Lecture 6 Radiation Transport

Summary figure EOS regimes – Pols Fig 3.4 Lecture 6 Radiation Transport Prialnik Chapter 5 Krumholtz and Glatzmaier 6 Pols Chapter 5 See Pols p 31 for discussion For an adiabatic expansion or compression Very roughly (do not use this for anything quantitative): dP Gm(r) regardless of equation of state (ideal, degenerate, = − ⇒ P ∝m2r −4 etc.) the first law of thermodynamics gives dm 4πr 4 ρ ∝ m / r 3 r −4 ∝ρ 4/3m−4/3 P = Kργ ad 2/3 4/3 The value of ad varies with the EOS. so for hydrostatic equilibrium, crudely, P ∝ m ρ . For an ideal gas, fully ionized, P 3 P If the global density changes due to expansion or contraction, u = φ = ρ 2 ρ leading to new pressure and density, P' and ρ', hydrostatic φ +1 equilibrum suggests γ = =5/3 ad 4/3 φ ⎛ P '⎞ ⎛ ρ '⎞ = ⎝⎜ P ⎠⎟ ⎝⎜ ρ ⎠⎟ For a relativistic gas =4/3. γ ad If the pressure increases more than this, In regions of partial ionization 4/3 ⎛ P '⎞ ⎛ ρ '⎞ (or pair production or photo i.e., > there will be a restoring ⎝⎜ P ⎠⎟ ⎝⎜ ρ ⎠⎟ disintegration) γ can be ad force that will lead to expansion. If it is less, the contraction suppressed. will continue and perhaps accelerate. Now consider an adiabatic compression: Radiation Transport P=Kργ So far we have descriptions of hydrostatic equilibrium and the equation of state. Still γ missing: ⎛ P '⎞ ⎛ ρ '⎞ So ⎜ ⎟ = ⎜ ⎟ ρ ' > ρ ⎝ P ⎠ ⎝ ρ ⎠ • How the temperature must vary in order to transport γ 4/3 ⎛ ρ '⎞ ⎛ ρ '⎞ the energy that is generated (by both radiative If, for a given mass, > one has stability, ⎝⎜ ρ ⎠⎟ ⎝⎜ ρ ⎠⎟ diffusion and convection). hydrostatic equilibrium is satisfied. If not, things are unstable. • A way of calculating the opacity of stellar matter 4 4 Thus if γ > the star is stable and if γ < it is not. 3 3 • An explicit description of the relevant nuclear This is a global analysis and doesn't necessarily apply and particle physics to small regions of the star but illustrates the importance A technique for solving the resulting differential of γ =4/3. We will return to this later. • equations Transport by radiative diffusion in a star From our prior discussions, for blackbody radiation, we have the Planck function which tells us everything Because of the high density and long time scales, hydrodynamic equilibrium is an excellent description. about the equilibrium radiation Locally, and frequently globally, thermal equilibrium 2hν 3 dν B (T, )d erg cm−2 s−1 Hz−1 Ster −1 ν Ω ν = 2 h /(k T ) is also a valid assumption. c e ν B −1 The total flux leaving a cm2 comes from integrating this over The radiation field in the optically thick stellar interior is thus to high accuracy, given by the Planck function. solid angle and frequency Spherical symmetry and isotropy can be assumed F= B dν dΩ =σT 4 erg cm−2 s−1 (except for rapid rotators). In “radiative” (as opposed to ∫ ∫ ν “convective”) regions, the heat is transported because a small excess of diffusion proceeds in one direction It will also be useful to define a monochromatic intensity, I ( ) ν Ω owing to the decreasing temperature in that direction, 2 1 1 1 with units like B (erg cm− s− Hz− Ster − ), but taken to but to good accuracy heat is diffusing both ways ν nearly equally. describe the energy flux of a beam of radiation with a single frequencymoving into a given solid angle. As radiation passes through Simplified derivation of temperature gradient equation: matter it can be absorbed or scattered. Two blackbody slabs separated I I +dI by a photon mean free path One way of quantifying this is by a typical “cross section” F T+dT right T Clayton 3-2 for an absorber or scatterer.# page 171 dT << T # Fleft The mean free path of a given photon is 1/(n)# 1 = where n is the number of κρ absorbers/scatterers per cm-3. 4 A related quantity is the "opacity" κ defined by dI = - Iκρ dr 4 ⎡⎛ dT ⎞ ⎤ net heat flux F F F ⎡ T dT T 4 ⎤ T 4 ⎢ 1 1⎥ = right − left = σ ( + ) − = σ ⎜ + ⎟ − so that I = I0 exp(−κρr).The mean free path, , is thus 1/κρ ⎣⎢ ⎦⎥ ⎝ T ⎠ ⎣⎢ ⎦⎥ since the radative flux declines by one e-fold in that distance. ⎡⎛ dT ⎞ ⎤ Thus nσ = κρ. In fact, σ and κ are both functions of the ≈ σT 4 ⎢ 1+ 4 −1⎥ = 4σT 3 dT ⎝⎜ T ⎠⎟ composition, temperature, density and frequency. ⎣⎢ ⎦⎥ The temperature difference between two layers It is perhaps better to think of the gas as continuous, is given by the temperature gradient times the but with a small temperature jump. Radiation emitted mean free path of a photon, from the hotter region typically goes a distance l dT =1/ κρ so in 1D dT≈ and before being absorbed. Similarly radiation moves ( ) dx from the cooler region the same distance back into dT 4σT 3 dT acT 3 dT F = 4σT 3dT = 4σT 3 = = the hotter region dx κρ dx κρ dx ac −15 −3 −4 since σ ≡ a=7.566 ×10 erg cm K 4 =7.566 ×10−15 dyne cm−2 K−4 Pols p.iv σ = 5.670×10−5 erg cm−2 K−4 One obtains the same 2 equations as on the Multiplying F by the area, 4πr , gives the luminosity T+dT T 2 3 4πr acT dT previous page L(r) = or κρ dr The accurate result is dT κρ ≈ L(r) ¾ times this dr 4πr 2acT 3 Fick’s Law J=-D n Fick’s Law ∇ In general Flux = nv Assume an isotropic distribution, v x = vy = vz On the average 1/3 of the particles at a boundary will be moving in the z direction and half of those up and half down 1 1 ⎡ dn ⎛ dn ⎞ ⎤ F ≈ v ⎡n − n ⎤ = v n + − n − z z ⎣ z− z+ ⎦ z ⎢ ⎜ ⎟ ⎥ 6 6 ⎣ dz ⎝ dz ⎠ ⎦ 1 dn = v 3 z dz dn n = D where D is the "diffusion coefficient" z− nz dz + 1 1 1 defined as v = v 3 mfp 3 n > n κρ z− z+ Pols Chapter 5, Kippenhahn and Weigert According to Fick's law the flux of something equals a In particular, an energy gradient ∇U in the internal constant times the rate at which something changes energy per cm3 gives rise to an energy flux (erg cm−2 s−1) in space (the gradient) ∂U Energy Flux = F = -D ∂r 1 ∂U ⎛ ∂U ⎞ ∂T ∂T J=-D∇n with D= v the "diffusion coefficient" Since = = C with C the "heat capacity" 3 ⎜ ⎟ V V ∂r ⎝ ∂T ⎠V ∂r ∂r per unit volume (at constant volume) and n, some quantity (like number density) that is ∂T 1 diffusing. J is the flux of n across a surface. Since the only F = -Κ with Κ the conductivity = D C = v C ∂r V 3 V ∂ net gradiant in a spherical star is in the r direction ∇ → r ∂ 2 −1 −3 −1 D has units cm s ; CV =cV ρ has units erg cm K , so Κ, The unts of D are cm2 s−1 and so the units of J the conductivity, has units erg cm−1 s−1 K−1. 2 1 are cm− s− , the flux of particles per unit area per unit time. The energy we are interested in here is blackbody radiation. 4 ⎛ ∂U ⎞ 3 L(r) For photons, v = c and U = aT , so C 4aT ⎜ ⎟ = V = and since F = (1D) ⎝ ∂T ⎠ rad 4πr 2 The mean free path comes from the definition of opacity L(r) 4 acT 3 dT dI −s/ ν = −κ ρ I ⇒ I = I e mfp = K ∇T = - ds ν ν ν 0 4πr 2 rad 3 κρ dr is the distance over which the intensity of a beam of radiation mfp dT 3κρ L(r) ¾ times what 1 = − decreases by e, hence = . Ignore the frequency dependence dr 16πacT 3 r 2 we got before κ ν ρ for the moment. In Lagrangian coordinates this is dT 3κ L(r) In terms of the opacity, the conductivity is then = dm 2 3 4 1 4 acT 3 64π acT r Κ = v C = rad 3 V 3 κρ This is our second major equation for stellar structure and the radiative energy flux is and evolution. It describes heat transport so long as the dT 4 acT 3 dT Frad = −Krad = - material is not convecting. dr 3 κρ dr Pols 5.17 and since ∞ dB d ∞ d σT 4 4σ ac ν dν = B dν = = T 3 = T 3 ∫ ∫ ν 0 dT dT 0 dT π π π The same approach can be used with a frequency-dependent flux , where we have used a result from lecture 2 that ∞ H , as is necessary to correctly define the opacity 4 υ π B dν =F = σT ∫ ν 1 dU c 1 d ⎧4π ⎫ 0 H = − D ∇U = − v ν = − ⎨ B T,Ω ⎬ ν ν ν 3 ν dr 3 dr c ν ( ) 4σ κ ν ρ ⎩ ⎭ a ≡ c 4π 1 dB dT = − ν the π comes from integration over solid angle 3ρ κ dT dr ν Defining a mean opacity κ (the Rosseland mean) by ∞ ∞ 4π dT 1 dB ∞ 1 dB F = H dν = − ν dν ν dν ∫ ν ∫ ∫ ∞ 0 3ρ dr 0 κ dT 1 κ dT π 1 dB ν = 0 ν = ν dν κ ∞ dB acT 3 ∫ κ dT ∫ ν dν 0 ν 0 dT ∞ 1 dB acT 3 so ν dν = ∫ κ dT πκ 0 ν 4π dT ∞ 1 dB 4 acT 3 dT A condition for the validity of this equation is gives F = − ν dν = - as before 3 dr ∫ dT 3 dr that the background conditions vary only a tiny bit ρ 0 κ ν κ ρ 1 over a typical scattering length = .

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