Physics 312 Introduction to Astrophysics Lecture 24

Transcription

1 Physics 32 Intoduction to Astophysics Lectue 24 James Buckley Lectue 24 Stella Stuctue Reading Assignment Read Chapte 5 and 8 by next Wed. Physics 25, J. Buckley

2 The Life Stoy of a Sta The Life Stoy of Sta HighaMass Stas Fo low mass stas Fusion in tempeatue neve eaches that Fo low mass stas equied fo C buning tempeatue neve thatneve Fo low mass staseaches tempeatue Fo highe mass stas, ash of eaches that equied fo C buning equied fo C buning each cycle at contaction Fo highe mass stas,fusion ash of coe, Fo highe mass stas, ash contaction of each cycle eachpoducts cycle atocoe, heats up shells suounding collects at the coe, contaction heatsheats up shells up shellssuounding suounding that egion causing shell-buning egion causing egion causingshell-buning shell buning Physics 26 J. Buckley Physics 26 J. Buckley Main Sequence Themostat µ 3/2 µv 2 R = n n2 v (v) exp d3 v 2 kt 2kT S(E) 2 Z Z2 e2 (E) = exp E v Tcoe const

3 Random Walk Photons ae absobed and e-emitted in the optically thick inteio of a sta, making a andom walk fom the coe to the opitically thin suface whee they escape. Random Walk James Joyce s challenge: Find a path acoss Dublin with out encounteing a pub It s as had fo a photon to make it out of the sun as it is fo a Iishman to make it acoss Dublin without encounteing a pub (James Joyce s challenge) - kind of like a pub cawl out of the sun!

4 Diffusion/Random Walk What is the x coodinate afte N time steps? Aveaging ove many paths gives x i+ = x i =0 (i.e., it is just as likely fo x to be positive o negative, to advance to the left o the ight) Now conside how the RMS deviation gows with time: x i+ = x i ± l x 2 i+ = x 2 i ± 2 lx i + l 2 i 2 lx i =2l x i =0 so x 2 = l 2, x 2 2 =2l 2,... x 2 N = Nl 2 R Random walk time t di = N = x(t)2 l 2 whee = l/v The aveage time fo photons to di use a distance R is given by t di = R2 lc R = lct di Scaling: Radiative Tanspot l R 3 In adiation zone : l mfp M R 2 Time to di use out of adiation zone : t di usion l mfp R Rate enegy tanspoted = amount of enegy in ad zone time fo enegy to di use out L ad at 4 adiation zone volume t di usion T 4 R 3 R 2 /l mfp RT 4 l mfp L fusion T lage powe T coe constant fo di eent stas L ad RT 4 coe R 3 M R 4 M Substituting the elationship R M we obtain L M 3 which, despite goss ovesimplification, is close to the coect empiical elationship: L M 3.5

5 Main Sequence Lifetime Nuclea fuel M H c 2 M L M 3.5 Main sequence lifetime t ms Total enegy available fo Hydogen fusion Rate enegy eleased t ms M M 3.5 t ms M 2.5 Moe massive stas bun much moe bightly, but live fo a much shote peiod of time This is known as the Kut Cobain e ect, o the live fast, die young phenomenon Anothe Look at the HR Diagam

6 Equations of Stella Stuctue Hydostatic Equilibium : dp Mass Continuity : dm() d () d =4 2 () = () GM() 2 Eqn. of State : P gas = nkt() = () µm u kt() P ad () = at ()4 3 Radiative tanspot equation : T = 3 6 ac L 2 T 3 dt Enegy dl() Poduction/Consevation= T dp = 4π 2 ρ()ϵ() d Convective Enegy Tanspot: P Equations of State Recall... Fo Radiation: u P = 3 at 4 = at 4 Fo an ideal gas: u gas = 3 2 nkt n = P gas = nkt i whee X i N 0 A i ( + Z i ) Combining the two equations fo adiation we have P V = 3 U Combining the two equations fo an ideal gas P gas V = 3 2 U gas Even fo moe complicated situations whee thee ae molecula degees of feedom, phase tansitions, etc, in some paticula ange of state paametes (P,V,T) in a single phase it is often possible to wite the equation of state in the geneal fom PV =( )U

7 Hydostatic Equilibium m+dm P(+d) dm m ds Mass of volume element: dm = d da P() F g Gavitational foce : F G = GM()dm 2 d In equilibium, this must be balanced by the pessue di eence on the top and bottom of the di eential element: [P () P ( + d)]da = GM() d da 2 dp d = whee M() = GM() 2 0 ( )4 2 d o dm() d =4 2 Scaling Laws: Hydostatic Equilibium Hydostatic Equilibium : dp d = GM 2 P suface 0 P coe P suface R GM (M /R 3 ) R 2 P coe P coe, gavity M 2 R 4 P coe, themal kt coe M R 3 In hydostatic equilibium, P gavitational = P themal and we obtain: R M 2 R 4 / M R 3 giving the final esult that the adius of a main sequence sta is oughly popotional to its mass: R / M

8 Radiative Tanspot Conside a ay bundle, with solid angle d about some mean cental ay diection (, ) and specific intensity I da 2 Cente ay, diection (, ) da d ds The specific intensity is a constant if thee is no emission o absoption The change in specific intensity due to scatteing out of the beam o absoption is popotional to the intensity, and given by di,abs = I ds The change in specific intensity due to spontaneous emission, o scatteing into the solid angle of the beam is given by di,emus = j ds Radiative Tanspot (Ryden & Peteson) P ad () = a 3 T 4 P ad ( + d) = a 3 (T + dt )4 = a 3 T 4 + dt T 4 a + dt 3 T The adiation foce acting on the shell between and + d is df ad =[P ad () P ad ( + d)] 4 2 df ad 6 3 a2 T 3 dt So a tempeatue gadient implies a net adiation foce on mateial (e.g., fee electons) inside the shell. We can calculate this foce anothe way, by looking at how momentum is tansfeed as photons ae absobed by mateial. In a shell of optical depth d = ()apple()d The ate photons (with momentum p = E/c) cay momentum though the shell is L()/c The pobability of inteacting is P = exp( d ) =d so the foce, which is the ate of momentum tansfeed by absoption is: df ad = L() d = L() ()apple()d Equating these two expessions: c dt d = c 3 6 ac apple L 2 T 3

9 Radiative Tanspot Fo isotopic emission, this equation educes to dp d = F c P ad () = a 3 T 4 dp d a 4T F ad () = L 4 2 Radiative tanspot equation : T = 3 L 6 ac 2 T 3 Equations of Stella Stuctue Hydostatic Equilibium : dp Mass Continuity : dm() d () d =4 2 () = () GM() 2 Eqn. of State : P gas = nkt() = () µm u kt() P ad () = at ()4 3 Radiative tanspot equation : T = 3 6 ac L 2 T 3 dt Enegy dl() Poduction/Consevation= T dp = 4π 2 ρ()ϵ() d Convective Enegy Tanspot: P

10 Convection Conside a gas bubble slightly hotte than suoundings if 0 i < 0 upwad buoyant foce ρ i,p i,t i ρ i,p i,t i ρ o ρo( + + ), ), P o Po( + + ), ), T o To( + + ) ) P i = P o,t i >T o ρ i < ρ o ρ i,p i,t i ρ o (), P o (), T o () Timescale fo bubble to expand to achieve pessue balance, o to ise is shot compaed to timescale fo heat to be exchanged at the bounday ) Q = 0, adiabatic expansion. Sola Ganulation 30 minute time lapse images of the sola ganulation made with the SVST (La Palma) (G. Schame, et al.). Hot gas ises in cente of bight egions, is diveted hoizontally (adiates) and sinks back into the Sun in the dake integanula lanes. The size of the ganules ange fom 250 km (esolution limit) to 2000~km with an aveage of 300 km.

11 Adiabatic Expansion Fist law of themodynamics (consevation of enegy): du = Q W PV =( )U Q =0) P dv = du d(pv)=p dv + VdP=( )du P dv + = V dp P dv = V dp dp P = dv V d(ln P )= d(ln V ) ln P = ln V + C ln P +ln V e = e C PV = const Adiabatic Expansion Q: Why do potatoes bun you mouth? Whee did the expession hot potato come fom? A: Potatoes have a high specific heat Specific Heat at Constant Volume C V how intenal enegy changes with change in tempeatue if volume is held constant. Specific Heat at Constant Pessue C P how intenal enegy changes with change in tempeatue if pessue is held constant. The adiabatic index is defined to be: C P /C V =( + )/ PV P = constant = const Substituting P = m kt gives PT P = const P T = const

12 Themal fom Atmosphee Cicula Escape acceleation We can undestand a lot about atmosphees, fom fist undestanding themal escape fom atmosphees. A gas atom of mass m and velocity v can escape fom a planet of adius and mass M if its kinetic enegy K is geate than the depth of its gavitational potential well U (K>U) o v>vesc v v a = t K= m v2 2 U= GM m GM m 2 m vesc = 2 2GM vesc = Physics 25, J. Buckley Physics 32 - Lectue p. 25/27 Themal Escape fom Atmosphee Cicula acceleation We know the aveage velocity of gas paticles in the atmosphee fom the tempeatue. By definition, the tempeatue in Kelvin, times Boltzmann s constant k is equal to the aveage Kinetic enegy o v a = v t 2 m vave = kt 2 2k T vave = m 2GM vesc = If vave > vesc we quickly loose all paticles to space. Physics 25, J. Buckley Physics 32 - Lectue p. 25/27

13 Themal Escape fom Moon Cicula acceleation Example: Do Hydogen atoms escape fom the Moon? Day time tempeatue of the moon is T=400 K Mass of Hydogen is m= kg. Boltzmann s constant is k= vave = 2k T m = e ve v av sc = 2G M 2( )400 = 2.6 km/sec vesc = 2.4 km/sec v a = t vave,h > vesc So Hydogen eadily escapes fom the suface of the moon! Physics 25, J. Buckley Physics 32 - Lectue p. 25/27 ave m Escape velocity vesc Numbe of atoms at that velocity Slow Evapoation of Atmosphees Cicula acceleation 2kT v = Velocity of atoms What about sodium atoms? The mass of sodium (Na) is about 23 times as lage as Hydogen, v so the aveage velocity is the squae oot of 23 smalle: a = tp p vave,na = vave,h / 23 = 2.6 km/sec/ 23 = 0.5 km/s This is smalle than the escape velocity of the moon (2.4 km/sec) but thee ae still a good faction of paticles above this enegy. These paticles escape and thee is a slow leakage of Na paticles into space, but if thee is a souce of paticles. Fo the moon, micometeoites libeate Na poviding this souce Physics 25, J. Buckley Physics 32 - Lectue p. 25/27

14 Scale Height of atmosphee Foce on top = P (z + z)a z z z Aea, A Foce on bottom = P (z)a slab of mass m = A [P (z) P (z + z)]a = mg =( A {z z } )g volume P = m m kt ) = P kt P (z + z) P (z) mg = P z kt P z = kt h P whee h mg Conside a slab of atmosphee, diffeence in pessue on top and bottom must povide an upwad foce that balances the weight Physics 32, J. Buckley Scale Height P z = h P whee h kt mg P = P 0 e z/h Planet Venus Eath Mas Jupite Satun Uanus Neptune Scale Height 5.9 km 8.5 km. km 27 km 59.5 km 27.7 km 20 km Physics 32, J. Buckley