Lecture Notes 5: the Sun As a Star Ii As Stated in the Last Lecture the Sun’S Structure Is a Result of Various Equilibria, Among Them Hydrostatic Equilibrium

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Lecture notes 5: The Sun as a Star ii As stated in the last lecture the Sun’s structure is a result of various equilibria, among them hydrostatic equilibrium. This, fact, along with estimates of the Sun’s composition and internal state allows us to calculate the pressure in the solar core.

Hydrostatic equilibrium In general we may consider the forces acting on cylindrical piece of the Sun with mass dm, volume Adr. Newton’s second law may be written

d2r dm = F + F + F dt2 g P,t P,b where Fg is the gravitational force acting on dm and FP are the pressure forces (gas pressure times area, P × A) acting on the top and bottom surface of dm (Why do the horizontal pressure forces cancel?). Writing FP,t = −(FP,b + dFP ) 2 and Fg = −GMrdm/r , we ﬁnd

d2r M dm dm = −G r − AdP, dt2 r2 where Mr is the mass of the Sun within r (remember Gauss’ formulation of Newton’s theory of gravity!) and we have used dFP = AdP . We may continue with dm = ρAdr where ρ is the density of dm. Inserting this and dividing by Adr we ﬁnd

d2r M ρ dP ρ = −G r − (1) dt2 r2 dr In hydrostatic equilibrium the acceleration d2r/dt2 = 0 and we see that the pressure gradient balances the gravitational forces.

The pressure at the core, Pc

A quick estimate of the core pressure Pc is found by noting that the change in pressure from the core to the surface is roughly Pc since the pressure at the surface is essentially zero. This change occurs over the solar radius R. The Sun’s density is roughly ρ = M/R3. Inserting these assumptions we see that equation 1 may be written as Pc = gµ, i.e. that the pressure in the core must be great enough to support the weight of a column of material with mass/area µ. The solar gravitational acceleration is g ≈ 2 2 2 12 2 GMS/RS = 272 m/s , the mass/area is µ ≈ MS/RS = 4 × 10 kg/m , which 15 2 together give a core pressure Pc = 10 N/m . A more accurate calculation gives a pressure 19 times greater.

1 Temperature and density in the core It is a good approximation to assume that the gas in the solar core is a perfect gas obeying the perfect gas law P = nkT . Proper models of the solar interior 5 3 give a density ρc = 1.5 × 10 kg/m . Assuming that the solar core is comprised of ionized hydrogen implies that the average mass per particle is roughly m = 1 −27 32 −3 2 mp ≈ 10 kg. (Why?) This gives nc = ρc/m ≈ 10 m and a core 6 temperature Tc ≈ 15 × 10 K. Armed with the pressure, temperature and density at the core, we are now capable of computing/deriving other aspects of solar core physics.

The opacity of the solar interior Radiation diﬀuses from the solar core on a time scale set by the mean free path of a photon l which in turn is determined by the opacity κ such that lκ = 1. The two most important sources of opacity in the solar interior are hydrogen free–free transitions and the scattering of photons on electrons. The former is due the fact that when a free electron passes a proton a transitory dipole moment arises allowing both absorption and emission of photons as long as the encounter lasts. Electron scattering is independent of photon wavelength λ with −25 2 a scattering cross section σe = 6.6 × 10 cm giving an opacity

κe = neσe.

The formula for free–free transitions is somewhat more complicated and depends on photon wavelength as well as the proton and electron densities and the plasma temperature T .

8 −3 −1/2 κν,ff = 3.7 × 10 ν T nenp [1 − exp(−hν/kT )] gIII(ν, T ) where gIII is a Gaunt factor (which we will assume is of order 1). 3 3 Let us use the average solar density ρav = 14 × 10 kg/m , an average solar temperature Tav = 4.5 MK and assume that the solar plasma is only comprised of protons and electrons such that ne = np = ρav/m (Why?) This gives number densities 1024 cm−3. We need an estimate of a typical photon frequency to proceed. A good guess should be νav = c/λav = cT/0.0029 mK. (Why?) The above gives κν,ff ≈ 5 /cm and κe ≈ 1 /cm, leading to l ≈ 0.5 cm or so. At the solar center a typical photon wavelength is 1.8 A,˚ at T = 4.5 MK it is 6.4 A,˚ i.e. photon radiation is in the form of x–rays and the material in the solar interior is quite opaque to x–rays.

Radiative transfer in the Sun As mentioned above radiative transfer in the solar interior proceeds by a random walk of photons with a step length l. Consider the expected average x-position

2 of a photon after N + 1 interactions with the plasma: the photon should have moved hxN+1i = 0. However, the expected square of the position is 1 1 hx2 i = h(x − l)2i + h(x + l)2i = hx2 i + l2 N+1 2 N 2 N N since it is equally likely that the photon go left or right and where we have used the fact that hxN i = 0. But this means that by induction we may prove

2 2 hxN i = Nl , √ and that the root mean squared distance covered is proportional to N. There- fore the number of steps required to cover the distance RS in steps of size 2 2 l in three dimensions is N = 3RS/l (Why?) and the total flight time is 2 11 t = Nl/c = 3RS/lc ≈ 9.8 × 10 s or 30 000 years. We are now in a position to set up an expression for the luminosity of the Sun which may be written as the product of the Sun’s volume and the average energy density contained in photons divided by the diffusion time calculated above (another way of seeing this is by computing the total radiative energy of the Sun as the product of the luminosity and the flight time, i.e. the energy loss rate times the time to lose energy).

4 3 4 3 πRS × aT 26 LS = 2 = 4.54 × 10 W 3RS/lc Which compares well with the observed luminosity 3.9 × 1026 W.

The source of energy of the Sun New photons must continuously be created in order to replace those that escape into the rest of the universe. A hot plasma such as the Sun will create photons 3 3 at the expense of the plasma’s energy content 2 nckT = 2 Pc. If we compare this energy content with that contained in the Sun’s photons we ﬁnd

3 2 Pc 4 ≈ 1000 aTc

16 2 6 with Pc = 2.1 × 10 N/m and Tc = 15 × 10 K. Thus, multiplying this factor by the photon leakage time, we ﬁnd an expected cooling time for the Solar plasma of 1000 × 30 000 yr= 3 × 107 yr. However, the age of the Earth is 4 × 109 yr and of life on Earth at least 3 × 109 yr indicating that the Sun has been stable for at least this period. We must therefore conclude that the Sun requires and energy source.

Gravitational potential energy of the Sun In discussions below we will need an expression for the potential energy of the Sun. This may be derived as followed: Consider the potential energy dUi of a

3 piece of solar material dmi at distance r from the Sun. M dm dU = −G r i i r

But the potential energy of any dmi at distance r from the Sun must be the same by symmetry so inserting dm = 4πr2ρdr we ﬁnd the potential energy for the spherical shell at distance r with thickness dr

M 4πr2ρ dU = −G r dr, r and integrating over the entire Sun we ﬁnd potential energy

Z RS U = −4πG Mrρrdr 0

4 3 4 3 Noting that Mr ≈ 3 πr ρav and ρ ≈ ρav = MS/( 3 πRS) we see that the poten- 2 tial energy is of order −GMS/RS. The exact value will depend on the mass distribution inside the Sun or star, for the Sun a good value is

M 2 U = −2G S RS

Gravitational contraction One early energy source propounded, among others, by Lord Kelvin in the late 19th century — before the geological and paleontological evidence was well un- derstood — was the energy given by gravitational contraction. As we show in the appendix to this lecture it is possible to prove the virial theorem that states that the total energy of the Sun — the sum of the thermal and gravitational potential energy — is, at any given time, equal to half of its gravitational potential energy. Thus 1 total energy of‘Sun = × potential energy 2 1 thermal + gravitational = × potential energy or 2 1 thermal = − × potential energy 2 As the Sun radiates it loses energy and the (negative) potential energy becomes greater in absolute value. The above equation shows that this is accompanied by an increase in thermal energy.

As the Sun contracts its core gets hotter!

If we assume that the Sun has been heated by gravitational contraction as 2 it has reached its present state and potential energy of WS = −2GMS/RS.

4 Assume in addition assume that the Sun’s luminosity has been more or less constant at its present value of LS. We the ﬁnd a time-scale 1 t = − W /L = 3 × 107 yr 2 S S

1 (the − 2 from the virial theorem) which is much too short to explain the presence of life on Earth.

The stability of the Sun In the last section we proved that the Suns energy source could not be gravitational contraction. We will later see that the source lies in the fusion of hydrogen to helium (no other elements will fuse at the temperatures and pressures we have derived for the Solar core). Let us for now simply note that the production of energy from fusion increases with increasing temperature and densities, since these imply increasing collision rates between protons. Consider the stability of the Sun in the face of perturbations such as those incurred through for example photon loss through the surface, random pressure changes, or whatever. A perturbation leading to increased energy production will cause the plasma to heat up. A hotter plasma will expand and cool and the energy production will decrease. On the other hand a decrease in the energy production will cause the plasma to cool. A cooler plasma will contract, which in turn will increase energy production towards its equilibrium value. Thus the solar radius RS is such that Tc and ρc are large enough to drive the solar luminosity LS by the leakage of photons. Any change in the temperature, density or radius of the Sun will be countered by either contraction or expansion so as to maintain this equilibrium between energy production and energy loss. This in turn implies that the radius and luminosity of a star such as the Sun (i.e. in quasi-equilibrium) can be derived once one is given the mass and the chemical composition.

Appendix: The virial theorem Let us start with a set of particles with momenta r and positions r. Now form P the quantity Q ≡ i pi · ri where the sum extends over all particles. We will consider time derivative dQ/dt:

dQ X dpi dri = ( · r + p · ) (2) dt dt i i dt i The left hand side of equation 2 may be written

dQ d X dri = m · r dt dt i dt i i

5 d X 1 d = (m r2) dt 2 dt i i i 1 d2I X = where I = m r2 2 dt2 i i i We recognize I as the moment of inertia. Split the right hand side of equation 2 into to two sums. The ﬁrst of these is then

X dri X X 1 p · = m v · v = 2 m v2 i dt i i i 2 i i i i i

This is obviously twice the total kinetic energy of the particles Ek. Rear- ranging and using Newtons 2. law F = dp/dt, we now have

1 d2I X − 2E = F · r 2 dt2 k i i i The term remaining on the right hand side has a name: Clausius virial. Noting that the total force on a particle is a result of the superposition principle, it i.e. the sum of the forces from all the other particles we write X X X Fi · ri = ( Fij) · ri. i i j

But according to Newtons 3. law Fij = −Fji, so continuing with Clausius virial

  1 X X = (F − F ) · r 2  ij ji  i i j 1 X X X X = ( F · r − F · r ) 2 ij i ji i i j i j 1 X X = F · (r − r ) 2 ij i j i j

The last transition is clear when one considers that both sums go over all P P P P particles so that i j Fji · ri may be rearranged into i j Fij · rj. Assume now that the interactions between the particles are gravitational:

mimj rj − ri Fij = G 2 er where er = rij rij

p 2 and rij is the distance (ri − rj) . We then have

6 X 1 X X mimj F · r = − G (r − r )2 i i 2 r3 i j i i j ij 1 X X mimj = − G 2 r i j ij

The term inside the double sum Gmimj/rij is the potential energy Uij between particle i and j. This term shows up twice for each combination i, j and we therefore ﬁnd Clausius virial X 1 X X F · r = U = U i i 2 ij i i j

We then ﬁnally have

1 d2I − 2E = U 2 dt2 k and noting that the time average

d2I 1 Z τ d2I 1 dI dI h 2 i ≡ 2 dt = ( |τ − |0) dt τ 0 dt τ dt dt is zero when τ → ∞ and dI/dt is bounded or if the system is periodic. Which proves the virial theorem in the form we need to use in this course: 1 hE i = − hUi. k 2 And if all this seems a bit abstract and hard to grasp!? Consider the much simpler system consisting of two particles: a satellite mass m in orbit around the much heavier (mass ME) and therefore stationary Earth, with a velocity v at a distance r from Earths center. Computing the kinetic and potential energies of the satellite gives the relation 1 1 mM mv2 = G E 2 2 r Recognize it? Use Newtons 2. law to check that this is right!

Exercises: The Sun as a Star

1. Assume that a 100 W bulb occupies 30 cm2 of area. Cover the Earth with such bulbs. What is the power requirement of these compared with the Sun?

7 2. Define the solar luminosity, the solar constant (at 1 AU) and the effective 26 11 temperature. Given that LS = 3.90×10 W, that r = 1.50×10 m, and that the angular diameter of the sun is θ = 32 arcmin, deduce the linear size of the Sun and its effective temperature. 3. Calculate the Sun’s mass by assuming a circular orbit and equating the Earth’s centripetal acceleration and the gravitational acceleration pro- duced by the Sun. Show at the same time that the total energy of the Earth’s orbit is one half the potential energy. 4. Use the equation of hydrostatic equilibrium M ρ dP ρG r = − r2 dr

to estimate the pressure and temperature in the solar core Pc. Clearly identify the assumptions that have been made in forming the estimate.

−25 −2 5. The scattering cross section for electrons is σe = 6.6 × 10 cm . The opacity is κ = neσe with ne the electron number density. The mean free path l for photons is given by 1 = lκ. (a) Use the estimates above to calculate the mean free path of photons in the solar core assuming that the main source of opacity is scattering oﬀ electrons. (b) What is a typical photon wavelength in the solar core? 6. Calculate the ratio of the plasma-pressure to the radiation-pressure at 16 2 7 Sun-center. Use Pc = 2.1 × 10 N/m and Tc = 1.5 × 10 K. 7. Calculate the ratio of the plasma-energy density to the radiation-energy 16 2 7 density at Sun-center. Use Pc = 2.1 × 10 N/m and Tc = 1.5 × 10 K. 8. The gravitational potential energy W of a self-gravitating sphere of mass M and radius R depends on the detailed distribution of mass within the 2 sphere, but is generally of order −GM /R. For the Sun WS = −2 · 2 GMS/RS. What is the time scale over which gravitational contraction could have supplied the power radiated by the Sun at its present rate?