Nearly everything that we know about the Sun (& other stars) comes from the photons it emits into space.

Transcription

1 THE SOLAR ATMOSPHERE Nearly everything that we know about the Sun (& other stars) comes from the photons it emits into space. In the interior, we saw the photon distribution had a dominant isotropic part I 0, and just a small perturbation I 1 to make the net flux flow outwards. At the surface of the star, we see a RAPID transition from this ( optically thick ) regime to a regime in which ALL photons are flowing out and don t undergo any more absorption or scattering ( optically thin ). thick... τ 1 photons are trapped & mostly isotropic thin... τ 1 photons escape freely We ll start by talking about the photosphere, i.e., this visible layer in which photons stop random walking and start running. (τ 1) Thus, we ll delve a bit into the field of radiative transfer, which is the study of how radiation is transported through gas (and emitted, absorbed, scattered). The photosphere is special because this layer is in radiative equilibrium. { } { Flux is conserved; radiative is equivalent to i.e., F = 0 heating = radiative cooling Later, we ll talk about the other interesting layers (above photosphere) that are NOT in radiative equilibrium... there are extra sources of heating! } 8.1

2 Overview of the radiative layers... Thin surface layers conceal major changes in T and ρ... Remember that R 700,000 km, so the above plot really is thin! Height z = 0 defined by τ 2/3 (optical depth unity ). 8.2

3 The solar spectrum I ν is close to a blackbody at T eff 5770 K, which is set by the surface flux... F rad = σt 4 eff = L 4πR 2 but there are discrepancies due to the ν-dependence of opacity: The Earth s atmosphere absorbs a lot of the Sun s spectrum, too: 8.3

4 The flow of radiation through the atmosphere depends on 2 equations both of which we ve seen (sort of) before: { } the equation of radiative transfer energy conservation The general quantity we want to solve for is I ν (ˆn,r,t) Why do we care? The angle dependence will let us predict the Sun s limb darkening, and use it as a probe of depth in the atmosphere. Solving these equations will also let us build a model of how T and ρ vary with depth in the atmosphere, and knowing that helps us predict the solar spectrum emitted by all those layers. The full form I ν (ˆn,r,t) depends on 7 scalar quantities (ˆn is a unit vector; only 2 angles). It pays to simplify the problem... Ignore time-dependence (i.e., all / t = 0). For some applications, we ll integrate over all photon frequencies, and work with a single gray equation of transfer. Assume spherical symmetry (i.e., spatial properties depend on r only). Assume plane-parallel geometry to reduce the ˆn vector down to just one angle... If the total extent of the atmosphere (z max z min ) is R, then we can forget about spherical stellar curvature. 8.4

5 Also, if there s no preference for the x or y transverse dimensions, the only important angle is θ, and we assume everything s symmetric in φ. Many equations will look nicer if we use µ = cosθ Thus, we want to solve for I ν (µ,z). [reduced 7 scalars to 3] Equation of Radiative Transfer It s really just the Boltzmann equation for photon f(p). f t +v f + a v f = {sources, sinks} Where we note that the system is time-steady and non-relativistic (i.e., there are no forces strong enough to bend rays of light). However, we re keeping the RHS general: there are likely to be lots of processes that can create/destroy photons. Remembering that I ν f and that v = cˆn, the equation simplifies to ˆn I ν = S net and the RHS has been boiled down to a single net rate of creation and/or destruction. What gives rise to S net 0? Photons can be absorbed by atoms & ions. Rate S depends on local properties of the gas (e.g., ρ, T) and on incoming I ν. Photons can be emitted spontaneously by atoms & ions. Rate S just depends on local properties. Photons can also be lost by being scattered out of a given direction ˆn or up/down in frequency ν. Rates depend on both local properties and on incoming I ν. Scattering can also create photons in the given direction ˆn and frequency ν, by coming from some other direction or frequency. Scattering in rates depend on local properties & on the I ν in all those other directions, but not on the given I ν (ˆn) that we re following. 8.5

6 The left-hand side above can be simplified in plane-parallel geometry. ˆn = (cosθ)ê z + (sinθ)ê x ( ) Iν However, I ν (zonly), so I ν = z Thus, the equation of radiative transfer is ê z. cosθ I ν z = S net which we can rewrite by breaking out the creation/destruction terms on the right side: µ I ν z = j ν χ ν I ν We saw this before, when discussing radiative diffusion in the interior. The opacity χ ν has units of 1/length, and we saw earlier that it depends on density ρ, so that we can break it down like χ ν = κ ν ρ = [κ ν (abs) + κ ν (scat)] ρ where the κ s are typically just called absorption coefficients (even if they arise from a combination of true absorption & scattering). Units of κ: (m 2 /kg); i.e., a cross section per unit mass of stuff that s doing the absorbing. The gain term j ν is often written as a product of χ ν times an intensity-like quantity called the source function µ I ν z = χ ν (S ν I ν ) but keep in mind that although I ν is a function of both µ and z, S ν depends only on z (and not angle). What is S ν? We ll get to that soon. 8.6

7 There s one more thing we can do to simplify the math of the radiative transfer equation: define a new depth coordinate: Optical depth τ ν goes up as you go deeper into the atmosphere; dτ ν = χ ν dz Note that it s frequency dependent. Some photons escape more easily than others. τ 1: optically thin regions where photons mainly flow freely; absorptions/scatterings are rare. τ 1: optically thick regions where photons are trapped. Absorptions/scatterings are so frequent that the distribution I ν (ˆn) is rapidly randomized to be nearly isotropic. τ 1: this defines the photosphere; i.e., the layer where MOST of the photons that we see are created! Since τ goes to zero as z + (high up), we can integrate to get the absolute optical depth scale as τ ν = τν 0 dτ ν = z + dz χ ν (z ) = + and, lastly, the equation of radiative transfer simplifies to µ di ν dτ ν = I ν S ν z dz κ ν (z )ρ(z ) where partial deriv s were replaced by d because we re following how the beams are processed vs. depth only (i.e., depth is the only coordinate that we ll differentiate with respect to) We ll just treat µ as a fixed parameter

8 However, earlier I said that solving for I ν (µ,z) requires solving two equations. We ve just looked at one so far. The other one is energy conservation. Recall from stellar interiors, dl r dr = 4πr2 ρǫ I ll show you that another way to write this is F rad = ρǫ Why? Write out the divergence of the flux in gory detail... F rad = 1 ( ) r 2 r r 2 L r = 1 L r 4π r 2 4πr 2 r = ρǫ and it s equivalent! However, far from the nuclear-burning core, the stellar atmosphere essentially has ǫ = 0. Thus, F = 0. Throughout much of the interior, this means r 2 F r = constant, i.e., F r 1 r 2 which we already know. However, in the tiny box of the plane-parallel atmosphere, we can be confident in approximating this as F = constant, and that constant is often defined as σt 4 eff. and this contains the assumption that the vector F is pointed in the radial (or +z) direction, so that F r = F = F. (This is a natural consequence of our plane-parallel symmetry!) Leaving off the ν subscripts for now, we essentially know two things: µ di dτ = I S df dτ = 0 Remember that the flux F is an angle-moment of I. Let s define these moments in a consistent way... J = 1 dωi(µ) = 1 +1 dµi(µ) U rad = 4πJ 4π 2 1 c H = 1 dωµi(µ) = 1 +1 dµµi(µ) F = 4πH 4π 2 1 K = 1 dωµ 2 I(µ) = 1 +1 dµµ 2 I(µ) 4π

9 Just like with the Vlasov/Boltzmann equation for fluids, it s possible to take moments of the equation of radiative transfer... i.e., multiply by some power of µ and integrate over all solid angle. Noting above, if we multiply each term by (1/2) and the power of µ, we can integrate over dµ from 1 to +1. Let s do it... but first we ll have to pause and think a bit more about the source function S.... Think about some limiting cases... (1) Deep in the high-ρ interior, I is close to isotropic, and τ 1 everywhere. There s (almost) no net transport; i.e., it s just about a perfect blackbody, in which emissions (the S term on RHS) are perfectly balanced by absorptions (the I term on RHS). When that s the case, I ν S ν, and both of them are roughly given by the equilibrium Planck function B ν (T). Note: the gray version: B = 0 dν B ν (T) = σt4 π. In most stellar atmospheres, it s safe to assume S B when the opacity is dominated by absorption/emission. (2) On the other hand, what if the only source of opacity was scattering? Here, the source function S tells us how photons get scattered into the direction ˆn that we re following with the radiative transfer equation. There are many physical mechanisms of scattering, and a lot of them are isotropic; i.e., the photon scatters into an essentially RANDOM angle, losing all memory of its incoming direction. (Think about the daytime sky...) When scattering is isotropic, it doesn t really matter where it came from, so S should depend on the angle-average of I over all directions. In other words, for pure scattering, S J. 8.9

10 In general, if the fraction of total κ ν that comes from scattering is called a (sometimes called albedo ), then S = (1 a)b + aj.... The equation of radiative transfer is thus given by µ di dτ = I (1 a)b aj and we can take the zeroth moment by just multiplying each term by 1/2 and integrating over dµ: ( d 1 +1 ) ( 1 +1 ) ( 1 +1 ) ( 1 +1 ) dµµi = dµi (1 a) dµb a dµj dτ In the 2 right-most terms, the factors of B and J are not functions of µ! They can be pulled out, and the integral is 2, which cancels out with the 1/2. Thus, dh dτ = J (1 a)b aj = (1 a)(j B) which is nice, but remember that H = constant, so the left-hand side = 0. Thus, for a plane-parallel atmosphere, we define the term local thermodynamic equilibrium (LTE) to mean that: No flux is created or destroyed (i.e., H = constant). J = B (i.e., the mean radiation field is the Planck function). Furthermore, this also means that S = J = B. This doesn t mean that I(µ) is the Planck function! How does this help us? Well, if we take the first moment of the equation of transfer, we multiply each term by µ/2 and integrate as before. Note that we can revert to earlier version µ di dτ = I S 8.10

11 because we know all about S = B = J by now. Thus, ( d 1 +1 ) ( 1 +1 ) ( 1 dµµ 2 I = dµµi dτ dk = H {zero!} dτ (noting that, after we pull out S, the rest is an odd function!) 1 ) dµµs Because H is a constant with depth, this is a very simple differential equation, which can be solved for K(τ) = Hτ + K 0 where K 0 is an integration constant that tells us the value of K at τ = 0. In the 1920s, Sir Arthur Stanley Eddington thought about this problem, and he made 2 key approximations that allow us to move forward to something useful. The Eddington approxmations aren t perfect (often wrong by 10% 20%), but they re not terrible. They involve how the moments relate to one another: 1st approx (diffusion approx): J = 3K. In radiative stellar interiors, we saw that this worked well for I I 0 +I 1 cosθ. It s best for the deep, optically thick layers. 2nd approx (surface approx): J = 2H. This is true at the optically thin top of the atmosphere. If all radiation is going up and none is coming down, then half of the sky is filled, and half is empty. Thus, the flux is half of the mean intensity. Deeper down, J H. Let s plug in those approximations, one at a time... J = 3Hτ + 3K 0 (using the 1st) = 3Hτ + J 0 = 3Hτ + 2H (using the 2nd) ( B = 3H τ + 2 ) (collecting terms & recalling S = B = J) 3 We ve seen that B = σt 4 /π and that F r = 4πH = L 4πR 2 = σt 4 eff (which defines T eff ) 8.11

12 Plugging those in, we get the classical gray atmosphere temperature stratification law: T(τ) 4 = 3 ( 4 T4 eff τ + 2 ) 3 Limiting values: As τ 0, T (1/2) 1/4 T eff 0.841T eff. We ll define this as T rad later. Note that the real Sun can go down to 0.7 T eff When τ = 2/3, T = T eff, and this height is traditionally called the optical surface or photosphere. For τ 1, T keeps growing. This is a close cousin to the greenhouse effect, in which more atmospheric absorption causes the atmosphere to heat up more and more. In fact, in the deep interior, this T(τ) gives same answer as ( dt dr ) rad = 3κρ 4acT 3 ( Lr 4πr 2 ). Note: The in-class lectures on this topic end here. The remaining pages will be summarized briefly

13 How does this depth dependence help us understand the radiation emerging from the Sun s atmosphere? We ll now work out how I(µ) produces the so-called limb darkening effect. In the homework, you ll use the integrating factor method to solve a simple version of the radiative transfer equation. Recall, for S = constant, µ di dτ = I S is solved by I(τ) = I 0e τ + S ( 1 e τ). The boundary condition I = I 0 was applied at the top of the atmosphere (τ = 0); i.e., we assumed that a flashlight-beam of intensity is begin projected into a gasesous slab. However, the Sun s atmosphere has no beam shining down from above. It generates its own light from within (i.e., mainly from the deep τ 1 layers)! Thus, let s assume I 0 = 0. For the general case of S(τ) not constant, the integrating-factor solution is an integral. Let s also restrict ourselves to µ > 0 (rays coming up from below): I(µ,τ) = 1 µ τ dt S(t) e (τ t)/µ For these upward-going rays, the integral is over all layers below the height τ at which we re computing I. The intensity is a weighted sum of the source function in all layers that can contribute to I at τ. Note: for S = constant, the above solution gives just I = S. The down from above version of the general solution (for µ < 0 and the integral going from 0 to τ) gives back the solution given above in the case of S = constant. Anyway, let s evaluate this integral for: The gray solution for S(τ), and The specific value τ = 0 (i.e., to solve for the emergent intensity coming from the top of the atmosphere). 8.13

14 For S(t) = 3H[t+(2/3)], I(µ,0) = 3H µ 0 dt ( t+ 2 ) ( e t/µ = = 3H µ+ 2 ) 3 3 This convenient conversion from τ-dependence to µ-dependence is called the Eddington-Barbier relation. (Not general for any S(τ).) As we look across an image of the Sun, we see rays that go from µ = 0 (limb) to µ = 1 (center) to µ = 0 again (other limb):. This relationship makes intuitive sense... the rays skimming the limb (µ 0) penetrate less deep (in the r direction) than the rays hitting disk-center. Photons that come from shallower (higher up) regions emerge from cooler plasma... and B ν (T) is lower for lower T. Thus, limb darkening is a good (albeit indirect) way to probe the depth dependence of gas properties in a stellar atmosphere. 8.14

15 FYI, the full angle & depth dependence of the gray I(µ,τ) can be plotted using polar coordinates

16 Another important question: How do we convert optical depth τ to actual atmospheric height z? Recall the definition of optical depth: τ ν = + z dz κ ν (z )ρ(z ) One way we can get this into a form where we can solve for z is to approximate: In upper layers of atmosphere, T constant. Although the absorption coefficient is often approximated as κ ν κ 0 ρ a T b, let s assume here that a = b = 0; i.e., that κ ν constant with depth, too. For an isothermal (constant T) atmosphere, we derived earlier that ( ρ(z) = ρ 0 exp z ) where H = k BT H µm H g constant. The expression for τ(z) boils down to an integral over e z, and you can show that τ ν (z) = κ ν ρ(z)h. From the above, it s clear that τ ν drops exponentially with scale height H, just like density. κ ν depends on photon frequency, but for a mean value of κ 0 we can solve for the photospheric density, κ 0 ρ 0 H 2 3 so ρ 0 2 3κ 0 H. However, κ ν really varies as a function of ν, so different frequencies see the photophere at different relative heights. i.e., τ ν = 2 3 ρ(z) = ρ 0 e z/h 2 3κ ν H e z/h means different things at different values of ν κ ν ρ 0 H κ 0 κ ν z Hln ( κν κ 0 ) 8.16

17 Thus, at some specific frequencies (with markedly HIGHER values of κ ν ), the photosphere occurs at a lower density & larger height: Next, we ll see the consequences of that. Limb darkening was one way of probing the depth dependence of what s going on in the Sun s atmosphere. There s another very useful way: spectral lines. We ve talked about the fact that the opacity κ ν has contributions from continuum processes and from narrow lines that only occupy narrow ranges of ν. Why? It s due to the fact that atomic energy levels are quantized. There are only specific discrete E values for jumps between the various levels. Photons of that exact energy ( E = hν) can be easily captured, and this boosts a bound e to a higher level. The higher levels are unstable, so often photons are eventually re-emitted, but the original ones were still removed from the incoming spectrum. 8.17

18 Note: Sometimes the photon s hν may exceed the ionization potential of the atom, and its capture causes photoionization; recall: hν +X X + +e But there s a whole continuous range of photon energies that can do this; it s a source of continuum opacity, not spectral lines. The opacity due to a spectral line transition is slightly broadened in frequency around its central quantized value. Why? Doppler broadening: The atoms are moving around randomly, in a Maxwell-Boltzmann distribution. Sometimes the right photon to excite an e is a little red-shifted or blue-shifted due to this motion. Natural broadening: The excited states have a finite lifetime, with a quantum mechanical spread. Heisenberg s uncertainty principle says that you can t know both an exact time interval AND an exact energy; You may have seen x p h but there s also E t h so there s a quantum fuzz around the exact energy level. Collisional broadening: Our atom is sitting in a bath of other atoms, ions, and free electrons. They can all undergo collisions with our atom, and give it random kicks of energy. Sometimes this can happen in tandem with photon absorption, so the amount of energy-level jump has another random component to it. Thus, the frequency dependence of opacity in the vicinity of a given spectral line looks like: 8.18

19 The big question: What does the solar atmosphere s spectrum I ν look like in the vicinity of these extra opacity peaks? It depends on T(τ), i.e., on how temperature varies with depth. Far away from the peaks, roughly a I ν is roughly a blackbody B ν (T eff ), which later we ll call B C (T C ) (for continuum). An instructive way of thinking about spectral lines was first formulated by Arthur Schuster (1905). We saw that the photosphere in a line is physically higher up in the atmosphere, at a place where the surrounding continuum has τ 1. So, let s think about the line-forming region as being physically separated from the rest of the atmosphere. Schuster postulated that the line is formed in a cloud of gas sitting above the main source of continuum radiation (B C ). Far from the central line frequency ν 0, the cloud is invisible; i.e., it has κ ν = 0, but there s a sharp peak inside the line. Assume the cloud has constant properties throughout its volume (i.e., constant ρ and κ ν versus height z). Thus, the frequency-dependent optical depth of the cloud is [ ( ) ] 2 ν ν0 τ ν = dz κ ν ρ = κ ν ρ z τ 0 exp ν where the line strength is essentially τ 0, and the parameters inside the exponential tell us about the central frequency & line width. If everything is constant inside the cloud, then its source function S ν is constant versus z, too! We know the solution for the emergent intensity, when there s an incoming beam I 0ν going through a SLAB with constant S ν : I ν (τ ν ) = I 0ν e τ ν + S ν ( 1 e τ ν ). 8.19

20 We can continue assuming simple things about the two intensity parameters above. The incoming beam comes from the continuum photosphere, I 0ν B C (T C ) where T C T eff. If the line-forming cloud is in LTE, too, then S ν B L (T L ) where T L is the temperature in the cloud. We d like to know the residual intensity r ν = I ν I 0ν i.e., the fractional dip or enhancement, relative to continuum. Divide everything in the I ν solution by the incoming intensity, and we get r ν = e τ ν + B L B C ( 1 e τ ν ). Roughly speaking, recall that B T 4, so... B L B C ( TL T C ) 4. The appearance of the line profile depends on whether T goes up or down as you go up from the photosphere. 8.20

21 Examples: For weak lines (τ 0 1), e τ (1 τ), so ( ) [ 4 (TL ) 4 TL r ν (1 τ ν ) + τ ν 1+ 1] T C An absorption line occurs if T L < T C (i.e., most of the Fraunhofer lines in the photospheric spectrum). An emission line occurs if T L > T C (lines formed in warmer chromosphere!). For strong lines (τ 0 1), the e τ term vanishes in the core of the line, and ( ) 4 TL r ν at line center T C i.e., for a cold overlying layer (T L T C ) it s saturated and black at line-center. If we look above the limb and our line-of-sight passes only through the cloud (and not through the underlying photosphere), we have to go back and not divide by I 0ν (since it s zero), With I 0ν = 0, I ν = B L ( 1 e τ ν ) τν B L (for weak lines with τ 0 1) and thus there will be an emission line on top of nothing (instead of on top of a continuum with r ν = 1) T C τ ν Observing a large number of lines helps put empirical constraints on the temperature structure T(z) of the solar atmosphere. 8.21

22 Irradiated (Planetary) Atmospheres The radiative transfer theory above allowed us to study how light from the Sun s interior is processed as it passes through the Sun s outer layers. That same theory tells us how this light is processed when it enters the top of the Earth s atmosphere and (some of it) makes it down to ground level. There are optical depth effects here, too! The Eddington/gray atmosphere had 2 boundary conditions, though we didn t tend to think about them as such: From beneath: F = σt 4 eff = constant > 0. From above: I(µ < 0) = I down = 0. If we move on to thinking about planetary atmospheres, both of those boundary conditions change. Ignoring internal heat generation (tides or radioactive decay!), the total F can be zero, but I down 0. Recall, of course, that these quantities are integrated over ν. In isolated parts of the spectrum, F ν 0. Spectrally, we often see that F = 0 is maintained as a balance between: { } Inward flux: from stellar irradiation ( short-wave ) Outward flux: thermal re-radiation ( long-wave ) For the Earth, these two sources of radiation occupy (essentially) separate parts of the spectrum: 8.22

23 When the short-wave & long-wave pieces of the flux cancel each other out, we have F SW = F LW. We use this flux balance to determine the atmospheric temperature structure of terrestrial planets. There are several ways to estimate F SW. First, examine instantaneous insolation on a spherical planet: At noon (θ = 0), F SW = L 4πr 2 (1 A) F max where r is the star planet distance D. (Strictly, r = D R p.) For us, the insolation flux (L /4πr1AU 2 ) is called the solar constant. 8.23

24 A = the short-wave Bond albedo of the planet: the fraction of total incident flux that is immediately reflected back into space. (Earth: A = 0.30, Venus: A = 0.77) Depends on cloud cover! At other locations, F SW = { Fmax cosθ, on the day side 0, on the night side. For a tidally locked planet, that s all one needs. For a rotating planet, the incoming flux is spread around in longitude [and a bit in latitude, if there s axis obliquity, but usually F SW (poles) F SW (equator)]. If we assume rapid redistribution of the incoming flux, we can compute the average over the globe: F SW = 1 dω F SW (θ,φ) = F 2π π/2 max dφ dθ sinθ cosθ = F max 4π 4π } 0{{}} 0 4 {{} 2π 1/2. This may not correspond exactly to any one point on the planet, but it lets us estimate surface-averaged quantities. If the planet re-emits the absorbed radiation as something close to a blackbody, the outgoing long-wave flux can be written as F LW = ǫσt 4 eff where ǫ is a fractional emissivity (ǫ = 1 for a perfect blackbody; Earth has ǫ 0.96). We can thus compute the planet-averaged equilibrium temperature F SW = F LW = T eff = Earth: T eff 257K 16 C. [ L (1 A) 16πǫσr 2 ] 1/4 Venus: T eff 229 K. Lower because of higher albedo!

25 Note: some books take an energy in = energy out approach: [ ] L (1 A) E in = πr 2 4πr 2 p, E out = ( ǫσteff) 4 4πR 2 p which gives the same answer as above.... This isn t the end of the story because terrestrial planets can have atmospheres. The deeper you go, the more opacity can trap photons... the greenhouse effect. When computing height dependence of air temperature, our standard gray/eddington result is still more or less valid: [ ( 3 T air (τ) = T eff τ + 2 )] 1/4 4 3 and at the infrared λ s of interest to the outgoing long-wave Planck function, the near-surface values are Earth: τ 1.4 T air 288K +15 C Venus: τ 140 T air 735K which makes sense, if we know that Venus surface pressure is roughly 100 times that of Earth. There are many other complications: Real atmospheres aren t gray or Eddington. For CO 2, strong-opacity bands are surrounded by transparent spectral windows. Even for a very rapidly rotating planet, computing T air as a function of latitude demands including a horizontal diffusion term (due mainly to atmospheric & ocean dynamics). Gerald North s classic (1975, JAtmSci, 32, 2033) energy-balance model reproduces lots of what we see... including the possibility of snowball Earth ice age phases. Computing the ground temperature of the solid surface takes extra care. It s often > T air, but simple models overestimate the discontinuity. 8.25