Today in Astronomy 142: stellar structure
Observations of stars which we seek to explain theoretically. Principles of stellar structure Hydrostatic equilibrium • Crude stellar interior models • Pressure, density and temperature in the center (where the luminosity is produced) Opacity of the Sun: diffusion of light from center to surface.
Rotational structure of the Sun: the different colors indicate flows slower (blue) and faster (red) than average. (SOHO/NASA).
29 January 2013 Astronomy 142, Spring 2013 1 Data on eclipsing binary stars
The most useful, ongoing, big compendium of detached, main-sequence, double-line eclipsing binary data is by Oleg Malkov and collaborators (1993, 1997, 2006, 2007). Much of the data come from many decades of work by Dan Popper. Those objects for which orbital velocities have been measured comprise the fundamental reference data on the dependences of stellar parameters – temperature, radius, luminosity on their masses. A strong trend emerges in plots of mass vs. any other quantity, involving all stars which are not outliers in the mass vs. radius plot. This trend, in whatever graph it appears, is called the main sequence, and its members are called the dwarf stars.
24 January 2013 Leftovers Astronomy 142, Spring 2013 2 Radii of eclipsing binary stars
100
Giants
10 ) R Radius ( Radius 1
Detached binaries
Semidetached/contact binaries 0.1 0.1 1 10 100
Mass (M)
24 January 2013 Leftovers Astronomy 142, Spring 2013 3 Luminosities of eclipsing binary stars
1000000
10000 ) L 100
1 Luminosity ( Luminosity
0.01 Detached binaries Semidetached/contact binaries
0.0001 0.1 1 10 100
Mass (M)
24 January 2013 Leftovers Astronomy 142, Spring 2013 4 Effective temperatures of eclipsing binary stars
50000 Detached binaries
40000 Semidetached/contact binaries
30000
20000 Temperature (K) Temperature
10000
0 0.1 1 10 100
Mass (M)
24 January 2013 Leftovers Astronomy 142, Spring 2013 5 The H-R diagram
We can’t, of course, measure the masses of single stars, or binary stars in orbits of unknown orientation. Unfortunately this includes more than 99% of the stars in the sky. We can usually measure luminosity (from bolometric magnitude) and temperature (from colors) though. The plot of this relation is called the Hertzsprung-Russell diagram. The H-R diagram for detached eclipsing binaries is very similar to that of stars in general: it corresponds to what is obviously the main sequence in other samples of stars. This correspondence allows us to infer the masses of single or non-eclipsing multiple stars.
24 January 2013 Leftovers Astronomy 142, Spring 2013 6 H-R diagram for eclipsing binary stars
Usually the T axis is plotted backwards in H-R diagrams.
1000000 Detached binaries
Semidetached/contact binaries 10000 ) L 100 Giants
1 Luminosity ( Luminosity
0.01
0.0001 10000 1000 Effective temperature (K)
24 January 2013 Leftovers Astronomy 142, Spring 2013 7 Eclipsing binaries vs. nearby stars
Stars within 25 pc: Gliese & Jahreiss 1991.
1000000 Stars within 25 pc of the Sun
10000 Detached binaries ) L 100
1 Luminosity ( Luminosity
0.01
0.0001 10000 1000 Effective temperature (K)
24 January 2013 Leftovers Astronomy 142, Spring 2013 8 Eclipsing binaries vs. young stars in clusters
X-ray selected Pleiades: Stauffer et al. 1994.
1000000 The Pleiades young open cluster
10000 Detached binaries ) L 100
1 Luminosity ( Luminosity
0.01
0.0001 10000 1000 Effective temperature (K)
24 January 2013 Leftovers Astronomy 142, Spring 2013 9 Eclipsing binaries vs. bright or very nearby stars
Bright/nearby: Allen’s Astrophysical Quantities (2000).
1000000
10000 ) L 100
1 Nearest and Luminosity ( Luminosity brightest stars 0.01 Detached binaries 0.0001 10000 1000 Effective temperature (K)
24 January 2013 Leftovers Astronomy 142, Spring 2013 10 Theoretical principles of stellar structure
Vogt-Russell “theorem:” the mass and composition of a star uniquely determine its radius, luminosity, internal structure, and subsequent evolution. Stars are spherical, to good approximation. Stable stars are in hydrostatic equilibrium: the weight of each infinitesimal piece of the star’s interior is balanced by the force from the pressure difference across the piece. • Most of the time the pressure is gas pressure, and is described well in terms of density and temperature by the ideal gas law, PV = NkT. • However, in very hot stars or giant stars, the pressure exerted by light – radiation pressure – can dominate.
29 January 2013 Astronomy 142, Spring 2013 11 Principles of stellar structure (continued)
Energy is transported from the inside to the outside, most of the time in the form of light. • The interiors of stars are opaque. Photons are absorbed and reemitted many many times on their way from the center to the surface: a random-walk process called diffusion. • The opacity depends upon the density, temperature and chemical composition. • Most stars have regions in their interiors in which the radial variations of temperature and pressure are such that hot bubbles of gas can “boil” up toward the surface. This process, called convection, is a very efficient energy-transport mechanism and can be more important than light diffusion in many cases.
29 January 2013 Astronomy 142, Spring 2013 12 The equations of stellar structure
Hydrostatic equilibrium: Equations of state dP GM ρ = − r Pressure: 2 dr r PP= (ρ , T ,composition) in general Mass conservation: ρσkT4 T 4 dM 2 = + throughout most normal stars r = 4πρr µ 3c dr Energy generation: Opacity: κ κρ dL = ( ,T ,composition) in general r = 4πr2 ρε dr Energy generation: ε= ερ ( ,T , composition) in general Energy transport: dT 3 κρ L = − r Boundary conditions: dr 16σ 22π rr4 Mr → 0 Adiabatic temperature gradient: as r → 0 Lr → 0 dT 1 µ GM =−−1 r T → 0 2 dr rad γ k r P → 0 as r → Rstar Convection occurs if: ρ → 0 T dP γ < Mr , Lr : mass or luminosity contained P dT γ − 1 within radius r.
29 January 2013 Astronomy 142, Spring 2013 13 The equations of stellar structure (continued)
There are no analytical solutions to the equations; usually stellar-interior models are generated by computer solution. Does that mean we get to go home now? No. Progress can be made by assuming a formula for one of the parameters, and solving for the rest. In this manner we can learn the workings of some of these differential equations, and establish some of the scaling relations useful in understanding the shapes of the empirical RM ( ) , T ee ( M ) , LM ( ) , and LT ( ) results. First it will be useful to derive the stellar-structure equation we’ll use most, the equation of hydrostatic equilibrium.
29 January 2013 Astronomy 142, Spring 2013 14 Hydrostatic equilibrium
In AST 111 we applied hydrostatic equilibrium to the structure of planetary atmospheres. Because atmospheres are thin compared to the radii of planets, it was sufficient there to approximate atmo- spheres as plane parallel slabs with constant gravitational acceleration. Thus we got a Cartesian one-dimensional differential equation, which we solved in various cases: dP = −ρ g . dz In stellar interiors we still get a one-D differential equation (because stars are spherically symmetric), but we can’t ignore the spherical shape or the radial dependence of gravitational acceleration.
29 January 2013 Astronomy 142, Spring 2013 15 Hydrostatic equilibrium (continued)
Consider a spherical shell of radius r and thickness ∆r r within a star with mass density ∆m (mass per unit volume) ρ(r). Its M weight is r ∆r GM∆ m ∆=−F r r r2 GM =−∆r 4πρrrr2 ( ) 2 R r M
=−∆4.πρGMr ( r) r Mr : mass contained (minus sign: force points inward) within radius r
29 January 2013 Astronomy 142, Spring 2013 16 Hydrostatic equilibrium (continued)
If the star is in hydrostatic equilibrium the weight is balanced by the pressure difference across the shell: 2 −∆F = Pr( ) 44ππ r2 − Pr( +∆ r) ( r +∆ r) (∆rr )
22dP (to first order ≅Pr( ) 44ππ r − Pr( ) +∆ r r dr in pressure) dP =−∆4,π rr2 dr dP∆ F 4πρGM( r)∆ r GM ρ( r) so ==−=rr−. dr 44ππrr22∆∆ rr r 2
29 January 2013 Astronomy 142, Spring 2013 17 Crude stellar models
(Not even complex enough to be called “simple”…) One can learn a surprisingly large amount about stellar interiors by starting with a formula for the mass density, ρ(r), and solving the equations of hydrostatic equilibrium and the equation of state self consistently for pressure and temperature. One learns most, of course, if the formula for density resembles the real thing. In AST 142 we will consider problems in which density of polynomial or exponential form to be within bounds. That is, all forms in which the hydrostatic equilibrium equation can be integrated straightforwardly.
29 January 2013 Astronomy 142, Spring 2013 18 Crudest approximation: the “uniform Sun”
Since we know its distance (from radar), mass (from Earth’s orbital period) and radius (from angular size and known distance): r =1 AU = 1.4960 × 1013 cm
33 M =1.98843 × 10 gm
= × 10 R 6.9599 10 cm we know the average mass density (mass per unit volume): MM3 ρ = = = 1.41 g cm-3 3 V 4π R = only 26% of Earth's. What would the pressure and temperature be, if the Sun had uniform density? 29 January 2013 Astronomy 142, Spring 2013 19 The “uniform Sun” (continued)
3 Let us therefore assume that ρρ ( r ) = = 3 MR 4, π and integrate away: 2 dP GM ρ G r3 33M GM =−=−r Mr =− dr 22 33 6 r rRR 44ππ R R 0 3GM2 dP = − rdr Surface has P = 0, if it doesn't mov. e ∫∫π 6 P(00) 4 R 22 2 3GM R 3 GM −=−PP(0) ⇒( 0.) = 6428π 4π RR 15 -2 9 PC ==P(0) 1.34×= 10 dyne cm 1.3× 10 atmospheres.
29 January 2013 Astronomy 142, Spring 2013 20 The “uniform Sun” (continued)
Thus, for the pressure elsewhere within the “uniform-density Sun,” Pr( ) r 33GM22GM dP′ = −r′′ dr = P r −=− P r2 ∫∫66( ) C 48ππRR PC 0 22 2 33GM GM 3 GM r2 Pr( ) =−=−r2 1 88ππR4 R 6 8 π RR 42 Now, the Sun is an ideal gas, through and through: PV= NkT (N= number of gas particles, n= number ρkT density: number of particles per unit P= nkT = µ volume, µ = avg. particle mass)
29 January 2013 Astronomy 142, Spring 2013 21 The “uniform Sun” (continued) so 322 µPr( ) µ 4π R3 GM r Tr( ) = = 1 − kρπ kM38RR42 1 µGM r2 = 1.− 2 kR 2 R
Suppose the average mass of the particles in this ideal gas is the same as the mass of the proton: µ =1.67 × 10−24 gm. Then = = × 7 TTC (0) 1.2 10 K . Note that in our approximations we wind up with T = 0 on the surface. We know the temperature is really more like 6000 K there, but indeed that’s much less than 12 MK.
29 January 2013 Astronomy 142, Spring 2013 22
Central pressure in a star
Return for a moment to the central pressure in a uniform star: 3 GM2 PC = . 8π R4 The only part of the equation which depends upon the functional form of the density is the dimensionless coefficient, 38 π . In this week’s homework, for example, you will use different density functions: 2 rr ρρ=−=−ρ (r) CC1 , or 1 , RR and obtain 5 GM2215 GM PC = , or = . 4ππRR4416
29 January 2013 Astronomy 142, Spring 2013 23 Central pressure in a star (continued)
Evidently this is a scaling relation: GM2 PC ∝ , R4 and the proportionality constant gets larger, the larger the central density is. Lo and behold, a complete calculation for stars of moderate to low mass (Astronomy 453 style) yields GM2 PC = 19 . R4
29 January 2013 Astronomy 142, Spring 2013 24 Central pressure in a star (continued)
For the Sun:
MR=×=×1.99 1033 g 6.96 1010 cm GM2 P ≅×19 =2.1 1017 dyne cm-2 C 4 R
> 1011 atmospheres
So, for other main sequence stars, to adequate approximation,
24− GM2 MR P ≅×19 =2.1 1017 dyne cm-2 . C 4 R MR
29 January 2013 Astronomy 142, Spring 2013 25 Central pressure in a star (continued)
That is, 24 GM22 GM MR P ≅=19 19 C 44 RRMR
2 24− GM MR = 19 4 MR R 2 ××−8 −− 1 2 33 24− (6.67 10 cm gm sec)( 1.99 10 gm) MR = 19 4 MR (6.96× 1010 cm)
24− MR = × 17 -2 2.1 10 dyne cm . MR
(You’ll need to get used to doing this. ) 29 January 2013 Astronomy 142, Spring 2013 26 Central density and temperature of the Sun
Central pressure is a little more than a factor of 100 larger than that of the Uniform Sun. Guess: central density 100 times higher than average? (That’s equivalent to guessing that the internal temperature doesn’t vary much with radius.) For the Sun, that’s not bad; the central density turns out to -3 be 110 times the average density, ρC = 150 gm cm . Thus, since ρ ∝ MR3 , C 3 MMR ρ =25 = 150 gm cm-3 C 3 R MR As we will see in a couple of weeks, the average gas- particle mass in the center of the Sun, considering its composition and the fact that the center is completely −24 ionized, is µ C = 1.5 × 10 gm .
29 January 2013 Astronomy 142, Spring 2013 27 Central density and temperature of the Sun (continued) But the material is still an ideal gas, so PV P Pµ T =C = C = CC =15.7 × 106 K. C Nk nkρ k CC C And, indeed, T doesn’t vary very much with radius. We can make a scaling relation out of this as well, to use in extrapolating to stars similar to the Sun but having different masses, sizes and composition: P µ GM23 R T =∝=×CC µ 15.7 106 K for the Sun; CC4 ρC kMR M R µ = × 6 TC 15.7 10 K . MR µ
29 January 2013 Astronomy 142, Spring 2013 28 Opacity and luminosity in stars
At the high densities and temperatures found on average in stellar interiors, matter is opaque. The mean free path, or average distance a photon can travel before being absorbed, is about = 0.5 cm for the Sun’s average density and temperature (given above; see also Homework #2, prob. 1). Photons produced in the center have to random-walk their way out, a process called diffusion. How many steps (= mean free paths) does it take for a photon to random-walk from center to surface? Suppose photon starts off at the center of the star, and has an equal chance to go right or left after each absorption and re-emission. Average value of position after N steps is xN = (x1 + x2 ++xN )/ N = 0 29 January 2013 Astronomy 142, Spring 2013 29 Opacity and luminosity in stars (continued)
However, the average value of the square of the position is not zero. Consider step N+1, assuming the chances of going left or right are equal: 2 1 2 1 2 xN+1 = bxN − g + bxN + g 2 2 1 1 = x2 − 2x + 2 + x2 + 2x + 2 2 N N 2 N N
= x2 + 2 . N 2 But if this is true for all N, then we can find x N by starting at zero and adding 2 , N times (i.e. using induction): 2 2 xN = N .
29 January 2013 Astronomy 142, Spring 2013 30 Opacity and luminosity in stars (continued)
2 Thus to random-walk a distance x N = L , the photon needs to take on the average
L2 N = steps. 2
So far, we have discussed only one dimension of a three- dimensional random walk. Three times as many steps need to be taken in this case, so to travel a distance R, the photon on the average needs to take 3R2 N = steps. 2
29 January 2013 Astronomy 142, Spring 2013 31 Opacity and luminosity in stars (continued)
For the Sun, and for a constant mean free path of 0.5 cm, 10 2 3e6.96 × 10 j N = = 5.81× 1022 steps. 2 a0.5f (Very opaque indeed.) Each step takes a time ∆ t = c , so the average time it takes for a photon to diffuse from the center of the Sun to the surface is 3R2 t= Nt ∆= =9.7 × 1011 s = 3.1 × 104 years. c Note that the same trip only takes Rc / = 2.3 s for a photon travelling in a straight line.
29 January 2013 Astronomy 142, Spring 2013 32