where v is the velocity of the particles. In this case the density of photons is given by the radiative energy density : U = at 4 (5.

Transcription

1 5.3. CONECIE RANSOR where v is the velocity of the particles. In this case the density of photons is given by the riative energy density : U = a 4 (5.11) where a is the riation density constant, and so the photon energy flux is: F R = 4 ac 3 = 4 ac 3 apple = (5.12) where we have used that a =4 /c where is Stefan Boltzmanns constant. We also know that the riative flux has to carry all the energy produced by the star, so we also know that F R = so we can solve for the temperature grient L 4 r 2 (5.13) L 16 4 r 2 = = 3apple L 1 64 r = 3appleL r 4 3 (5.14) In principle Eq is only true when all the energy is transported by riation. he grient in temperature is the grient needed for the photons to be able to move all the energy out through the sun, so if some other mechanism would set in that was more e cient than energy transport by riation, the temperature grient from Eq would not be correct. 5.3 Convective transport Convective transport is the transport of energy by motions. It depends on the stability of the gas for small perturbations. Convection in stars is similar to what happens when you heat water in a pot on a stove. he stove heats the pot and the water at the bottom of the pot. Water is convectively unstable under normal circumstances, and the e ect on the water is most obvious just before it boils. Large cells of water well up from below where it has been heated, when it reaches the surface, it cools when coming in contact with the colder air, and then flows back down to the bottom of the 40

2 CHAER 5. ENERGY RANSOR pot, usually along the cool sides of the pot. In stars the same e ect happens, and in the sun, the e ect is almost identical. he gas is being heated from below, and the gas becomes convectively unstable, and the same large cells of upwelling gas can be seen on the solar surface, but since there are no sides on the sun, the cells are surrounded by cooler downflow lanes as can be seen in 5.1. o find out if the gas in a medium is convectively unstable, meaning that it will start to move even at the smallest perturbation, a number of thermodynamic principles must be understood. hermodynamics is a very large subject, and only the absolute minimum needed to understand convective transport will be included in these notes. Figure 5.1: he convective pattern on the solar surface, showing the bubbles of hot gas sticking up above the cool and dark downflow valleys. he observation is me with the Swedish 1-m Solar elescope on the Canary Islands. ick marks are arcseconds corresponding to roughly 725 km on the solar surface. he suns surface is seen under an angle of hermodynamics he behavior of a gas can be explained with the help of a number of constants and principle laws. he first relation is the equation for ded heat per unit mass: dq = du + d (5.15) 41

3 5.3. CONECIE RANSOR and the internal energy u and the specific volume =1/. It is important to notice that du is a total di erential form, so that for instance: du = d + d hese expressions are rather useless for our purposes so we need to reform them into something more useful. We want to get an expression for what happens to the gas if we d just a small amount of energy dq, and that involves both that we know something about the gas itself and how it changes its properties when it is being perturbed by outside disturbances. In order to understand these details we need to be as general as possible. o that end we assume a very general equation of state (, ) and u(, ), which usually is also dependent on the chemical composition, but we assume here the chemical composition is constant or at least that the evolution of the chemical composition is so slow that it has no consequence. Be aware that the independent variables ( and for (, ) and and for u(, ) is chosen specifically but could have been chosen di erently as long as two of the three, and are kept as primary variables and the last as a derived quantity. In the following we will use the independent variables mentioned above. A number of derivatives will be needed later on and to simplify things they will be renamed by the following ln = = ln ln = = he subscripts, and later means that the derivatives are to be taken at constant temperature, pressure and specific volume respectively. We can now write the equation of state as d = d d (5.19) Exercise 5.1: Show Eq using Equations , alternatively show Eq using Eq and Eq repeating what is done in Eq We also need the specific heats of which there are two. Specific heats are the amount of energy needed to raise the temperature of a unit mass of the 42

4 CHAER 5. ENERGY RANSOR gas by one Kelvin, at constant pressure(eq. 5.20) and at constant specific volume(eq. 5.21): dq c = = + and c = dq = (5.21) Eq can now be written dq = d + d + apple = c d + + d (5.22) If we choose a case where the pressure is kept constant, then we can find a change in temperature that only changes the volume and we can then use the derivative of the volume with temperature to get dq = c d + apple d (5.23) At the same time a change in energy dq at constant pressure due to a change in temperature is the definition of heat capacity, so in this case so we finally get apple c d = c d + + apple (c c ) d = apple c c = dq = c d (5.24) d (5.25) Even if we have an equation of state, u is still not a very useful variable to use, so in order to get rid of the derivative of u we can use the equation for the entropy ds = dq = 1 (du + d) (5.26) 43

5 5.3. CONECIE RANSOR and if we insert the full derivative for u in Eq. 5.26, we can find ds = 1 du + 1 d = 1 apple d + d + d = 1 d + 1 apple + d We compare that with the expression for the full derivative ds = d + d and then we can see = 1 apple + (5.29) (5.30) we can take the derivative with respect to of the left expression in Eq then that should be the same as the derivative with respect to of the right expression, so we apple 2 apple 1 1 = u 2 @ 0 = We now have an expression for the derivative of u we needed in Eq. 5.25, and by inserting we c c and since we can transform the @ = (5.33) 44

6 CHAER 5. ENERGY RANSOR we finally arrive at c c = 2 (5.34) which for an ideal gas transforms into c c = nk B (5.35) where k B is Boltzmanns constant and n is the number density of atoms or molecules. Exercise 5.2: Using the first law of thermodynamics (Eq. 5.15) and assuming a process where the volume does not change, so that dq = c d = du find an expression for the c for an ideal gas. Exercise 5.3: Deduce the expression for c of state and Eq c (Eq. 5.35) by using the ideal equation Exercise 5.4: Using the answers to Exerc. 5.2 and Exerc. 5.3, deduce an expression for c Eq can now be transformed dq = du + d apple = d + d + = d + c d + d apple = c d + + d (5.36) 45

7 5.3. CONECIE RANSOR using Eq and Eq we get apple dq = c d + = c d apple d d = c d + d (5.37) we know need to convert d and that can be done using Eq for the general equation of state: d = d 1 2 d = d d d d = d + d (5.38) inserting that expression back into Eq gives dq = c d + d + d = c d d + 2 d = c + 2 d d (5.39) and finally using Eq dq = c + 2 d d = c d d (5.40) We are now almost done. In our case we are interested in how the temperature, density and pressure would have to change inside the star in order for it to be stable to small perturbations. If a small perturbation moves a parcel of gas up or down, we need to know how its pressure, temperature and density changes. If the changes are such that it now gets the same temperature, density and pressure as its new surroundings, the star would be very stable, and nothing much would move. hat same would be true if 46

8 CHAER 5. ENERGY RANSOR the parcel would start to move back towards its initial position after being moved up or down. We will look at how the temperature structure looks for a star where the gas can be perturbed up or down without becoming unstable. In such a case the temperature, density and pressure is exactly the same as the surroundings after it has moved, so there is no exchange of energy from the parcel to the surroundings. Such a process is iabatic, and so we can use the definition of the iabatic temperature grient to ln r = ln and using Eq and the assumption that there is no exchange of energy between the package of gas we are looking at and its surroundings we now have ds =0=dq = c d d (5.42) or d = d s c (5.43) so that the iabatic temperature grient can now be rewritten as d r = = d c (5.44) s s Exercise 5.5: Show that using the equation of state for an ideal gas (Eq. 4.3), Eq can be simplified to Eq Exercise 5.6: By inserting the expression for the pressure of an ideal gas into Eq and find a value for r for an ideal gas Convective instability Convection can easily be explained as a stratified gas or a fluid that unstable to small changes in position. Imagine a small parcel or volume of gas is displaced slightly up towards the stellar surface. Assume that the movement is fast enough for the parcel not to exchange heat with its surroundings, while at the same time having time to get into pressure equilibrium. hese assumptions might not seem very realistic, so we will see if they are realistic 47

9 5.3. CONECIE RANSOR or not later in this chapter. In certain circumstances the small parcel of gas becomes buoyant (the density is lower than its surroundings) as an e ect of that movement. If those circumstances are present, then the gas is convectively unstable, as the gas will continue to rise through the star until the circumstances change. he assumptions we have me in this example are that the timescale for a pressure perturbation to travel across the volume of the gas parcel, is smaller than the timescale for a photon to cross the parcel taking into account the times it is absorbed and reemitted. hat makes it obvious that we are talking about a parcel of gas that is many times larger than the mean free path of a photon, otherwise the photons would make the parcel the same temperature as the gas surrounding it in a time that is on the order of the photon flight time from one side of the parcel to the other. A pressure perturbation travels with sound speed. he sound speed for an ideal gas is c 2 s = (5.45) where is the ratio between the specific heats and can also be written as = c c (5.46) =1+ 2 f (5.47) where f is the number of degrees of freedom. A pressure perturbation will travel across its rius r in a time t r t s = r (5.48) A photon will have a mean free path through the volume of = 1 apple and the number of steps it will have to take is (see Sec. 5.2) (5.49) and finally the time it would take would be N = r2 2 (5.50) t r = N c + N t em (5.51) 48

10 CHAER 5. ENERGY RANSOR where c is the speed of light, and t em is the time it on average takes between an absorption and before the photon is reemitted. In order for our assumption to be correct t s < t r r < N + N t em c s c r < N c s c + t em r < r2 2 c s c + t em 1 < r 2 c s c + t em 2 c r > c s + c t em r > c c s + c t em (5.52) Here we can see that the r in units of the mean free path of the photons needs to be roughly c/c s which is clearly large. he expression in the parantheses on the right hand side can then decrease the requirement for r if the distance a photon can travel during the time it is absorbed by an atom is on the order of or larger than the mean free path. It is hard to estimate if this is realistic. It turns out that inside stars, the sound speed is very high so it makes this requirement less stringent. Exercise 5.7: Calculate the requirement for r at the solar core, the bottom of the convection zone and in the photosphere, using an ideal equation of state, so that the sound speed can be expressed as c 2 s = / where =5/3. Use the values found in the appendices and assume that the opacity apple =1m 2 kg Onset of convection Imagine now a parcel of gas that starts out at a certain height r, where it has a density 0,pressure 0 and temperature 0. he gas of the star around the parcel has a density 0,pressure 0 and temperature 0 which are equal to inside the parcel, i.e. 0 = 0, 0 = 0 and 0 = 0 (5.53) 49

11 5.3. CONECIE RANSOR he parcel is now displaced upward in the stably stratified stellar interior by Figure 5.2: A convectively unstable parcel of gas inside a star a distance r from a position r to r + r. During the iabatic displacement, the density 0 of the parcel changes to while the density outside changes to. he pressure inside the parcel justs to the same level as the pressure outside the parcel. If the density of parcel after the displacement is lower than the surroundings, the parcel is buoyant and will move. So in order for the stratification to be convectively unstable, the density contrast can be written as apple 0 > = r apple = r = r 0 r 0 r 0 r 0 (5.54) where the iabatic grient is part of the assumption of the parcel displacement being iabatic. If we assume an ideal gas where the mean molecular weight µ can vary then = m u k B " µ = m apple u k B 1 + µ@ (5.55) 50

12 CHAER 5. ENERGY RANSOR and then inserting into Eq gives apple apple mu 0 > r k B 1 + µ@ apple 1 + µ@ apple 0 > 1 + µ@ 0 > apple @µ + (5.56) Using that the pressure at the new height r+dr inside and outside the parcel is the same so that the grient in pressure inside and outside the parcel is also the same, so finally we get 0 @ > + µ @µ (5.57) (5.58) (5.59) he instability criterion says that if the temperature grient is lower than the iabatic temperature grient, then we have a convective instability, and the gas will begin to move. here are two terms on the right, that can influence this criterion. If we assume that the mean molecular weight does not change under iabatic expansion or contraction (@µ/) = 0 and we are dealing with the sun, then the mean molecular weight decreases with rius 1, so that the term (@µ/) is stabilizing the temperature grient 1 For a gas consisting of only hydrogen and helium, both fully ionized, the mean molecular weight µ =1/(2X +3/4 Y ) which for the sun decreases with rius because at the center of the sun, the mass fraction of Helium is much larger than at the surface. here are though regions in the outer parts of the sun where He goes from being twice ionized to only ionized once, which means that at that rius the mean molecular weight actually increases. 51

13 5.3. CONECIE RANSOR can now be larger than the iabatic temperature grient. he instability criterion is often recast into r > r (5.60) where r is defined by Eq and r is then the same double logarithmic grient, but for the stellar stratification being investigated. It is equivalent to Eq except for the terms depending on the mean molecular weight, @ ln (5.61) (5.62) because the parcel is supposed to be rising iabatically and have the same pressure as outside, so the two grients are identical. Exercise 5.8: Why does the inequality sign change from Eq to Eq. 5.62? Convective energy flux In general the energy flux at any point in a star, outside its energy producing core, has to obey F R + F C = L 4 r 2 (5.63) where the F R is the riative flux found in Eq. 5.12, F C is the convective flux and L is the luminosity of the star. We have here implicitly assumed that heat conduction is not important. he convective flux can from first principles be written as F C = c v (5.64) which just states that the convective flux, is the excess energy c a parcel of density can move, with the velocity v. he trouble is of course to estimate what the velocity and temperature di erence is. In the following we will try to estimate what the temperature di erence between the parcel and the surroundings ( ) is and what the convective velocity (v) is. 52

14 CHAER 5. ENERGY RANSOR he temperature di erence We imagine as in Sec that we have a rising parcel of gas, which is continuously in pressure equilibrium with its surroundings, but here we cannot assume that it rises iabatically, because the time it takes for the parcel to rise might be long, and so it would be able to exchange energy with its surroundings. o estimate the temperature di erence, we write = p r =(r r p ) r H (5.65) where a subscript p is for the parcel, while a superscript is for the star. We have used the definition of the pressure scale height. H to change the grient of r to a grient in pressure: r ln ln ln = H = (5.66) (5.67) Exercise 5.9: Assume that we are looking at a static atmosphere where gravity, and pressure are the only two forces present. Assume also that gravity can be represented by the gravitational acceleration and is constant and has the value g. Assume furthermore that gravity is aligned along r. Write down the force balance, and use it to find an equation for as a function of r. What is the pressure scale height, for an ideal gas, when the pressure scale height is defined as the height over which the pressure drops by a factor 1/e? We need to assume a value for the distance travelled r by a parcel of gas in order to get further with Eq For historical reasons the distance usually assumed is half a mixing length r = 1 2 l m (5.68) 53

15 5.3. CONECIE RANSOR Inserting this directly into Eq we get: F C = c v l m 2H (r r p ) (5.69) If we split this into two terms by introducing r which is the temperature grient for an iabatically rising parcel we get: F C = c v l m 2H (r r )+ c v l m 2H (r r p ) (5.70) hese two terms can be understood quite easily. he first term is just the convective flux we would have if the parcels were rising iabatically. he second term is the modification needed when taking into account the riative losses of the rising parcel. he second term is just the riative loss of the parcel so by using the riative flux(eq. 5.12) and assuming that the grient in temperature from the outside to inside of the parcel is just twice the temperature di erence divided by the diameter d of the parcel: f R = apple 2 d = apple H l m d (r r p ) (5.71) he flux is the riative losses per unit area so we need to multiply it with the surface of the parcel S in order to get the total energy loss by the parcel. Since the second term of Eq is the di erence in flux of the gas and the gas only can move up, then that term needs to be multiplied with the surface Q normal to the velocity v. By multiplying with those geometric factors we get: Sl m apple H d (r r p ) = Q c v l m (r r p ) (5.72) 2H (r p r ) = 32 3 S 3apple 2 c v dq (r r p ) (5.73) We now finally just need to figure out how fast the parcel rises so that we can find the convective flux. Exercise 5.10: Calculate the geometric factor S/(d Q) for a spherical parcel with rius r p he convective velocity After having found the temperature di erence between the convective parcel we need to find its rise-velocity in order to be able to use Eq

16 CHAER 5. ENERGY RANSOR If we look at a buoyant parcel, then the buoyancy force working on it is just the gravitational force working on the di erence in density: f B = g (5.74) and taking into account that the expression is per unit mass, so we need to divede by the density. he work the buoyancy force produces is just the buoyancy force times the distance over which it works. he parcel we are looking at cannot turn all the work it is subjected to into acceleration, since it also needs to push gas in front of it away while it rises. If we assume that half the work produced by the buoyancy force is used to accelerate the parcel and since we alrey have assumed that it moves a distance r, then 1 / 2 f B r = g 2 r = g 2 r = g 2 ( r) H (r r p )( r)(5.75) = g 2H (r r p )( r) 2 (5.76) where Eq and the fact that there is no change in pressure compared to the outside (d = 0) has been used. hat amount of work is then equal to the kinetic energy of the parcel (remember again that all of these expressions are per unit mass): 1 / 2 v 2 = g (r r p )( r) 2 (5.77) 2H r g v = (r r p ) 1 / 2 ( r) H s g lm = 2 (r r p ) 1 / 2 (5.78) 4H We can insert that into Eq to get s (r p r ) = H S 3apple 2 c g dql m s = 64 3 H S 3apple 2 c g dql m so finally inserting into Eq. 5.69, we get (r r p ) 1 / 2 (r r p ) 1 / 2 (5.79) r g l m F C = c H 2 (r r p ) l m (r r p ) 1 / 2 2H = c p 2 g H 3 2 lm (r r p ) 3 2 (5.80) 2 55

17 5.3. CONECIE RANSOR So now we can find the energy flux and through that the temperature grient our star has in the convection zone by using that the total flux is F R + F C = apple H r stable (5.81) where r stable is the temperature grient needed for all the energy to be carried by riation. In reality the riative flux is F R = apple H r (5.82) where the r now is the actual temperature grient. hese two equations combined with Eq and Eq the temperature grient of the star can be found. So far we have encountered four di erent temperature grients: r p, r stable, r and r. he relations between these grients have so far been quite involved, but we can say something about their relative sizes. ressure and temperature both decreases with rius so the grients are all positive. If we furthermore look at a convectively unstable layer, then we alrey know from Eq that r p > r. Since for a convective gas, some of the energy is now moved through convection, the temperature grient does not need to be as steep as in the case where all the energy has to moved by riation, so r < r stable. For a rising parcel of gas we know that it cools both by iabatic expansion and by riative losses, so the temperature must decrease faster than for a parcel which only cools by iabatic expansion alone, i.e. r p > r. Finally we know that for a rising parcel, the temperature has to decrease slower than the surroundings, otherwise it would become unable to rise very quickly, so r p < r. Combining these we get r stable > r > r p > r (5.83) Exercise 5.11: Insert Eq and Eq into Eq to get an expression that contains an expression for (r r p ) as a function of r and r stable. Hint: If you can insert the expression for the pressure scale height(you found it in Exerc. 5.9) and end up with only known quantities for the gas, i.e.,, apple,, c,g and in the expression, as well as the r s. 56

18 CHAER 5. ENERGY RANSOR Exercise 5.12: Insert Eq into (r p r )=(r r ) (r r p ) and use Eq. 5.78, to get a second order equation for (r r p ) 1 / 2.Express the solution to the resulting equation as a function of r, r, U and the geometrical constant in front of U in the equation above, where s U = 64 3 H 3apple 2 c g Argue why this equation has only one viable solution. Exercise 5.13: Use your solution to Exerc into your solution of Exerc to eliminate r except in the form of used in Exerc here is still one free parameter left, which we have not determined: l m. One can give physical arguments for one value or another, but all in all, it is a free parameter that can be tuned to any value that suits the application. Generally it is assumed that l m = H (5.84) where is a parameter between 1 / 2 and 2, again this is a free parameter, but at least slightly better constrained than l m. hat is not satisfying in the strictest sense, but a complete treatment of convection takes full compressible hydrodynamics simulations for some cases even magneto hydrodynamic simulations, which are computationally very expensive to complete. he theory behind hydrodynamics and magneto hydrodynamics will be treated later in these notes. Example 5.1: Inserting values for the middle of the convection zone, assuming that 57

19 5.3. CONECIE RANSOR the gas is ideal and using the following values: µ = 0.6 = K = 55.9kgm 3 R = 0.84 R M(R) = 0.99M apple = 3.98 m 2 kg 1 = 1 then for purely riative transport the following values can be found r stable = 3.26 r = 2 / 5 so the gas is unstable, so the value for r stable (the temperature grient which would be necessary for all the energy to be carried by riation) cannot be used. Calculating the convective and riative energy flux ratios then gives: H = 32.5Mm U = = r = v = m s 1 F C F C + F R = 0.88 F R F C + F R = 0.12 A further check you should make is that Eq hold. In this case I have: 2 5 r < r p < r < r stable < < <

20 CHAER 5. ENERGY RANSOR In a physical sense we cannot write up this sort of expression, with such a precision, since most of the constants we use are not this precise, but in this case we use it only for internal consistency and as a mathematical check. 59