MSCA&A_L11_L14.pdf

Introduction to Astronomy and Astrophysics (I & II, 4 credits) The Sun Prof. Anupam Bhardwaj IUCAA The Sun A star Not just another star in the cosmos!! N...

Introduction to Astronomy and Astrophysics (I & II, 4 credits) The Sun Prof. Anupam Bhardwaj IUCAA The Sun A star Not just another star in the cosmos!! Nearest and the only star in our planetary system Provides almost all of our energy Only star on which we can resolve the spatial scales where fundamental processes take place provides us with a unique laboratory in which to learn about various branches of physics Life giving star….  The Sun What we want to study? Estimate values of the basic parameters of the Sun Describe different layers of the Sun’s structure describe different layers of the Sun’s atmosphere; Understand solar interior and its atmosphere describe some of the observed features such as sunspot, prominence and solar Describe flare associatedsome of theactivity; with solar observed features such as sunspot, prominence andthe explain solar roleflare of theassociated with solar Sun’s magnetic activity field in solar activity; derive the basic Explain theresults role ofofthe solar magnetohydrodynamics; Sun’s magnetic field in solar andactivity The explain the seismology of Solar System describe Thethe Sun. Solar different layers of the Sun’s atmosphere; System describe different layers of the Sun’s atmosphere; Explain the seismology and Stars of and Stars the Sun. describe some of the observed features such as sunspot, prominence and solar describe some of the observed features such as sunspo flare associated with solar activity; 5.2 SOLAR PARAMETERS flare associated with solar activity; explain the role of the Sun’s magnetic field in solar activity; explain the role of the Sun’s magnetic field in solar ac The basic parameters of the Sun are the derive its mass (MΘ), of basic results radius (RΘ), luminosity (LΘ) and solar magnetohydrodynamics; and derive the basic results of solar magnetohydrodynamic effective surface temperature (T eff). Inthe explain theseismology followingofdiscussion, the Sun. you will learn to estimate these parameters. explain the seismology of the Sun. 5.2 SOLAR PARAMETERS Mass: You know ar System that the mean distance describe different layersof ofthe the Sun 5.2the Earth Sun’sfrom SOLAR atmosphere; × 1011m. It is is 1.5PARAMETERS rs called the mean solar distance, a. Inparameters The basic astronomy of we measure the Sun are itsmean distances mass (M Θ), radiusin(Rterms Θ), luminosity (LΘ) and the describe some of the observed AUfeatures such as distance). sunspot, prominence and aresolar of a and it defines Astronomical Unit effective surface (1 (TThe = mean temperature basic solar eff). In the parameters following To of obtain the Sun discussion, an you its willmass learn(M toΘ), radius The expression Sun for the flare associated estimate mass of the with solar Sun,these activity; parameters. we may effective surface temperature (Teff). In the following discus use Kepler’s third law under the estimate these parameters. assumption that theexplain mass ofthethe role of thecan planet Sun’s be magnetic neglectedfield in solar activity; in comparison with the mass Mass: You know that the mean distance of the Sun from the Earth is 1.5 × 1011m. It is Basic of the Sun. parameters This assumption derive the is valid basic for the results Sun-Earth of solar systemYou andknow magnetohydrodynamics; Mass: we can andwrite: that the mean distance of the Sun from t called the mean solar distance, a. In astronomy we measure mean distances in terms of aseismology and it defines calledUnitthe (1 meanAU solar distance, a. In astronomy weanmeasu 2 3explain the of the the Astronomical Sun. = mean solar distance). To obtain 4 π Using Kepler’s a thirdexpression law (neglecting for the mass theofmass of the Sun, a of weand themay it defines earth) the use Kepler’s Astronomical third law under the = mean so Unit (1 AU = GM Θ (5.1) Kepler’s t f 2 P5.2 assumption that the mass of the expression planet can be forneglected the mass in of comparison the Sun, we with may the usemass SOLAR PARAMETERS assumption that the of the Sun. This assumption is valid for the Sun-Earth system and we can write: mass of the planet can be neglected in where a, P,Solar Mass G and Θ areparameters the mean of solar distance, of the Sun. This assumption is valid for the Sun-Earth syst TheMbasic the 2Sun3 are its orbital mass (Mperiod (~ 365 days) of the Θ), radius (RΘ), luminosity (LΘ) and 4π a Earth, gravitational For the Sun-Earth constant effective and mass of the Sun, systemtemperature2(Teff=).GM surface In respectively. Θ With the following discussion, 4π 2the a 3 values you will oflearn the to (5.1) orbital period and You may recall from the mean Unit 6 estimate of thesesolar parameters. P distance available at present, the value = GM of GMΘΘ is the physics elective course You may12 recall 3 −2from Unit 6 of P2 estimated to be PHE-01 entitled 132712438where Elementary ×the10physics m s and P,elective a,−11 G. Since course MΘ−2are thethelaboratory mean solarmeasurements distance, orbital for G (~ 365 days) of the period 3 −1 Mechanics that Mass: gives a value equal to 6.672Earth, Kepler’s third You knowPHE-01 that m kg aconstant × 10gravitational the entitled mean s = ,mean Elementary andsolar we obtain: distance of massdistance the where Sun a, from of P, theGSun, and=1 the MAU Earth Θ are = the mean is respectively. 1.5 1011 ×With m. values solar the Itdistance, is of the orbital law is given by: called the mean Mechanics solar that distance,Kepler’sa. third In astronomyEarth, we measure gravitational mean distances constant and in massterms of the Sun, respecti 30orbital period law is given by:and the mean solar12distance available at present, the value of GMΘ is 2 3MΘof 4π a ≈ a2 × and 10it kg. defines estimated the Astronomical to be P 132712438 = Orbital Unit× (1 10 period AU orbital m 3 =−2 s (365 mean period. Since d)thethe solar and distance). mean solar laboratory To distance obtain anavailable measurements for G at pres T2 = 2 3 −11 Kepler’s 3 to −1 be −2 third × 12 3 −2 expression for the mass gives Ta2value 4π of the Sun, we may estimated use equal to 6.672 × 10 m kg s , we obtain: a 132712438 law under 10 them s. Since the laborat This valueGMis taken as the mass assumption that the mass of the = Sun as GMof theGplanet it exists today. = Gravitational can gives In fact, constant a value be neglected solar equal in mass =to 6.672 with comparison −11 3 −1 −2 × 10 themmass kg s , we obtain where T, a, Gcontinuously decreases and M are since the Sun continuously 30 emits radiation and particles which of the respectively the time period, Sun. This assumption M Θ ≈ is 2 × valid 10 forkg.the Sun-Earth system and we30 can write: where T, a, G and M are carry withaxis, semi-major them some mass.This gravitational However, respectively the the time total period, mass loss during the ≈ 2 × 10estimated MΘ Sun’s kg. 10 2 value 3 is taken as the 27 mass of the Sun as it exists today. In fact, solar mass life timeand(~mass constant 10 of the yrs)Sun.is found πtoabe less 4semi-major decreases axis,than 10 since gravitational continuously kg. This theThis Sunvalue value isis much continuously taken emits aslessthethan massthe radiation of the andSun as it exists particles whichtoday error in ~ 99.86% of measurement mass the of solar the constant = mass GM solar and massΘsystem and of the is, Sun. therefore, negligible. This method can (5.1) y recall from Unit 6 of carry 2 P with them decreases continuously since some mass. However, the total mass loss during the Sun’s estimated the Sun continuously emits r 10 27 alsocourse sics elective be used to estimate thelife masses time (~of10theyrs) is found tocarry satellites/Moons be less with of thethem than 10 some planets inmass. kg. This our However, solar value is much the lesstotal thanmass the loss entitled Elementary ~ 330,000 where times mass and of the Earth lifemass timeand (~ 10 10 yrs) is (~ found 365to be less than 27 10 kg.canThis v system. How about a, P, G solving error an SAQ Θ are inMmeasurement the mean of this solar of the nature? distance, solar orbital period is, therefore, days) negligible. of the This method ics that Kepler’s third Earth, gravitationalalso be used constantto estimate and mass theof error masses the ofinthe Sun, measurement satellites/Moons respectively. of thethe With ofsolar the massofand planets values is, therefore, in our the solar n iven by: orbital period system. and the How mean about solar solving distance also be an SAQ used of this available to estimate at nature? present, the the value masses of GM ofΘthe is satellites/Moons nd 2 SAQ 1 12 3 −2 system. How about solving an SAQ of this nature? 4π a 3 estimated to be 132712438 × 10 m s. Since the laboratory measurements for G  error in measurement of the solar mass and is, 1therefore, AU negligible. Sun This method can SAQ nd also 1 SAQbe1used to estimate lar System describe the masses of the satellites/Moons of the planets in our solar different layers of the Sun’s atmosphere; in. system. How about solving an SAQ of this nature? ars 32´ One of the four Galilean Galilean satellites of the theobserved planet Jupiter issuch Io. as Itssunspot, orbitalprominence period is and solar One ofThe the fourSun describe some of satellites of the planet features Jupiter 10 is Io. Its orbital period is nd 1.77 SAQdays. 1 TheThe semi-major axis of its flare associated withorbit 4.22 × 10 cm.10Calculate the mass of solarisactivity; 1.77 days. semi-major axis of its orbit is 4.22 × 10 cm. Calculate the mass of en. Jupiter under the assumption that the Jupiter is too massive in comparison to Io. JupiterBasic the assumption that the Jupiter is too field underparametersexplain the role of the Sun’s magnetic massive in solar in activity; comparison to Io. nd One of the four Galilean satellites derive the basic results ofof the planet Astronomical Jupiter observations solar magnetohydrodynamics; is Io. indicateIts orbital that the solarperiod radius is notis constant; rather, its value changes slowly. Over 10 a period of ~ 109and years, the average change is about 1.77 days. Radius: The semi-major The radius of the Sunaxis can of beits orbit estimated isper 4.22 Astronomicalifyr.we Earth ×know 10radius observations cm. the indicate Calculate values the6angular ofradius that the solar its ismass of rather, its not constant; θ) ofis5.1). 2.4 cmchanges Further, of the Earth 9is 6.4 × 10 m. Thus, the Sun’s radius is Radius: The JupiterIfθunder we radius andknow the mean of explain the the angular assumption the Sun can seismology diameter that be estimated of the a ((Fig. value the Jupiter Sun.the too if we Sun: slowly. know Over massive a period the of values ~ 10 of years, the its average angular change is about ds, diameter, the solar distance, 2.4 cm per yr. The Further, angular radius of in comparison diameter the Earth is 6.4 × 106ofm. totheIo.the Thus, SunSun’s radius is diameter, θ θ, is 32′ and mean solar distance is 1.5 × 10 m.11Thus, with the help of Fig. 5.1, we the Sun and the mean solar distance,11 a (Fig. 5.1). The angular diameter of obtain the and is 32′ Radius: mean Solar value The of5.2 radius solar radius the of SOLAR distance solarthe radius Sun canRΘ1.5as:× 10 m. Thus, with the help 1of isPARAMETERS be estimated if we know the values AU Fig. 5.1, we of its angular Sun s, obtain the value of the solar radius RΘ as: 1 AU Sun θ, diameter, θ andThe thebasic meanparameters solar distance, of the Sun a (Fig. are its5.1). massThe (MΘangular ), radius (R diameter Θ), luminosity of the(LSun 32´ Θ) and Fig.5.1: When is 32′ and mean 1 viewed 1/2(11(θ solar RΘ =effective [ [.distance 2 =thelarger × 10 in from 11 radians) 5surface (1.5 ×than m) the is111.5 10 that Earth, × 10 × ( 32 temperature m) the 11 angular m.× × 2(T.9eff ×as: Thus, ). 10 32 ×Earth. (the − 4 In therad) 2.9 ×To diameter with following 10 −get4 rad) ] ] theof the help Sun of discussion, is Fig. 32´ approximately you5.1, willwe learn to 32′′ obtain almost 100Restimate the value Θof times 2 these solar parameters. radius RΘof an idea of the relative sizes of es the Sun and the Earth, refer to Fig. 5.2. 8 know Earth 11 that the meana distance = meanof solar distance = 1Earth AU =is 1.5 × 10 m. It is if d. =Mass: RΘ 6.7 1× You == 6.7 called 2 [ 10 m. (1×mean the.510 8 × 10 11 solarm) m. × ( 32 ×a.2In distance,.Sun − × 10 rad) 9 astronomy 4 the Sun ] from the Earth we measure mean distances in terms es of a and it defines the Astronomical 1 arc-sec Unit = (14.84 AU =x mean 10-6 radians solar distance). To obtain an f expression for the mass of the Fig.5.1: Sun, we Fig.5.1:When may When viewed use Kepler’s fromthe theEarth, Earth,thethe third angular law ofunder diameter of the the SunSun 32′′ 32′′ is approximately 8 viewed from angular diameter the is approximately d. = 6.7 × 10 assumption thatm.the mass of the planet Moon’s can orbit be neglected in comparison with the mass almost almost100 100 times times larger thanthat larger than thatofofthe theEarth. Earth. ToTogetget an an ideaidea of the of the relative relative sizessizes of of of the Sun. This assumptiontheis valid theSun andfor Sunand the the Earth,Sun-Earth the Earth, refertotoFig. refer system and we can write: Fig.5.2. 5.2. Sun Sun 4π 2 a 3 ~ two times the Earth-Moon distance = GM Θ (5.1) 2 ay recall from Unit 6 of P Moon’s orbit sics elective course 100 times larger than the Earth’s radius Moon’s orbit Earth 1 entitled Elementary where a, P, G and MΘ are the mean solar distance, orbital period (~ 365 days) of the nics that Kepler’s third Earth, gravitational constant and mass of the Sun, respectively. With the values of the Earth iven by: Earth orbital period and the mean solar distance available at present, the value of GMΘ is 4π2 a 3 estimated to be 132712438 × 1012 m3s−2. Since the laboratory measurements for G = −11 3 −1 −2 GM gives a value equal to 6.672 × 10 m kg s , we obtain: T, a, G and M are The Sun 30 Fig.5.2: The size of the Sun is so big that it can contain the Earth as well as the orbit of the Moon! olar System Fig.5.2: The size of the Sun describe Θis≈so 2 big Mdifferent × 10 that it of kg. layers canthe Fig.5.2: contain The Sun’s the Earth as well as the orbit of the Moon! size of theatmosphere; Sun is so big that it can contain the Earth as well as the orbit of the Moon! vely the time period, Luminosity: The solar luminosity, LΘ, is defined as the total energy radiated by the tars Basic parameters ajor axis, gravitational Luminosity:This value The is taken solar describe asofthe luminosity, some the LSun, of mass per is Luminosity: observed Θ unit time the The in Sun defined features the asas solar form it such the of total exists electromagnetic luminosity, as today. LΘ,energy radiation. Inprominence fact, is defined sunspot, radiated as the To estimate solar total mass the by solar energy and the value ofby the radiated t and mass of the Sun. luminosity Sun per unitoftime the Sun, in theletform us imagine a sphere with radiation. of electromagnetic the Sun at its Tocentre (Fig.the estimate 5.3). Theof value Sun per unitdecreasesin continuously timeflare associated the form ofwith since solartheactivity; Sun electromagnetic radius of this luminosity continuously radiation. of imaginary the Sun, sphere let isemits a, To us imagine radiation estimate theamean distance sphere and with the the particles Sun value between atthe itsSun ofwhich and centre the Earth. (Fig. 5.3). The carry with them some mass. However, Now, each unit the area total A of the mass sphere loss receivesduring energy the equal Sun’s to S, estimated called the solar If we know the luminosity of the Solarthe Sun, explain constant let us imagine—Sun’s roleisoffound the S = 1370 a radiussphere of this W/m with imaginary 2 the Sun sphere is a, at theits meancentre distance (Fig. between 5.3). the SunThe and the Earth. life time (~ 1010 yrs) Now, bemagnetic constant. to each Therefore, less unit area field thanluminosity A 10 27 incan of thekg. solar sphereThis activity; be expressed value receives as: is much energy less equal to thanthe S, called thesolar radius of this imaginary sphere is a,constant. the mean distance Therefore, 2 = 4πaluminosity between the Sun can be negligible. expressed and the Earth. in error measurement derive of the solar the basic results of solarmass LΘand is,S therefore, magnetohydrodynamics; andas: This method (5.2) can Now,Solar each also unit area luminosity A of the sphere receives be used to estimate the masses ofLthe explain constant. Therefore, the seismology luminosity can beof expressed energy the Sun. as: 2 equal to S, called the solar satellites/Moons Θ = 4πa S of the planets in our solar(5.2) system. How about solving an SAQ of this nature? Since the solar radiation is absorbed in the Earth’s atmosphere, it ais obvious that S 1m should be measured above 2the atmosphere. S has now been measured accurately using Spend SAQLΘ = 5.2 1 4πa S SOLAR PARAMETERS Sun a Earth (5.2) satellites and its value is 1370 Wm−2. Substituting the value of S and the mean solar A 1m1m 5 min. 11 Earth distance, a = 1.5The × 10 basicm parameters in Eq. (5.2),ofwe theget: Sun are its mass Sun (M ), radius 1m One of the four Galilean satellites of the planet Jupiter Θ is Io. ItsΘ),orbital (R luminosity period(LAisΘ) and onomy, we encounter effective 10 1.77 days. surface temperature The26semi-major axis(T ). In ofeffits theisfollowing orbit × 10discussion, 1 AU 4.22 cm. Calculateyou will the learn mass to of ly small angles (of the = 3.86 ×these LΘ estimate 10 parameters. W. a that the Jupiter is too massive in comparison to Io. f minute (′) and second Jupiter under the assumption 1 AU 1m Fig.5.3: Imaginary sphere of radius a surrounding the Sun where a is its mean solar distance from r which simple the Earth imations areTemperature: used for The Mass: temperature You know of thethe that Sun meanataits = surface mean Earth distance and the its ofsolar interior distance Sun from the regions = 1 AU Earth =is 1.5 × 1011m. It is are different. Radius:temperature The radiusSun of the Sun can beFig.5.3: be estimated Imaginary if we sphere of radius know the a surrounding values the Sun 1mof where itsmean a is its angular solar distance from metric functions. Thus,The surface called the mean solar candistance, estimated the In a. using astronomy Earth Stephan-Boltzmann we measure mean A law distances in terms all values ofwhich the angle diameter, θ, studied θ and the mean solar distance, a (Fig. 5.1). The angular diameter of the Sun you of ainand ouritcourse definesontheThermodynamics Astronomical Unit and 11 (1 Statistical AU = mean Mechanics solar distance). To obtain an is 32′ and mean solar distance is 1.5 × 10 m. Thus, with the help of Fig. 5.1, we (PHE-06). anθ = θ = sin θ We leave this asfor expression anthe exercise mass of forthe you in the Sun, formuse we may of an SAQ. third law under the Kepler’s Energyobtain the value radiated perof unit the solar time radius Θ as: Relectromagnetic assumption that the mass of in theall planet can be neglected bands in comparison with the mass SAQ 2 11 AU of the Sun. This assumption is valid for the Sun-Earth system and we can write: pend osθ = 1 min. S is measured above R Θ = 2 2 the 3 (1. 5 [ atmosphere × 10 11 m) × ( 32 × 2.9 × 10 − 4 rad) ] Assume er, as you know, that the Sun radiates4like angles π a a black body at temperature T. Calculate T using e measuredStephan-Boltzmann inFig.5.3: radian if Imaginary law. sphereTake =a GM Stephan of radius constant Θ surrounding σ the = 5.67 10−8 aWm Sun×where −2 −4 (5.1) K. solar distance is its mean from may recall ormulae arefrom Unit to be 6 of valid. P 26.7 × 108m. = hysics elective course the Earth The Sun Basic parameters Solar radius = 695,990 km = 432,470 mi = 109 Earth radii Solar mass = 1.989 X 1030 kg = 4.376 X 1030 lb = 333,000 MEarth Solar luminosity (energy output of the Sun) = 3.846 X 1033 erg/s Surface temperature = 5770 K = 10,400 ºF Central temperature = 15,600,000 K = 28,000,000 ºF Surface density = 2.07 X 10-7 g/cm3 = 1.6 X 10-4 Air density Central density = 150 g/cm3 = 8 × Gold density Surface composition = 70% H, 28% He, 2% (C, N, O,...) by mass Central composition = 35% H, 63% He, 2% (C, N, O,...) by mass The Sun Basic parameters Solar age = 4.57 X 109 yr (from meteorite isotopes and moon rocks) — G2V main sequence star Rotation period = 25.4 days at the equator, with the period of rotation increasing to about 36 days near the pole Its axis of rotation is inclined 8° to the ecliptic (the plane in which the planets orbit) Surface gravitational acceleration = 274 m/s2 1 arc sec = 722±12 km on solar surface, at centre of solar disc (elliptical Earth orbit) The Sun Solar mass loss The solar mass decreases continuously since the Sun continuously emits radiation and particles which carry with them some mass. Sun sends out about 4 x 1033 ergs/sec of energy (Solar Luminosity) Einstein's mass-energy relation is E = mc2. Thus the equivalent mass loss of all this energy loss is: m = 4 x 1033 ergs/sec / (3 x 1010 cm/sec)2 or a mass loss of ~4 x 1012 grams/sec Since the mass of the Sun = 2 X 1033 gms, find the total mass loss in Sun’s current lifetime? The Sun Solar mass loss Q. Since the mass of the Sun = 2 X 1033 gms, find the total percentage of mass loss in Sun’s current lifetime? mass loss of ~4 x 1012 grams/sec The Sun Solar mass loss Q. Since the mass of the Sun = 2 X 1033 gms, find the total percentage of mass loss in Sun’s current lifetime? mass loss of ~4 x 1012 grams/sec age = 4.57 X 109 yr The Sun The lifetime of the Sun (nuclear lifetime) The current age of the sun is around 5 billion years. This number is determined from radioactive dating of objects in the solar system which are known to have formed around the same time as the sun. lifetime = (energy) / (rate [energy/time] at which sun emits energy) The Sun shines via nuclear reactions in the core that transforms four hydrogen atoms into one helium atom. If you look at a periodic table, you will see that one helium atom has a little less mass than four hydrogen atoms combined; about 0.7% of the original mass has "disappeared". This missing mass gets transformed into energy that causes the Sun to shine. The Sun The lifetime of the Sun (nuclear lifetime) The current age of the sun is around 5 billion years. E = 0.007 x Msun c2 But only on the order of 10% of the sun's mass is in the central part, hot enough to undergo nuclear reactions. E = 0.007 x 0.1 x Msun c2 = 0.0126 x 1053 ergs. Sun sends out about 4 x 1033 ergs/sec of energy. So the nuclear lifetime of the Sun = (12.6 x 1050 ergs)/(4 x 1033 ergs/sec) = 3.15 x 1017 sec = 1010 years The Sun should last another 5 billion years. The Sun The lifetime of the Sun (nuclear lifetime) Since becoming a main-sequence star, the Sun’s luminosity has increased nearly 48% (from 0.677 Lsun) The radius has increased 15% (from 0.869 Rsun) Temperature has also increased from 5620K to 5777K The Sun The Solar Composition We can learn a great deal about composition of the Sun from the pattern of absorption lines in its spectrum (the Fraunhofer lines). The pattern of these lines serves as a set of fingerprints for the elements that are present in the surface of the Sun, and their intensity serves as a measure of the concentration of these elements. High Resolution Solar Spectrum Showing Absorption Lines (created to mimic an echelle spectrum). Each of the 50 slices covers 60 angstrom between 4000 to 7000 angstroms — credit NOAO. The Sun The Solar abundance Approximately 60 elements have been identified in the solar spectrum. The most abundant are listed in the table below, both with respect to the number of atoms or ions present, and with respect to the total mass of the atoms or ions. The Sun is clearly mostly hydrogen and helium, with only a trace of heavier elements. Solar Elemental Abundances Element Number % Mass % Hydrogen 92.0 73.4 Helium 7.8 25.0 Carbon 0.02 0.20 Nitrogen 0.008 0.09 Oxygen 0.06 0.8 Neon 0.01 0.16 Magnesium 0.003 0.06 Silicon 0.004 0.09 Sulfur 0.002 0.05 Iron 0.003 0.14 The Sun The Solar abundance The element helium is the second most abundant in both the Sun and the Universe, but it is very difficult to trace it on the Earth. In fact, helium was discovered in the spectrum of the Sun (the name helium derives from helios, which is the Greek name for the Sun). It was postulated that a set of spectral lines observed in the Solar emission spectrum that could not be associated with any known element belonged to a new element (the Sun is too cool to ionize helium appreciably, so absorption lines associated with helium are very weak). The Sun The Solar abundance The first evidence of helium was observed on August 18, 1868 as a bright yellow line with a wavelength of 587.49 nanometers in the spectrum of the chromosphere of the Sun. The line was detected by French astronomer Pierre Janssen during a total solar eclipse in Guntur, India. Only after this was helium discovered on the Earth and this hypothesis confirmed (helium occurs in certain very deep gas wells on the Earth). The Sun The Solar abundance The Sun is classified as a typical main-sequence star of spectral class G2, which contains mostly hydrogen (X = 0.74), some helium (Y = 0.24), and a little metal (Z = 0.02) by mass. Using a solar model, it is found that the hydrogen mass fraction in the core has decreased from 0.71 to 0.34 and the helium mass fraction increased from 0.27 to 0.64 over the Sun’s lifetime. At the surface the hydrogen mass fraction increased by 0.03, while the helium mass fraction decreased by 0.03. The Sun The Solar neutrinos Sun gains practically all its energy from the reaction 4p → α + 2e+ + 2ν ↔ 4H → 4He+ 2e+ + 2ν p-p chain: yields about 99.2% of energy in Sun CNO cycle : 0.8% of energy released in present day (dominant form of energy release in hotter stars) Both chains yield a total energy Q of 26.7 MeV per He nucleus, mainly in form of γ-radiation Qγ (which is absorbed and heats the gas) and neutrinos Qν (Which escapes from the Sun) The Sun The Solar neutrinos Neutrinos, ν, are produced at various stages of the pp-chain. In 1970, Raymond Davis began measuring the neutrino flux from the Sun. The neutrino has a very low cross section for interaction with other matter, which makes it difficult to detect. The main reaction that accounted for 77% of the neutrinos detected: This Homestake 37Cl experiment gave a value of 2.1 ± 0.3 SNU (Solar neutrino units; 1SNU =1/1036 target atoms) The Sun The Solar neutrinos The standard solar model predicted a capture rate of 7.9 SNU - inconsistency with the observations - The solar neutrino problem Other experiments have confirmed the discrepancy including Japan’s underground Super-Kamiokande observatory. This observatory detects the Cerenkov light produced when neutrinos scatter electrons. The deficit of neutrinos suggested that either (a) some fundamental physical process in the solar model is incorrect or (b) something happens to the neutrinos in-transit (i.e., on their way from the Sun’s core to the Earth) The Sun The Solar neutrinos The Mikheyev-Smirnov-Wolfenstein effect involves transformation of neutrinos between the three types. This idea is an extension of the electroweak theory of particle physics. Confirmation of this idea came in 1998 when Super-Kamiokande was used to detect atmospheric neutrinos produced in Earth’s upper atmosphere. Cosmic rays are capable of producing only electron and muon neutrinos. Sudbury Neutrino Observatory) in Sudbury, Canada uses D2O and can detect not just the electron neutrino, but 𝜈𝜇 and 𝜈𝜏 as well. The neutrinos aren’t missing!!! The Sun The Solar neutrinos 𝜈e neutrinos produced in the Sun just convert into 𝜈𝜇 and 𝜈𝜏 For neutrino mixing to occur, the neutrino could not be a massless particle The neutrino has a small rest mass (10-8 me), which allows it oscillate between the three flavours. The confirmation by measuring anti-neutrinos from power plant (with Super- kamiokande) Nobel Prize 2015 for the discovery of neutrino oscillations, which shows that neutrinos have mass The Sun’s structure Six regions The core Radiative zone Convection zone Photosphere (the visible surface) Chromosphere (the lower atmosphere) Corona NASA (the upper atmosphere) The Sun’s structure Solar interior The Sun’s Structure Below the Sun’s (optical) surface Solar interior: Hydrogen-burning core Everything below the Radiation zone Sun’s (optical) surface Convection zone Divided into hydrogen-burning Solar atmosphere core, radiation and convection zones Directly observable part of the Sun Solar atmosphere: Photosphere Directly observable Chromospherepart of the Sun Corona Divided into photosphere, Heliosphere chromosphere, corona, heliosphere  The Sun’s structure Six regions 5.4 SOLAR ATMOSPHERE The temperature and density of the Sun change from its interior to the surface. The Sun’s atmosphere is divided into two layers namely, the chromosphere and the corona. A schematic diagram of these layers is shown in Fig. 5.7. Corona 2600 km This picture summari