You ll see in the final homework (also Cravens, chapter 6.1.1) that a purely hydrostatic corona doesn t make sense. P is way too big.

Transcription

1 THE SOLAR WIND You ll see in the final homework (also Cravens, chapter 6.1.1) that a purely hydrostatic corona doesn t make sense. P is way too big. We now know that the outer corona requires a dynamic solution with u 0. This is the outflowing solar wind. In the early part of the 20th century, other observational hints began to emerge that hinted at the presence of outflowing charged particles in the solar system. Ludwig Biermann began collecting these pieces of evidence together in the early 1950s... Some comets have 2 tails: a dust tail traces the orbit, an ion tail points radially away from the Sun. Sometimes tails break apart, and pieces flow away from the Sun at 100 km/s speeds. 2 3 days after big solar flares, things often begin happening on Earth: geomagnetic storms, brightenings of aurora borealis, blown circuits in long conductors (telephone & telegraph lines, pipelines). = (1 AU)/(2 3 days) a few 100 km/s. In 1958, Eugene Parker realized there is a connection between this kind of outflow & the existence of a hot corona! Let s look at Parker s initial hydrodynamic solution: Time steady (all / t = 0), but we must allow for u 0. Spherically symmetric (i.e., stretched-out split-monopole magnetic field) Isothermal: T 10 6 K, and assume conduction keeps it constant. Since u points radially, let s simplify our notation by writing u instead of u r. 11.1

2 Parker solved conservation equations... Mass conservation: Momentum conservation: 1 d ( r 2 ρu ) = 0 r 2 dr u du dr = 1 ρ dp dr GM r 2 For a time-steady system, mass conservation is simple: r 2 ρu = constant. By convention, we define M = 4πρur 2 and Ṁ is the total mass loss rate of the wind (in kg/s), integrated around all 4π steradians of solid angle. Note: Parker s theory doesn t predict Ṁ. For that, we would need to go back to coronal heating models like RTV. It predicts the base pressure (at the TR), i.e., how much plasma is there at the base of the corona. Parker s wind model just draws out that plasma. We can see the basic idea already: if u r eventually becomes constant, then ρ 1/r 2. If T = constant, then P 1/r 2. (P 0, not like hydrostatic!) But this makes for a gas pressure gradient that produces an outward force.... So, if T = constant, then dp dr = ( kb T µm H ) dρ dr = c2 i dρ dr where c i is known as the isothermal sound speed (constant). Recall, though, that the mass conservation equation lets us write dρ/dr in terms of du/dr: u dρ dr + ρdu dr + 2ρu = 0, r so we can write dp dr = c2 i ( ρ u du dr 2ρ ) r. 11.2

3 Parker then took this and inserted it back into the momentum equation, and rearranged some terms, to get the classic isothermal equation of motion: ( ) ( u c2 i du 2c 2 u dr = i r GM ) r 2 There s lots to say about this equation. The right-hand side is a sum of imposed accelerations: 1st term is due to P in spherical geometry: it s an outward force, and it s stronger when plasma is hotter (i.e., when c i is larger). 2nd term is due to gravity: it s an inward force, The left-hand side is kind of counter-intuitive. Usual intuition says that pushing outward on a parcel (RHS > 0) will make it accelerate as it flows out. i.e., we expect that (du/dr) RHS. Our intuition makes sense for supersonic flows (u c i ). In this case, RHS > 0 means that du/dr > 0. However, for subsonic flows (u c i ), we have c2 i u and if RHS > 0, we get deceleration with increasing distance! du dr RHS Aside: This makes sense if we think of the subsonic flow as being similar to hydrostatic equilibrium (i.e., the flow isn t very important to the main P ρg balance). If there s a positive force on the RHS, the effects of gravity are partially cancelled out. Thus, hydrostatic balance can be maintained with a flatter (shallower) P. This means ρ(r) falls off more flatly/shallowly. Lastly, mass conservation implies that u(r) increases more flatly/shallowly. Thus, an extra outward force means this subsonic (nearly hydrostatic) wind is accelerated less than it was before! Lesson: fully-compressible hydrodynamics with flows on either side of the sound barrier can be weird. Cravens discusses this in the context of gas flow through a nozzle or wind tunnel. 11.3

4 But what does the actual Sun do? In general, the equation of motion can be written like this: du dr = A B C D = A(r) B(r) C(u) D(u) If we know A,B,C,D then we can solve for the slope du/dr. Note that there are singular points when: A = B, but C D: du/dr = 0 (flat) A B, but C = D: du/dr (vertical; unphysical?) Both A = B and C = D: du/dr = 0/0 (Parker s critical point!) Who remembers L Hopital s rule? It s possible that there are finite values of du/dr at the critical point.. one must take derivatives of the numerator & denominator, then take the limits (to 0/0) simultaneously. Note also: A & B are functions of r only: A = B defines a critical radius. C & D are functions of u only: C = D defines a critical velocity. The critical radius (sometimes called the Parker radius ) is given by r = r c = GM 2c 2 i and the critical velocity is given by u = u c = c i (sometimes called the transonic velocity ). When c i is larger, r c is smaller. For a hotter corona, P beats gravity sooner! Note: the full Parker critical point occurs when u = u c at r = r c. 11.4

5 For a fixed value of c i, we can compute du/dr for every point in (r,u) space, and then connect the little bits of slope by continuous curves: Which solutions are physically realistic? We should start at the lower-left: Why? When we re close to the solar surface (r R ), we know already that there are no hugely fast outflows. Putting aside granulation & waves, the photosphere & chromosphere are static. We can safely ignore the solutions that double back on themselves. A realistic solution for u(r) should have only one unique value of u at each radial distance r. What s left? Solar breeze solutions that are always subsonic (u < c i ). The transonic solar wind solution that starts subsonic, goes through the critical point (r c,u c ), then is supersonic far from the Sun. Parker (1958) suggested the transonic/accelerating solution should be what the Sun naturally exhibits. In 1952, Bondi examined the decelerating accretion solution for the case of u < 0 (i.e., infalling parcels speed up in u ). Despite Bondi also having studied a solution that passes through the critical point, there was widespread skepticism about Parker s idea. WHY should the system select only this one delicate solution, which somehow knows to pass through (r c,u c ), over all others? 11.5

6 There are 3 answers to that: 1. The breeze solutions all have u 0 as r, so they become hydrostatic at large distances. Thus, they have the same problem (P P ISM ). 2. Breeze solutions are unstable! Marco Velli (2001) showed that any small perturbation to a breeze will kick it, causing it to collapse back onto the transonic solution (which is the ONLY stable, realistic solution). 3. Parker published his analysis at the dawn of the Space Age. He had to only wait 4 years for Mariner 2 to leave the Earth s magnetosphere in 1962 and directly probe the solar wind plasma. u 1AU > c i! Thus, Parker was vindicated.... How does the transonic solution behave? du dr = A B C D = (2c2 i /r) (GM /r 2 ) u (c 2 i /u) is always > 0 Below critical point, u < c i, so gravity must win in the numerator. Above critical point, u > c i, so the pressure gradient wins. i.e., the outward force that accelerates the wind dials up gradually, so that low heights remain hydrostatic. (That s the only way a wind can really be time-steady, and not totally evaporate the solar surface!) Once we take L Hopital s rule to find (du/dr) c, we can integrate the equation up and down from r c to get u(r). It s separable: all u terms on one side, all r terms on other: ( ) ( ) ( u (u 2 c 2 2 r 1 i ) c2 i ln = 4c 2 i ln +2GM r c r 1 ) r c c 2 i Problem: one can t solve equations like this (e.g., lnx+ax = B) analytically! 11.6

7 Math-geek solution: mathematicians have created something called the Lambert W function that lets you do it, as long as you don t mind the solutions being in the form of some new function. (We re used to it with exp, ln, sin, cos,...) Usually, solar physicists integrate equations like this numerically. However, if we just care about how it behaves far from the Sun (r r c, u u c ), the largest terms dominate: u 2 4c 2 i ln ( r r c ) i.e., u lnr, which is increasing very slowly as r. In reality, we see that u constant u pretty much by the orbit of Mercury (r 70R ), and it remains constant from there on out. u 250 to 900 km/s (average: 450 km/s) Why? Really, T isn t isothermal! T as r, so the RHS acceleration term decays away. So does gravity, too, so eventually the equation of motion is just u du 0 so u constant coasting speed. dr Can we predict values for u? We know that it should go up as the coronal c i goes up. Parker s initial models showed that: 11.7

8 Cravens assumes a polytropic kind of variation with r, i.e., P ρ γ, where our isothermal case is γ = 1. If the wind obeys P ρ γ far from the Sun (where u constant), then we know that ρ 1, so T ρ γ 1 1 r 2 r. 2(γ 1) For a polytropic (pseudo-adiabatic?) solar wind, one can integrate the full energy conservation equation to obtain Bernoulli s equation, which says that the total energy of the corona should be conserved: E = u2 2 GM r + ( γ γ 1 ) c 2 i = constant. (Note this doesn t work for γ = 1; the last term blows up!) Working it out for r c and r, we get ( ) E c = u2 c 5 3γ E = u2 5 3γ i.e., u = u c 2 γ 1 2 γ 1 What should we use for γ? We know γ = 1 is unrealistic (constant T, also u ). We also can see that the adiabatic result (γ = 5/3) doesn t work, either. It gives u = 0 in the above equation! Numerical models also show that u > u c (i.e., the wind keeps accelerating above the critical point). That requires: 1 < γ < 1.5. The polytrope law gave T r δ, with δ = 2(γ 1). However, actual spacecraft measurements give δ 0.2 to 0.8. Thus, the effective γ is

9 Does it make sense? Given above range of γ, we get u 1.4 4u c. For coronal temperatures of 1 2 MK, u c km/s. Thus, this model predicts u km/s which is pretty much the range we observe! A perfectly adiabatic γ = 5/3 would mean T r 4/3, which is a much steeper drop-off than we ever see. γ < 5/3 means that there must be sufficient continual heating of the plasma in the extended corona, over many solar radii, to keep T high enough, for long enough, so the RHS acceleration term has a chance to beat gravity. Thus, we really need to solve the coronal heating problem in order to fully understand solar wind acceleration.... A few other things about the 1D radially outflowing wind... Some other outward forces have been suggested to help Parker s P acceleration. If Alfvén waves are propagating out radially, they can exert a time-averaged wave pressure gradient force, too. Wang & Sheeley (1990) noticed an interesting trend. When reconstructing the PFSS magnetic field and tracing field lines out to 1 AU, one can compare the degree of flux-tube expansion to the wind speed. There s an anti-correlation... It s used for space weather prediction, but it s not great (correlation coefficient never gets better than 50%). 11.9

10 Coronal holes fill the majority of the heliosphere volume with fast wind. However, we didn t really know that until the 1990s, when the Ulysses spacecraft got a gravitational assist from Jupiter and was kicked into a polar orbit. This was the first exploration far outside the ecliptic plane! 11.10

11 The Interplanetary Magnetic Field (IMF) How does the solar wind (and magnetic field) behave far from the Sun? Right above source surface (r 2.5R ), both u & B are pointed radially. Further out, though, it gets a bit more complicated. The Sun & the inner corona rotate quasi-rigidly (with a roughly 27-day period), and the footpoints of magnetic field lines are fixed onto the rotating surface. Magnetic tension wants to keep them rigid. In ideal MHD, B is frozen-in to the motion of plasma parcels, but it s NOT true that u B. One counterexample that we ve discussed: a collapsing gas cloud: For this cylindrical ( toothpaste tube ) case, u is to B, but B is still frozen in.... What about parcels of solar wind? Above the source surface β 1. Thus, the field gets dragged along passively with the parcels. We ll look at this in 2 ways: (1) The garden sprinkler analogy. Consider just 1 longitude that rotates with the Sun & spits out wind parcels (each with u = u r ê r ): 11.11

12 Even though u is radial (in the nonrotating frame), B becomes stretched out into an Archimedian spiral: the Parker spiral. (2) Cravens uses coordinate transformations to show that if we write the diffusionless induction equation B t = (u B)...in a frame rotating with the Sun & inner corona, then there ought to be NO time variations. Transform: u = u r ê r (Ωr sinθ)ê φ B = B (i.e., same in any frame, as long as u c) We ll think mostly about the ecliptic plane: sinθ = 1. ( ) B Thus, = (u B) = 0 t rot frame and a trivial way to make this true is: u to B. This gives the same result as the garden sprinkler model... i.e., u B means u φ u r = B φ B r So, outside the source surface, ( rss ) ( ) 2 Ωr sinθ B r = B ss and B φ = B r r Up at these distances, u r constant, so B φ /B r increases linearly with r. u r WedefinetheParkerspiralangleφ p as the angle between u and B: tanφ p = B φ B r = Ωr u r in the ecliptic. It s less tightly wound at higher latitudes

13 Using the average u r 450 km/s in the heliosphere, At r = r ss = 2.5R, Ωr = 4.7 km/s φ p 0.6 At Mercury, r = 70R, Ωr = 130 km/s φ p 16 At Earth, r = 215R, Ωr = 400 km/s φ p 42 At Pluto, r = R, Ωr = 16,000 km/s φ p 88 At 1 AU, we re at the coincidental point where φ p is always around 45 ; i.e., B φ B r.... Of course, u r isn t always constant... it typically varies between about 250 and 900 km/s. Different streams of parcels have different speeds. Thus, different streams have different values of φ p, and the outcomes depend on what precedes what: These kinds of co-rotating interaction regions (CIRs) are seen very frequently (typically 3 4 times/month) at 1 AU. CIR shocks are a source of energetic particles that can lead to damaging space weather. (At solar max, it s mostly CMEs, but at solar min, CMEs are rare and CIRs are the dominant source of damage.) 11.13

14 In the outer heliosphere, where φ p approaches 90, multiple CIRs wrap around to become global/merged interaction regions (GMIRs). These end up being huge pulses that slam into the edge of the heliosphere and cause long-term modulations in the cosmic rays (galactic energetic particles) that make it into the heliosphere. The Large-scale 3D Heliosphere Early in situ measurements at 1 AU showed that the sign of B r flipped back and forth either 2 or 4 times in a solar rotation. These magnetic sectors turned out to be crossings of the current sheet that divides polarities at the source surface. The dipole component can be tilted! The combination of a tilted split-monopole and the rotating frame turns the current sheet into a warped ballerina skirt