More on stellar structure s equations

Transcription

1 University of Naples Federico II, Academic Year Istituzioni di Astrofisica, read by prof. Massimo Capaccioli Lecture 7 More on stellar structure s equations Two hours of Introduction to Astrophysics per day is too much!!! Arthur Stanley Eddington ( )

2 Learning outcomes The student will : consider the equation of energy production; examine which are the sources of energy production and when they are at work; conclude that only nuclear energy fits the requirements (but gravity may also help); discuss how energy is transported inside the star; derive the condition for convection to win on radiative transport.

3 Equation of energy production The third equation of stellar structure establishes the relation between energy release and the rate of energy transport. Consider a spherically symmetric star in which energy transport is radial and time variations are unimportant. r r +δ r The rate of energy flow is: L r L( r) = across sphere of radius r; L( r + δ r) = across sphere of radius r + δ r. Lr + δ r Because shell is thin: δv ( r) = 4 π r δ r and δ m( r) = 4π r ρ( r) δ r 3

4 Equation of energy production 1 We define ε ( r) = energy release per unit mass [ W kg ]. Hence energy release in shell is written : π r ρ r) δ r ε ( r). Conservation of energy implies: ( 4 ( ) L( r + δ r) L( r) δ ( π ρ δ ) ε π ρ ε δ r L( r + r) = L( r) + 4 r ( r) r ( r) = 4 r ( r) ( r dl( r) and for δ r 0 : = 4π r ρ( r) ε ( r) dr This is the equation of energy production. We now have three of the equations of stellar structure. However we have five unk nowns : P( r), M ( r), L( r), ρ( r), ε ( r). 3 equations 5 unknowns In order to progress further, the energy transport in stars has to be considered. 4 )

5 Energy generation in stars So far, we have only considered the dynamical properties of the star, and the (average) state of the stellar material. We need to consider the source of the stellar energy. Let us consider the origin of the energy, i.e. the conversion of energy from some form in which it is not immediately available into some other that the star can radiate. In order the establish the amount of energy we are speaking of, let s ask how much energy does the Sun need to generate in order to shine with its measured flux for its lifetime? Solar flux L = 4 10 W = 4 10 J s 9 Sun has not changed flux in 10 yr Sun has radiated J in yr. 7 [ 1 yr = 3 10 s] 17 Since E = mc [ m 10 kg/j] the equivalent mass lost by the Sun is: M lost 10 M [ ] kg M =.0 10 kg 5

6 Sources of energy generation Where does the Solar /stellar energy come from? Possibilities: 1. Gravitational potential energy (energy is released as star contracts/cools). Chemical energy (energy released when atoms combine; burning) 3. Nuclear energy (energy released when nuclei form/split)

7 1. Cooling and contraction Cooling and contraction are closely related, so we consider them together. Suppose that: 1a. the radiative energy of our star is due to the Sun being much hotter when it was formed, and since then it has been cooling down. We can test how plausible this is; or that 1b.the Sun is slowly contracting with consequent release of gravitational potential energy which is converted into radiation. 7

8 Gravitational potential The gravitational potential energy for a system of two particles of masses M and m: Mm U = G r As M and m are brought closer together, the potential energy becomes more negative. Assume the Sun was originally much larger than it is today, and contracted. How much gravitational potential energy has been 41 released? [ U 10 J] Since lim U r [ remember black holes] =, how close can two masses go?

9 Gravitational potential at spherical symmetry dr r R r Assume spherical symmetry: M ( r) = 4 π s ρ( s) ds. 0 The potential energy out to the sphere radius R is: ( 4 π ρ( ) ) R M ( r) r r dr U = G = G r ( 4π ) 0 0 R ( r s ρ( s) ds)( r ρ( r) dr) 0 r.

10 Gravitational potential at spherical symmetry Assume tha t ρ( r) = ρ r 0 α. ( 4π ) ( r ρ0 )( ρ0 ) ( r )( ) α α α α s s ds r r dr s ds r dr R 0 R 0 U = G ( 4π ) = G ( 4π ) ρ 0 0 r = 0 r r 3 α α ( r dr ) 5 α 3 α G ( 4π ) ρ r dr G ( 4π ) ρ R R R α ρ = G = = r dr r 3 α r 3 α U ( 4π ) ρ0 5 α 9 R ( 3 α )( 5 α ) ( 3 α )( 5 α ) G G M = = + R 1 α 16π 3 M For a homogeneous sphere, ρ( r) ρ, a α =0: U = G ρ R = G 15 5 R 5 0 nd 0.

11 Cooling and contraction for ideal gas kt In an ideal gas, the thermal energy of a particle, n f, since the number of 3kT degrees of freedom is n f = 3, amounts to:. If n is the number of particles per unit volum e, the total thermal energy per du 3knT unit volum is: =. dv Assuming the stellar material to be an ideal gas so that Pr is negligicle and the V s overall pressure reduces to P = nkt, the Virial Theorem: 3 PdV + Ω = 0, 0 writes: Vs 3 nktdv + Ω = 0 U + Ω = 0 0 where U = integral over volume of the thermal energy per unit vo lume. 11

12 Cooling and contraction U + Ω = 0 The negative gravitational energy of a star is equal to twice its th ermal energy. This means that the time for which the present thermal energy of the Sun can supply its radiation is one half of the time for which the past release of gravitational potential energy could have supplied its present rate of radiation. We can estimate the latter. Negative gravitational potential energy of a star is related by the inequality: GM s 3 GM s Ω > for a homogeneous sphere: Ω = rs 5 rs GM s As an approximation let us assume that: Ω ~. r s 1

13 The total release of gravitational potential energy would have been sufficient to provide radiant energy at a rate given by the luminosity of the star Ls, for a time t, known as thermal (or Kelvin-Helmholtz ) timescale: th Cooling and contraction t th GM L r s ~ s s. Hermann von Helmholtz ( ) William Thomson, Lord Kelvin ( ) 7 ( t ) = For the Sun: th 3 10 yr. Hence, either contraction or cooling would have been sufficient to leave the Sun unchanged for no more than 30 million years: a factor of ~10 too short to account for the constraints on age of the Sun imposed by fossil and geological records. Lergely insufficient. 13

14 Chemical energy Chemical reactions are based on the of orbital electrons in atoms. interactions Chemical reactions such as the combustion of fossil fue ls 10 release ~ 5 10 of the rest mass energy of the fuel. Class task: Show that the above rate is correct knowing that combustion of methane 7 releases 50.1 MJ kg ( 5 10 J/kg). Note that 1 BTU 1055 J. Order of magnited estimate: one atom of H can release a maximum of 13.6 ev (its ionization potential) Since 1 ev = J, EH =. 10 J mh = EH c = 10 kg. 7 8 Since the mass of the proton m = kg m m 10. p H p 14

15 Chemical energy in stars Chemical energy is quickly ruled out as a possible source of energy for the Sun. Typical energy differences between atomic are ~10 ev. orbitals For instance, assume the Sun is pure hydrogen. 30 M Sun The total number of atoms is therefore: n = = = m If each atom releases 10 ev of energy due to chemical reactions, the total amount of chemical energy available is: 10 ev = 10 J. This figure is ~10 less than the gravitational potential energy available, and 5 it would be radiated away in only 10 years at the present solar luminosity 6 ( W). H 1 electron volt = joules

16 Atomic nuclei Helium The mass of nuclei is usually expressed in atomic 1 mass units u, defined to be 1 1 the mass of C (the 1 is the mass number A; 1 C has 6 protons, 6 neutrons, and 6 electrons): 7 1 u = kg. m m m p n e = u = u = u The mass of 1 proton + 1 electron is u. Note: 6p + 6n + 6e = Does this make sense? It should be 1.0!

17 The equation Einstein showed that mass and energy are equivalent ( but not the same thing), and related by: E = mc. Thus atomic masses can be expressed as energies: 1 u = MeV c 1 GeV c When equating rest masses to energies, it is customary to omit the factor c and leave it implicitly assumed.

18 Binding energy We have seen that the mass of an atom (protons+neutrons+electrons) is not equal to the cumulative mass of its particles. 8 For instance, the hydrogen ato m has u less mass than the sum of proton's and electron's masses. This mass difference corresponds to an energy: E = 13.6 ev. This energy is the binding energy: it is the energy released when an electron and a proton are brought together ( also the ionization potential). Recall: Carbon-1: 6p + 6n + 6e = The mass difference is u, equivalent to 9. MeV! 1 This is the binding energy of the C atom.

19 Nuclear reactions A way of producing sufficiently large amounts of energy is through nuclear reactions. There are two types of nuclear reactions, fission and fusion. Fission reactions, such as those that occur in nuclear reactors or atomic weapons, can release ~ of rest mass energy by breaking heavy nuclei (uranium or plutonium). Class task: Show that the fusion reactions can release enough energy to feasibly power a star. [Assume atomic weight of H = and He 4 = ]. 19

20 Nuclear reactions: which is best? Simple calculations have shown that both fusion and fission could in principle power the Sun. Question # 1: which is the more likely? As in the Solar System light elements are much more abundant than heavy ones, we would expect nuclear fusion to be the dominant source. Question # : given the limits on P(r) and T(r) that we have just obtained, are the central conditions suitable for fusion? 0

21 Nuclear energy: fusion Making a larger nucleus out of smaller ones is a process known as fusion. For example: H + H + H + H He + low mass remnants. The mass of 4 H atoms is u. The mass of He is u. The mass difference is u, equivalent to 6.71 MeV. As we will see, the energy of the typical low-mass remnants amounts to only ~ 1 MeV. C onclusion: ~ 0.7% of the H mass is converted into energy. Show it! Thus, nuclear reactions (which change one type of nucleus into another) typically release energies in the MeV range, 6 10 times higher than chemical reacti ons.

22 Nuclear energy: doubts 1. Nuclear fusion is sufficient to sustain the luminosity of the Sun;. But, we shall see, fusion is difficult to achieve! 3. Can it actually occur naturally in the Sun? Sir Arthur Eddinghton

23 Tricky business: Coulomb repulsion The repulsive force between like-charged particles turns in a potential barrier that gets stronger as the particles get closer The strong nuclear force becomes dominant on very smal l scales 15, 10 m. 1 Z Z e 1 : UC =. 4πε 0 r What temperature is required to overcome the Coulomb barrier? Charles Augustin de Coulomb ( )

24 9 Since the Coulomb constant is = N m C, for: Tricky business : Coulomb repulsion 1 15 =, with r = 10 m, 0 1 4πε the Boltzman constant k = the electron charge e = C, 3 1 kt 4πε Z Z e r 10 the temper ature is T 10 Z Z K. 1 0 J K 3 1, big problem indeed Charles Augustin de Coulomb ( )

25 Recap of statistical mechanics If the gas is in thermal equilibrium with temperature T, the atoms have a range of velocities described by the Maxwell-Boltzmann distribution function: the number density of gas particles with speed between v and v + dv is: 3/ mv m kt v 4 v v nvd = n e π d π kt Most probable veloci Average k kt m 1 3 ty: v =. inetic energy: m v = kt.

26 Quantum tunneling The solution to the problem, arosen by Eddington in the 0's, lies in quantum physics. The uncertainty principle states that momentum and position are not precisely defined: x px ħ. The uncertainty in the position means that if two protons can get close enough to each other, there is some probability that they will be found within the Coulomb barrier. This effect is known a s tunneling. The effectiveness of this process depends on the momentum of the particle.

27 Quantum tunneling In 194 Louis de Bloglie, in his Ph.D. thesis, suggested that the wave-particle duality should apply to matter too and proposed to associate a wave of lenght: h h h h λdebroglie = = = = p µ v µ to a particle of momentum p. µ v 3kT µ. Louis-Victor Pierre de Broglie ( ) Approximately, tunneling is possible if the wavelength of each other. protons come within one debroglie

28 Quantum tunneling Which temperature is required for two protons to come within one de Broglie wavelength, λ debroglie h =, of each other? 3kT µ 3 1 Z Z e h λ = 4πε r 3kT µ 1 It must be kt, where the distance r =. 0 7 µ Fusion is thus possible at T Z1 Z K. mh 7 For two protons, this implies ~ 10 K. T 15 So the protons do not need to get anywhere near 10 m before they can begin to tunnel past the barrier. Without this quantum effect, fusion would not be possible in the Sun and such high luminosities could never be achieved.

29 Nuclear reactions: another problem So what are the specific reactions we are talking of? In order to form a He atom one needs four H atoms. But the probability that four H atoms will collide at once is exceedingly small. Even this simple fusion reaction must occur via a number of steps.

30 Nuclear energy: rate of collisions The energy production rate must be related to the rate of collisions. σ ( E) ds = v( E) dt Since the full rate equation can be very complex, it is common to approximate it as a power-law over a particular temperature range: 1 rix r0 X i X xρ +α T β. i X i and X are the mass fractions of the incident and target particles. x i For two-body interactions ( i.e. p + p collisions), α ~ 1. i The parameter β can have a wide range of values. 30

31 Nuclear energy production If each reaction releases an energy L, the mass ε ε ρ + 1 is just: ix = 0 X i X α x T β. amount of energy released per unit The sum over all reactions gives the nuclear reaction contribution to ε in our dl 5-th fundamental equation : r = 4π r ρε. dr 31

32 Methods of energy transport in stars There are three ways energy can be transported in stars: 1. Convection: energy transport by mass motions of the gas;. Conduction: energy transport by exchange of energy during collisions of gas particles (usually e ); 3. Radiation: energy transport by the emission and absorption of photons. Conduction and radiation are similar processes; they both involve transfer of energy by direct interaction, either among particles or between photons and particles. Which is the process dominating in stars? 3

33 Conduction vs. radiation The energy carried by a typical particle, ~ 3kT/, is comparable to the energy carried by a typical photon, ~ hc/λ. [to show this, use Wien s displacement law λt = b, k = , h = , and b = MKS]. But number density of particles is much greater than that of photons. Show! This would imply conduction is more important than radiation. But: mean free path of photon ~ 10 m; mean free path of particle ~ m. Photons can move across temperature gradients more easily, hence a larger transport of energy. Conclusion: conduction is negligible, radiation transport is dominant. 33

34 Energy transport: some numbers Let's look at the Solar light flux as a starting place: the Solar output is the rate of energy release by the Sun = erg/sec; from spherical geometry, the energy density at the Earth's orbit is the Solar Constant = erg/cm /sec; 40% of the Sun's energy is visible light = erg/cm /sec; the energy of a typical visible light photon is: E = hν = erg; so the visible solar flux is: photon/cm /sec; 70% of visible light gets through the atmosphere when viewed at the Zenith and is equal to photon/cm /sec; therefore 100 thousand million million photons blast away at every square centimeter of the surface of the Earth every second. This partially explains sunburn. 34

35 Solar surface from Swedish Solar Telescope Resolution ~ 100 km Granule size ~ 1000 km 35

36 Convection is the mass motion of gas elements; it occurs only when the temperature gradient exceeds some critical value. Let us derive an expression for this. Consider a convective element at a dista nce r from center of the star. The element is in equilibrium with the surroundings. Convection r +δ r Convective element of stellar material Element P δ P ρ δρ Surrounding P P ρ ρ g Now let's suppose it rises to r + δ r. It expands P( r) and δ ( r) are reduced to P δ P and ρ δρ. r P ρ P ρ But these may not be the same as the new surrounding gas conditions. Define those as P P and ρ ρ. If gas element is denser than surroundings at r + δ r then it will sink. If it is less dense, then it will keep on rising convectively unstable. 36

37 Convection The condition for instability is therefore: ρ δρ < ρ ρ Whether or not this condition is satisfied depends on two things: i. the rate at which the element expands due to decreasing pressure; ii. the rate at which the density of the surroundings decreases with height. Let's make two assumptions: 1. the element rises adiabatically;. the element rises at a speed much less than the sound speed. During motion, sound waves have time to smooth out the pressure differences between the element and the surroundings. Hence δ P = P at all times. The first assumption means that the element must obey the adiabatic relation γ between pressure and volume: PV = constant, where γ = cp cv is the ration of the specific heat ( i. e. the energy in J to raise temperature of 1 kg of material by 1 K) at constant pressure to that at constant volume. 37

38 Convection γ Given that V is inversely proportional to ρ, from PV = constant we can write: P = constant. γ ρ Hence eq uating the term at r and r + δ r: P δ P P =. γ γ ( ρ δρ) ρ γ If δρ is small we can expand ( ρ δρ) using the binomial theorem [review] as follows: γ γ γ 1 ( ρ- δρ) ~ ρ γρ δρ, and, combining the last two expressions: ρ δρ = δ P. γ P Now we need to evaluate the change in density of the surroundings, ρ. d ρ Let's consider an infinitesimal rise of δ r ρ = δ r. dr 38

39 Convection Let's substitute the condition for convective instability derived above [ δρ < ρ] : ρ d ρ δ P < δ r γ P dr nd This can be rewritten by recalling our assumption that the element will remain at the same pressure as it surroundings, so that in the limit δ r 0, δ P dp =,it is : δ r dr ρ dp d ρ < γ P dr dr ρ d ρ the expressions for δρ and ρ δρ = δ P; ρ = δ r into γ P dr Density gradient in the surroundings if there an adiabatic relation between ρ and P holds. Actual density in the surroundings. We can convert this to a more useful expression by first dividing both sides by dp dr. Note that dp dr is negative, hence the inequality sign must change. 39

40 Convection ρ dρ dp ρ d ρ P dρ 1 < > dr dr < γ P γ P dp ρ dp γ ρkt For an ideal gas in which radiation pressure is negligible: P =, that is: m ln P = ln ρ + lnt + constan t, (where m is the mean mass of particles in the stellar material). Once differentiated, the equation gives: dp d ρ dt = +. P ρ T Combining this with the equation above gives the condition P dt γ 1 dt γ 1 T > or >. T dp γ dp γ P for convection to occur: 40

41 The adiabatic gradient The adiabatic gradient is: dt dp ad = γ 1 T. γ P γ dp P dv From PV = constant = γ. dt ad V dt ad By deriving the equation of state: PV T, we obtain: dv 1 dp = 1 V dt P dt By combining this with the other above: dp P 1 dp 1 dp = γ 1 V = γ dt ad V P dt ad V dt ad dp P dp = γ dt ad T dt ad dp P dt γ 1 T ( γ 1 ) = γ =. dt T dp γ P ad ad 41

42 Condition for occurrence of convection Remembering our result: P dt T dp γ 1 > γ which is then the condition for the occu rrence of convection temperature gradient)? (in terms of the The answer is: a gas is convectively unstable if the actual temperature gradient is dt γ 1 T steeper than the adiabatic gradient =. dp γ P dt dt If the condition > is satisfied, then large scale rising and falling dp dp ad motions transport energy upwards. The criterion can be satisfied in two ways: i the ratio of specific heats γ is close to unity i the temperature gradient is very steep. ad γ 1 0, or γ For example, if a large amount of energy is released at the centre of a star, it may require a large temperature gradient to carry the energy away. Hence, where nuclear energy is being released, convection may occur. 4

43 Condition for occurrence of convection Alternatively in the cool outer layers of a star, gas may only be partially ionised, hence much of the heat used to raise the temperature of the gas goes into ionisation and hence the specific heat of the gas at constant V is nearly the same as the specific heat at constant P, and γ ~ 1. In such a case, a star can have a cool outer convective layer. Convection is an extremely complicated subject and it is true to say that the lack of a good theory of convection is one of the worst defects in our present studies of stellar structure and evolution. We know the conditions under which convection is likely to occur but don t know how much energy is carried by convection. Fortunately we will see that we can often find occasions where we can manage without this knowledge. 43

44 Convection movie: A-type star 44

45 Convection movie: F-type star 45

46 Convection movie: K-type star 46

47 The real Sun 47