PHYS 633: Introduction to Stellar Astrophysics

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1 PHYS 6: Introduction to Stellar Astrophysics Spring Semester 2006 Rich Townsend Convective nergy Transport From our derivation of the radiative diffusion equation, we find that in regions of a star where energy is transported by radiation alone, the physical temperature gradient satisfies = rad, (1 where the radiative temperature gradient is defined in terms of the local flux F (r by rad κρh P 4ac T 4 F (r, (2 and H P is the pressure scale height, H P P r P = P r2 Gmρ. ( If we wish to include the energy transported due to conduction (usually unimportant except in degenerate stars, we can replace the κ term in the definition of rad with κ + κ con, where κ con is the conductive opacity. So, in a radiative region, rad tells us what the physical temperature gradient will be. But what is its interpretation in a convective region? nergy is still transported by radiative diffusion when convection is occurring, but the convection itself also transports energy. In such cases, rad provides an indicator of the temperature gradient that would ensue if the convection could artificially be switched off, leaving radiation to carry all of the flux. Hence, the local flux F (r can be expressed as F (r = F C (r + F R (r = 4ac T 4 κρh P rad, (4 where F C is the convective energy flux, and F R is the radiative energy flux, We can write F R in the form, F R (r = 4ac F R (r = T 4 κρh P. (5 rad F (r, (6 which allows us to express the physical temperature gradient at each radius in the form [ = rad 1 F ] C(r. (7 F (r 1

2 Clearly, for a positive convective flux, we have < rad ; so the effect of convection is to make the temperature gradient less steep than it would be if energy was transported by radiation/conduction alone. In fact, as the convection becomes more and more efficient (we ll see what that means further on, the physical temperature gradient approaches the adiabatic gradient ad. Due to the Schwarzschild stability criterion, the above expression only holds in regions where rad itself is greater than the adiabatic gradient ad ( ln T ln P. (8 s Otherwise, radiation is able to transport all of the flux, with a temperature gradient = rad < ad that gives a stable hydrostatic equilibrium (i.e., no convection, F C (r = 0. Given eqn. (7, we can calculate the physical gradient throughout a convective region, if we know both the total flux F (r and the convective flux F C (r. We obtain the former from the local luminosity l(r; but how do we go about deriving the convective flux? Here lies one of the biggest problems in present-day stellar astrophysics: there is no comprehensive, physically-realistic theory for convective transport of energy. None. Nada. This topic has received an inordinately-large amount of attention over the past 50 years or so, but still remains extremely problematical. Is all lost, then? Fortunately not. Although there are no complete models, there do exist very useful parameterizations of convection, which turn out to produce workable results. The most simple of these is the mixing-length theory developed originally by Ludwig Prandtl, and first applied in the field of astrophysics by rika Böhm- Vitense.. The theory is based on the assumption that the rising and falling fluid elements responsible for the convective energy transport retain their identity over a typical radial distance l m (the mixing length, before they dissolve into their new surroundings. Let s focus on an average fluid element at radius r, which we assume has already traveled a distance l m /2 since it began moving (see Fig. 1. Adopting the same notation as for the convective stability analysis, let T be the change in the temperature of the element during its motion, and let T A be the corresponding change in the ambient fluid surrounding the element. We may write these two quantities as and T = T l m (9 T A = T l m (10 where ( ln T (11 ln P is the temperature gradient (w.r.t. pressure experienced by the fluid element. If during its motion the element remains completely insulated from its surroundings, then = ad. But, more realistically, the element will lose some of its 2

3 Figure 1: The path followed by the average fluid element. excess heat to its surroundings, so the general inequality for the four nablas in convective regions ( ad < rad is ad < < < rad. (12 In terms of the temperature changes above, the excess heat per unit mass of the element over its surroundings is given by (T s T s A = c P ( T T A (1 = T c P ( l m. (14 Hence, the convective flux being the amount of excess heat transported by

4 the element through radius r, per unit area, per second is just F C (r = ρv(t s T s A = ρvt c P ( l m, (15 where v is the suitably-averaged velocity of the element. The next step in our derivation is to obtain values for v and. Let s focus first on the convective velocity. This quantity is of course non-zero because the element has been accelerated by buoyancy forces acting upon it. As with the stability analysis, the buoyancy force per unit mass (i.e., the acceleration is given by g b = g ρ ( ρ ρ A, (16 where ρ and ρ A are just the density changes in the element and its surroundings. With a little algebra, that takes advantage of the fact that the element remains in pressure equilibrium with its surroundings, we can write these density changes in terms of the nablas as and ρ = ρ(α δ l m (17 ρ A = ρ(α δ l m, (18 where α and δ are the usual isothermal compressibility and isobaric thermal expansion coefficients. Hence, the buoyancy force per unit mass is g b = gδ( l m. (19 This expression gives the force when the element reaches radius r; assume that the average force between r l m /2 and r is then just half this value. We further assume that drag forces on the element due to having to push surrounding fluid out of the way will mean that only half of the average force is able to accelerate the element. Hence, we write the (assumed to be uniform acceleration of the element as a = g b /4 = gδ( l m 8H P. (20 Given this acceleration, we can evaluate the element s velocity at r as v 2 = 2a l m 2 = gδ( l2 m 8H P. (21 So where have we got to? With reference to eqn. (15, we have eliminated the convective velocity v, so the only outstanding unknown quantity (apart from the mixing length l m is the element s temperature gradient. To get a 4

5 handle on this quantity, we need to think about how rapidly the element cools off as it rises. We can write the rate-of-change of temperature as ( ( ( ( ( T T P T s = +. (22 P s s p = T P P ad + T ( s (2 c P The first term on the right-hand side comes from adiabatic expansion, and the second represents heat losses at constant pressure to the element s surroundings. Since the element moves a distance v dt in time dt, we can rewrite this equation in terms of the spatial temperature gradient experienced by the element, ( T r = 1 v T = T P ad ( P r + T c P v ( s (24 Multiplying through by the pressure scale height, we therefore find after a little algebra ( ln T = = ad H ( P s. (25 ln P c P v We re further along now; we have an expression for the elemental temperature gradient, in terms of ad and the rate at which the element exchanges heat with its surroundings. In a twist of irony, this heat loss is actually caused by radiative transport of energy. If the element has an approximate dimension d, then the radiative flux through its surface can be expressed using the diffusion equation, f r 4acT κρ 4acT 4 κρ ( T T A, (26 d/2 ( d l m H P. (27 Here, we have used the temperature difference between the element and the ambient fluid to write the temperature gradient term. In terms of f r, the rate of heat exchange per unit mass of the element will be T ( s = f ra Bρ, (28 (note the minus sign, indicating a heat loss, where A is the surface area of the element, and B its volume. Substituting this into the expression (25 for, and eliminating f r using eqn. (26, we obtain = ad + H P f r A T c P vbρ, (29 = ad + 4ac T κρ 2 c P v ( l ma d B. (0 5

6 The right-most fraction in this expression contains information about the geometry of the fluid elements. For a spherical element with radius equal to l m, it would equal 6/l m ; but most authors (including Kippenhahn & Weigert take l m A d B = 9/2, (1 l m in order to achieve agreement with the original derivations by Böhm-Vitense. With this substitution, we have = ad + 6acT κρ 2 c P vl m (. (2 With this final expression, we now have four equations that describe energy transport in a convection zone. To recap: [ = rad 1 F C ] F, F C = ρvt c P ( lm, v 2 = gδ( l2 m 8H P, = ad + 6acT κρ 2 c P vl m (. There are four unknowns in these equations: F C,, v and (we assume that we know l m somehow. So, we have enough information in principle to calculate the physical temperature gradient. In practice, it is useful to eliminate F C, and v, to obtain a single expression for the physical temperature gradient: where ξ is the (single real root of the cubic equation = ad + ξ 2 U 2, ( (ξ U + 8U 9 (ξ2 U 2 W = 0, (4 and the parameters U and W are defined by U act 8H P κρ 2 c P l 2 m gδ, W = rad ad. (5 What is the interpretation of these parameters? Well, W simply measures how much the radiative temperature gradient exceeds the adiabatic temperature gradient. Since we can write the convective flux in the form F C (r = F (r rad rad (6 (see eqn. 7, and since > ad in a convection zone, we can see that W gives the upper limit on the total flux that will be transported by convection: F C (r < F (r W = 4ac W T 4. (7 rad κρh P 6

7 To understand the significance of U, let s look at the limiting cases where this parameter is very small or very large. In the U 0 case, eqn. (4 has the solution ξ = 0. Hence, ad, and we see that the convection is very efficient; the temperature gradient need only be marginally above the adiabatic gradient for convection to transport the stellar flux. In the opposite case of very large U case, it is less clear how to solve eqn. (4. Let us first introduce the parameter x ξ 2 U 2. (8 Since is bounded between ad and rad, we can see from eqn ( that x must remain small even when U becomes large. Hence, we can make the approximation ξ U + x/2u (9 in the cubic equation (4, to obtain x 8U + 8U 9 (x W = 0. (40 For U 1, this becomes 8U (x W = 0, (41 9 giving us the solution x = W, or ξ 2 U 2 = W. (42 Substituting this back into eqn. (, we obtain the result rad (4 in the limit U. Based on the foregoing analysis, we can interpret U as a measure of the efficiency of the convection: the U 0 case corresponds to efficient convection ( ad, and the U case corresponds to inefficient convection ( rad. 7