Presentation on theme: "PH507 The Hydrostatic Equilibrium Equations for Stars"— Presentation transcript:
1 PH507 The Hydrostatic Equilibrium Equations for Stars Dr J. Miao
2 We will cover the following topics: Equation of Mass distributionEquation of hydrostatic support- Virial Theorem-Centre pressure and mean temperature of a star-Different time scales for stars-How do we know we are right?3. Equation of Energy generation4. Equation of Energy transportation
3 The Equations for stars How does a star exist?Internal pressure gradientForce of gravitationStars are spherical and symmetric about their centersStars are in hydrostatic equilibriumTwo fundamental assumptions:Four equations of stellar structureStars are held together by the force of gravitation of all the stellar material.This gravitational force is resisted by an internal pressure gradient due to the thermal motion of the particles which make up the stellar material.If these two assumptions are made, the structure of a star is governed by a set of equations in which all the physical quantities depend on the distance from the centre of the star alone.Firstly, we assume that the density distribution inside a star is , then the mass element ….1. Equation of mass distribution
4 2. Equation of hydrostatic support The gravitational mass m(r) situated at the centre gives rise to an inward gravitational acceleration equal to :Newton’s second law:Note that to oppose gravity the pressure must increase towards the centre
5 m0 2.1 What can we know from the equation of hydrostatic support i) What will happen if there is no pressure gradient to oppose the gravity?Each spherical shell of matter converges on the centre free fall of the starm0r0The mass of the thin shell is M with an initial radius of r0rThe mass included in the sphere of radius r0 is m0,When the thin shell collapses to the distance r: KE = GEIt follows that the free fall time to the centre of the sphere is given by(See Appendix)For the sun, tff ~ 2000sTo find the inward velocity of the shell when its radius is r, we assume that the shell is initially at rest at a radius r0, and that it encloses a mass m0 which remains constant during collapse. The inward velocity can than be found from the conservation of energy equationIn fact, collapse under gravity is never completely unopposed. During the process, released gravitational energy is usually dissipated into random thermal motion of the constituents, thereby creating a pressure which opposes further collapse The internal pressure will rise and slow down the rate of collapse. The cloud will then approach hydrostatic equilibrium
6 ii) What will happen if a star is in hydrostatic equilibrium state? an element of matter at a distance r from the centre will be in hydrostatic equilibrium if the pressure gradient at r is (d2r/dt2 =0 in eq.1.3)The whole system is in equilibrium if this equation is valid at all radii.* Eq. (1.4) implies that the pressure gradient must be negative, or in other words, pressure decreases from the inner central region to the outer region* The three quantities m, r, are not independent2.2 From Hydrostatic equilibrium equation to Virial TheoremIf m is chosen as the independent space variable rather than r,Divide [4r2] into two sides of (1.4)Spherical symmetry implies that each spherical shell of matter converges on the centre .
7 Which is gravitational potential energy of the star Using the symbol: Integrating the left-hand side of above by parts, the equation can be written:Which is gravitational potential energy of the starUsing the symbol:Note: Vc=0, and dm = dVIf the star were surrounded by a vacuum, its surface pressure would be zeroThe integrated part vanishes at the lower limit of integration because Vc=0.The term on the right-hand side of above is the negative gravitational potential energy of the star(i.e., apart from the minus sign it is the energy released in forming the star from its component parts dispersed to infinity)and we denote this by the symbol …If the star were surrounded by a vacuum, its surface pressure would be zero and the left-hand side of it could be put equal to zero. In fact thesurface pressure of a star will not be zero but it will be many orders of magnitude smaller than the central pressure or the mean pressure in the interior, which means that the term on the left-hand side is very small compared to either of the terms on the right-hand and it can usually be neglected so we haveThis is the general, global form of the Virial Theorem and will be used very often later on. It relates the gravitational energy of a star to its thermal energy.
8 2.3 What can Virial theorem tell us for classical idea gas system ? the equation of state of a classical gas is known asthe internal energy per unit mass isthe system is stable and bound at all pointsThis is also the Virial theorem in another formBecause the total energy (binding energy) E = + U = - / 2 = / 2 , and E are always negative, if a star is in stable.Virial theorem tells that in a contracting gas system (protostar ),This is also the Virial theorem in another form, which implies that the system is stable and bound at all points, because the total energy (binding energy)It is instructive to consider what happens to these terms in a contracting gas (protostar).the energy for radiation is provided by half of the decreased gravitational potential energy: E = / 2.When there is no energy from contraction, the radiation of a star is provided by thermonuclear reactions.
9 The pressure at the centre of the Sun exceeds 450 million atmospheres 2.4 Estimate the minimum pressure at the centre of a star:Integrating eq.(1.5) from the centre to the surface of the starOn the right-hand side we may replace r by the stellar radius to obtain a lower limit for the central pressure: i.e, 1/r > 1/RThe pressure at the centre of the Sun exceeds 450 million atmospheres
10 2.5 Estimate the minimum mean temperature of a star: We can use this result to estimate the average internal temperature of a starIn the gravitational potential energy expression, r is less than R everywhereAt such a high temperature, the average kinetic energy of the particles is higher than the energy required to remove many bound electrons from atoms and the gas will thus be highly ionised. Because the gas is ionised, the distances between particles are much greater than their sizes and corrections to ideal gas law are small. between two stars of the same mass, the denser one is also hotter.For the Sun, Eq. (1.9) gives us T > 4 106 K if the gas is assumed to be atomic hydrogen
11 2.6 Estimate the importance of the radiation pressure: the corresponding expression for radiation pressure iswith T =4 106 K and( =1.4 103 kgm-3, a =7.5510-16 Jm-3K-4),We have:Therefore it certainly appears that radiation pressure is unimportant at an average point in the Sun!This is not true of all stars, however. We shall see later that radiation pressure is of importance in some stars, and some stars are much denser than the Sun and hence correction to the idea gas are very important.
12 If t > 3*1016s < (tff / t)2 < 10-27 2.7 How accurate is the Hydrostatic Assumption?FromSuppose:i.e:If a mass element starts from rest with this acceleration, its inward displacement s after a time t :if we allow the element to fall all the way to the centre of the star, we can replace s in the above equation by r and then substituteThe time t is that it would take a star to collapse if the forces are out of balance by a factor In deriving the equation of hydrostatic equilibrium, it has been assumed that the gravity and pressure forces are balanced in a star.we ask: what will happen if the two terms on the right-hand side of the above equation is not zero?Suppose that the sum of the right side is a small fraction of the gravitational term, i.e., the inward acceleration a=g,But fossil and geological records indicate that the properties of the Sun have not changed significantly for at least 109 years(31016s)If t > 3*1016s < (tff / t)2 <most stars are like the sun and so we may conclude that: the equation of hydrostatic equilibrium must be true to a very high degree of accuracy !
13 ~ 2 10-5 2. 8 How valid is the spherical symmetry assumption? Departure from spherical symmetry may be caused by rotation of the star.~ 2 10-5 of the Sun is about 2.5 10 -6Departures from spherical symmetry due to rotation can be neglected.This statement is true for the vast majority of stars. There are some stars which rotate much more rapidly than the Sun. For these stars, the rotation-distorted shape of the star must be accounted for in the equations of stellar structure. r (r, , )
14 3. Energy generation in stars 3.1 Gravitational potential energyIt is a likely source of the stellar energy and has the formThe total energy of the system :Assuming a constant density distributionthe gravitational potential energy:We all know that the rate of energy output from stars is very large. But where is the energy source coming from? The first candidate isthe total mechanical energy of the star is:What can this tell us?
15 Is the Kelvin-Helmholz time scale Assuming that the Sun were originally much larger than it is todayhow much energy would have been liberated in its gravitational collapse?If its original radius was Ri, where Ri >> R, then the energy radiated away during collapse would beFurther assuming Lsun is a constant throughout its lifetime, then it would emit energy at that rate for approximatelyIs the Kelvin-Helmholz time scaleWe have already noted that fossil and geological records indicate that the properties of the Sun have not changed significantly for at least 109 years (3 × 1016 s)But the Sun has actually lost energy: L * 3 * 1016 = 1.2 × 1043 JGravitational potential energy alone cannot account for the Sun’s luminosity throughout its lifetime !
16 td << tth << tn. (1.9) 3.2 Nuclear reactionthe total energy equivalent of the mass of the Sun, .If all this energy could be converted to radiation, the Sun could continue shining at its present rate for as long asis called nuclear timescaleThe Sun just have consumed its mass:Hence, for most stars at most stages in their evolution, the following inequalities are truetd << tth << tn (1.9)
17 3.3 How do we include the energy source? Define luminosity L (r) as the energy flow across any sphere of radius r. The change in L across the shell dr is provided by the energy generated in the shell:rr+drwhere (r) is the density; (r) is the energy production rate per unit massWe now have three of the equations of stellar structure. This is not yet a complete set of equations as we have five unknowns (P, M, , L and ). In order to make further progress, we need to considerThis is usually called the energy-generation equationThe energy generation rate depends on the physical conditions of the material at the given radius.
18 4. How is energy generated transported from center to outside? 4. 1 Convection.Energy transport by conduction (and radiation ) occurs whenever a temperature gradient is maintained in any bodyBut convection is the mass motion of elements of gas, only occurs when.Consider a convective element of stellar material a distance r from the centre of the starr+rrT+TP+P+T, P, T, P, T+T, P+P, +define P, as the change in pressure and density of the elementConvection - energy transport by mass motions of elements of the gas.Conduction - energy transport by exchange of energy during collisions of gas particles (usually electrons).Radiation - energy transport by the emission and re absorption of photons generated in the gas.Radiation neutrinosHow to quantify them?fig1.6, when fluid bulb get heated to a certain temperature, it rises up in the centre and the cold fluid sinks down from two sides.Let us suppose that the element rises a distance r towards the surface of the star, expanding (and hence reducing in pressure and density) as it does so.The pressure and density of the element at r+r may or may not be the same as its surroundings,If the blob is denser than its surroundings at r+r it will tend to sink back to its original position and the gas is said to be stable against convection. however, ifP, , as the change in pressure and density of the surroundings
19 The condition for this instability is therefore: If the blob is less dense than its surroundings at r+r then it will keep on rising and the gas is said to be convectively unstable.The condition for this instability is therefore:Whether or not this condition is satisfied depends on two things:a) the rate at which the element expands (and hence decreases in density) due to the decreasing pressure exerted on itb) the rate at which the density of the surroundings decreases with height.We can make two assumptions about the motion of the elementThe element rises adiabatically, i.e. it moves fast enough to ensure that there is no exchange of heat with its surroundings;PV =constantHaving obtained a condition for the occurrence of convection, we now need to rewrite it in terms of a temperature gradient for compatibility with the stellar structure equations, we can do this by making two assumptions about the motion of the element2. The element rises with a speed < the speed of sound..This means that, during the motion, sound waves have plenty of time to smooth out the pressure differences between the element and its surroundings and hence P = P at all times
20 By using P/ =constant By the second assumption P = P :For an ideal gas in which radiation pressure is negligible, we have:P = kT / m, log P = log+ log T + constant.This can be differentiated to give relationship of the changes between P, T and of the surrounding:Substitute (1.12) and (1.13) into (1.11)Given that V is inversely proportional to the density , we can equivalently write for the element:The critical temperature gradient for convection is given by
21 Eq.(1.15) can also be written as: Note that the temperature and the pressure gradients are both negative in this equation, we can use modulus sigh to express their magnitudes:Eq.(1.15) can also be written as:Convection will occur if temperature gradient exceeds a certain multiple of the pressure gradient.The criterion for convection derived above can be satisfied in two ways :a) The ratio of specific heats, , is close to unityIn the cool outer layers of a star, the gas is only partially ionized, much of the heat used to raise the temperature of the gas goes into ionization and hence cv and cp are nearly same ~ A star can have an outer convective layerWe will return to the subject of convective cores and outer convective layers when we discuss the structure of main-sequence starConvection 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 evolutionIn particular, although we know the conditions under which convection is likely to occur, we do not know how much energy is carried by convection. Fortunately, as we shall see later, there are occasions when we can manage without this knowledge.b) the temperature gradient is very steepa large amount of energy is released in a small volume at the centre of a star, it may require a large temperature gradient to carry the energy away A star can have convective core.
22 ~ > << 4.2 Conduction and radiation Conduction and radiation are similar processes because they both involve the transfer of energy by direct interaction,the flux of energy flowWhich of the two - conduction and radiation - is the more dominant in stars to transport energy?particlesphotonsEnergy:~Number density:nparti>nphotonThis means that the energy density in the form of particles is much greater than that in the form of photons, so it might be expected that conduction is a more important mechanism of energy transport in stars than radiation.However, the smaller number of photons is far outweighed by their much larger mean free path between collisions: a photon at a typical point inside a star travels about 10-2 m before being absorbed or scattered, whereas a particle only travels around m.mean free path:parti ~ mphoton~ 10-2m<<Photons can walk more easily from a point where the temperature is high to one where it is significantly lower before colliding and transferring energy, resulting in a higher transport of energy.
23 4.3 Equation of Radiative transport Conduction is therefore negligible in nearly all main sequence stars and radiation is the dominant energy transport mechanism over conduction in most stars.4.3 Equation of Radiative transportIf we assume for the moment that the conditions for the occurrence of convection is not satisfiedwe can write down the fourth equation of stellar structure,The energy carried by radiation in the flux Frad, can be expressed in terms of the dT/dr and a coefficient of radiative conductivity, rad,we can write down the fourth equation of stellar structure, which expresses the rate of change of temperature with radius in a star assuming that all energy is transported by radiation (i.e. ignoring the effects of convection and conduction).where the minus sign indicates that heat flows down the temperature gradient.The radiative conductivity measures the readiness of heat to flow
24 Astronomers generally prefer to work with an inverse of the conductivity, known as the opacity, which measures the resistance of the material to the flow of heat. Detailed arguments (see Appendix 2 of Taylor’s book ‘stellar evolutions’) show that the opacitywhere a is the radiation density constant and c is the speed of lightCombining the above equations we obtain:Recalling that flux and luminosity are related bythe equation of radiative transport
25 It is the temperature gradient that would arise in a star if all the energy were transported by radiationIt should be noted that the above equation also holds if a significant fraction of energy transport is due to conduction, but in this case, Lr Lr + Lcond.Then (1.22) can be written asClearly, the flow of energy by radiation/conduction can only be determined if an expression for is available.
26 4. 4 Radiation of Neutrinos In massive stars late in their lives, the amount of energy that must be transported is sometimes larger than either radiation of photons or convection can account for .In these cases, significant amounts of energy may be transported from the center to space by the radiation of neutrinos.This is the dominant method of cooling of stars in advanced burning stages, which also plays a central role in events like supernovae associated with the death of massive stars.
27 Summary: 1. Based on two fundamental assumptions: we derived the four equations of stellar structureThere are four primary variables M(r ), P(r), L(r ), T(r ) in these equations, all as a function of radiusWe also have three auxiliary equations P: equation of state, P=P(,T, Xi) : opacity (,T,Xi) : nuclear fusion rate, (,T,Xi).These are three key pieces of physics and we will discuss them in detail
28 td << tth << tn. (1.9) 2. From the most important hydrostatic equilibrium equation:-- Drive the global form of Viral theorem.With the gravitational potential energy of a statIf the density of the system is a constant,Drive another form of Viral theorem:Which tells us: a star in hydrostatic equilibrium is stable and bound at all points- E = U = - / –only half of the released potential energy can be used as radiation during the collapse process inside a star!-- estimate the minimum center pressure in a star :-- estimate the minimum mean temperature of a star:3. Criteria for convection:4. Three important time scale:td << tth << tn (1.9)
29 5. Show that the radiation pressure is not important in Sun-like stars 6. Radiation is more efficient way to transport energy from place to place than conduction
30 Appendix:This may be simplified by introducing the parameterto give