Welcome Back to PHYS 3368

Principles of Astrophysics and Cosmology

Henry Norris Russell October 25, 1877– February, 18 1957

Principles of Astrophysics & Cosmology - Professor Jodi Cooley Announcements

- Reading Assignments: Chapter 3.3 - 3.12. - Problem Set 4 is due Wednesday, February 18th, 2015. - Next lab is Monday, February 23rd. Be sure to report to FOSC 032 that day.

Principles of Astrophysics & Cosmology - Professor Jodi Cooley Goals for this Class

1. Calculate how long it takes a photon to travel from the center of the sun and emerge at its surface. 2. Estimate the luminosity of the sun. 3. Derive equations for the gas and radiation pressure inside a star.

Principles of Astrophysics & Cosmology - Professor Jodi Cooley basicastro4 October 26, 2006

STELLAR PHYSICS 37

A density profile, ρ(r), that falls with radius will give a somewhat more negative 2 value of Egr. Taking a characteristic Egr GM /r, the mean pressure in the Sun is ∼− 2 2 1 GM GM 15 2 9 P¯ ⊙ = ⊙ 10 dyne cm− = 10 Atm. (3.30) ⊙ 4 3 4 ∼ 3 3 πr r 4πr ≈ ⊙ ⊙ ⊙ To find a typical temperature, which we will call the virial temperature, let us assume again a classical nonrelativistic ideal gas, with particles of mean mass m¯ . Equation 3.27 then applies, and 3 1 GM 2 1 GM Nm¯ NkT ⊙ = ⊙ . (3.31) From Last Time:2 vir ∼ 2 r 2 r ⊙ ⊙ The mass of an electron is negligibly small, only 1/2000 compared to the mass Mean mass m of a proton. For an ionized hydrogen gas, consisting≈ of an equal number of protons and electrons, the mean mass m¯ , m + m m m¯ = e p = H , (3.32) 2 2 is therefore close to one-half the mass of the proton or exactly one-half of the 24 hydrogen atom, m =1.7 10− g. The typical thermal energy is then H × 8 33 24 TheGM meanmH mass6 .is7 1/210 the− cgs mass2 of the10 hydrogeng 1.7 atom.10− g kTvir ⊙ = × × × × × (3.33) ∼Since,6r the me << mp, the mean6 mass7 10 of10 hydrogencm is ⊙ × × close to 1/2 the mass of a10 proton. =5.4 10− erg = 0.34 keV. × 16 1 5 1 With k =1.4 10− erg K− =8.6 10− eV K− , this gives a virial temperature of about 4× 106 K. As we will see,× at temperatures of this order of magnitude, nuclear reactions× can take place, and thus replenish the thermal energy that the star radiates away, halting the gravitational collapse (if only temporarily). Of course, in reality, just like P (r), the density ρ(r) and the temperature T (r) are also functions of radius and they grow toward the center of a star. To find them,Principles we of Astrophysics will need & Cosmology to define - Professor additionalJodi Cooley equations. We will see that the equation of hydrostatic equilibrium is one of four coupled differential equations that determine stellar structure.

3.2 MASS CONTINUITY

In the hydrostatic equilibrium equation (Eq. 3.19), we have M(r) and ρ(r), which are easily related to each other: dM(r)=ρ(r)4πr2dr, (3.34) or dM(r) =4πr2ρ(r). (3.35) dr Although this is, in essence, merely the deﬁnition of density, in the context of stellar structure this equation is often referred to as the equation of mass continuity or the equation of mass conservation. basicastro4 October 26, 2006

38 CHAPTER 3

Figure 3.2 A volume element in a ﬁeld of targets as viewed in perspective (left). The target number density is n, and each target presents a cross section σ. From the base of the cylindrical volume, ndx targets per unit area are seen in projection (right). A straight line along the length of the volume will therefore intercept, on average, nσdx targets.

3.3 RADIATIVE ENERGY TRANSPORT

The radial gradient in P (r) that supports a star is produced by a gradient in ρ(r) and T (r). In much of the volume of most stars, T (r) is determined by the rate at which radiative energy flows in and out through every radius, i.e., the luminosity L(r). To find theRadiative equationbasicastro4basicastro4 that October determines 26, OctoberEnergy 2006 26,T 2006(r), we Transport will need to study some of the basics of “radiative transfer”, the passage of radiation through matter. In some of the volumeThe of radial some gradient stars, in thepressure energy that transportsupports a star mechanism is produced that by a dominatesgradient in is convection,38 ratherdensity than and temperature. radiative transport. Pressure, density We will andCHAPTER discuss temperature 3 convection are functions is Section of radius. 3.12. 38 CHAPTER 3 Energy transport by means of conduction plays a role only in dense stellar remnants – white dwarfs and neutron stars – which are discussed in Chapter 4. σ = effective cross Photons in stars can be absorbed or scattered out of a beamsection via interactions for absorption with molecules, with atoms (either neutral or ions), and with electrons.or scattering. If a photon traverses a path dx filled with “targets” with a number density n (i.e., the number of targets per unit volume), then the projected number of targets per unit area lying in the pathFigure of 3.2 the A volume photon element in is a fieldndx of targets(see as viewed Fig. in perspective3.2). If (left). each The target target poses an effective “cross A numberphoton density transverses is n, and each target a presentspath adx, cross section σ. FromProjected the base of number of targets section”2 Figureσthefor cylindrical 3.2 absorption A volume volume, elementndx targets inor a fieldper scattering, unit of area targets are as seen viewed in then projection in perspectiveper the (right). unit fraction A (left). area The lying of target the in the area path covered by filledstraight with linenumber along “targets” densitythe length is ofn ,with the and volume each a target will therefore presents intercept, a cross section on average,σ. From the base of targets is σnumberndxnσdx targets..the Thus,density. cylindrical the volume, numberndx targets of per targets unit area are thatof seen the will in projection photon typically (right). is given A be intersectedas ndx. by a straight line along the length of the volume will therefore intercept, on average, straight3.3 RADIATIVE line traversing ENERGYnσdx targets. TRANSPORT the path dx, or in other words, the number of interactions the photonFor undergoes, a straight willline along be the length of the path The radial3.3 gradient RADIATIVE in P (r) that ENERGY supports TRANSPORT a star is produced by a gradient in ρ(r) and T (r). In much of the volume of# most of stars, interactionsT (r) is determined by = then rateσdx. at (3.36) which radiative energy flows in and out through every radius, i.e., the luminosity The radial gradient in P (r) that supports a star is produced by a gradient in ρ(r) L(r). To find the equation that determines T (r), we will need to study some of the EquationPrinciples 3.36 of Astrophysics defines & Cosmology the - Professor concept Jodi Cooley of cross section. (Cross section can be defined basics ofand “radiativeT (r). In transfer”, much of the the passage volume of radiation of most stars,throughT matter.(r) is determined In some of by the rate at equivalentlythe volumewhich of as some radiative the stars, ratio energy the energy flows between transport in and mechanism out the through interaction that every dominates radius, is i.e., rate con- the per luminosity target particle and the vection,L rather(r). To than find radiative the equation transport. that We determines will discussT convection(r), we will is Section need to 3.12. study some of the incomingEnergy transportbasics flux of of by “radiative means projectiles.) of conduction transfer”, plays the Settingpassage a role only of inradiation the dense left-hand stellar through remnants matter. side In some equal of to 1, the typical distance– white athe dwarfs photon volume and neutron of will some stars travel stars, – which the between energy are discussed transport interactions in Chapter mechanism 4. that is dominates called is the con- “mean free path”: Photonsvection, in stars rather can be than absorbed radiative or scattered transport. out We of a will beam discuss via interactions convection with is Section 3.12. molecules,Energy with transport atoms (either by means neutral of or conduction ions), and plays with electrons. a role1 only If in a dense photon stellar remnants traverses– a white path dx dwarfsfilled withand neutron “targets” stars with a – number which aredensityl = discussedn (i.e.,. the in Chapter number of 4. (3.37) targets per unit volume), then the projected number of targetsn perσ unit area lying in Photons in stars can be absorbed or scattered out of a beam via interactions with the path of the photon is ndx (see Fig. 3.2). If each target poses an effective “cross section”molecules,2 σ for absorption with atoms or scattering, (either then neutral the fraction or ions), of and the area with covered electrons. by If a photon traverses a path dx filled with “targets” with a number density n (i.e., the number of 2 targets is σndx. Thus, the number of targets that will typically be intersected by a Thestraight crosstargets line section traversing per unit of the a volume), particle, path dx, then or which in the other projected has words, units thenumber number of area, of targets of interactions quantifies per unit area the lying degree in to which the particle is liable tothe take photonthe part undergoes, path in of a the particular will photon be is ndx interaction(see Fig. 3.2). (e.g., If a each collision target poses or a an reaction) effective “cross with some other particle. 2 section” σ for absorption# of interactions or scattering, = nσdx. then the fraction of(3.36) the area covered by Equationtargets 3.36 defines is σndx the. Thus, concept the of number cross section. of targets (Cross that section will typically can be defined be intersected by a equivalentlystraight as the line ratio traversing between the the path interactiondx, or rate in other per target words, particle the number and the of interactions incomingthe flux photon of projectiles.) undergoes, Setting will be the left-hand side equal to 1, the typical distance a photon will travel between# interactions of interactions is called = then “σmeandx. free path”: (3.36) 1 l = . (3.37) Equation 3.36 defines the conceptnσ of cross section. (Cross section can be defined equivalently as the ratio between the interaction rate per target particle and the incoming flux of projectiles.) Setting the left-hand side equal to 1, the typical 2The cross section of a particle, which has units of area, quantifies the degree to which the particle is liable to takedistance part in a a particular photon interaction will travel (e.g., betweena collision or interactions a reaction) with is some called other the particle. “mean free path”: 1 l = . (3.37) nσ

2The cross section of a particle, which has units of area, quantiﬁes the degree to which the particle is liable to take part in a particular interaction (e.g., a collision or a reaction) with some other particle. basicastro4 October 26, 2006

38 CHAPTER 3

3.3 RADIATIVE ENERGY TRANSPORT

3Note the inverse square dependence of the Thomson cross section on the electron mass. For this reason, protons and nuclei, which are much heavier, are much-less-effective photon scatterers. Similarly, the relevant mass for electrons bound in atoms is the mass of the entire atom, and hence bound electrons pose a very small Thomson scattering cross section. basicastro4 October 26, 2006

STELLAR PHYSICS 39

basicastro4 October 26, 2006

Figure 3.3 Thomson scattering of a photon on a free electron. STELLAR PHYSICS 39

More generally, the stellar matter will consist of a variety of absorbers and scatterers, each with its own density n and cross section σ . Thus i i basicastro4 October 26, 2006 1 1 l = , (3.38) Thomsonniσi ≡ ρκScattering where we have used the fact that allP the particle densities will be proportional to the mass density ρ, to define the “opacityFigure” 3.3κ. Thomson The opacity scattering obviously of a photon has on a cgs free units electron. 2 1 STELLAR PHYSICS 39 of cm g− , and will depend on the local density, temperature, and element abundance.In Thomson ScatteringMore generally, a photon the stellar is matter will consist of a variety of absorbers and scatter- Wescattered will return off later a free toers, the electron. each various with processes its own density thatn producei and cross opacity. section σi However,. Thus to get an idea of the magnitude of the scattering process, let1 us consider1 one of the l = , (3.38) importantThe cross interactions section – Thomson for this scatteringprocess isof given photons ni onσi ≡ freeρκ electrons (see Fig. 3.3). where we have used the fact that allP the particle densities will be proportional to Theby: Thomson cross sectionthe mass is density ρ, to define the “opacity” κ. The opacity obviously has cgs units 2 1 of cm g− , and2 will depend on the local density, temperature, and element abun- 2 Figure 3.3 Thomson scattering of a photon on a free electron. dance.8π e Note:25 Thomson2 cross section is independent σ = =6.7 10− cm . (3.39) T 2 of temperature and photon energy. We3 willme returnc later to the× various processes that produce opacity. However, to get anµ idea of the∂ magnitudeMore generally, of the scattering the stellar process, matter will let consist us consider of a variety one of of the absorbers and scatter- It is independent of temperature and photon energy.3 In the hot interiors of stars, the important interactionsers, – Thomson each with its scattering own densityofn photonsi and cross on section free electronsσi. Thus (see gas isCalculate fully ionized the and Thomson thereforeFig. 3.3). free cross electrons section are abundant.(using cgs If we units): approximate,1 1 l = , (3.38) for simplicity, that the gas isThe all Thomson hydrogen, cross then section there is is one electron per atomniσi of≡ ρκ 10 2 2 mass mH , and 8⇡ (4.8 10 esu)8π e2 = ( where we have used)2 the fact that25 allP the2 particle densities will be proportional to T 28 ⇥ σT =10 1 22 =6.7 10− cm . (3.39) 3 9.1 10 g (3 ρthe10 masscm3 density sm e)c ρ, to define× the “opacity” κ. The opacity obviously has cgs units n . 2 µ1 ∂ (3.40) ⇥ e⇥ ⇥of cm g− , and will depend on the local density, temperature, and element abun- It is independent≈ ofm temperatureH and photon energy.3 In the hot interiors of stars, the dance. The meanNote: free path 1 esu for = electron gas1cm/sqrt(dyne) is fully scattering ionized is and thenWe therefore = willg1/2 return free cm electrons later3/2 tos-1 the are various abundant. processes If we that approximate, produce opacity. However, to for simplicity, that theget gas an24 is idea all hydrogen, of the magnitude then there of the is onescattering electron process, per atom let us of consider one of the 1 mH -11 .7 10− g 2 -2 1 dyne mass= ergmH cm, and ; 1 ergimportant = g interactionscm s – Thomson scattering of photons on free electrons (see les = 3 × 25 2 2 cm, (3.41) neσT ≈ ρσT ≈ 1.4 g cm− Fig.6.7 3.3).10− cm ρ≈ × × ne . (3.40) wherePrinciples of we Astrophysics have used& Cosmology the - mean Professor mass Jodi Cooley density ofThe the Thomson Sun calculated cross≈ m sectionH previously. is In The mean free path for electron scattering is then 2 2 reality, the density of the Sun is higher than average in regions8π wheree electron 25 2 σT = 24 2 =6.7 10− cm . (3.39) scattering is the dominant source of opacity,1 m whileH in other1 regions.7 10− other3 g m processes,ec × les = × µ ∂ 2 cm, (3.41) n σ ≈ ρσ ≈ 1.4 g cm 3 6.7 10 25 cm2 ≈ 3 apart from electron scattering, are important.e T It As isT independent a result, the of− temperature× actual× typical and− photon photon energy. In the hot interiors of stars, the mean free path is even smaller,where weand have is l used1 themm.gas mean is fully mass ionized density and therefore of the Sun free calculated electrons previously. are abundant. In If we approximate, reality, the density≈ offor the simplicity, Sun is higher that the than gas average is all hydrogen, in regions then where there electron is one electron per atom of scattering is the dominantmass sourcemH , and of opacity, while in other regions other processes, 3 apart from electron scattering, are important. As a result, the actualρ typical photon Note the inverse square dependence of the Thomson cross section on the electron mass.ne For this. (3.40) mean free path is even smaller, and is l 1 mm. ≈ mH reason, protons and nuclei, which are much heavier, are much-less-effective≈ photon scatterers. Similarly, the relevant mass for electrons bound in atoms is the massThe of mean the entire free atom, path for and electron hence bound scattering electrons is then pose a very small Thomson scattering cross section. 24 3Note the inverse square dependence of1 the ThomsonmH cross section on1. the7 electron10− mass.g For this les = 3 × 25 2 2 cm, (3.41) reason, protons and nuclei, which are muchne heavier,σT ≈ areρσ much-less-effectiveT ≈ 1.4 g cm− photon6. scatterers.7 10− Similarly,cm ≈ the relevant mass for electrons bound in atoms is the mass of the entire atom, and× hence× bound electrons pose a very small Thomsonwhere scattering we cross have section. used the mean mass density of the Sun calculated previously. In reality, the density of the Sun is higher than average in regions where electron scattering is the dominant source of opacity, while in other regions other processes, apart from electron scattering, are important. As a result, the actual typical photon mean free path is even smaller, and is l 1 mm. ≈

basicastro4 October 26, 2006

STELLAR PHYSICS 39 basicastro4 October 26, 2006 STELLAR PHYSICS basicastro4 October 26, 2006 39

STELLAR PHYSICS 39 basicastro4 October 26, 2006

STELLAR PHYSICS 39 STELLAR PHYSICS 39

STELLAR PHYSICS 39 Figure 3.3 Thomson scattering of a photon on a free electron. Figure 3.3 Thomson scattering of a photon on a free electron.

More generally, theMore stellarFigure generally, matter 3.3 Thomson will the consist scattering stellar of a mattervariety of a photon of will absorbers on consist a free and electron. of scatter- a variety of absorbers and scatterers, each with its own density n and cross section σ . Thus ers, each withi its own densityi ni and cross section σi. Thus More generally, the stellar matter1 will consist1 of a variety of absorbers and scatter- Figure 3.3 Thomson scattering of a photon on a freeFigure electron. 3.3l Thomson= scattering of, a photon on a free1 electron. 1(3.38) ers, each with its own density nniiσandi ≡ crossρκ section σi. Thus Figure 3.3 Thomson scattering of a photonl = on a free electron. , (3.38) niσi ≡ ρκ where we haveMore used generally, the fact the stellar that all matter the particle will consist1 densities of a variety1 will beof absorbers proportional and scatter- to More generally, the stellar matter will consist of a variety of absorbersP andl scatter-= , (3.38) the mass densityers, eachρwhere, with to define its we own the have density “opacity usedn and” κ the. cross Then factσ section opacity thatρκσ obviously all. Thus the particle has cgs units densities will be proportional to More generally, the stellar matteri will consisti i of≡ a varietyi of absorbers and scatterers, each with its own density ni and cross2 section1 σi. Thus P of cm g− , and willthe depend mass density on the localρ, to density, define1 temperature, the1 “opacity and element” κ. The abun- opacity obviously has cgs units whereers, we each have with used its own the density factn thatli and= allP cross the section particleσ,i. densities Thus will be proportional(3.38) to dance.1 1 2 1 n σ ≡ ρκ l = the mass, densityof cm ρg, to− define, and the will “(3.38)opacity depend1 i i ” κ on1. The the opacity local obviously density, temperature, has cgs units and element abun- Weniσ willi ≡ returnρκ2 later1 to the various processesl = that produce, opacity. However,(3.38) to of cmwhereg− wedance., have and willused depend the fact that on the allP thelocalniσ particlei ≡ density,ρκ densities temperature, will be proportional and element to abun- where we have used the fact that allget the an particle ideadance.the ofwhere mass the densities we magnitude density have will usedρ, to be of the define proportional the fact scattering the that “opacity all the to process, particle” κ. The densities let opacity us consider will obviously be proportional one has of cgs the units to P 2 1We will return laterP to the various processes that produce opacity. However, to the mass density ρ, to define the “opacityimportant” κ.We interactionsof Thethe cm will mass opacityg− return density, and – obviouslyThomson later willρ, to depend to define the has scattering the on various cgs “ theopacity units local processesof” density,κ photons. The opacity that temperature, on produce free obviously electrons and opacity. has element cgs (see unitsHowever, abun- to 2 1 Fig. 3.3). dance. 2 1 of cm g− , and will depend on the local density,get anof cm idea temperature,getg− of, an the and idea willmagnitude and depend of element the on of magnitude the the abun- local scattering density, of temperature, process, the scattering let and us element consider process, abun- one let of the us consider one of the dance.We will return later to the various processes that produce opacity. However, to dance. The Thomsonimportant crossimportant interactions section isinteractions – Thomson scattering – Thomsonof photons scattering on freeof electrons photons (see on free electrons (see getWe an idea will return of the later magnitude to the various of2 the scattering processes that process, produce let opacity.us consider However, one of to the We will return later to the various processesFig. 3.3). that produce opacity.8π e2 However, to importantget anFig. ideaσ interactions of 3.3).= the magnitude – Thomson of the=6 scattering scattering.7 10 process,25ofcm photons2. let us on consider free electrons one(3.39) of the (see get an idea of the magnitude of the scattering process, letT us consider2 one of the − TheFig.important Thomson3.3). The interactions cross Thomson3 section –mThomsonec cross is scattering section× of is photons on free electrons (see important interactions – Thomson scattering of photons on freeµ electrons∂ (see Fig.The 3.3). Thomson cross section is 2 2 3 It is independent of temperature and8 photonπ e energy. In the hot2 interiors252 2 of stars, the Fig. 3.3). Let’s simplify.The Thomson Assume crossσ section= all the is gas is2 hydrogen8=6π .7 e10 andcm that. there is one(3.39) T 8π e2 2 − 25 2 gas is fully ionized and therefore free3 electronsmσ2eTc =2 are abundant.× 25 If we2 =6 approximate,.7 10− cm . (3.39) The Thomson cross section is electron per atom of massσT = 8 mπ µ. e ∂=6.7 10− 2cm . (3.39) for simplicity, that the gas is allσ hydrogen,= 3 Hm c then2 =6 there3.7 is×10 oneme25c electroncm2. per atom×(3.39) of 2 2 T e 2 −3 8π e It is independent of temperature3 µm andec photon∂ energy.×µ In∂ the hot interiors of stars, the mass mH , and 25 2 µ ∂ 3 3 σT = 2 =6gasIt.7 is is fully independent10It− is ionizedcm independent of. and temperature therefore of and temperature(3.39) free photon electrons energy. and3 areIn the photon abundant. hot interiors energy. If we of stars, approximate,In the the hot interiors of stars, the 3 mec It× is independent of temperature andρ photon energy. In the hot interiors of stars, the µ ∂ forgas simplicity, is fully ionized that the and gas thereforen is all hydrogen, free. electrons then are there abundant. is one If we electron approximate,(3.40) per atom of gas is3 gas fully isionized fully and ionized thereforee andfree electrons therefore are abundant. free electrons If we approximate, are abundant. If we approximate, It is independent of temperature and photon energy.for simplicity,In the hot that interiors the gas is of all≈ stars, hydrogen,mH the then there is one electron per atom of massform simplicity,H , and that the gas is all hydrogen, then there is one electron per atom of gas is fully ionized and therefore freeThe electrons mean freemassmass are pathmm abundant.forH , for, and and simplicity, electron If we scattering approximate, that the is then gas is all hydrogen, then there is one electron per atom of Calculate theH mean free path for electronρ ρ scattering in this star. for simplicity, that the gas is all hydrogen, then theremass is onem electronH , and per atomnn ofe 24ρ . . (3.40)(3.40) 1 mH 1.7 ne10e −≈ mg. (3.40) ≈ mmH H -3 mass mH , and Assumeles = the mean density to be3 × that≈ ofH the25 sun,2 1.42 cmg cmρ, . (3.41) Then meaneσT ≈ freeρσ pathT ≈ for1.4 electron g cm− scattering6.7 10 is− thencm ≈ ρ TheThe mean mean free free path path for for electron electron scattering scattering× × is is then then ne . (3.40) wherene we have. used the mean mass density(3.40) of the Sun2424 calculated24 ≈ previously.mH In 111 mmmH 11..717.7 101010− gg g ≈ mH l = HH − − 2 cm, (3.41) reality, the densityles es=les = of the Sun is higher than average33 ××× in regions2525 2 2 where2 cm, electron2 cm(3.41), (3.41) Thenne meanσσT ≈≈ ρσρσ freeT ≈ path1..44 g g for cm cm electron3 66.7.7 10 scattering10 cmcm25 ≈≈ is2 then The mean free path for electron scattering is then neσeT T≈ ρσTT ≈ 1.4 g cm−−−×× 6×.×7 −−10− cm ≈ scattering iswherewhere the dominant we we have have used source used the the of mean opacity, mass mass while density density in× of of other the the× Sun regions Sun calculated calculated other previously. processes, previously.24 In In 1 m 1.7 where10 24 weg have used the mean1 massmH density of the Sun1. calculated7 10− previously.g In H apart fromreality, electronreality,− the the scattering, density densityl = of of are the the important. Sun is is higher higher As thana than result, average average the inactual in regions regions typical where where photon electron electron 2 cm, (3.41) les = 3 ×reality, the25 densityes2 of2 the cm, Sun is(3.41) higher than average in3 regions× where25 electron2 neσT ≈ ρσT ≈ 1.4 gmean cm− free6 path.scattering7scattering is10 even− is is thecm smaller, the dominant dominant≈ andneσ source isTl ≈ of of1ρσ opacity, opacity,mm.T ≈ while while1.4 in in otherg other cm regions regions other6 other.7 processes, processes,10 cm ≈ ×scattering× is the dominant source≈ of opacity, while in other− regions other processes,− where we have used the mean mass density ofapartapart the from from Sun electron electron calculated scattering, scattering, previously. are are important. important. In As As a a result, result, the the actual actual× typical typical× photon photon apart fromwhere electron we scattering, have used are important. the mean As mass a result, density the actual of the typical Sun photon calculated previously. In reality, the density of the Sun is higherIn thanreality,meanmean average freethere free path inpath regionsare is is even even regions smaller, smaller, where where and andelectron is isll electron11mm.mm. scattering is more 3Note themean inverse free squarereality, path dependence is even the smaller, density of the Thomson and of is the cross≈l≈ Sun1 sectionmm. is on higher the electron than mass. average For this in regions where electron scattering is the dominant source ofreason, opacity,prominent protons while and in nuclei, otherand which regionsregions are much other where heavier, processes, are it much-less-effective is less≈ important. photon scatterers. As a Similarly,result, the 3Notescattering the inverse square is dependence the dominant of the Thomson source cross section of opacity, on the electron while mass. For in this other regions other processes, apart from electron scattering, are important.the relevanttypical mass As 3photonNote for a result, electrons the inverse themean bound actualsquare in free atoms dependence typical path is the photon of mass theis of1Thomson themm. entire cross atom, section and on hence the electron bound electrons mass. For this pose a very smallreason, Thomson protons scattering and nuclei, cross which section. are much heavier, are much-less-effective photon scatterers. Similarly, mean free path is even smaller, and is l 1 mm.3reason,Notethe relevant the protonsapart inverse mass and fromfor nuclei, square electrons which electron dependence bound are much in atoms scattering, of heavier, the is the Thomson aremass much-less-effective of the are cross entire important. section atom, and photon on hence the scatterers. boundAselectron a electrons Similarly, result, mass. For the this actual typical photon ≈ reason,thepose relevant protons a very mass small and for nuclei, Thomson electrons which scattering bound are muchin cross atoms section. heavier, is the mass are much-less-effective of the entire atom, and photon hence bound scatterers. electrons Similarly, In whichpose aareas verymean small of the Thomson free star path scatteringdo you is crosseven suppose section. smaller, electron and scattering is l 1 ismm. more dominant? the relevant mass for electrons bound in atoms is the mass of the entire≈ atom, and hence bound electrons 3Note the inverse square dependence of the Thomsonpose cross a very section small Thomson on the electron scattering mass. cross For section. this reason, protons and nuclei, which are much heavier,Principles are of Astrophysics much-less-effective & Cosmology - photon Professor scatterers. Jodi Cooley Similarly, the relevant mass for electrons bound in atoms is the mass of the entire3 atom,Note and the hence inverse bound square electrons dependence of the Thomson cross section on the electron mass. For this pose a very small Thomson scattering cross section. reason, protons and nuclei, which are much heavier, are much-less-effective photon scatterers. Similarly, the relevant mass for electrons bound in atoms is the mass of the entire atom, and hence bound electrons pose a very small Thomson scattering cross section. basicastro4 October 26, 2006

basicastro4basicastro4 October October 26, 2006 26, 2006

40 basicastro4 October 26, 2006 CHAPTER 3

40 CHAPTER 3 40 CHAPTER 3

40 CHAPTER 3

basicastro4 October 26, 2006

Figure 3.4 The net advance of a photon performing a random walk consisting of N steps,

Figure 3.4 The net advance of a photon performingeach decribed a random by40 a walkvector consistingli, is the of vectorN steps,D. CHAPTER 3 eachSo, decribed we can by a see vector thatli, is the the vector photonsD. only travel a

tinyFigure distance 3.4 TheThus, before net advance photons being of a can photonscattered travel performing only or a a randomtiny distance walk consisting inside of theN steps, Sun before being scat- Thus, photons can travel only a tiny distance inside the Sun before being scat- Figure 3.4 The net advanceabsorbed of a photon andteredeach performing remitted decribed or absorbed by a ain vector random a and newli, re-emitted is walk direction. the vector consistingD in. a new of N direction.steps, Since the new direction is tered or absorbed and re-emitted in a new direction. Since the new direction is each decribed by a vector l , is the vector D. random, the emergence ofrandom,i photons the from emergence the Sun is necessarily of photons a fromrandom the walk Sun is necessarily a random walk process. The vectorThus,Thus,D photonsdescribingprocess. photons in the can the The change travel sun vector only infollow positionD a tinydescribing a distance ofrandom a photon the inside change after theN Sunsteps, in before position being of a scat- photon after N steps, tered or absorbed and re-emitted in a new direction. Since the new direction is each described by a vectoreachl having described length byl and a vector randoml orientationi having length (see Fig.l and 3.4), random orientation (see Fig. 3.4), Thus, photons can travelwalkrandom, only process. a the tinyi emergence distance of inside photons the from Sun the before Sun is necessarily being scat- a random walk is is tered or absorbed and re-emittedprocess. The in a vector newD direction.describing the Since change the in newposition direction of a photon is after N steps, D = l1 + l2 + l3 + ... + lN.D = l + l + l (3.42)+ ... + l . (3.42) random, the emergence ofeach photons described from by a the vector Sunli having is necessarily length l and1 a randomrandom2 orientation3 walk (seeN Fig. 3.4), The square of theis linear distanceThe square covered of is the linear distance covered is process. The vector D2 describing2 the2 change in2 position of a photon after N steps, D =Thel1 square+ l2 + of... the+ l Nlinear2 +D 2( distancel=12 l l2++l l21 +coveredll3 ++...... )+ ,isl 2. (3.43) (3.42) | | | | |D | = l1 ·+1 l22 +· 3... + lNN + 2(l1 l2 + l1 l3 + ...), (3.43) each describedand by its expectation a vector Theli valuehaving square is oflength the linearl and distance| random| covered| orientation|Figure is 3.4 The| (see net| advance Fig. of 3.4), a· photon performing· a random walk consisting of N steps, and its expectation value is each decribed by a vector li, is the vector D. is D2 = l D2 +2 l= Nl2 +2.... + l 2 + 2(l l + l l (3.44)+ ...), (3.43) | 1| | 2| | N| 1 · D2 2 =1 ·Nl3 2. (3.44) The expectation valueDand= its ofl expectation1 the+ terml2 + inl value3 parentheses+ ... is + lN in. Eq. 3.43Thus, is photons zero because can travel(3.42) it is only a a tiny distance inside the Sun before being scat- 2 This2 term approaches zero sum over many vector dotThe products, expectation each with value a random ofD thetered angle,= termNl or and absorbed. in hence parentheses and with re-emitted both in Eq. in a 3.43 new(3.44) direction. is zero because Since the it new is a direction is The square of the linear distance covered is for a large number of steps. 2 positive2 and negative2The cosines expectationsum contributing over2 value many of equally.the vector term The in dot parenthesesrandom, linear products, distance the in each emergence Eq. covered 3.43 with is of a in zero random photons a because from angle, it the is and Sun a is hence necessarily with a bothrandom walk D random= l1 walk+ l is2Thus, thereforesum+ ... over we+positive many canlN vectorwrite+ and 2( dot negativel1 products,l2 + cosinesl1 eachprocess.l3 with+ contributing... a The random), vector angle,D equally.describing(3.43) and hence The the linear with change both distance in position covered of a photon in after a N steps, | | | | | |2 1/2 · · and its expectation value ispositive andrandom negativeD walk cosines= D is= therefore contributing√Nl. each equally. described The by linear a vector(3.45) distanceli having covered length l inand a random orientation (see Fig. 3.4), is random walkD2 is therefore= Nl2. and D2 1/2 = D =(3.44)√Nl. (3.45) To gain some intuition, it is instructive to calculateD2 1/2 = howD= long√Nl. it takes a photonD = l1 + l2 +(3.45)l3 + ... + lN. (3.42) The expectation value of the termexpectation in parentheses value in Eq. 3.43The islinear square zero distance of because the linear covered distance it is in a random covered walk is to travel from the center of the Sun, where most of the energy is2 produced,2 to2 the 2 4 To gain some intuition, it is instructiveD = 2l1 2+ tol2 calculate+ ... + lN how+ long 2(l1 l it2 + takesl1 l3 a+ photon...), (3.43) sum over manysurface. vectorFromPrinciples dot Eq. products, of AstrophysicsTo 3.45, gain traveling & someeach Cosmology intuition, with a - Professor distance a random Jodi it isCooleyr instructivewill angle, require to and calculateN hence= r| /l| how withsteps| long| both in it takes| a| photon · · to travelto from travel the center from of the the center⊙ Sun, where ofand the its most expectation Sun, of where the⊙ energy value most is is produced,of the energy to the is produced, to the positive andthe negative random cosines walk. Each contributing step requires equally. a time l/c The. Thus, linear the total distance time for covered the photon in a 2 2 4 4 2 D2 = Nl . 2 2 (3.44) to emerge from thesurface. Sun issurface.From Eq.From 3.45, traveling Eq. 3.45, a distance travelingr will a distance require Nr =willr /l requiresteps inN = r /l steps in random walk is therefore ⊙ ⊙ 2 2the random walk. Each10 step2 requires a timeThe expectationl/c. Thus, value the total of the time⊙ term for in the parentheses photon in Eq.⊙ 3.43 is zero because it is a l r r 2 the1/2(7 random10 walk.cm) Each step requires12 a time l/c. Thus, the total time for the photon ⊙ ⊙ D = D = √Nl. sum over many vector dot(3.45) products, each with a random angle, and hence with both τrw 2 = to= emerge1 from× the Sun is10 1 =1.6 10 s = 52, 000 yr. ≈ c l lc 10−to2cm emerge23 from10 thecm s Sun− is10positive×2 and negative cosines contributing equally. The linear distance covered in a l r ×r ×2 2(7 10 cm) (3.46)12 ⊙ ⊙ random walk is10 therefore2 τrw 2 = l r= 1r × 10(7 101 =1cm).6 10 s = 52, 000 yr. 12 ≈ c τl lc ⊙10=− cm⊙ = 3 10 cm s− × 2 =11/2 .6 10√ s = 52, 000 yr. To gain some intuition, it is instructiverw to2 calculate× how× 1 long× it takes10 a photonD1 = D = Nl. (3.45) ≈ c l lc 10− cm 3 10 cm s− (3.46)× to travel from4 theIn reality, center of course, of the it is Sun,not the same where photon most that travels of the this energy path. In every is× produced, interaction,× the photonto the (3.46) 4 can transfer energy to the particle it scatters on, or distribute its originalTo energy gain among some2 several intuition,2 photons it is instructive to calculate how long it takes a photon surface. From Eq. 3.45, traveling4 a distance r will require N = r /l steps in that emerge from the interaction.In reality, Hence of course, the photon it is not⊙ energy the same is stronglyphotonto that travel “degraded” travels from this during⊙ the path. center theIn every passage of theinteraction, Sun, where the photon most of the energy is produced, to the the random walk.through the Each Sun. stepcan requires transfer energy a time to thel/c particle. Thus, it scatters the on, total orsurface. distribute time4 From its for original Eq. the 3.45, energy photon traveling among several a distance photonsr will require N = r2 /l2 steps in 4In reality, of course, it is not the same photon that travels this path. In every interaction, the photon that emerge from the interaction. Hence the photonthe random energy is walk. strongly Each “degraded” step requires during a the time passagel/c.⊙ Thus, the total time for⊙ the photon to emerge from the Sun is through thecan Sun. transfer energy to the particle it scatters on, or distribute its original energy among several photons to emerge from the Sun is 2 2 that10 emerge from2 the interaction. Hence the photon energy is strongly “degraded” during the passage l r r (7 10 cm) 2 2 10 2 12l r r (7 10 cm) 12 τrw ⊙ = ⊙ = ×through the Sun. =1.6 10 ⊙s =⊙ 52, 000 yr. 2 1 10 1 τrw 2 = = 1 × 10 1 =1.6 10 s = 52, 000 yr. ≈ c l lc 10− cm 3 10 cm s− × ≈ c l lc 10− cm 3 10 cm s− × × × (3.46)× × (3.46)

4In reality, of course, it is not the same photon that travels this path. In every interaction, the photon 4In reality, of course, it is not the same photon that travels this path. Incan every transfer interaction, energy to the theparticle photon it scatters on, or distribute its original energy among several photons that emerge from the interaction. Hence the photon energy is strongly “degraded” during the passage can transfer energy to the particle it scatters on, or distribute its original energythrough the among Sun. several photons that emerge from the interaction. Hence the photon energy is strongly “degraded” during the passage through the Sun. Photon Travel Time

Calculate how long it takes for a photon to travel from the center of the sun to the surface. The time it takes for a photon to reach the surface is the time it takes in each step, times the number of steps in the random walk.

⌧rw = tstepN The time for each step is given by ` t = step c And N is given by our equation for linear distance covered by random walk.

D2 r2 D = pN` N = = sun ! `2 `2 Put it all together and simplify. 2 2 10 2 ` rsun rsun (7 10 cm) 12 ⌧rw = = = 10⇥ 1 10 =1.6 10 s = 52, 000 years 2 (3 10 cm s )(10 cm) ⇥ c ` c` ⇥

Principles of Astrophysics & Cosmology - Professor Jodi Cooley basicastro4 October 26, 2006

basicastro4 October 26, 2006

STELLAR PHYSICS 41

Figure 3.5 Radiative diffusion of energy between volume shells in a star, driven by the gra- Figure 3.5dient Radiative in the thermal diffusion energy of energy density. between volume shells in a star, driven by the gra- dientEquation in the thermal energy density.of Radiativebasicastro4 Transport October 26, 2006 Thus, if the nuclear reactions powering the Sun were to suddenly switch off, we wouldThus, not if notice the nuclear5 anything reactions unusual powering for 50,000 the Sun years. were to suddenly switch off, we Inside the sun, it is a good approximation that every volume element radiates as a Withwould this not background, notice5 anything we can unusual now for derive 50,000 the years. equation that relates the temperature profile,Withblackbody. thisT (r background,), to However, the flow we of can there radiative now is derive still energy a the net equation throughflow of that aradiation star. relates The theoutward. small tempera- mean This implies freeture paththat profile, of photons thereT (r is), inside a to higher the the flow Sundensity of radiativeand STELLAR theat smaller very energy PHYSICS numerous radii. through scatterings, a star. The small absorptions, mean 41 andfree re-emissions path of photons every inside photon the undergoes Sun and the reaffirms very numerous that locally, scatterings, every absorptions, volume ele- mentand inside re-emissionsL(r) the = net Sun flow every radiates of photon radiation as a undergoes blackbody through reaffirms to a a very that good locally, approximation. every volume How- ele- ever,ment theremass inside is a shell net the flow Sun at radius radiates of radiation r. as a blackbody energy outwards, to a very meaning good approximation. there is some How- small anisotropyever, there (a preferred is a net flow direction), of radiation and energy implying outwards, there is meaning a higher there energy is some density, small at anisotropy (a preferredexcess direction), energy and implying there is a higher energy density, at smaller radii thanL( atr)= larger radii. The net flow of radiation energy through a mass smaller radii than at largertransit radii. time TheFigure net flow 3.5 of Radiative radiation diffusion energy of energy through between a mass volume shells in a star, driven by the gra- shell at radius r, per unit time, is just L(r) (seedient Fig. in the 3.5). thermal This energy must density. equal the shell at radius ∆r,u per = excess unit time, energy is just densityL(r) (see Fig. 3.5). This must equal the excessexcess energy energy in the in the shell, shell, compared compared to to a a shell shell at at larger larger radius, divided divided by by the the time time it takes this excess energyr2 to flow across the shell’s width ∆r. The excess energy it takes this excess⌧rw energy= to= flowtransit across Thus,time the for if shell’s the photons nuclear width reactionsto∆ crossr. The powering shell excess theenergy Sun were to suddenly switch off, we 5 densitydensity is ∆ isu∆, whichu, which multiplied multipliedc` by by the thewould shell’s shell’s not volume volume notice gives givesanything the the unusual total total excess excess for 50,000 energy energy years. 2 of theof the shell,Thus, shell,4πr 4weπ∆r2 rcan∆∆ru∆ write:.u The. The time time for for the theWith photons photons this background, to to cross cross the thewe canshell shell now in in a derive a random random the equation that relates the tempera- 2 walkwalk is (∆ is r(∆) r/lc)2/lc. Thus. Thus ture profile, T (r), to the flow of radiative energy through a star. The small mean free path of photons inside the Sun and the very numerous scatterings, absorptions, 4πr2∆2 r∆u ∆u 4πr ∆andr∆ re-emissionsu 22 every∆u photon undergoes reaffirms that locally, every volume ele- L(Lr)(r) 2 == 44ππrr lclc .. WARNING:(3.47)(3.47) this is an approximation ≈−≈−(∆(∆r)r)ment/lc2/lc inside−− the Sun∆ radiatesrr as a blackbody to a very good approximation. How- A moreA more rigorous rigorous derivation derivation of of this this equation equationever, adds there adds isa a factor a factor net flow 1/3 of on on radiation the the right right energy hand hand side, outwards, side, meaning there is some small whichwhich comes comes about about from from an an integration integration ofanisotropy ofcoscos22θθover (aover preferred solid solid angle angle direction), (see (see the and the derivation implyingderivation there is a higher energy density, at of theof the equation equation of of radiation radiation pressure, pressure, Eq.smaller Eq. 3.74, 3.74, radii below). below). than at larger Including radii. this this The factor net factor flow and and of radiation energy through a mass Principles of Astrophysics & Cosmology - Professor Jodi Cooley replacingreplacing the the differences differences with with differentials, differentials,shell we at we radius obtain obtainr, per unit time, is just L(r) (see Fig. 3.5). This must equal the excess energy in the shell, compared to a shell at larger radius, divided by the time LL(r()r) clcldudu 2it= takes this excess. energy to flow across the(3.48) shell’s width ∆r. The excess energy 4π2r = − 3 dr. (3.48) Note that this is, in effect, a “diffusion4πr density equation− 3 isdr∆u”,, which describing multiplied the outward by the shell’s flow ofvolume gives the total excess energy Note that this is, in effect, a “diffusion equation”, describing2 the outward flow of energy. The left-hand side is the energyof the flux. shell, On4 theπr right-hand∆r∆u. The side, timedu/dr for theis photons the to cross the shell in a random energy. The left-hand side is the energywalk flux. is ( On∆r) the2/lc right-hand. Thus side, du/dr is the gradient in energy density, and cl/3 is a diffusion coefficient that sets the pro- gradient in energy density, and cl/3 is a diffusion coefficient that4πr sets2∆r∆ theu pro- ∆u portionality relating the energy flow− and the energy density gradient. The opacity, 2 − L(r) 2 = 4πr lc . (3.47) portionalityas reflected relating in the the mean energy free path flowl, and controls the energy the flow density of radiation gradient.≈− through(∆ Ther) /lc the opacity, star.− ∆r A more rigorous derivation of this equation adds a factor 1/3 on the right hand side, as reflectedFor low opacityin the mean (large freel), the path flowl, controls will be relatively the flow of unobstructed, radiation through and hence the the star. which comes about from an integration of cos2 θ over solid angle (see the derivation Forluminosity low opacity will (large be high,l), andthe flowvice versa. will be relatively unobstructed, and hence the luminosity will be high, and vice versa.of the equation of radiation pressure, Eq. 3.74, below). Including this factor and Since at every radius the energy densityreplacing is close the differences to that of with blackbody differentials, radiation, we obtain Sincethen (Eq. at every 2.9) radius the energy density is close to that of blackbodyL(r) radiation,cl du 4 = . (3.48) then (Eq. 2.9) u = aT , 4πr2 −(3.49)3 dr u =NoteaT that4, this is, in effect, a “diffusion equation(3.49)”, describing the outward flow of energy. The left-hand side is the energy flux. On the right-hand side, du/dr is the 5 In fact, even then, nothing dramatic would happen.gradient As in we energy will see density, in Section and 3.9, a slowcl/3 contractionis a diffusion coefficient that sets the pro- of the Sun would begin, with a timescale of 107 yr. Over 105 yr, the Solar radius− would only 5In fact, even then, nothing dramatic would happen.portionality As we will relating see in the Section energy 3.9, flow a slow and contraction the energy density gradient. The opacity, shrink by 1%, which is small but discernible.∼ ∼ of the Sun would∼ begin, with a timescale of as107 reflectedyr. Over in the10 mean5 yr, the free Solar path radiusl, controls would the only flow of radiation through the star. shrink by 1%, which is small but discernible.∼ ∼ ∼ For low opacity (large l), the flow will be relatively unobstructed, and hence the luminosity will be high, and vice versa. Since at every radius the energy density is close to that of blackbody radiation, then (Eq. 2.9) u = aT 4, (3.49)

5In fact, even then, nothing dramatic would happen. As we will see in Section 3.9, a slow contraction of the Sun would begin, with a timescale of 107 yr. Over 105 yr, the Solar radius would only shrink by 1%, which is small but discernible.∼ ∼ ∼ basicastro4 October 26, 2006

STELLAR PHYSICS 41

Figure 3.5 Radiative diffusion of energy between volume shells in a star, driven by the gradient in the thermal energy density.

Principles of Astrophysics & Cosmology - Professor Jodi Cooley basicastro4 October 26, 2006

STELLAR PHYSICS 41

Figure 3.5 Radiative diffusion of energy between volume shells in a star, driven by the gradient in the thermal energy density.

dT (r) 3L(r)κ(r)ρ5In(r fact,) even then, nothing dramatic would happen. As we will see in Section 3.9, a slow contraction dT (r) 3L(r)κ(r)ρ(r) 7 5 = 2 of3 the Sun. would begin, with a timescale(3.51) of 10 yr. Over 10 yr, the Solar radius would only dr =− 4πr 4acT2 shrink(r by)3 1%., which is small but discernible.∼ (3.51) ∼ dr − 4πr 4acT (∼r) From Eqns. 3.48 and 3.49, together with an estimate of the mean free path, we Fromcan make Eqns. an 3.48 order-of-magnitudeThis and is the 3.49, Equation together prediction of with Radiative of an the estimate Sun’s Energy luminosity. of Transport the mean Approximat-. free path, we caning makedu/dr an order-of-magnitudewith u/r = aT 4/r prediction, we have of the Sun’s luminosity. Approximat- − ∼ ⊙ 4 ⊙ ing du/dr with u/r = aT /r , wecl haveaT 4 − ∼ ⊙ L ⊙4πr2 . (3.52) ⊙ ∼ ⊙ 3 r 4 2 cl⊙aT Principles of Astrophysics & Cosmology - LProfessor Jodi 4Cooleyπr . (3.52) Based on the virial theorem and the⊙ Sun’s∼ mass⊙ 3 andr radius, we obtained in Eq. 3.33 6 an estimate of the Sun’s internal temperature, Tvir ⊙ 4 10 K. Using this as a Basedtypical on the temperature virial theorem and taking andl =0 the. Sun’s1 cm, we mass ﬁnd and∼ radius,× we obtained in Eq. 3.33 6 an estimate of the Sun’s internal temperature, Tvir 4 10 K. Using this as a 4π 10 10 1 1 15 6 4 L 7 10 cm 3 10 cm s− 10− 7.6 ∼10− ×cgs (4 10 K) typical⊙ ∼ temperature3 × and× taking× l =0.1 cm,× we× ﬁnd× × ×

4π 10 10 331 1 1 15 6 4 L 7 10 cm 3 10=2cm10 s− ergs10−−, 7.6 10− cgs (4(3.53)10 K) ⊙ ∼ 3 × × × × × × × × × 33 1 in reasonable agreement with the observed Solar luminosity, L =3.8 10 ergs− . 33 1 ⊙ × (We have abbreviated above the=2 units of10 the “radiationergs− , constant”, a, as “cgs”.) The (3.53) above estimate can also be used to argue× that, based on its observed luminosity, the 33 1 in reasonableSun must be agreement composed with primarily the observed of ionized Solar hydrogen. luminosity, If the SunL were=3 composed.8 10 ergs− . ⊙ × (Weof, have say, abbreviated ionized carbon, above the mean the units particle of mass the “radiation would be m¯ constant”,12mH /a7, as2 “cgs”.)mH , The ≈ ≈ aboverather estimate than m canH /2 also. Equation be used 3.33 to wouldargue thenthat, give based a virial on its temperature observed thatluminosity, is 4 the Suntimes must as be high, composed resulting primarily in a luminosity of ionized prediction hydrogen. in Eq. 3.52 If the that Sun is too were high composed by two orders of magnitude. of, say, ionized carbon, the mean particle mass would be m¯ 12m /7 2m , ≈ H ≈ H rather than mH /2. Equation 3.33 would then give a virial temperature that is 4 times3.4 as ENERGY high, resulting CONSERVATION in a luminosity prediction in Eq. 3.52 that is too high by two orders of magnitude. We will see that the luminosity of a star is produced by nuclear reactions, with output energies that depend on the local conditions (density and temperature) and hence on r. Let us deﬁne (r) as the power produced per unit mass of stellar 3.4 ENERGY CONSERVATION material. Energy conservation means that the addition to a star’s luminosity due to the energy production in a thin shell at radius r is We will see that the luminosity of a star is produced by nuclear reactions, with dL = dm = ρ4πr2dr, (3.54) output energies that depend on the local conditions (density and temperature) and henceor on r. Let us deﬁne (r) as the power produced per unit mass of stellar material. Energy conservationdL means(r) that the addition to a star’s luminosity due to =4πr2ρ(r)(r), (3.55) the energy production in a thindr shell at radius r is which is the equation of energydL conservation= dm = ρ. 4πr2dr, (3.54) or dL(r) =4πr2ρ(r)(r), (3.55) dr which is the equation of energy conservation. basicastro4 October 26, 2006

STELLAR PHYSICS 41

Figure 3.5 Radiative diffusion of energy between volume shells in a star, driven by the gradient in the thermal energy density.

Thus, if the nuclear reactions powering the Sun were to suddenly switch off, we would not notice5 anything unusual for 50,000 years. With this background, we can now derive the equation that relates the temperature profile, T (r), to the flow of radiative energy through a star. The small mean free path of photons inside the Sun and the very numerous scatterings, absorptions, and re-emissions every photon undergoes reaffirms that locally, every volume element inside the Sun radiates as a blackbody to a very good approximation. How- ever,basicastro4 there is a October net flow of 26, radiation 2006 energy outwards, meaning there is some small anisotropy (a preferred direction), and implying there is a higher energy density, at smaller radii than at larger radii. The net flow of radiation energy through a mass shell at radius r, per unit time, is just L(r)basicastro4(see Fig. 3.5). October This 26, must 2006 equal the excess energy in the shell, compared to a shell at larger radius, divided by the time basicastro4it takes this October excess energy 26, 2006basicastro4 to flow across October the shell’s 26, 2006 width ∆r. The excess energy 42 density is ∆u, which multiplied by the shell’s volumeCHAPTER gives 3 the total excess energy of the shell, 4πr2∆r∆u. The time for the photons to cross the shell in a random and 2 walk is (∆r) /lc. Thus42 CHAPTER 3 du du dTThe dTSun’s2 Luminosity = =4aT 3 .4πr ∆r∆u 2(3.50)∆u Land(r) 2 = 4πr lc . (3.47) 42dr dT dr ≈−dr (∆r) /lc − ∆r CHAPTER 3 1 CHAPTERdu du dT 3 3 dT 42 Substituting in eq. 3.48,A and more expressing rigorous derivationl as (κρ) of− this, we equation obtain adds the equation a factor= 1/3 of=4 on theaT right. hand side, (3.50) whichUsingand comes the about diffusion from an integrationequation of togethercos2 θdrover soliddTwithdr angle the (see meandr the derivation free path radiative energy transport, 1 and Substituting in eq. 3.48, and expressing l as (κρ)− , we obtain the equation of ofestimate the equation we of can radiation estimate pressure,du duthe Eq.dT luminosity 3.74, below).3 dT of Including the sun. this factor and radiative energy= transport=4aT, . (3.50) dureplacingdT du(r) thedT differences3L(r)κ3 (dTr with)ρ(rdr differentials,) dT dr we obtaindr = = =4aT2 3. . (3.50)1 (3.51) dr SubstitutingdrdT dr− in4π eq.r 4 3.48,acTdr and(r) expressingL(r) lclasdudT(κρ(r))− , we3L obtain(r)κ(r the)ρ(requation) of = . = . (3.48) (3.51) radiative energy transport1 , 4πr2 − 3 drdr − 4πr24acT 3(r) SubstitutingFrom in Eqns. eq. 3.48 3.48, and andNote 3.49, expressing that together thisl is, withas in( effect,κρ an) estimate− a, “wediffusion obtain of the equation meanthe equation free”, describingpath, of we the outward flow of radiativecan make energy an transport order-of-magnitude,energy. The prediction left-hand of side theFrom is Sun’sdT the Eqns.(r energy) luminosity. 3.48 and3 flux.L( 3.49,r) Onκ Approximat-(r together) theρ(r right-hand) with an side, estimatedu/dr of theis mean the free path, we Approximate = 2 3 . (3.51) ing du/dr with u/rgradient= aT 4 in/r energy, we have density,can make anddr an order-of-magnitudecl/3−is4π ar diffusion4acT ( predictionr) coefficient of the that Sun’s sets luminosity. the pro- Approximat- − ∼ dT⊙(r) 3⊙L(r)κ(r)ingρ(r)du/dr−with u/r = aT 4/r , we haveRecall: portionality relating the energy flow4 and the⊙ energy density⊙ gradient. The opacity,4 From= Eqns.du2 3.48 and3u −4 3.49,. aT together∼ with an estimate(3.51) of the mean free4 path,u weaT dr − 4πr 42acTcl aT(r) u = 4aT = as reflectedL in4 theπr mean free.= path l, controls the flow of(3.52) radiation2 cl aT through! r the star.r can make andr order-of-magnitude⇠ rsun rsun prediction of theL Sun’s4π luminosity.r . Approximat- (3.52) ⊙ ∼ ⊙ 3 r 4 ⊙ Foring lowdu/dr opacitywith (largeu/rl),⊙ the= flowaT /r will, we be have relatively∼ unobstructed,⊙ 3 r and hence the From Eqns. 3.48 and 3.49, together− with an estimate∼ ⊙ of the mean⊙ free path, we ⊙ Based on the virial theoremluminosity and the will Sun’s be mass high,Based and and on radius, vice the versa. virial we theorem obtained and in4 the Eq. Sun’s 3.33 mass and radius, we obtained in Eq. 3.33 can make an order-of-magnitude prediction of the Sun’s luminosity.6 2 Approximat-cl aT 6 an estimate of the Sun’s internalThus,Since4 at temperature, we every can radius writeanT the estimatevir energy4L of the density104 Sun’sπKr . is internal Using close. to this temperature, that as of a blackbodyTvir 4 radiation,10(3.52)K. Using this as a ing du/dr with u/r = aT /r , we have ∼ ×⊙ ∼ ⊙ 3 r ∼ × −typical temperature∼ ⊙ and takingthen (Eq.l⊙=0 2.9).1 cm, wetypical find temperature and taking⊙l =0.1 cm, we find Based on the virialcl aT theorem4 4π and theu Sun’s= aT mass4, and radius, we obtained in Eq.(3.49) 3.33 4π 2 10 10 1 6 1 15 6 4 10 Lan estimate10 4πr of1 the Sun’sL.1= internal7 10 temperature,15cm 3 10Tvir(3.52)cm64 s− 104 10K− . Using7.6 10 this− ascgs a (4 10 K) L 7 10 cm 3 ⊙10∼ cm s⊙−3 r10−⊙ ∼ 73.6 ×10− cgs× ×(4 10∼ K)× × × × × × ⊙ ∼ 3 × × ×typical temperature× ⊙ and× taking×l =0.1 cm, we× find× 5 33 1 Based on the virial theorem andIn the fact, Sun’s even then, mass nothing and dramatic radius, would we obtained happen. As in we Eq. will=2 see3.33 in10 Sectionergs 3.9,− , a slow contraction (3.53) 4π 3310 1 10 17 1 5 15 6 4 of theL Sun would7 begin,10 withcm a timescale3 10 ofcm6 s10− yr.10 Over− 733.106× yr,10-1 − the Solarcgs radius(4 would10 K) only =2 10 ergs− , (compared∼ to 3.8 x 10∼ ergs(3.53)) 33 1 an estimate of the Sun’s internalshrink⊙ temperature, by∼ 1%3×, which× isTvir smallin reasonable× but×4 discernible.10 agreementK. Using× with the this× observed as× a Solar luminosity,× × L =3.8 10 ergs− . ∼ ∼ × 33 1 ⊙ × typicalin temperature reasonable agreement and taking withl =0 the.1 observedcm, we Solar find(We luminosity, have abbreviatedL =333 above.8 the101 unitsergs of the− “radiation. constant”, a, as “cgs”.) The Principles of Astrophysics & Cosmology - Professor Jodi=2 Cooley⊙10 ergs×− , (3.53) (We4π have abbreviated above the units of the “radiationabove estimate constant”, can× alsoa be, as used “cgs”.) to argue The that, based on its observed luminosity, the 10 10 1 1 Sun must be composed15 primarily6 of ionized4 hydrogen. If33 the Sun1 were composed L above7 estimate10 cm can also3 10 be usedincm reasonable to s− argue10 agreement that,− based7. with6 on10 the its− observed observedcgs Solar(4 luminosity, luminosity,10 K) theL =3.8 10 ergs− . ⊙ ∼ 3 × × × × ×of, say,× ionized carbon,× the× mean particle mass⊙ would× be m¯ 12m /7 2m , Sun must be composed primarily(We have of abbreviated ionized hydrogen. above the If units the of Sun the were “radiation composed constant”, a, as “cgs”.)≈ The H ≈ H above estimate33 can alsorather1 be than usedm toH argue/2. Equation that, based 3.33 on would its observed then give luminosity, a virial temperature the that is 4 of, say, ionized carbon, the=2 mean10 particleergs mass− , would be m¯ 12mH /7 (3.53)2mH , Sun× must be composedtimes primarily as high, ofresulting ionized≈ in hydrogen. a luminosity≈ If the prediction Sun were in Eq. composed 3.52 that is too high by rather than mH /2. Equation 3.33 would thentwo give orders a ofvirial magnitude. temperature33 that is1 4 in reasonable agreement with theof, observed say, ionized Solar carbon, luminosity, the meanL particle=3.8 mass10 wouldergs be−m¯. 12mH /7 2mH , times as high, resulting in a luminosity prediction in Eq.⊙ 3.52 that× is too high by≈ ≈ (We have abbreviated above therather units than of themH “radiation/2. Equation constant”, 3.33 woulda, as then “cgs”.) give a The virial temperature that is 4 two orders of magnitude. times as high, resulting in a luminosity prediction in Eq. 3.52 that is too high by above estimate can also be used to argue that, based3.4 on ENERGY its observed CONSERVATION luminosity, the two orders of magnitude. Sun must be composed primarily of ionized hydrogen. If the Sun were composed We will see that the luminosity of a star is produced by nuclear reactions, with of, say,3.4 ionized ENERGY carbon, CONSERVATION the mean particle mass would be m¯ 12mH /7 2mH , output energies≈ that depend on≈ the local conditions (density and temperature) and rather than mH /2. Equation 3.333.4 ENERGY would then CONSERVATION give a virial temperature that is 4 hence on r. Let us define (r) as the power produced per unit mass of stellar timesWe as high, will see resulting that the in luminosity a luminosity of a prediction star ismaterial. produced in Eq. Energy 3.52 by nuclear conservationthat is reactions, too high means by withthat the addition to a star’s luminosity due to two ordersoutput of energies magnitude. that dependWe on will the seelocal that conditions thethe luminosity energy (density production of a and star in temperature) is a thinproduced shell at by and radius nuclearr is reactions, with output energies that depend on the local conditions (density and temperature) and hence on r. Let us define (r) as the power produced per unit mass of stellar 2 hence on r. Let us define (r) as the powerdL produced= dm per= ρ unit4πr massdr, of stellar (3.54) material. Energy conservation means that the addition to a star’s luminosity due to material. Energy conservation means that the addition to a star’s luminosity due to 3.4 ENERGY CONSERVATION or the energy production in a thinthe energy shell at production radius r is in a thin shell at radius r is dL(r) 2 dL = dm = ρ4πr2dr, 2 (3.54)=4πr ρ(r)(r), (3.55) We will see that the luminosity of a star is produced bydL nuclear= dm = reactions,ρ4πrdrdr, with (3.54) outputor energies that depend onor the local conditionswhich (density is the equation and temperature) of energy conservation and . hence on r. Let us define (r) as the power produced per unit mass of stellar dL(r) dL(r) 2 material. Energy conservation means that=4 theπr addition2ρ(r)(r to), a star’s=4 luminosityπr ρ(r)(r due), (3.55) to (3.55) dr dr the energy production in a thin shell at radius r is which is the equation of energywhich conservation is the equation. of energy conservation. dL = dm = ρ4πr2dr, (3.54) or dL(r) =4πr2ρ(r)(r), (3.55) dr which is the equation of energy conservation. basicastro4 October 26, 2006

basicastro4 October 26, 2006

42 CHAPTER 3 and42 CHAPTER 3 du du dT dT and = =4aT 3 . (3.50) dr dT dr dr du du dT dT = =4aT 3 1. (3.50) Substituting in eq. 3.48, anddr expressingdT dr l as (κρdr)− , we obtain the equation of radiative energy transport, 1 Substituting in eq. 3.48, and expressing l as (κρ)− , we obtain the equation of radiative energy transportdT, (r) 3L(r)κ(r)ρ(r) = 2 3 . (3.51) dTdr(r) −34Lπ(rr)4κacT(r)ρ((rr)) = . (3.51) From Eqns. 3.48 and 3.49,dr together− 4 withπr24 anacT estimate3(r) of the mean free path, we can make an order-of-magnitude prediction of the Sun’s luminosity. Approximat- From Eqns. 3.48 and 3.49, together with an estimate of the mean free path, we ing du/dr with u/r = aT 4/r , we have can− make an order-of-magnitude∼ ⊙ prediction⊙ of the Sun’s luminosity. Approximat- ing du/dr with u/r = aT 4/r , we havecl aT 4 − ∼ ⊙ L ⊙ 4πr2 . (3.52) ⊙ ∼ ⊙cl3aTr 4 L 4πr2 ⊙ . (3.52) Based on the virial theorem and the⊙ ∼ Sun’s⊙ mass3 r and radius, we obtained in Eq. 3.33 ⊙ an estimate of the Sun’s internal temperature, T 4 106 K. Using this as a Based on the virial theorem and the Sun’s mass andvir radius,∼ × we obtained in Eq. 3.33 typicalan estimate temperature of the Sun’s and taking internall =0 temperature,.1 cm, weT ﬁnd 4 106 K. Using this as a vir ∼ × typical4π temperature10 and taking 10l =0.1 cm,1 we ﬁnd1 15 6 4 L 7 10 cm 3 10 cm s− 10− 7.6 10− cgs (4 10 K) ⊙ ∼ 34π × 10 × × 10 1 × 1 × × 15 × × 6 4 L 7 10 cm 3 10 cm s− 10− 7.6 10− cgs (4 10 K) ⊙ ∼ 3 × × × 33× ×1 × × × =2 10 ergs− , (3.53) × 33 1 =2 10 ergs− , 33(3.53) 1 in reasonable agreement with the observed× Solar luminosity, L =3.8 10 ergs− . ⊙ × 33 1 (Wein reasonable have abbreviated agreement above with the the units observed of the Solar “radiation luminosity, constant”,L =3a.8, as10 “cgs”.)ergs The− . above(We have estimate abbreviated can also above be used the to units argue of the that, “radiation based on constant”, its observed⊙ a, as luminosity,× “cgs”.) The the Sunabove must estimate be composed can also primarily be used to of argue ionized that, hydrogen. based on its If observed the Sun luminosity, were composed the of,Sun say, must ionized be composed carbon, the primarily mean of particle ionized mass hydrogen. would beIf them¯ Sun12 werem / composed7 2m , ≈ H ≈ H ratherof, say, than ionizedm / carbon,2. Equation the mean 3.33 particle would mass then would give a be virialm¯ temperature12mH /7 that2mH is, 4 H ≈ ≈ timesrather as than high,mH resulting/2. Equation in a luminosity 3.33 would prediction then give in a virial Eq. 3.52 temperature that is too that high is 4 by twotimes orders as high, of magnitude. resulting in a luminosity prediction in Eq. 3.52 that is too high by two orders of magnitude.

3.4 ENERGY CONSERVATION 3.4 ENERGY CONSERVATIONEnergy Conservation We will see that the luminosity of a star is produced by nuclear reactions, with outputWe will energies see that that the depend luminosity on the of local a star conditions is produced (density by nuclear and temperature) reactions, with and henceoutput on energiesr.The Let that luminosity us depend deﬁne of on (ar star the) as is local theproduced conditions power by producednuclear (density reactions. per and unit temperature) Let mass of stellar and ε(r) be the power per unit mass of stellar material. The star’s material.hence on Energyr. Let conservation us deﬁne (r means) as the that power the addition produced to per a star’s unit luminosity mass of stellar due to material. Energyluminosity conservation due to energy means production that the addition in a thin toshell a star’s at radius luminosity r is due to the energy production in a thin shell at radius r is the energy productiongiven by in a thin shell at radius r is dL = dm = ρ4πr2dr, (3.54) dL = dm = ρ4πr2dr, (3.54) or or rearranging terms dL(r) dL(r) 2 =4=4ππrr2ρρ((rr))((rr),), (3.55)(3.55) drdr whichwhich is is the theequationequation of of energy energy conservation conservation.. This is the Equation of Energy Conservation.

Principles of Astrophysics & Cosmology - Professor Jodi Cooley basicastro4 October 26, 2006

STELLAR PHYSICS 43

3.5Equations THE EQUATIONS OF STELLAR of Stellar STRUCTURE Structure

We have derived four coupled ﬁrst-order differential equations describing stellar structure.We Letnow us re-writehave all them 4 of here. the equations for stellar structure!

dP (r) GM(r)ρ(r) = , (3.56) dr − r2

dM(r) =4πr2ρ(r), (3.57) dr

dT (r) 3L(r)κ(r)ρ(r) = , (3.58) dr − 4πr24acT (r)3

dL(r) =4πr2ρ(r)(r). (3.59) dr We can deﬁne four boundary conditions for these equations. For example: And it Mhardly(r = 0) hurt = 0 at all …. (3.60)

L(r = 0) = 0 (3.61) Principles of Astrophysics & Cosmology - Professor Jodi Cooley P (r = r )=0 (3.62) ∗ M(r = r )=M (3.63) ∗ ∗ where M is the total mass of the star. (In reality, at the radius r of the photosphere of the star,∗ P does not really go completely to zero, nor do ∗T and ρ, and more sophisticated boundary conditions are required, which account for the processes in the photosphere.) To these four differential equations we need to add three equations connecting the pressure, the opacity, and the energy production rate of the gas with its density, temperature, and composition: P = P (ρ,T,composition); (3.64)

κ = κ(ρ,T,composition); (3.65)

= (ρ,T,composition). (3.66) P (ρ,T) is usually called the equation of state. Each of these three functions will depend on the composition through the element abundances and the ionization states of each element in the gas. It is common in astronomy to parametrize the mass abundances of hydrogen, helium, and the heavier elements (the latter are often referred to collectively by the term “metals”) as ρ ρ ρ X H ,Y He ,Z metals . (3.67) ≡ ρ ≡ ρ ≡ ρ basicastro4 October 26, 2006

STELLAR PHYSICS 43

3.5 THE EQUATIONS OF STELLAR STRUCTURE

We have derived four coupled ﬁrst-order differential equations describing stellar structure. Let us re-write them here.

dP (r) GM(r)ρ(r) = , (3.56) dr − r2

dM(r) =4πr2ρ(r), (3.57) dr

dT (r) 3L(r)κ(r)ρ(r) = , (3.58) dr − 4πr24acT (r)3

dL(r) =4πr2ρ(r)(r). (3.59) dr To solve these equations, we need to deﬁne boundary conditions. We can deﬁne four boundary conditions for these equations. For example: M(r = 0) = 0 (3.60)

L(r = 0) = 0 (3.61)

P (r = r )=0 (3.62) ∗ M(r = r )=M (3.63) ∗ ∗ where M is the total mass of the star. (In reality, at the radius r of the photosphere of the star,∗ P does not really go completely to zero, nor do ∗T and ρ, and more sophisticatedVogt-Russell boundary Conjecture: conditions are required, which account for the processes in the photosphere.)Properties and evolution of a star are fully determined by its To these four differential equations we need to add three equations connecting initial mass and its chemical composition. the pressure, the opacity, and the energy production rate of the gas with its density, temperature, and composition: From this we can determine a stars observable parameters: surface temperature,P =radiusP (ρ,T, andcomposition); luminosity. (3.64)

κ = κ(ρ,T,composition); (3.65)

Principles of Astrophysics & Cosmology - Professor = Jodi Cooley(ρ,T,composition). (3.66) P (ρ,T) is usually called the equation of state. Each of these three functions will depend on the composition through the element abundances and the ionization states of each element in the gas. It is common in astronomy to parametrize the mass abundances of hydrogen, helium, and the heavier elements (the latter are often referred to collectively by the term “metals”) as ρ ρ ρ X H ,Y He ,Z metals . (3.67) ≡ ρ ≡ ρ ≡ ρ basicastro4 October 26, 2006 basicastro4 October 26, 2006

STELLAR PHYSICSSTELLAR PHYSICS 43 43

3.5 THE EQUATIONS3.5 THE EQUATIONS OF STELLAR OF STRUCTURE STELLAR STRUCTURE

We have derivedWe four have coupled derived fourﬁrst-order coupled differential ﬁrst-order equations differential describing equations describingstellar stellar structure. Let usstructure. re-write Let them us re-write here. them here.

dP (r) GMdP ((rr))ρ(r)GM(r)ρ(r) = = , 2 , (3.56) (3.56) dr − drr2 − r

dM(r) dM(r) 2 =4πr2ρ(r),=4πr ρ(r), (3.57) (3.57) dr dr

dT (r) 3L(r)κ(r)ρ(r) dT (r) 3L(r)κ(=r)ρ(r) , (3.58) = dr − 4πr, 24acT (r)3 (3.58) dr − 4πr24acT (r)3

dL(r) dL(r) =4πr2ρ(r)(r). (3.59) =4πr2ρdr(r)(r). (3.59) dr We can deﬁne four boundary conditions for these equations. For example: We can deﬁne four boundary conditions for these equations. For example: M(r = 0) = 0 (3.60) M(r = 0) = 0 (3.60) L(r = 0) = 0 (3.61) L(r = 0) = 0 (3.61) P (r = r )=0 (3.62) ∗ P (r = r )=0 (3.62) ∗ M(r = r )=M (3.63) ∗ ∗ where M is the totalM( massr = ofr )= the star.M (In reality, at the radius r of(3.63) the photosphere ∗ ∗ ∗ ∗ where M is theof total the star, massP ofdoes theAncillary star. not really (In reality, go completely at the radiusEquations tor zero,of the nor photosphere do T and ρ, and more of the star,∗ P sophisticateddoes not really boundary go completely conditions to are zero, required, nor do which∗T and accountρ, and for more the processes in sophisticated boundarytheThe photosphere.) Equations conditions of are Stellar required, Structure which accountdo not give for the us processes enough ininformation To these four differential equations we need to add three equations connecting the photosphere.) to understand how the gas in the sun behaves. So, we add to the the pressure, the opacity, and the energy production rate of the gas with its density, To these four differential equations we need to add three equations connecting temperature,following and equations composition: to connect pressure, opacity, energy generations, the pressure, the opacity,density and and the temperature. energy production rate of the gas with its density, temperature, and composition: P = P (ρ,T,composition); (3.64) “equation of state” P = P (ρ,T,composition);κ = κ(ρ,T,composition); (3.64) (3.65) “opacity” κ = κ(ρ,T,composition); = (ρ,T,composition). (3.65) (3.66) P (ρ,T) is usually called the equation of state. Each of“energy these threegeneration” functions will = (ρ,T,composition). (3.66) depend on the composition through the element abundances and the ionization P (ρ,T) is usuallystates called of eachthe element equation in ofthe state gas.. It Each is common of these in three astronomy functions to parametrize will the depend on themass compositionThe abundances mass abundances through of hydrogen, the element of He, helium, H abundances and and heavier the heavierand elements the elements ionization are (the latter are states of each elementoftenparameterized referred in the to gas. collectively by It is common by the term in astronomy “metals”) as to parametrize the mass abundances of hydrogen, helium,ρ andH the heavierρHe elementsρmetals (the latter are X ,Y ,Z . (3.67) often referred to collectively by the term≡ “metals”)ρ ≡ asρ ≡ ρ ρ ρ ρ Principles of AstrophysicsX &H Cosmology,Y - ProfessorHe Jodi,Z Cooley metals . (3.67) ≡ ρ ≡ ρ ≡ ρ basicastro4 October 26, 2006

basicastro4 October 26, 2006

44 CHAPTER 3

We have thus ended up with seven coupled equations defining the seven unknown functions: Pbasicastro4(r), M(r), ρ October(r), T (r 26,), κ 2006(r), L(r), and (r). As there are four boundary conditions for the four first-order differential equations, if there is a solution, it 44 CHAPTER 3 is unique. This is usually expressed in the form of the “Vogt-Russell conjecture”, We have thus endedwhich up with states seven that the coupled properties equations and evolution defining of an the isolated seven star unknown are fully determined by its initial mass and its chemical abundances. These determine the star’s functions: P (r), Mobservable(r), ρ(r), parameters:T (r), κ(r) its, L surface(r), and temperature,(r). As radius, there areand fourluminosity. bound- Two vari- ary conditions44 for theables four which first-order we have differential neglected in this equations, treatment, if and there whichCHAPTER is a have solution, minor 3 it influence is unique. This is usuallyon stellar expressed structure, are in stellar the form rotation of the and “Vogt-Russell magnetic fields. To conjecture”, proceed, we need to which statesWe that have the thusdefine properties ended the up three with and functions, seven evolution coupledP , κ, equations and of an. isolated defining the star seven are unknown fully deter- functions: P (r), M(r), ρ(r), T (r), κ(r), L(r), and (r). As there are four bound- mined byary its conditions initial mass for the and four its first-order chemical differential abundances. equations, These if there determine is a solution, the it star’s observableis unique. parameters: This3.6 is THE its usually surface EQUATION expressed temperature, in OF the STATE form radius, of the “Vogt-Russell and luminosity. conjecture”, Two vari- ables whichwhich we states have that neglected the properties in this and treatment, evolution of and an isolated which star have are minor fully deter- influence on stellarmined structure, by itsDifferent are initial stellar mass equations rotation and its chemical of and state magneticP abundances.(ρ,T,X,Y,Z fields. These) apply To determine proceed, for different the we star’s ranges need of to gas den- observable parameters: its surface temperature, radius, and luminosity. Two vari- sity, temperature, and abundance. Under the conditions in most normal stars, the define theables three which functions, we haveP neglected, κ, and in. this treatment, and which have minor influence equation of state of a classical, nonrelativistic, ideal gas, provides a good descrip- on stellar structure, are stellar rotation and magnetic fields. To proceed, we need to tion. Consider, for example, such a gas, composed of three different kinds of par- define the three functions, P , κ, and . ticles, each with itsEquation own mass mi and density of nStatei. The mean particle mass will 3.6 THE EQUATIONbe OF STATE 3.6 THE EQUATION OF STATE n1m1 + n2m2 + n3m3 ρ In most normal stars,m¯ = the classical, nonrelativistic= .ideal gas law is a(3.68) Different equations of state P (ρ,T,X,Y,Z) applyn1 + forn2 different+ n3 rangesn of gas density, temperature,Different equations andThe abundance.good gas of pressure stateapproximationP ( willρ Under,T,X,Y,Z then the befor) the conditionsapply equation for different in of moststate: ranges normal of gas den- stars, the equation ofsity, state temperature, of a classical, and abundance. nonrelativistic, Under the conditions ideal gas, in providesρ most normal a good stars, descrip-the P = nkT = kT. (3.69) tion. Consider,equation for of state example, of a classical, such a nonrelativistic, gas, composedg ideal of gas, three providesm¯ different a good kinds descrip- of par- tion. Consider,The for meanThe example, number mass such will density depend a gas, composedn on is the related chemical of threeto the abundance different mean mass. kinds and ionization ofConsider par- state gas of the ticles, each with its own mass mi and density ni. The mean particle mass will ticles, each withgas. As itsof ownthreewe have mass kinds alreadym iofand particles. seen, density for completelyni. The mean ionized particle pure mass hydrogen, will be be mH n1mn1 m+ n+2nmm2 ++nn3mmm¯ 3= ρρ, (3.70) m¯ = m¯ = 1 1 2 2 3 3 ==2 . . (3.68)(3.68) n +n1 n+ n+2 +nn3 nn and therefore m/m¯ 1H =02.5. 3 The gas pressureThe gas pressure willMore then will be generally, then be the number densities of hydrogen, helium, or an element of ρ atomicWhich mass numbermeansPg =weAnkT(i.e., can= anwriteρ elementkT. with a total of A protons(3.69) and neutrons in each atomic nucleus)Pg = nkT will be= m¯kT. This is the kinetic (3.69)gas pressure The mean mass will depend on the chemicalm¯ abundance⇢ and ionization state of the XρPg = kTY ρ ZAρ The meangas. mass As we will have depend already on seen, the for chemicaln completelyH = abundance,n ionizedHem¯= pure and hydrogen,,n ionizationA = state, of the (3.71) mmH 4mH AmH gas. As we have already seen, for completelym¯ = H ionized, pure hydrogen, (3.70) wherePrinciples ofZ AstrophysicsA is the & Cosmology mass - abundanceProfessor Jodi2 Cooley of an element of atomic mass number A. Com- plete ionization of hydrogenmH results in two particles (an electron and a proton); of and therefore m/m¯ H =0.5. m¯ = , (3.70) More generally,helium, the three number particles densities (two2 electrons of hydrogen, and helium, a nucleus); or an and element of an atom of with atomic number (i.e., with protons or electrons), +1particles, which for heavy and thereforeatomicm/m¯ massH number=0.5ZA. (i.e., an elementZ with a total of A protonsZ and neutrons in More generally,each atomic nucleus)theenough number atoms will be is densities always close of to hydrogen,A/2. Thus, helium, for an ionized or an gas element we will have of Xρ Y ρ A Z ρρ 3 1 atomic mass number A (i.e.,n =n an=2 elementn,n+3=n with+ a,n totaln= of=AA protons, 2X and+ Y neutrons(3.71)+ Z in H m H He He4m 2A A Amm 4 2 each atomic nucleus) will be H H HH µ ∂ where ZA is the mass abundance of an elementX of atomic mass number A. Com- Xρ Y ρ ρ ZYAρ plete ionizationn of= hydrogen,n results= in two= particles,n3 (an=X + electron+1, and, a proton); of(3.71) (3.72) H He 2m A 2 helium, three particlesm (twoH electrons and4m aH nucleus);H µ andAm of anH atom∂ with atomic number (i.e., with protons or electrons), +1particles, which for heavy where ZA is the mass abundance of an element of atomic mass number A. Com- enough atomsZ is alwaysZ close to A/2. Thus, forZ an ionized gas we will have plete ionization of hydrogen results in two particles (an electron and a proton); of A ρ 3 1 helium, three particlesn =2 (twonH +3 electronsnHe + andnA a= nucleus);2X and+ ofY + anZ atom with atomic 2 mH 4 2 number (i.e., with protons orX electrons), µ+1particles, which∂ for heavy Z Z ρ Y Z enough atoms is always close to=A/2. Thus,3X + for an+1 ionized, gas we will have(3.72) 2m 2 AH µ ρ ∂ 3 1 n =2nH +3nHe + nA = 2X + Y + Z 2 mH 4 2 X µ ∂ ρ Y = 3X + +1 , (3.72) 2m 2 H µ ∂ basicastro4 October 26, 2006 basicastro4 October 26, 2006

basicastro4 October 26, 2006

44 CHAPTER 3 44 CHAPTER 3 We have thus ended up with seven coupled equations defining the seven unknown We have thus ended up with seven coupled equations defining the seven unknown functions: P (r), M(r), ρ(r), T (r), κ(r), L(r), and (r). As there are four bound- functions: P (r), M(r), ρ(r), T (r), κ(r), L(r), and (r). As there are four bound- 44ary conditions for the four first-order differential equations, if thereCHAPTERbasicastro4 is a solution, October 3 26, 2006 it ary conditions for the four first-order differential equations, if there is a solution, it is unique. This is usually expressed in the form of the “Vogt-Russell conjecture”, We have thus endedis unique. up with seven This coupled is usually equations expressed defining in the the form seven of unknown the “Vogt-Russell conjecture”, which states that the properties and evolution of an isolated star are fully deter- functions: P (r), Mwhich(r), ρ(r states), T (r) that, κ(r the), L properties(r), and (r and). As evolution there are four of an bound- isolated star are fully determined by its initial mass and its chemical abundances.STELLAR PHYSICS These determine the star’s 43 ary conditions for themined four first-orderby its initial differential mass and equations, its chemical if there abundances. is a solution, These it determine the star’s isobservable unique. This parameters: is usuallyobservable expressed its surface parameters: in the temperature, form its surface of the radius, “Vogt-Russell3.5 temperature, THE EQUATIONS and luminosity. conjecture”, OFradius, STELLAR and Two STRUCTURE luminosity. vari- Two vari- whichables states which that we the have properties neglected and in evolution this treatment, of an isolatedWe and have which derived star arefour have coupled fully minor first-order deter- influencedifferential equations describing stellar ables which we have neglected instructure. this treatment, Let us re-write them and here. which have minor influence minedon stellar by its structure, initialon mass stellar are and stellar its structure, chemical rotation are abundances. and stellar magnetic rotation These fields. and determine magnetic To proceed, the fields. star’s we To need proceed, to we need to observabledefine the parameters: three functions, its surfaceP , κ temperature,, and . radius, and luminosity. TwodP (r) vari-GM(r)ρ(r) define the three functions, P , κ, and . = , (3.56) ables which we have neglected in this treatment, and which have minor influencedr − r2 on stellar structure, are stellar rotation and magnetic fields. To proceed, we needdM(r) to =4πr2ρ(r), (3.57) define the three functions, P , κ, and . dr 3.6 THE EQUATION3.6 THE OF EQUATION STATE OF STATE dT (r) 3L(r)κ(r)ρ(r) = , (3.58) dr − 4πr24acT (r)3 Different equationsDifferent of state equationsP (ρ,T,X,Y,Z of state P) (applyρ,T,X,Y,Z for different) apply ranges for different of gas den- ranges of gas den- 3.6 THE EQUATION OF STATE dL(r) =4πr2ρ(r)(r). (3.59) sity, temperature,sity, and temperature, abundance. and Under abundance. the conditions Under in the most conditions normaldr in stars, most the normal stars, the Differentequation equations of stateequation of of state a classical,P of(ρ,T,X,Y,Z state nonrelativistic, of a classical,) apply for nonrelativistic, ideal differentWe can gas, define ranges provides four boundary ideal of gas a conditions gas, good den- provides for descrip- these equations. a good For example: descrip- sity,tion. temperature, Consider, and fortion. abundance. example, Consider, such Under for a example, gas, the conditions composed such a in gas, of most three composed normal different stars, ofM threekinds(r the= 0) = different of 0 par- kinds of(3.60) par- equationticles, each of state with of a its classical, own mass nonrelativistic,m and density ideal gas,n provides. The mean a good particle descrip-L(r = 0)mass = 0 will (3.61) ticles, each with itsi own mass mii and density ni. The mean particle mass will tion. Consider, for example, such a gas, composed of three different kinds ofP ( par-r = r )=0 (3.62) be be ∗ ticles, each with its own mass mni andm density+ n mni+. Then m mean particleρ massM(r will= r )=M (3.63) 1 1 2 2 n1m31 +3 n2m2 + n3m3 ρ∗ ∗ be m¯ = m¯ = where M=is the total. mass of the star.= (In reality,. (3.68) at the radius r of the photosphere(3.68) n + n + n ∗ n ∗ n m + n m1 + n2 m 3 ofnρ1 the+ star,nP2 does+ n not3 really go completelyn to zero, nor do T and ρ, and more mbasicastro4¯ = 1 1 October2 2 26,3 3 2006=sophisticated. boundary conditions(3.68) are required, which account for the processes in The gas pressure will then be the photosphere.) The gas pressuren1 + willn2 + thenn3 be n ρ To these four differential equations we need to add three equations connecting The gas pressure will then be the pressure, the opacity,ρ and the energy production rate of the gas with its density, Pg = nkT = PkT.g = nkT = kT. (3.69) (3.69) ρ m¯ temperature, and composition:m¯ P = nkT = kT. P = P(3.69)(ρ,T,composition); (3.64) The mean massThe will mean depend massg on the will chemical dependm¯ on abundance the chemical and abundance ionization andstate ionization of the state of the κ = κ(ρ,T,composition); (3.65) Thegas. mean As we mass have willgas. already depend As we on seen, the have for chemical already completely abundance seen, ionized for completely and ionization pure hydrogen, ionized state of pure the hydrogen, = (ρ,T,composition). (3.66) gas. As we have already seen, for completely ionizedmH pure hydrogen,m STELLAR PHYSICS m¯ = , P (ρm¯,T)=is usuallyH called, the equation of45 state.(3.70) Each of these three functions(3.70) will mH 2 depend on the composition through the element abundances and the ionization m¯ = , states of each element2 in the gas.(3.70) It is common in astronomy to parametrize the 2 mass abundances of hydrogen, helium, and the heavier elements (the latter are and therefore m/m¯ andH therefore=0.5. m/m¯ H =0.5. and therefore m/m¯ =0.5. often referred to collectively by the term “metals”) as H ρ ρ ρ More generally,More the number generally, densities the number of hydrogen, densities helium, of hydrogen,X orH ,Y an helium, elementHe ,Z metals or of. an element(3.67) of More generally, the number densities of hydrogen, helium, or an elementρ ofρ ρ Weatomic can more mass numbergenerallyatomicA mass(i.e., write number an the element numberA (i.e., with an adensities element total of A withprotons a total≡ and of≡A neutronsprotons≡ in and neutrons in atomic mass number A (i.e., an element with a total of A protons and neutrons in ofeach eachH, atomic He atomic and nucleus) nucleus) metals.each will atomic will be be nucleus) will be Xρ YXρρ ZY ρρ Z ρ Xρ Y ρ ZAρ A A nH n=H = ,nHe,n=HenH==,nA =,n,nHeA ==, ,n, A (3.71)= (3.71), (3.71) mH mH 4mH 4mH AmH Am4mHH AmH wherewhereZAZisA theis the masswhere mass abundance abundanceZA is the of an mass of element an abundance element of atomic of of atomic mass an element number mass of numberA. atomic Com-A mass. Com- number A. Com- Howpleteplete ionizationmany ionization particles of hydrogenplete of hydrogen ionizationresults results results from in of two hydrogen inthe particles two complete particles results (an electron in (anionization two electron and particles a proton); andof (an a of electronproton); and of a proton); of Figurehelium,helium, 3.6 Calculation three three particles particles ofhelium, radiation (two (two three electrons pressure. electrons particles and A a beam and nucleus);(two a of electrons nucleus); photons and of with and an and atom blackbodya of nucleus); an with atom atomic intensity and with of atomic an atom with atomic hydrogen?B strikes Helium? the wall of a container at an angle θ to its perpendicular. The projected number (i.e., withnumberprotons(i.e., or electrons), with protons+1particles, or electrons), which for+1 heavyparticles, which for heavy numberareaZ of the(i.e., beam, withZdA, isprotons increased or by electrons),1/ cosZθ, and therefore+1particles, the power which reaching for heavy Thus,enough for atomsZ an is ionized alwaysenough close Zgas: atomsZ to A/ is2. always Thus,Z for close an ionized to ZA/2 gas. Thus, we will for haveZ an ionized gas we will have enoughthe wall atoms per is unit always area close is decreased to A/2 to. Thus,B cos forθ. an Since ionized a photon’s gas we momentum, will have A ρ 3 1 p, is itsn =2 energynH +3 dividednHe + by c, thenA momentumA= 2 fluxρX + inA theY Here+ beam3 weZ isρusedB/c 1the. fact Every3 1 n =2nH +3nHe + nA = 2X + Y + Z reflectionn of=2 a photonnH +3 transfersnHe + twice2 itsn perpendicularmAH= 42 componentXX+ + 2Y +Y Z of=+ 1 momentum,Z 2 µm 2 4 m∂H 2 4 2 X H µ µ ∂ ∂ and therefore the momentumρ transferX perY unit timeX and per unit area, i.e., the pressure, is 2B cos2 θ=/c. The total3X pressure+ +1 is obtained,ρ by integratingY (3.72) over all ⇢ 2m 3 ρ 1 2 Y angles ofn the= beams(2 thatX approach+=H Yµ + the wall.3ZX) += ∂ +1 3,X + +1 , (3.72) (3.72) m 42m 2 22mH 2 H H µ ∂µ ∂ Recall, m = ρ/n where we have used the fact that X + Y + Z =1. Thus, m¯ ρ 2 = = (3.73) mH nmH 1+3X +0.5Y for a totally ionized gas. For Solar abundances, X =0.71, Y =0.27, Z =0.02, andPrinciples therefore of Astrophysicsm/m¯ & CosmologyH =0 - Professor.61. Jodi In Cooley the central regions of the Sun, about half of the hydrogen has already been converted into helium by nuclear reactions, and as a result X =0.34, Y =0.64, and Z =0.02, giving m/m¯ H =0.85. In addition to the kinetic gas pressure, the photons in a star exert radiation pressure. Let us digress briefly, and derive the equation of state for this kind of pressure. Consider photons inside a blackbody radiator with an intensity given by the Planck function, Iν = Bν , which, when integrated over wavelength, we denoted as B. As illustrated in Fig. 3.6, the energy arriving at the surface of the radiator per unit time, per unit area, at some angle θ to the perpendicular to the surface, is B cos θ, because the area of the beam, when projected onto the wall of the radiator, is increased by 1/ cos θ. Now, consider the photons in the beam, which strike the fully reflective surface of the radiator at the angle θ. Every photon of energy E has momentum p = E/c. When reflected, it transmits to the surface a momentum ∆p =(2E/c) cos θ. Therefore, there is a second factor of cos θ that must be ap- plied to the incoming beam. The rate of momentum tranfer per unit area will be obtained by integrating over all angles at which the photons hit the surface. But, the rate of momentum transfer (i.e., the force) per unit area is, by definition, the pressure. Thus, F dp/dt 2 π 4π 1 P = = = B cos2 θ sin θdθdφ = B = u, (3.74) A A c 3c 3 Zπ/2 where in the last equality we have used the previously found relation (Eq. 2.4) between intensity and energy density. Note that the derivation above applies not basicastro4 October 26, 2006

basicastro4 October 26, 2006

STELLAR PHYSICS 45 STELLAR PHYSICS 45

Figure 3.6 Calculation of radiation pressure.Figure A beam 3.6 Calculation of photons of with radiation blackbody pressure. intensity A beam of photons with blackbody intensity B strikes the wall of a container at an angleB θstrikesto its the perpendicular. wall of a container The projected at an angle θ to its perpendicular. The projected area of the beam, dA, is increased by 1/ cos θ, and therefore the power reaching area of the beam, dA, is increased by 1/ costheθ wall, and per therefore unit area the is power decreased reaching to B cos θ. Since a photon’s momentum, the wall per unit area is decreased to Bpcos, isθ its. energy Since adivided photon’s by c, momentum, the momentum flux in the beam is B/c. Every p, is its energy divided by c, the momentumreflection flux of in a photon the beam transfers is B/c twice. Every its perpendicular component of momentum, reflection of a photon transfers twice its perpendicularand therefore component the momentum of momentum, transfer per unit time and per unit area, i.e., the and therefore the momentum transfer perpressure, unit time is 2 andB cos per2 θ/c unit. The area, total i.e., pressure the is obtained by integrating over all pressure, is 2B cos2 θ/c. The total pressureangles is ofobtained the beams by that integrating approach over the wall. all m¯ 2 angles of the beams that approach the wall. = The mean mass to hydrogen mass ratio for mH 1+3X +0.5Y pure hydrogenwhere = 1/2. we (X have = 1, usedY = 0, the Z = fact 0). that X + Y + Z =1. Thus, m¯ ρ 2 where we have used the fact that X + Y + Z =1. Thus, = = (3.73) For solar abundances, X = 0.71, Y = 0.27m andH Z nm= 0.02.H Calculate1+3X +0the .5Y m¯ ρ 2 mean= to hydrogen=for a mass totally ratio. ionized gas. For Solar abundances,(3.73)X =0.71, Y =0.27, Z =0.02, m nm 1+3X +0.5Y H H and therefore m/m¯ H =0.61. In the central regions of the Sun, about half of the for a totally ionized gas. For Solar abundances,hydrogen hasX already=0.71 been, Y =0 converted.27, Z into=0 helium.02, by nuclear reactions, and as a In the core of the sun, X = 0.34, Y = 0.64 and Z = 0.02. Calculate the and therefore m/m¯ =0.61. In the centralresult X regions=0.34, ofY the=0. Sun,64, and aboutZ =0 half.02 of, giving the m/m¯ H =0.85. H mean to hydrogen mass ratio. hydrogen has already been converted intoIn addition helium to by the nuclear kinetic gasreactions, pressure, and the as photons a in a star exert radiation pressure. Let us digress briefly, and derive the equation of state for this kind of pres- result X =0.34, Y =0.64, and Z =0sure..02, giving Considerm/m¯ photonsH =0 inside.85. a blackbody radiator with an intensity given by the In addition to the kinetic gas pressure, the photons in a star exert radiation pres- Planck function, Iν = Bν , which, when integrated over wavelength, we denoted sure. Let us digress briefly,Notice: and deriveX + Yas + the ZB =. equation 1. As illustrated of state in Fig. for 3.6, this the kind energy of pres- arriving at the surface of the radiator sure. Consider photons inside a blackbodyper unit radiator time, per with unit an area, intensity at some given angle byθ to the the perpendicular to the surface, is Planck function, Iν = Bν , which, whenB cos integratedθ, because over the area wavelength, of the beam, we when denoted projected onto the wall of the radiator, as B. As illustrated in Fig. 3.6, the energyis increased arriving by 1 at/ cos theθ surface. Now, consider of the radiator the photons in the beam, which strike the per unit time, per unit area, at some anglefullyθ reflectiveto the perpendicular surface of the to radiator the surface, at the angle is θ. Every photon of energy E Principles of Astrophysics & Cosmologyhas momentum - Professor Jodi Cooley p = E/c. When reflected, it transmits to the surface a momentum B cos θ, because the area of the beam, when projected onto the wall of the radiator, ∆p =(2E/c) cos θ. Therefore, there is a second factor of cos θ that must be ap- is increased by 1/ cos θ. Now, considerplied the to photons the incoming in the beam. beam, The which rate strike of momentum the tranfer per unit area will be fully reflective surface of the radiatorobtained at the angle by integratingθ. Every over photon all angles of energy at whichE the photons hit the surface. But, has momentum p = E/c. When reflected,the rate it transmits of momentum to the transfer surface (i.e., a momentum the force) per unit area is, by definition, the ∆p =(2E/c) cos θ. Therefore, therepressure. is a second Thus, factor of cos θ that must be ap- plied to the incoming beam. The rate of momentum tranfer per unitπ area will be F dp/dt 2 2 4π 1 obtained by integrating over all angles at whichP = the= photons= hit the surface.B cos θ sin But,θdθdφ = B = u, (3.74) A A c π/2 3c 3 the rate of momentum transfer (i.e., the force) per unit area is, byZ definition, the where in the last equality we have used the previously found relation (Eq. 2.4) pressure. Thus, between intensity and energy density. Note that the derivation above applies not F dp/dt 2 π 4π 1 P = = = B cos2 θ sin θdθdφ = B = u, (3.74) A A c 3c 3 Zπ/2 where in the last equality we have used the previously found relation (Eq. 2.4) between intensity and energy density. Note that the derivation above applies not Radiation Pressure

Since the gas in a star locally emits as a blackbody, the blackbody photons will produce radiation pressure. The intensity of this radiation (B) is integrated over all frequencies and directions. basicastro4Reminder: October 26, 2006 Intensity = power/area B = Id Power = energy/time Z

STELLAR PHYSICS 45 The projected area of the beam is increased by 1/cosθ, thus the energy per unit time (power) arriving at the surface is Bcosθ.

Figure 3.6 Calculation of radiation pressure. A beam of photons with blackbody intensity B strikes the wall of a container at an angle θ to its perpendicular. The projected Principles of Astrophysics & Cosmology - Professor Jodi Cooley area of the beam, dA, is increased by 1/ cos θ, and therefore the power reaching the wall per unit area is decreased to B cos θ. Since a photon’s momentum, p, is its energy divided by c, the momentum ﬂux in the beam is B/c. Every reﬂection of a photon transfers twice its perpendicular component of momentum, and therefore the momentum transfer per unit time and per unit area, i.e., the pressure, is 2B cos2 θ/c. The total pressure is obtained by integrating over all angles of the beams that approach the wall.

where we have used the fact that X + Y + Z =1. Thus, m¯ ρ 2 = = (3.73) mH nmH 1+3X +0.5Y for a totally ionized gas. For Solar abundances, X =0.71, Y =0.27, Z =0.02, and therefore m/m¯ H =0.61. In the central regions of the Sun, about half of the hydrogen has already been converted into helium by nuclear reactions, and as a result X =0.34, Y =0.64, and Z =0.02, giving m/m¯ H =0.85. In addition to the kinetic gas pressure, the photons in a star exert radiation pressure. Let us digress briefly, and derive the equation of state for this kind of pressure. Consider photons inside a blackbody radiator with an intensity given by the Planck function, Iν = Bν , which, when integrated over wavelength, we denoted as B. As illustrated in Fig. 3.6, the energy arriving at the surface of the radiator per unit time, per unit area, at some angle θ to the perpendicular to the surface, is B cos θ, because the area of the beam, when projected onto the wall of the radiator, is increased by 1/ cos θ. Now, consider the photons in the beam, which strike the fully reflective surface of the radiator at the angle θ. Every photon of energy E has momentum p = E/c. When reflected, it transmits to the surface a momentum ∆p =(2E/c) cos θ. Therefore, there is a second factor of cos θ that must be ap- plied to the incoming beam. The rate of momentum tranfer per unit area will be obtained by integrating over all angles at which the photons hit the surface. But, the rate of momentum transfer (i.e., the force) per unit area is, by definition, the pressure. Thus, F dp/dt 2 π 4π 1 P = = = B cos2 θ sin θdθdφ = B = u, (3.74) A A c 3c 3 Zπ/2 where in the last equality we have used the previously found relation (Eq. 2.4) between intensity and energy density. Note that the derivation above applies not basicastro4 October 26, 2006

STELLAR PHYSICS 45

basicastro4 October 26, 2006

STELLAR PHYSICS 45 Figure 3.6 Calculation of radiation pressure. A beam of photons with blackbody intensity B strikes the wall of a container at an angle θ to its perpendicular. The projected area of the beam, dA, is increased by 1/ cos θ, and therefore the power reaching the wall per unitSTELLAR area PHYSICS is decreased to B cos θ. Since a photon’s momentum, 45 basicastro4p, is its energy October divided 26, 2006 by c, the momentum flux in the beam is B/c. Every reflection of a photon transfers twice its perpendicular component of momentum, and therefore the momentum transfer per unit time and per unit area, i.e., the 2 Figure 3.6pressure, Calculation is of2 radiationB cos pressure.θ/c. The A beam total of pressure photons with is obtained blackbody byintensity integrating over all anglesB strikes of the the wall beams of a container that approach at an angle theθ to wall. its perpendicular. The projected STELLAR PHYSICS area of the beam, dA, is increased by 1/ cos θ, and therefore45 the power reaching the wall per unit area is decreased to B cos θ. Since a photon’s momentum, p, is its energy divided by c, the momentum flux in the beam is B/c. Every where we havereflection used of a the photon fact transfers that twiceX + itsY perpendicular+ Z =1 component. Thus, of momentum, and therefore theFigure momentum 3.6 Calculation transfer per of unit radiation time and pressure. per unit Aarea, beam i.e., of the photons with blackbody intensity 2m¯ ρ 2 pressure, is 2B cos θ/c.= TheB totalstrikes pressure= the wall is obtained of a container by integrating at an angle overθ allto its perpendicular.(3.73) The projected angles of the beamsm thatH approachareanm of theH the wall. beam,1+3dAX, is+0 increased.5Y by 1/ cos θ, and therefore the power reaching for a totally ionized gas. For Solarthe wall abundances, per unit areaX is decreased=0.71, toY B=0cos θ.27. Since, Z =0 a photon’s.02, momentum, p, is its energy divided by c, the momentum flux in the beam is B/c. Every andwhere therefore we havem/m¯ used theH fact=0 that.61X.+ InY the+ Z central=1. Thus, regions of the Sun, about half of the Figure 3.6 Calculation of radiation pressure. A beam of photonsreflection with blackbody of a photon intensity transfersbasicastro4 twice its perpendicular October component 26, 2006 of momentum, hydrogen has already beenm¯ convertedρ into2 helium by nuclear reactions, and as a B strikes the wall of a container at an angle=θ to itsand perpendicular.= therefore The the projected momentum transfer per(3.73) unit time and per unit area, i.e., the 2 arearesult of the beam,X =0dA, is.34 increased, Y =0 bym1H./64cos,θ andnm, andHpressure, thereforeZ =01+3 the. is02X2 powerB,+0 givingcos. reaching5Yθ/cm/m.¯ The totalH =0 pressure.85. is obtained by integrating over all the wall per unit area is decreased to B cos θ. Sinceangles a of photon’s the beams momentum, that approach the wall. p, is itsInfor energy addition a totally divided ionized to by thec, the gas. kinetic momentum For Solar gas flux pressure, abundances, in the beam the isXB/c=0 photons..71 Every, Y in=0 a.27 star, Z exert=0.02radiation, pres- reflectionsureand of. thereforea Let photon us transfers digressm/m¯ H twice=0 briefly, its.61 perpendicular. In and the central derive component regions the of momentum,equation of the Sun, of about state half for of this the kind of pres- and thereforehydrogen the momentum has already transfer been per converted unit time into and per helium unit area, by nuclear i.e., the reactions, and as a sure. Consider2 photons inside a blackbody radiator with an intensity given by the pressure,result is 2BXcos=0θ/c.34., TheY =0 total.64 pressurewhere, and we isZ obtained=0 have.02 used, by giving integrating them/m¯ fact overH that=0 allX.85+.Y + Z =1. Thus, anglesPlanck of the beams function, that approachI = the wall.B , which,STELLAR when PHYSICS integrated over wavelength, we denoted 45 In addition to theν kinetic gasν pressure, the photonsm in¯ a star exertρ radiation pres-2 Foras sureB a. .photon, As Let illustrated us digress p=E/c, briefly, in Fig. and incident 3.6, derive the the energy equation arriving of state= for at this the= kind surface of pres- of the radiator (3.73) m nm 1+3X +0.5Y sure. Consider photons inside a blackbody radiator withH an intensityH given by the where we haveatper used angle unit the fact time, θ that, the perX + photon unitY + Z area,=1 .surface at Thus, some angle θ to the perpendicular to the surface, is Planck function, Iν = Bνfor, which, a totally when ionized integrated gas. over For Solar wavelength, abundances, we denotedX =0.71, Y =0.27, Z =0.02, B cos θ, becausem¯ ρ the area of2 the beam, when projected onto the wall of the radiator, densityas B. As is illustrated =reduced= in Fig. andby 3.6, thereforea factor the energym/m¯ arrivingH =0 at.61 the(3.73). In surface the central of the regionsradiator of the Sun, about half of the is increasedper unitm time,H by pernm1/H unitcos area,1+3θhydrogen. at Now,X some+0.5 consider has angleY alreadyθ to the been perpendicular photons converted in to the into the beam, surface, helium which is by nuclear strike reactions, the and as a for a totallycos ionizedBθcos. gas. Theθ, Forbecause Solarmomentum the abundances, area of theX beam,transfer=0.71 when, Y =0 projected.27, Z onto=0.02 the, wall of the radiator, fully reflective surfaceresult of theX radiator=0.34, Y at=0 the.64 angle, and Zθ=0. Every.02, giving photonm/m¯ ofH energy=0.85. E and thereforeishasm/m¯ thenis momentum increasedH =0 .61. by In1 thep/ cos= centralθE/c. Now, regionsIn. When addition consider of the reflected, to the Sun, the photons about kinetic it half in transmits gas the of pressure, the beam, to which the the photonsstrike surface the in a a momentum star exert radiation pres- hydrogen has already been converted into helium by nuclear reactions, and as a ∆pfully=(2 reflectiveE/c) cossurfaceθ. of Therefore,sure the. radiator Let us there at digress the is angle briefly, a secondθ. Every and factor derive photon the of ofcos equation energyθ thatE of must state for be this ap- kind of pres- result X =0.34,hasY =0 momentum.64,E andcosZp✓=0= E/c.02,E. giving Whencos ✓m/m¯ reflected,H2E=0 it.85 transmits. to the surface a momentum p =[ ( sure. Consider)] = cos photons✓ inside a blackbody radiator with an intensity given by the In additionplied to the∆p kinetic to=(2 theE/c gas incomingc pressure,) cosθ. Therefore, the beam. photonsc there in The a star isc rate a exert second ofradiation momentum factor pres- of cos θ tranferthat must per be unitap- area will be Figure 3.6 Calculation of radiation pressure. A beam of photons with blackbody intensity sure. Let us digressobtainedplied briefly, to bythe and incoming integrating derive the beam.Planck equation over The function, all of rate state angles of formomentumIν this at= kind whichBν , tranfer of which, pres- the per photons when unit area integrated hit will the be over surface. wavelength, But, we denoted B strikes the wall of a container at an angle θ to its perpendicular. The projected sure. Consider photonsobtained inside by integrating a blackbody overas radiatorB all. As angles with illustrated an at intensity which in the given Fig. photons by 3.6, the the hit theenergy surface. arriving But, at the surface of the radiator Planck function,theI rate= B of, which, momentum when integrated transfer over (i.e., wavelength, the force) we denoted per unit area is, by definition, the So theweν rate canν of momentumwrite transferper unit (i.e., time, the per force) unitarea per area, of unit the at area beam, some is,dA angleby, definition, is increasedθ to the the perpendicular by 1/ cos θ, and to therefore the surface, the power is reaching as B. As illustratedpressure. in Fig. Thus, 3.6, the energy arriving at the surface of the radiator pressure. Thus, B cos θ, because thethe area wall of the per beam, unit area when is decreased projected to ontoB cos theθ wall. Since of the a radiator, photon’s momentum, per unit time, per unit area, at some angle θ to the perpendicular top, the is its surface, energy is divided by c, the momentum flux in the beam is B/c. Every is increasedπ 2⇡ π by⇡1/ cos θ. Now, consider the photons⇡ in the beam, which strike the B cos θ, because the area of theFF beam,dp/dtdp/dt when projected2 2 onto the2 2 wallreflection2 of the2 radiator, of4π a photon1 transfers4π4⇡ twice1 its2 perpendicular component of momentum, is increased by 1/ cosPθ.P Now,== consider== the=fully photons== reflective inB thecos beam,B surfaceθcossinB whichθcosdθθ ofdsinφ strike✓ the=sinθd radiator theθ✓Bddφ✓=d= atu,= theB angleB=(3.74)θu,.cos Every✓ sin photon(3.74)✓d✓ of energy E AA AA c πc/⇡2=0 ⇡ c and therefore3c the momentum3 3c c transfer3⇡ per unit time and per unit area, i.e., the fully reflective surface of the radiator at thehas angle momentumZ θ.π Every/22 p photon= E/c of. energy WhenE reflected, it transmitsZ 2 to the surface a momentum Z Z Z pressure, is 2B cos2 θ/c. The total pressure is obtained by integrating over all has momentum pwhere= E/c in. When the last reflected, equality it transmits we have to used the surface the previously a momentum found relation (Eq. 2.4) whereRecall: in the The energy last equality per unit∆p time=(2 we (power)E/c have) arriving cos usedθangles. Therefore,at the the of surface previously the beams there is Bcos thatisθ found a. approach second relation factor the wall. of (Eq.cos θ 2.4)that must be ap- ∆p =(2E/c) cosbetweenθ. Therefore, intensity there and is aenergy secondplied density. to factor the of incoming Notecos θ thatthat beam. the must derivation be The ap- rate above of momentum applies not tranfer per unit area will be plied to the incomingbetween beam. intensity The rate and of momentum energy density. tranfer per unit Note area that will the be derivation above applies not obtainedcos byn+1 integratingax over all angles at which the photons hit the surface. But, obtained by integrating overcos allax anglesn sin axdx at which= the photons hit the surface. But, Hint: the rate ( ofnwhere+ momentum 1)a we have transfer used the (i.e., fact the that force)X + perY + unitZ =1 area. is, Thus, by definition, the the rate of momentumZ transfer (i.e., the force) per unit area is, by definition, the pressure. Thus, pressure. Thus, Wherem¯ we haveρ used 2 = = (3.73) F dp/dt 2 π F4π dp/dt1 2 theπ relationship: 4π 1 P = = = B cos2 θ sin θdθdφ = B = u, (3.74)mH 2 nmH 1+3X +0.5Y A A c P = 3c = 3 = B cos θcsin θdθdφ = B = u, (3.74) Zπ/2 A A c π/2 B = u 3c 3 for a totally ionizedZ gas. For Solar4⇡ abundances, X =0.71, Y =0.27, Z =0.02, where in the last equality we have used the previously found relation (Eq. 2.4) where inand the therefore last equalitym/m¯ weH have=0. used61. In the the previously central regions found relationof the Sun, (Eq. about 2.4) half of the between intensity and energy density. Note that the derivation above applies not Principles of Astrophysics & Cosmology -between Professor Jodihydrogen intensity Cooley and has energy already density. been converted Note that into the derivation helium by above nuclear applies reactions, not and as a result X =0.34, Y =0.64, and Z =0.02, giving m/m¯ H =0.85. In addition to the kinetic gas pressure, the photons in a star exert radiation pressure. Let us digress briefly, and derive the equation of state for this kind of pressure. Consider photons inside a blackbody radiator with an intensity given by the Planck function, Iν = Bν , which, when integrated over wavelength, we denoted as B. As illustrated in Fig. 3.6, the energy arriving at the surface of the radiator per unit time, per unit area, at some angle θ to the perpendicular to the surface, is B cos θ, because the area of the beam, when projected onto the wall of the radiator, is increased by 1/ cos θ. Now, consider the photons in the beam, which strike the fully reflective surface of the radiator at the angle θ. Every photon of energy E has momentum p = E/c. When reflected, it transmits to the surface a momentum ∆p =(2E/c) cos θ. Therefore, there is a second factor of cos θ that must be ap- plied to the incoming beam. The rate of momentum tranfer per unit area will be obtained by integrating over all angles at which the photons hit the surface. But, the rate of momentum transfer (i.e., the force) per unit area is, by definition, the pressure. Thus, F dp/dt 2 π 4π 1 P = = = B cos2 θ sin θdθdφ = B = u, (3.74) A A c 3c 3 Zπ/2 where in the last equality we have used the previously found relation (Eq. 2.4) between intensity and energy density. Note that the derivation above applies not basicastro4 October 26, 2006

STELLAR PHYSICS 45 basicastro4 October 26, 2006

STELLAR PHYSICS 45

Figure 3.6 Calculation of radiation pressure. A beam of photons with blackbody intensity B strikes the wall of a container at an angle θ to its perpendicular. The projected area of the beam, dA, is increased by 1/ cos θ, and therefore the power reaching the wall per unit area is decreased to B cos θ. Since a photon’s momentum, p, is its energy divided by c, the momentum flux in the beam is B/c. Every Figure 3.6 Calculation of radiation pressure. A beam of photonsreflection with blackbody of a photon intensity transfers twice its perpendicular component of momentum, B strikes the wall of a container at an angle θ to itsand perpendicular. therefore The the projected momentum transfer per unit time and per unit area, i.e., the area of the beam, dA, is increased by 1/ cos θ, andpressure, therefore the is power2B cos reaching2 θ/c. The total pressure is obtained by integrating over all the wall per unit area is decreased to B cos θ. Sinceangles a photon’sof the beams momentum, that approach the wall. p, is its energy divided by c, the momentum flux in the beam is B/c. Every reflection of a photon transfers twice its perpendicular component of momentum, and therefore the momentum transfer per unit time and per unit area, i.e., the pressure, is 2B cos2 θ/c. The total pressurewhere we is obtained have used by integrating the fact over that allX + Y + Z =1. Thus, angles of the beams that approach the wall. m¯ ρ 2 = = (3.73) mH nmH 1+3X +0.5Y where we have used the fact that X + Y + Z =1. Thus, for a totally ionized gas. For Solar abundances, X =0.71, Y =0.27, Z =0.02, m¯ ρ 2 = = and therefore m/m¯ H =0.61(3.73). In the central regions of the Sun, about half of the mH nmH 1+3hydrogenX +0.5 hasY already been converted into helium by nuclear reactions, and as a for a totally ionized gas. For Solar abundances, X =0.71, Y =0.27, Z =0.02, result X =0.34, Y =0.64, and Z =0.02, giving m/m¯ H =0.85. and therefore m/m¯ H =0.61. In the central regionsIn addition of the to Sun, the about kinetic half gas of the pressure, the photons in a star exert radiation pres- hydrogen has already been converted into helium by nuclear reactions, and as a sure. Let us digress briefly, and derive the equation of state for this kind of pres- result X =0.34, Y =0.64, and Z =0.02, giving m/m¯ H =0.85. In addition to the kinetic gas pressure, thesure. photons Consider in a star exert photonsradiation inside pres- a blackbody radiator with an intensity given by the sure. Let us digress briefly, and derive thePlanck equation function, of state forI thisν = kindBν of, which, pres- when integrated over wavelength, we denoted sure. Consider photons inside a blackbodyas radiatorB. As with illustrated an intensity in given Fig. by 3.6, the the energy arriving at the surface of the radiator basicastro4 OctoberPlanck 26, 2006 function, Iν = Bν , which, when integratedper unit over time, wavelength, per unit area, we denoted at some angle θ to the perpendicular to the surface, is as B. As illustrated in Fig. 3.6, the energyB arrivingcos θ, atbecause the surface the ofarea the of radiator the beam, when projected onto the wall of the radiator, basicastro4 Octoberper unit time, 26, 2006 per unit area, at some angle θ to the perpendicular to the surface, is B cos θ, because the area of the beam, whenis projected increased onto by the1 wall/ cos ofθ the. Now, radiator, consider the photons in the beam, which strike the is increased by 1/ cos θ. Now, consider thefully photons reflective in the beam, surface which of strike the theradiator at the angle θ. Every photon of energy E fully reflective surface of the radiator at thehas angle momentumθ. Every photonp = E/c of energy. WhenE reflected, it transmits to the surface a momentum has momentum p = E/c. When reflected, it∆ transmitsp =(2E/c to the) cos surfaceθ. Therefore,a momentum there is a second factor of cos θ that must be ap- 46 ∆p =(2E/c) cos θ. Therefore,CHAPTER there is a 3 secondplied to factor the of incomingcos θ that beam. must be The ap- rate of momentum tranfer per unit area will be plied to the incoming beam. The rate of momentumobtained tranfer by integrating per unit area over will all be angles at which the photons hit the surface. But, only46 to photons, but to any particles with an isotropicobtained velocity by integrating distribution, over all angles and withCHAPTER at which the 3 photons hit the surface. But, the rate of momentum transfer (i.e., the force)the rate per unit of momentumarea is, by definition, transfer the (i.e., the force) per unit area is, by definition, the kinetic energies large compared to their rest-masspressure. energies, Thus, so that the relation p pressure. Thus, E/conly(which to photons, is exact but for to photons) any particles is a good with approximation. an isotropic velocity Thus, the distribution, equation of≈ and with F dp/dt 2 π F4π dp/dt1 2 π 4π 1 statekinetic in which energies the radiation large compared pressure equals to their one-third rest-massP of= the energies,= thermal so= energy that theB densitycos relation2 θ sin θdpθPdφ== B== u, = (3.74) B cos2 θ sin θdθdφ = B = u, (3.74) We can relate this to the energy densityA andA thec π/2 ≈ A3c 3A c 3c 3 holdsE/c for(which any ultra-relativistic is exact for photons) gas. is a good approximation. Thus,Z the equation of Zπ/2 Returningstatetemperature in which to the the case radiationusing of the relations pressure pressure due equals for towhere ablackbody one-third thermal in the last photon of equality the radiation. gas thermal we inside have energy a used star, thewhere density previously in the found last relation equality (Eq. we 2.4) have used the previously found relation (Eq. 2.4) between intensity and energy density. Note that the derivation above applies not weholds can write for any ultra-relativistic gas. between intensity and energy density. Note that the derivation above applies not Returning to the case of the pressure1 1 due to a thermal photon gas inside a star, P = u = aT 4, (3.75) we can write rad 3 3 which under some circumstances can become1 important1 or4 dominant (see Problems Prad = u = aT , (3.75) 2 andThe 3). The total full equationpressure of state is then for normal the3 sum stars3 will of thereforethe gas be pressure and the which under some circumstances can becomeρkT important1 or dominant (see Problems radiation pressure.P = P + P = + aT 4. (3.76) 2 and 3). The full equationg of staterad for normalm¯ 3 stars will therefore be We will see in Chapter 4 that the conditions in whiteρkT dwarfs1 and in neutron stars P = P + P = + aT 4. (3.76) dictate equations of state that are veryg differentrad fromm¯ this form.3 We will see in Chapter 4 that the conditions in white dwarfs and in neutron stars 3.7dictate OPACITY equations of state that are very different from this form. In normal starts, the gas (“kinetic”) pressure usually dominates. Like theCan equation you ofthink state, of the when opacity, thisκ, at everymight radius not in be the the star willcase? depend on the3.7 density, OPACITY the temperature, and the chemical composition at that radius. We have already mentioned one important source of opacity, Thomson scattering of photons offLike free electrons. the equation Let us of now state, calculate the opacity, correctlyκ, at the every electron radius density in the for star an will ionized depend on gasthe of arbitrary density, the abundance temperature, (rather and than the pure chemical hydrogen, composition as before): at that radius. We have A ρ 2 1 ρ nalready= n +2 mentionedn + onen important= sourceX + ofY opacity,+ Z Thomson= (1+ scatteringX), (3.77) of photons e H He 2 A m 4 2 2m offPrinciples free of electrons.Astrophysics & Cosmology Let us - now Professor calculateH Jodiµ Cooley correctly the∂ electronH density for an ionized wheregas weof arbitrary have again abundanceX assumed (rather that the than number pure of hydrogen, electrons as in before): an atom of mass number A is A/2. Therefore,A ρ 2 1 ρ n = n +2n + n = X + Y + Z = (1+X), (3.77) e H HeneσT σTA 2 1 κes = = 2 (1 +mXH )=(1+4X)0.22 cm g− 2.mH (3.78) ρ 2mH µ ∂ where we have againX assumed that the number of electrons in an atom of mass In regions of a star with relatively low temperatures, such that some or all of the number A is A/2. Therefore, electrons are still bound to their atoms, three additional processes that are important sources of opacity are bound-boundneσT σT, bound-free (also called photoionization2 1 ), κes = = (1 + X)=(1+X)0.2 cm g− . (3.78) and free-free absorption.ρ In bound-bound2mH and bound-free absorption, which we haveIn regions already of discussed a star with in the relatively context low of photospheric temperatures, absorption such that features, some or an all of the atomelectrons or ion is are excited still bound to a higher to their energy atoms, level, three or additional ionized to processes a higher that degree are ofimportant ionization,sources by of absorbing opacity are a photon.bound-bound Free-free, bound-free absorption is(also the calledinversephotoionization process of ), free-freeand free-free emission,absorption. often called Bremsstrahlung In bound-bound(“braking and bound-free radiation” in absorption, German). which In free-free emission (see Fig. 3.7), a free electron is accelerated by the electric we have already discussed in the context of photospheric absorption features, an potential of an ion, and as a result radiates. Thus, in free-free absorption, a photon atom or ion is excited to a higher energy level, or ionized to a higher degree of is absorbed by a free electron and an ion, which share the photon’s momentum andionization, energy. All by three absorbing processes a photon. depend on Free-free photon wavelength, absorption inis theaddition inverse to gas process of temperature,free-free emission, density, and often composition. called Bremsstrahlung (“braking radiation” in German). In free-free emission (see Fig. 3.7), a free electron is accelerated by the electric potential of an ion, and as a result radiates. Thus, in free-free absorption, a photon is absorbed by a free electron and an ion, which share the photon’s momentum and energy. All three processes depend on photon wavelength, in addition to gas temperature, density, and composition. Stay Tuned!

Principles of Astrophysics & Cosmology - Professor Jodi Cooley