energy time is of the same order as the Kelvin-Helmholtz time. So neither cooling of an already hot body nor gravitational collapse can sustain the solar radiation output for 4.6 × 109 years.

  28. Now the Sun’s total energy, E, is given by: Since then So a decrease in  leads to an increase inU. The gravitational energy released as the Sun contracted (a) was radiatedand (b) heated the gas in equal measures. So, in summary, the Virial Theorem shows that: 1. The source of the Sun’s energy cannot be either thermal or gravitational energy; The Sun heats up as it contracts: as the Sun contracts, its potential energy decreases, with half of this amount appearing as thermal energy, the other half as radiation.

  29. Other possible energy sources Chemical reactions like those involved in the burning of fossil fuels. But they are only capable of releasing up to 5x10-10 of the rest mass energy, whereas it is known that the Sun has used up 3×10-4 of its rest mass in its lifetime. Chemical reactions could only power the Sun for a few thousand years, so are ruled out. Again, fission (i.e. breaking up of heavy elements such as Fe or Ni) is possible as such reactions are exothermic (i.e. release energy). But the abundances of Fe etc. are much too small (typically 10-4 that of H) to explain the Sun’s energy.

  30. Fusion as solar energy source Fusion of light elements is capable of releasing 1% of the rest mass energy – light elements (H, He) are very abundant. Thus, fusion is the most probable source of the Sun’s energy. The nuclear timescale is given by: where 0.1 is the fraction of the mass used up before the star leaves the main sequence and  ≈ 0.7% is the fraction of the rest mass converted to thermal energy by protons fusing into He nuclei ( particles). So:

  31. Equations of solar structure 1. Hydrostatic equilibrium: (3) 2. Pressure in the solar interior– given by the ideal gas law: Pr = nr kB Tror (8) where  is the molecular weight and hence the number density n =  / ( mp). Note Pr means P(r) etc. This equation is the Equation of State.

  32. 3. Mass as function of radius: (4) Also, we can form an equation relating the rate of energy release and rate of energy transport through the Sun. Assume that energy flows across a sphere of radius r at a rate Lr (in W). Then equate the difference between the energy crossing a sphere of radius r + r and a sphere or radius r to the energy released in the spherical shell. Hence (9) where r is the rate of nuclear energy production per unit mass (W kg-1) at radius r

  33. Meaning of molecular weight μ It does not imply existence of molecules! (For the solar interior, there are only atoms, ions and electrons.) μ is the mean mass of a particle in atomic mass units (mass of a proton = mp = 1). (Note electron mass me = mp/1836 which we neglect.) Let X = fraction of material by mass due to H, Y = fraction by mass due to He (we neglect heavier elements). The Sun was made from “primordial” gas made from the Big Bang, enriched by first-generation stars: X = 0.75, Y = 0.25. Gas at the Sun’s core (T=15 MK) is fully ionized, so: for every H atom, there are 2 particles (1 proton + 1 electron) per unit mass (mp = 1); for every He atom, there is ¾ particle (there are 3 particles, 1 nucleus + 2 e’s, and the mass is 4 units).

  34. Meaning of μ (contd.) In a volume 1 m3, there is a mass Xρ due to H, Yρ due to He. So no. of particles in 1 m3 is 2Xρ/mpdue to H, 3Yρ/4mpdue to He. Total no. = n = 2 Xρ/mp + 3Yρ/4mp or n = (ρ/4mp) × (8X + 3Y). Now from before we had n =  / ( mp), i.e. µ = ρ / (n mp). So This is not quite right as we neglected heavy elements. Adding them makes μ just over 0.6.

  35. Energy transport in solar interior • How is energy transported outwards through the Sun? • Three possible mechanisms: • Conduction • Radiation • Convection • Role of conduction: gas pressure >> radiation pressure in the Sun, • so, since the thermal energy of the electrons is greater than that • of the photons, we might expect thermal conduction to be important. • But the mean free path of the electrons is tiny compared with the • dimensions of the solar interior. Thus, thermal conduction is • negligible in the Sun – at most it makes a very small contribution at the solar core (inner 0.2 Rʘ where Rʘ = solar radius).

  36. Radiation and convection transport in the solar interior Of the 2 remaining transport processes, energy is transported by radiation in the inner part of the solar interior (0.25Rʘ < r < 0.71 Rʘ) and convection in the outer part (r > 0.71 Rʘ). We now calculate how the temperature drops off – the temperature gradient dT/dr – with distance r from the Sun’s centre assuming first radiative energy transport then derive conditions for when convection takes over as the energy transport mechanism.

  37. Role of Radiation energy transfer:Photon transport in the solar interior The time for a photon to travel from the solar core to the solar photosphere can be estimated from a random walk formula. The photons are continually stopped by scattering off free electrons in the solar interior – Thomson scattering. Random walk process: End point Startingpoint

  38. Photon mean free path Mean free path for photon is given by λ= 1 / (σT <Ne> ) where σT = Thomson scattering cross section = 6.65 × 10-29 m2 and <Ne> is the mean electron number density. Now <Ne> ~ mean proton number density ~ Mʘ / (Vʘmp) = 2.1030 / ( 1.4.1027 × 1.67.10-27) = 8.6 × 1029 m-3. So λ = 0.018 m. The total number of random walk steps for a photon to get to the solar surface is (Rʘ / λ)2 = (7.108 / .018)2 = 1.5 × 1021. The time taken for a photon to travel a distance = λ / c = 6 × 10-11 s.

  39. Time for a photon to travel to the solar surface So the time for a typical photon to reach the surface is ttravel = (Rʘ/λ)2 × (λ/c) = 1.5.1021 × 6.10-11 s = 9 × 1010 s = 3000 years.

  40. Radiative temperature gradient We now derive dT/dr for radiative equilibrium (note dT/dr is negative, we will obtain |dT/dr| ). The (specific) intensity of radiation I(r,s) at a point inside the Sun (assumed in equilibrium) is defined as the radiative energy dE(J m-2 s-1 ster-1) in the frequency range  to  + d flowing across area da in the direction of vector s within the solid angle d = sin θdθdφ (See diagram next slide.) In the cases we consider, the intensity is symmetric about n so its value in the direction s is fully described by . dEν = Iν(r,θ) da cos θdωdν Radial direction r dφ See Böhm-Vitense vol. 2, p.18 Now consider a beam of radiation of intensity I() traversing a medium with mass absorption coefficient . We will consider how radiation is absorbed later.

  41. Definition of elementary solid angle dω θ = co-latitude dω = sin θdφ ×dθ (radius =1) dθ φ = azimuthal angle

  42. Then an element of intensity is absorbed along ds. Now, ds = sec  drso that: Eliminating the  dependence by rearranging and multiplying through by cos2: and integrating over solid angle dωgives:

  43. In solar interior, conditions are close to thermodynamic equilibrium. This means that I = B(T) = thePlanck function / π, which is independent of direction (i.e. θ and φ). (This only applies to deep interior of Sun.) Also, ∫I cos  d = F(r), the net outwardflux passing through the surface with radius r inside the Sun in the frequency range  to +d. Hence: Since dω = sin θdθdφ, Rewrite So Eq. (10) becomes (10) (11)

  44. If we now integrate over frequency νthen (11a) where the flux F(r) = Lr / 4  r2(unit: kW m-2) and Lr = total rate of energy flow (unit: kW) across a surface of radius r inside the Sun. Let’s see what the absorption coefficient κ is in detail in the solar interior.

  45. Absorption processes in the solar interior Let’s consider the sources of absorption (opacity) in the solar interior. 1) κbb: Absorption by bound-bound transitions (i.e. spectral lines): material is nearly entirely fully ionized H and He with very small amounts of heavier elements which are also highly ionized, so there is very little absorption by bound-bound transitions. κbb depends on ν. 2) κbf, κff: Bound-free (photon absorbed by an ion which releases an electron) or free-free (photon absorbed by an electron passing by an ion) continuum processes are the most important in the solar interior. κbf, κff both depend on ν. 3) κTS: Thomson scattering of photons by free electrons: where Ne = electron density, ρ= density, σT = Thomson scattering cross section (6.65 × 10-29 m2). This is very small in the solar interior. (Note: κTS does not depend on ν.)

  46. Definition of Rosseland mean opacity (absorption coefficient) We define a mean absorption coefficient – the Rosseland mean absorption coefficient: (11b)

  47. Calculated Rosseland means for bound-free & free-free absorption coefficients Y=.2500 Temperatures in 106 K. (Solar composition) Cox & Tabor 1976 ApJS

  48. L.H.S. of Eq. (11b), sinceʃBν(T) dν= integrated Planck function / π= σT 4 / π (whereσ is the Stefan-Boltzmann constant = 5.7 × 10-8 SI units). From Eq. (11a): or (12) -- an equation for the radiative temperature gradient. So if the energy transfer is by radiation, for a given radial distance, if Tr, Lr, and the Rosseland absorption coefficient are known, the radiative temperature gradient can be found. In the Sun, energy transfer is radiative for 0.25 R < r < 0.71 R

  49. Radiative & convective T gradients in the Sun Core (nuclear reactions) Radiative zone Conv. zone Sun’s surface Centre

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