Basic concepts in stellar physics Anthony Brown brown@strw.leidenuniv.nl 1.2.21 Abstract. These notes provide extra material on the basics of stellar physics. The aim is to provide an overview of the nature of stars using only elementary physics. Some concepts that are fundamental to stellar physics are introduced and the order of magnitude of astrophysical quantities is fixed through simple estimates. Most subjects treated in this note will be discussed in more detail using the chapters on stellar structure and evolution from Ostlie & Carroll (27). The material below is extracted from the first chapter in Phillips (1998). Revision History Issue Rev. No. Date Author Comments 3 9.2.29 AB Typos and minor errors fixed. 2 9.2.29 AB Typos and minor errors fixed. 1 26.2.28 AB Document put on blackboard site. 24.2.28 AB Document created. Contents 1 The composition of stars 3 2 Gravity and hydrostatic equilibrium 3 2.1 Gravitational contraction..................................... 3 2.1.1 Free fall.......................................... 4 2.2 Hydrostatic equilibrium...................................... 5 2.2.1 Equilibrium for a non-relativistic gas.......................... 6 2.2.2 Equilibrium for an ultra-relativistic gas......................... 7 3 Star formation 7 3.1 Conditions for gravitation collapse................................ 7 3.2 Contraction of a protostar..................................... 8 3.3 Conditions for being a star.................................... 9 4 The sun 1 4.1 Solar radiation........................................... 1 4.2 Nuclear fusion in the sun..................................... 11 5 Nucleosynthesis 12 6 The main sequence 12 7 Problems 14 8 Hints to exercises 15 1
REFERENCES 2 References Ostlie D.A., Carroll B.W., 27, An Introduction to Modern Stellar Astrophysics, Second Edition, Addison- Wesley, San Francisco Phillips A.C., 1998, The Physics of Stars, second edition, The Manchester Physics Series, John Wiley & Sons ltd, Chichester
2.1 Gravitational contraction 3 R r r P A g M (P + P ) A Figure 1: Spherically symmetric system of mass M and radius R. The arrows on the right hand side indicate the forces acting on a small element, with volume r A located at distance r, due to gravity and pressure. 1 The composition of stars Right after the big bang the process of nucleosynthesis led to the creation of the first elements in roughly the following ratio (by mass): 75% hydrogen, 25% helium and a tiny amount of lithium-7. These elements are the raw materials from which the first stars formed. The ratio of elements present in a star is referred to as its chemical composition. All stars convert hydrogen to helium in their interiors and helium is subsequently converted into the heavier elements such carbon, nitrogen, oxygen, iron, etc; the elements from which the earth and ourselves have been built up. The elements of atomic number higher than that of helium are referred to as metals in astronomical jargon. Stars that have formed after the first generation of stars still consist of roughly 75% hydrogen and 25% helium but also contain a small admixture of metals produced by the first stars. The second and later generation stars are said to have a non-zero metallicity. 2 Gravity and hydrostatic equilibrium The most important driving force in the formation and evolution of stars is that of gravity which leads to: 1. contraction of interstellar gas-clouds and the subsequent formation of stars; 2. conditions in the stellar interior which will allow nuclear fusion to proceed; 3. the fusion of hydrogen to helium is usually followed by further contraction of the star and the conversion of helium (and heavier elements) to elements such as C, O, Fe, etc. 2.1 Gravitational contraction We start by studying a number of basic properties of the compression of matter due to gravity. Consider a spherically symmetric system of mass M and radius R as illustrated in figure 1. The only forces acting on the matter in this sphere are its self-gravity and the internal pressure. The density and pressure at a distance r from the centre are ρ(r) and P (r). The acceleration of a mass element located at a distance r from the centre of the spherical system can be derived as follows. The mass inside the spherical shell with radius r is: m(r) = r ρ(r )4πr 2 dr, (1) which acts as a gravitational mass located at the centre and gives rise to an inward acceleration: g(r) = Gm(r) r 2. (2)
2.1 Gravitational contraction 4 In general there will also be a force due to the pressure gradient. Consider a small volume element located between radii r and r + r, with surface area A and volume r A, as in figure 1. There will be a net force acting on the volume element if the pressure on the inside surface is different from that on the outside surface. The inward force due to the pressure gradient is: [ P (r) + dp r P (r) dr ] A = dp r A. (3) dr With M = ρ(r) r A we can derive the acceleration of the volume element as: d 2 ( r dt 2 = g(r) + 1 ) dp. (4) ρ(r) dr The acceleration is inward and thus to compensate the force of gravity the pressure gradient has to be negative. That is, the pressure has to increase towards the centre in order to counteract the self-gravity of the spherical system. 2.1.1 Free fall If we now assume that there is no internal pressure we can derive how long it takes before all the mass in the system is concentrated at the centre. From symmetry considerations it follows that every spherical shell of matter will converge on the centre with acceleration Gm /r 2, where m is matter enclosed by the shell. The shell is assumed to be at rest at a radius r and the enclosed mass is assumed to remain constant during the collapse. From the conservation of energy (gravitation potential energy is converted to kinetic energy) it follows that the velocity of the shell after the collapse from r to r is given by: ( ) 1 dr 2 = Gm Gm. (5) 2 dt r r The time is takes for a free fall towards the centre of the sphere (the free-fall time is then given by: tend dt ( t ff = dt = dr dr = 2Gm 2Gm ) 1/2 dr. (6) r r r Where we take into account that dr/dt < (velocity vector is directed inward). The expression between brackets in the integral can also be written as: 2Gm 2Gm ( ) r r = 2Gm = 2Gm ( ) 1 r/r. (7) r r r r r r/r With the substitution x = r/r the free-fall time can be written as: ( ) r 3 1/2 1 ( ) t ff = x 1/2 dx. (8) 2Gm 1 x The last integral can be solved through the substitution x = sin 2 θ and evaluates to π/2. Hence the free-fall time depends on m /r 3 and thus on the average density within the shell with radius r. For a spherical system with uniform density ρ we can then write: ( ) 3π 1/2 t ff =. (9) 32Gρ In practice gravity will always be opposed to some extent because the energy released during free fall will be converted to random thermal motions thereby building up a pressure. The free fall approximation is however relevant if the released energy can be radiated away or used up in the excitation or dissociation of the gas particles in the system. An example relevant to the formation of stars is the collapse of a cloud of molecular hydrogen. It can collapse in free fall as long as it is transparent to its own radiation, or if the hydrogen molecules are being dissociated, or while the hydrogen atoms are being ionized. As soon as the gravitational energy is being released in a non-transparent cloud of ionized hydrogen the internal pressure will rise and the cloud will slowly reach hydrostatic equilibrium. r
2.2 Hydrostatic equilibrium 5 2.2 Hydrostatic equilibrium From figure 1 and equation (4) it follows that a volume element at distance r from the centre of the spherical system is in hydrostatic equilibrium if the pressure gradient at radius r is given by: dp dr = Gm(r)ρ(r) r 2. (1) If this equation holds at all radii then our entire spherical system is in hydrostatic equilibrium and in that case we can find an expression which relates the average internal pressure to the gravitational energy. We multiply both sides of the equation for hydrostatic equilibrium by 4πr 3 and we integrate from r = to r = R: R 4πr 3 dp R dr dr = Gm(r)ρ(r)4πr 2 dr. (11) r Both sides of the resulting equation have a physical meaning. De right-hand side is simply the gravitational potential energy of the system: m=m Gm(r) E gr = dm, (12) m= r where dm is the mass between r and r + dr, i.e. ρ(r)4πr 2 dr. The left-hand side can be evaluated through partial integration: [ P (r)4πr 3 ] R R 3 P (r)4πr 2 dr. (13) The first term in this expression is zero if we assume that the pressure outside of radius r = R is equal to zero. De second term is equal to 3 P V, where V is the volume of the system and P the volumeaveraged internal pressure. Thus the average pressure needed to support a spherical system with volume V and gravitational energy E gr is given by: P = 1 E gr 3 V. (14) In other words the average internal pressure is one-third of the density of the stored gravitational energy. This expression for the average pressure needed to support a self-gravitating system is one form of the so-called virial theorem. To examine this result further we will now consider the pressure generated by a classical gas due to the translational motion of the gas particles. Consider a gas of N particles confined to a cubical box with volume L 3 with its edges oriented along the x, y and z axes. Take one gas particle with velocity v = (v x, v y, v z ) and momentum p = (p x, p y, p z ). As this particle moves about in the box it bounces off the walls at regular intervals. The rate at which it strikes one of the sides perpendicular to the z-axis is v z /2L. Upon bouncing off the wall the particle will impart a momentum 2p z (as follows from momentum conservation with the wall being at rest). So the rate of momentum transfer to a unit area of the wall is p z v z /L. If now consider all particles in the box then the pressure generated on a wall of area L 2 perpendicular to the z-axis is: P = N L 3 p zv z, (15) where the brackets denote an average over all particles. If we assume the gas to be isotropic then all directions of motion for the particles are equally likely and: where Hence the pressure of each side of the box is the same and given by: p x v x = p y v y = p z v z = 1 p v, (16) 3 p v = p x v x + p y v y + p z v z. (17) P = n p v, (18) 3
2.2 Hydrostatic equilibrium 6 where n is the number of particles per unit volume. Note that this expression is valid for classical gases and for gases in which quantum effects are important. In addition it is valid for both non-relativistic and relativistic gases. We now consider the pressure for two types of ideal gases, one with non-relativistic particles and one with ultra-relativistic particles. The general expression for the energy ɛ p for a particle of mass m with momentum p is: ɛ 2 p = p 2 c 2 + m 2 c 4, (19) where the velocity of the particle is v = pc 2 /ɛ p. De non-relativistic limit follows from the assumption p mc for which ɛ p = mc 2 + p 2 /2m and v = p/m. The ultra-relativistic limit follows from p mc so that ɛ p = pc and v = c. The general expression for the pressure in an ideal gas derived above has the following form for these two limits: for a non-relativistic gas p v = mv 2 and the pressure is: or 2/3 of the translational kinetic energy density. for an ultra-relativistic gas p v = pc and the pressure is: or 1/3 of the translational kinetic energy density. 2.2.1 Equilibrium for a non-relativistic gas P = 2 1 3 n 2 mv2, (2) P = 1 n pc, (21) 3 For a non-relativistic self-gravitating ideal gas with volume V we can write for the average pressure: P = 2 E kin 3 V, (22) where E kin is the total translational kinetic energy of the gas particles. A comparison with the virial theorem shows that for a non-relativistic gas in hydrostatic equilibrium the following condition holds: 2E kin + E gr =. (23) The total energy E tot is equal to E kin +E gr if the particles have no internal degrees of freedom (i.e. they cannot rotate, vibrate, etc). In that case we can write: E tot = E kin and E tot = 1 2 E gr. (24) These equations are very important in astrophysics and the describe the consequences of the virial theorem for a system of non-relativistic gas particles that are held together by gravity and are in hydrostatic equilibrium. A couple of remarks: For a bound system we have E tot < and the stronger the system is bound the more negative the value of E tot is. This can be seen be rewriting equation (23) as E gr = 2E kin, or E gr > E kin, which means that E kin + E gr <. The quantity E tot is the binding energy of the system. The binding energy of the system, E tot, is equal to the total kinetic energy. Hence strongly bound clouds of gas are hot! If the system slowly changes but stays near hydrostatic equilibrium then the changes in the kinetic and gravitational energies are related in a simple way to the change in total energy. A decrease of E tot by 1% leads to a decrease of E gr by 2% and in increase of E kin by 1% (so that equation 23 is still satisfied).
3.1 Conditions for gravitation collapse 7 This kind of behaviour often occurs in astrophysical systems. Consider a gas cloud which loses energy from its surface by radiation. If this energy loss is supplied by the release of gravitational energy then the gravitational energy decreases while the kinetic energy increases, that is the cloud contracts and heats up. If this contraction occurs near hydrostatic equilibrium half the released gravitational energy is lost through radiation and the other half is used to heat up the gas. This heating up in turn causes an increase in pressure which is to compensate the increased gravity. However, the continuing energy loss will cause the cloud to keep contracting. The cloud contraction can be halted if the energy loss at its surface is compensated by, for example, energy released in nuclear fusion processes in its interior. This is what the sun does. If more energy is released through nuclear fusion that is radiated away the cloud will expand and cool (the total energy increases thus E kin decreases and E gr increases). Conversely, if the nuclear reaction cost energy the energy loss of the system will increase (energy is lost by radiation at the surface and absorbed by the nuclear reactions) and therefore the system will contract and the gas will heat up. Energy loss for stars thus leads to a heating up of the interior and an energy increase leads to cooling of the interior. This means that the specific heat capacity of stars is negative! 2.2.2 Equilibrium for an ultra-relativistic gas For an ideal gas with ultra-relativistic particles the pressure is 1/3 of the kinetic energy density and we can write: E kin + E gr =. (25) This means that hydrostatic equilibrium is only possible if the total energy is zero. Such a system is between bound and unbound and can easily be disrupted. This kind of instability occurs in stars for which the pressure is generated primarily by photons (an ultra-relativistic gas). This instability can also occur in stars with a degenerate electron-gas in the interior when the electrons become very energetic. 3 Star formation Stars are formed from clouds of molecular hydrogen (H 2 ) and most stars from in groups, named clusters: There are broadly speaking three types of clusters: Globular clusters consisting of many thousands to hundreds of thousands of stars. They are generally very old (more than a few Gyr) and the stars contain very little metals (i.e. they are metal-poor ). Open clusters consisting of 5 1 stars which are relatively young (< 1 Gyr) and metal-rich. Associations very loose groupings of stars which are very young (< 5 Myr) en often still surrounded by the remnants of the clouds from which they formed. In the rest of this section we will consider very global properties of star formation from elementary physical considerations from we will see that it is plausible that stars should form in groups. 3.1 Conditions for gravitation collapse Under which conditions can a cloud of gas collapse by its own gravity in order to form stars? The cloud should be compact enough for its mass such that gravity will overcome the internal pressure, or E gr > E kin. Consider a cloud of mass M and radius R consisting of N particles of average mass m. The cloud has a uniform temperature T. The gravitational energy can be calculated using equation (12) and equals: E gr = f GM 2 R, (26) where the factor f depends on the density distribution in the cloud. For a spherical cloud with uniform density it can be shown that f = 3/5. For the rough estimates that follow we will use f = 1. The kinetic energy can be written as: E kin = 3 NkT. (27) 2
3.2 Contraction of a protostar 8 The condition for the collapse of the cloud is E gr > E kin from which it follows that a cloud of radius R has to have a mass larger than: M J = 3kT 2Gm R. (28) We can also write this as a density criterion for a cloud of mass M. The condition for the radius is: R < GM 2 2Gm = NkT 3kT M, (29) from which it follows (with ρ = 3M/(4πR 3 )) that the density has to be larger than: 3 2 ρ J = 3 ( ) 3kT 3 4πM 2. (3) 2Gm The subscripts J in these expressions indicate that the criteria define the so-called Jeans mass and density. The expression for the Jeans density shows that it has to be low in order to make it easy for a cloud to collapse. This is easier for a cloud with a large mass. For example, a cloud with a mass of 2 1 33 kg (1 M ) can collapse if the density becomes larger than 1 22 kg m 3, about 1 5 molecules per cubic meter. For a cloud of 1 M the Jeans density is one million times higher! From these considerations it follows that the process of star formation may proceed as follows. First a large and massive cloud will start to contract by its own gravity. As the density becomes high enough smaller pieces of the cloud in the interior can start to collapse independently. The cloud will thus fragment into smaller clouds from which eventually protostars can form. This gives a qualitative explanation as to why we can expect stars to form in groups. 3.2 Contraction of a protostar The expression for the Jeans density indicates that if a cloud at a temperature of 2 K can reach densities of about 1 16 kg m 3 it can fragment into smaller clouds of 1 M which can contract independently. Such a fragment will constitute a protostar with a radius of about 1 15 m, 1 6 R. As long as the gravitational energy being released is not converted into thermal motion of the molecules (pressure) the fragment will collapse in free fall. This can go on as long as the dissociation of H 2 and the ionization of H absorb the released potential energy. The dissociation and ionization energies are ɛ D = 4.5 ev and ɛ I = 13.6 ev. So the energy required to convert all the H 2 to HII is roughly: M 2m H ɛ D + M m H ɛ I, (31) where m H is the mass of the hydrogen atom. This energy is released during the contraction of the protostar from a radius R 1 to a radius R 2, hence: GM 2 R 2 GM 2 R 1 M 2m H ɛ D + M m H ɛ I. (32) In this example (protostar of 1 M ) the required energy is 3 1 39 J and the protostar will contract in free fall from R 1 1 15 m to R 2 1 11 m, or about 1 R. The timescale is given by the expression for t ff (equation 9) with ρ 1 16 kg m 3 and is about 2 yr. After the complete ionization of all hydrogen (in the interior) the contraction of the protostar will proceed until the gas becomes opaque to its own radiation and then the released gravitational energy will be converted into random thermal energy of the electrons and protons. The pressure will rise in this process and the protostar will slowly reach hydrostatic equilibrium. With the aid of the virial theorem we can estimate the internal temperature at the moment the contraction starts to slow down. De kinetic energy of the electron and protons in the gas is: E kin M m H 3kT, (33)
3.3 Conditions for being a star 9 where we have used m.5m H. Because R 1 R 2 we can write for the gravitational energy: E gr GM 2 ( M ɛ D + M ) ɛ I. (34) R 2 2m H m H According to the virial theorem we have 2E kin + E gr = thus for the temperature we can write: kt 1 12 (ɛ D + 2ɛ I ) 2.6 ev. (35) This amounts to a temperature of approximately 3 K. This estimate is independent of the mass of the protostar. The protostar will now continue to contract slowly thereby maintaining approximate hydrostatic equilibrium. According to the virial theorem half the released potential energy is converted to internal kinetic energy and the rest is lost as radiation at the surface. The internal pressure and temperature now will continue to increase until the conditions are suitable for the thermonuclear fusion of hydrogen to helium. The nuclear reactions will release energy which will compensate the energy loss at the surface and then the contraction of the protostar stops and it becomes a real star. 3.3 Conditions for being a star Not all contracting protostars will in the end become stars. De pressure needed to stop the contraction can also be generated by a cold, dense gas of degenerate electrons. In such a gas the behaviour of electrons is governed by the laws of quantum mechanics and the electrons will occupy the lowest possible energy states in accordance with the Pauli exclusion principle. Such a gas generates pressure not by the increasing random thermal energy of the electrons but because the total kinetic energy of the electrons has a minimum which increases as the density rises. In fact the temperature no longer rises. An electron gas becomes degenerate if the average distance between the electrons is comparable to the the Broglie wavelength λ = h/p. The momentum of electrons in a classical electron gas at temperature T is roughly (m e kt ) 1/2 (kinetic energy of kt ) and then the typical the Broglie wavelength is: λ = h. (36) (m e kt ) 1/2 The electron gas is a classical gas as long as the wave functions of the electrons do not overlap, i.e. the average mutual distance between the electrons must be much larger than λ. The density of the gas then has to satisfy the following condition: ρ m λ 3 m(m ekt ) 3/2 h 3, (37) where m =.5m H for a gas with an equal amount of electrons and protons. As long as the gas is classical the temperature will rise as the density in the protostar increases. Using the approximate expression for E gr, the classical expression for E kin, and the condition for hydrostatic equilibrium (equations 26, 27 and 23) we find: kt GM 2 3NR = GMm 3R GmM 2/3 ρ 1/3, (38) where we have used M/3R M 2/3 ρ 1/3. Thus while the density remains below the critical value the temperature will rise with ρ 1/3. The temperature at which the electrons become degenerate can be estimated by substituting the critical density in the expression for kt above: or kt GmM 2/3 m 1/3 (m ekt ) 1/2, (39) h kt ( G 2 m 8/3 m e h 2 ) M 4/3. (4)
4.1 Solar radiation 1 Hence the mass of the protostar determines the internal temperature that is reached before the electron gas becomes degenerate! For 1 M the temperature reached is kt 1 kev. In other words if a protostar with a mass of 1 M keep contracting by its self-gravity eventually an average internal temperature of about 1 million K is reached, with a central temperature which is even higher. This is more than sufficient to trigger nuclear reactions and the conversion of hydrogen to helium. For less massive stars lower temperatures are reached (making it more difficult to ignite nuclear reactions). Detailed calculations show that in objects less massive than.8 M nuclear reactions will not be triggered. These are protostars that evolve toward a state in which gravity is opposed by the pressure of a degenerate electron gas, the so called brown dwarfs. At the other end of the mass spectrum we find the very massive stars for which the internal pressure is dominated by photons and we have already seen that for such an ultra-relativistic gas the equilibrium is unstable. Above 5 1 M stars are expected to easily disrupt and these are indeed very rare. 4 The sun Because our sun is the best known star for which we have the most accurate information we will proceed to apply to it a number of the estimates we have derived about the interiors of stars. The sun has a mass of about 2 1 3 kg and a radius of about 7 1 8 m. Hence the average density ρ is 1.4 1 3 kg m 3, comparable to the density of water. The free-fall time then is only a half hour according to equation (9). We can thus safely conclude that the sun is not in a state of free fall and that the internal pressure must be enough to oppose gravity. The age of the sun is about 4.5 billion years and we have no evidence that during its lifetime any drastic changes occurred (e.g., from the geological record). Hence the sun has remained in hydrostatic equilibrium during all that time which means we can determine the average internal pressure using the virial theorem: P = 1 E gr 3 V GM 2 4πR 4 1 14 Pa, (41) which is a billion times the pressure in the earth s atmosphere. Despite this high pressure we can still treat the solar interior as a classical ideal gas and that case the average internal pressure is also given by: P = ρ m kt in, (42) where T in is the typical internal temperature and m =.5 amu (atomic mass unit) for a gas of ionized hydrogen. The standard model of the sun in fact has a composition of 71% H, 27% He, and 2% heavier elements for which m =.61 amu. We can now estimate the internal temperature as: kt in GM m 3R.5 kev or T in 4.6 1 6 K. (43) The real central density, temperature and pressure are: 1.48 1 5 kg m 3, 15.6 1 6 K, 2.29 1 16 Pa. So our estimates are pretty good! 4.1 Solar radiation The sun radiates about 4 1 26 W of energy an has a corresponding effective temperature of about 6 K. Because kt eff.5 ev most of the radiation comes out in the visual. Hence the temperature at the surface of the sun is about 1 times lower than in the interior. If the radiation could leave the solar interior unhindered its luminosity would be: L 4πR 2 σt 4 in, (44) in which case the sun would radiate primarily in the X-ray regime (kt in.5 kev). Luckily this is not the case but what happens to the energy produced in the interior? The radiation from the interior of the sun is constantly scattered, absorbed and emitted by the electrons and ions in the gas. A temperature gradient is set up and the radiation slowly diffuses to the surface. The photons
4.2 Nuclear fusion in the sun 11 diffuse according to the so-called random walk. If l is the average free path length for a photon then the displacement after N interactions is with the gas is: D = l 1 + l 2 + + l N, (45) where we assume that l is constant over the whole solar interior. The value of D 2 is: D 2 = l 2 1 + l 2 2 + l 2 N + 2(l 1 l 2 + l 1 l 3 + ) (46) For a random walk the second term averages to zero so the average of the square of the displacement after N interactions is Nl 2. To escape the sun an average photon has to bridge a distance comparable to the solar radius and this takes about R 2 /l 2 steps. The time for each step is about l/c so the random walk escape time is: t rw R2 cl (47) The time it takes to escape the sun directly is R /c which is a factor l/r lower than t rw. Hence the rate at which energy escapes the sun is lower by the same factor as a consequence of the diffusion process. This means that the luminosity is also lower by a factor l/r compared to when the radiation can leave the sun freely. So for the effective temperature we have: ( ) l 1/4 T eff T in. (48) R From this result we can conclude that the mean free path for a photon in the sun is about 2 mm. The sun is very opaque. According to the equation for t rw it takes a photon 3 years to diffuse from the centre to the surface! Using the relation between T eff and T in and the relation between the internal temperature and the mass and radius of the sun (eq. 43), we can write: L 4πR σt 2 in 4 l (4π)2 σ R 3 5 k 4 G4 m 4 ρ lm 3. (49) Thus we expect that for stars like the sun the luminosity steeply increases with mass! The diffusion process for photons restricts the flow of energy and prevents the sun from catastrophically losing energy. This process determines the amount of energy that has to be released in nuclear reactions to compensate the energy loss at the surface. 4.2 Nuclear fusion in the sun The nuclear fusion process in the sun proceeds according to the so-called proton-proton chain: p + p d + e + + ν e, (5) p + d 3 He + γ, (51) 3 He + 3 He 4 He + p + p, (52) where d indicates the deuteron. Each of these reactions is exothermic and in total an energy of 26 MeV is released for each 4 He nucleus formed. All these reactions are in principle hindered by the Coulomb barrier and can only proceed due to quantum mechanical tunnelling. The first reaction is the slowest and on average it takes 5 billion years for a proton to fuse with another proton to a deuteron. The deuteron is converted to 3 He within 1 second and the time required to convert two 3 He nuclei to 4 He is about 3 years. So the first reaction determines the rate at which energy is generated within the sun. The luminosity of the sun is enormous but on average every kilo in the sun only produces.2 mw, which is about 1 times less than the heat we produce per kilo in our bodies!
6 The main sequence 12 Table 1: The most important stages of thermonuclear fusion in stars. The ashes of one stage of burning form the fuel for the next stage as long as the contracting star is massive enough to reach the temperatures indicated. Process Fuel Products Ignition temperature (K, approximate) Hydrogen burning Hydrogen Helium 1 1 7 Helium burning Helium Carbon, oxygen 1 1 8 Carbon burning Carbon Oxygen, neon, sodium, magnesium 5 1 8 Neon burning Neon Oxygen, magnesium 1 1 9 Oxygen burning Oxygen Magnesium to sulphur 2 1 9 Silicon burning Silicon Iron and nearby elements 3 1 9 To produce one 4 He 4 protons are used and 26 MeV of energy is released. Hence the rate at which protons are used up is: (4 4 1 26 )/(26 1.6 1 13 ) = 4 1 38 protons per second. (53) In this process at least 2 1 38 neutrinos are released which can leave the sun unhindered. The nuclear reactions in the sun s interior act as a kind of thermostat. If the temperature in the interior rises more nuclear reactions will occur and the total energy will increase (the nuclear reactions producing more energy than can escape the sun). From the virial theorem we know that the sun will then expand and cool. Conversely, a lowering of the internal temperature will lead to a contraction and heating up of the sun. This thermostat has functioned for the past 4.5 billion years and will do so until there is no longer enough hydrogen at the centre of the sun to keep the proton-proton chain going. There are about 7 1 56 protons in the sun and about 1% of those will be converted to 4 He in the next 6 billion years. The total hydrogen burning phase for the sun lasts about 1 billion years. After that time the core will contract and heat up until the pressure and temperature are high enough for the burning of helium to start. At that moment the outer layers of the sun will expand and cool and the sun becomes a red giant. 5 Nucleosynthesis Stellar evolution is a process whereby gravitational energy is released by the contraction of the star. This contraction temporarily stops if enough energy is added to the system by nuclear reactions so that the energy loss at the surface is compensated. The ashes of one nuclear burning stage form the fuel for the next stage. There is in fact a sequence of thermonuclear burning stages at ever increasing temperatures. This is described in table 1. The process of nuclear fusion can only continue as long as the formation of a new nucleus produces net energy. In figure 1.9 from Ostlie & Carroll (27) the binding energy per nucleon is plotted. Starting from hydrogen we see that as long as the binding energy increases nuclear fusion will produce net energy. This stops as soon as iron is reached. The elements near iron are formed (Cr, Mn, Fe, Co, Ni) but constitute an ash which cannot be burned any further and thus they constitute the end stages of nuclear fusion processes in the interiors of stars. Keep in mind that the particular nuclear burning stage that can be reached by a star depends on the central temperatures reached which in turn depend on the mass of the star. Only the most massive stars will eventually produce iron in their cores. Elements heavier than iron can only be formed through the capture of neutrons and β-decay. This only occurs during the very last stages of stellar evolution. 6 The main sequence Stars that convert hydrogen to helium in their interiors are in the phase of their evolution in which they are located on the main sequence in the Hertzsprung-Russell diagram. We have seen that the luminosity of a star is a steeply increasing function of its mass (eq. 49). If we assume that the mean free path l is inversely
6 The main sequence 13 1 6 Main sequence mass luminosity relation 1 4 Luminosity (L Sun ) 1 2 1 1 2.1 1. 1. 1. Mass (M Sun ) Figure 2: The main sequence mass-luminosity relation according to the data listed in appendix G of Ostlie & Carroll (27). The line is the estimated relation L M 3. proportional to the average density we expect that for stars on the main sequence L M 3. Figure 2 shows the actual relation between mass and luminosity for stars on the main sequence according to the data in appendix G of Ostlie & Carroll (27). The real relation is close to our estimate. In reality the relation L M α holds, with α 3 for massive stars and α 3.5 for stars less massive than the sun. De rapid increase of luminosity with mass means that more massive stars burn their nuclear fuel faster than less massive stars and thus have shorter main sequence lifetimes. For massive stars the main sequence lifetime is proportional to M 2 while for less massive stars the lifetime is proportional to M 2.5. The sun has a main sequence lifetime of 1 billion years so a star of 1 M only spends about 1 million years on the main sequence while a star of.5 M is on the main sequence for more than 5 billion years! The mass of a star is by far the most important parameter determining its evolution.
7 Problems 14 7 Problems The problems below are taken from the book by Phillips (1998). Try to solve the problems first without consulting the hints! 1. Consider a sphere of mass M and radius R. a. Calculate the gravitational potential energy of the sphere assuming (A) a density which is independent of the distance from the centre, and (B) a density which increases towards the centre according to: ρ(r) = ρ c (1 r/r). b. In both cases, (A) and (B), write down the average internal pressure needed for hydrostatic equilibrium, and c. determine how the pressure in the sphere depends on the distance from the centre. 2. As the sun evolved towards the main sequence, it contracted under gravity while remaining close to hydrostatic equilibrium, and its internal temperature changed from about 3 K, eq. (35) to about 4.6 1 6 K, eq. (43). (This stage of stellar evolution is called the Kelvin-Helmholtz stage.) a. Find the total energy radiated during this contraction. Assume that the luminosity during this contraction is comparable to the present luminosity of the sun and estimate the time taken to reach the main sequence. b. Explain how it is possible that while loosing energy the contracting sun heats up. 3. The main sequence of the Pleiades cluster of stars consists of stars with mass less than 6 M ; the more massive stars have already evolved off the main sequence. a. Estimate the age of the Pleiades cluster. b. The real age is 12 Myr. Explain why the estimated age is different. 4. Useful bounds can be set on the pressure at the centre of a star without detailed stellar structure calculations. Consider a star of mass M and radius R. Let P (r) be the pressure at distance r from the centre and m(r) be the mass enclosed by a sphere of radius r. Show that in hydrostatic equilibrium the function P (r) + Gm(r)2 8πr 4 decreases with r. Hence show that the central pressure satisfies the inequality where ρ is the average density. P c > 1 6 ( ) 4π 1/3 G ρ 4/3 M 2/3, 3 If you assume that the density ρ(r) decreases with r, it is possible to derive a tighter lower bound and, in addition, a useful upper bound for the central density. Show that P c > 1 2 ( ) 4π 1/3 G ρ 4/3 M 2/3. 3 In addition show that where ρ c is the central density. P c < 1 2 ( ) 4π 1/3 Gρ 4/3 c M 2/3, 3
8 Hints to exercises 15 8 Hints to exercises 1. Use equations (12), (14), and (1). 3. Bear in mind that the luminosity of a star of mass M is proportional to M α with α between 3 and 3.5. 4. Let f(r) = P (r) + Gm(r)2 df 8πr 4 and show that dr <. The first lower bound on P c is given by the condition f() > f(r). The second lower bound and the upper bound on P c can be obtained by noting that m(r) > 4π 3 ρ r3, m(r) < 4π 3 ρ cr 3, and using df dr + Gm(r)2 2πr 5 =.