The relevant physics: Problem 4.2 introduced the kinds of stars we observe. The physical parameters deduced directly from observations are luminosity and surface temperature, and indirectly the size of a star. These refer to the radiative properties of the surfaces of stars. We encounter a variety of other physics when we try to explain the luminosities and sizes of stars. The problems are arranged to isolate various aspects of physics:

• Hydrostatic equilibrium : The isothermal atmosphere

• Energy conservation: Conversion of gravitational energy into heat and radiation

• Energy conservation including E = mc²: The conversion of matter into energy by nuclear fusion

• The equation of hydrostatic equilibrium: a simple solution and an integral limit for a star

• The diffusion equation applied to radiation: a simple solution

• The diffusion of radiation: the Thomson scattering cross section

• Quantum theory: Degenerate electrons and their pressure

The astronomical setting: The several physics problems each treat one episode within the lives of the stars. A framework for these problems is provided by the following outline how the Sun was born, how it lives, and how it will die. The L-T diagram, used in problem 4.2 to describe existing stars, can also be used to represent the life of the Sun.

Physics introduction - The isothermal atmosphere: For a plane gaseous atmosphere with uniform gravity g, each layer of gas is supported against gravity by the pressure gradient: the pressure p at the bottom of a layer is slightly larger than at the top. Mathematically, dp/dz = -g, where is the gas density. For an ideal gas, p = nkT, where n is number density of particles. For stellar physics, we relate n to by = nm_p/, where m_p is the mass of a proton. Clearly, =1 for neutral hydrogen. This applies to the surfaces of cool stars like the Sun and Betelgeuse. For ionized hydrogen, each proton contributes mass, but protons and electrons contribute to the pressure. Therefore, =2 for ionized hydrogen. This applies generally for T > 10⁴K, throughout the interiors of normal stars, in the surfaces of the hotter stars, and in stellar coronae, here specifically the solar corona. Assuming uniform temperature and gravity, dp/dz = - p/H, where the scale height is H = kT/gm_p. Therefore, p is proportional to exp(-z/H),

The setting: This problem seeks to estimate how long the Sun's "youth" lasts, simplifying the track in the L-T diagram (see introduction to section V) into two straight lines. On the vertical track, the temperature T 3x10^3oK is determined by the atomic physics of hydrogen : The degree of ionization of the hydrogen determines how radiative energy can escape.

The solution:

1. Order of magnitude: The gravitational energy of the Sun is about GMo²/Ro. The rate of energy loss is about Lo. Therefore, the appropriate time scale is o = GMo²/RoLo = 10¹⁵s = 3x10⁷years.

2. Exact integration at constant T: We use L = 1/2 |d |/dt = -3/4 (GM²/R²)dR/dt = 4R²T⁴. This is a differential equation for R(t). But if we integrate in this form, with appearing explicitly, it is easy to make a mistake and difficult to see the physics. We eliminate via Lo = 4Ro²To⁴. The resulting dimensionless differential equation is oRo³R^-4dR/dt = -4/3 (T/To) ⁴, which integrates to Ro³ (R_f^-3-R_i^-3) = 4(T/To) ⁴ (t_f-t_i)/o, where subscript i and f stand for initial and final values. This is to be evaluated for initial and final luminosities LLo and Lo, respectively. Consequently, R_i/Ro 1, so that R_i^-3 may be neglected, and R_f/Ro = (To/T) ². The desired time interval becomes 1/4 (T/To) ², that is, o = 1/16 o, about 2 million years.

3. Exact integration at constant L: Now we have RooR^-2dR/dt = -4/3 or Ro(R_f^-1-R_i^-1) = 4/3 (t_f-t_i)/o, where now R_i/Ro = (To/T) ² and R_f/Ro = 1. Therefore, this interval is 9/16 o, about 17 million years.

4. The entire contraction time, 5/8 o is about 19 million years. It is dominated by the last contraction stage. The order of magnitude estimate, o, is quite satisfactory.

Interpretation: The Sun starts shining as a protostar at L 10²Lo, T 3x10^3oK. At high luminosity the contraction is fast. More time is spent once the Sun has reached lower L.

   Clearly, this computation oversimplifies the actual evolution of the Sun, which must be evaluated by detailed computer programs. For instance, the small wiggle in the Sun's track, just before reaching the main sequence, is caused by the nuclear fusion of deuterium, which "turns on" at a lower temperature than proton fusion. Since there is only little deuterium in the solar gases, this nuclear fuel is soon exhausted.

   Stars are born all the time. But we have only a small chance to catch them in very brief stages of their lives, such as when they first contract at high luminosity. (If you could search your entire city, how many babies less than one week old would you find, relative to the whole population?) The later stages of contraction take longer. We have a better chance to observe stars in these stages. Indeed, stars in the later contraction stages are now observed. Hubble Space Telescope observed that many contracting stars are surrounded by a disk of gas and dust, a suitable place to form planets.

Didactics:

1. Check that students have written the formula for o correctly before they proceed to construct the differential equations. For an order-of-magnitude estimate, factors like 3/5 are irrelevant.

2. Students find it difficult to eliminate and will keep it for several steps before simplifying the algebra. They need guidance to look for ways to make the algebra simpler and more understandable. The teacher's effort is worthwhile because it is a very useful habit for students to look for mathematical simplicity.

3. At any one time, the star has the gravitational energy = -[GM(r)/r]dM(r). With dM(r) = 4r²dr and uniform density , the result is = -3/5 GM²/R. The coefficient 3/2 used here represents the central condensation of gas and the deeper potential well in real stars.

4. There is a very basic reason that stars without nuclear energy must gradually shrink and must radiate. These stars, being in hydrostatic equilibrium at any one time, must satisfy the Virial Theorem = -2K , where for stars K is the total thermal energy within the star (see the next paragraph). As gravitational energy is gradually released, only half can go into thermal energy. The formula L = 1/2 d ||/dt represents the other half of the energy, which must be lost as radiation. Conversely, given that a protostar radiates, it must shrink. All this changes when nuclear fusion provides energy and d ||/dt = 0.

      Proof of the Virial Theorem for a star supported by gas pressure, that is, by thermal motions. Assume a spherically symmetric gas (such as the Sun) satisfying the equation of hydrostatic equilibrium. We chose to evaluate r (dp/dr) 4r²dr = -g(r)4r³dr. Evaluate the integral on the right side using g(r) = GM(r)/r² and dM(r) = 4r²dr. The integral becomes GM(r)/r]dM(r) = . Evaluate the integral on the left side using an integration by parts (with zero for the terms evaluated at r = o and r = R) and then use p = n < p_xv_x > = n < 1/3 mv² >, where the sum extends over electrons and protons and the 1/3 comes from averaging over a nearly isotropic velocity distribution. The integral becomes - nm < v ² > 4 r²dr = -2K . A general proof of the Virial Theorem is in the Appendix.

5. Students may question why we can have the star in hydrostatic equilibrium and yet say that it is gradually contracting. If a star is not in hydrostatic equilibrium, pressures that are not balanced will create sound waves. The sound waves can cross the star in a matter of hours or days. Equilibrium is re-established when the sound waves have adjusted the pressures throughout the star and have been dissipated by viscosity. Hours or days are a short time compared to the contraction time o. (An exception are stars that pulsate. Their periods of pulsation are of the order of the time needed for soundwaves to cross the star.)

6. Students' problem: A "brown dwarf" (a star which never reaches fusion temperature in the center) with M = 0.08Mo contracts for 1.2x10¹⁰years with constant T=0.5To. What is its L at the end of that time? (L1.1x10^-4Lo) Finding such a star and measuring its L might tell us the age of the Milky Way Galaxy.

7. Students' problem: How long does it take a massive protostar, M=20Mo, to contract at constant L = 10⁵Lo from T = 3x10^3oK to T = 3.3x10^4oK? (3x10^-4o 10⁴yr)

The problem: Given Lo = 4x10²⁶w and Einstein's relation E = mc², how much mass does the Sun convert to energy and lose per second ? At this rate, how many years would it take to lose all of the Sun's mass, Mo=2x10³⁰kg?

   Given Lo, how much mass of hydrogen is converted into helium per second? (The resulting helium has 0.7% less mass than the original hydrogen.) At this rate, how many years would it take to convert one solar mass of hydrogen into helium?

   Actual stars convert 10% of their hydrogen into helium during their lifetime. How many years can one expect for the Sun's lifetime?

The setting: On the time scale of a human lifetime, stars appear to last "forever". Of course, stars must change with time because they exhaust their nuclear fuels. But, fortunately for the evolution of life on Earth, the Sun is very nearly constant for some 10¹⁰years. This is the period during which its luminosity is derived from the conversion of hydrogen to helium. The most common reaction chain is: p+pd+e⁺+ (1.44 MeV) and d+p³He+ (5.49 MeV); after both reactions have occurred twice, then ³He+³He⁴He+2p (12.9 MeV). Here p = proton, d = deuteron (proton plus one neutron), and the energy release is expressed in MeV = 1.6x10^-13j. The most difficult step is the first step, the pp fusion, because the two protons repel each other by their electrical charges and yet they must be near each other long enough for a weak interaction to take place. The interaction requires quantum-mechanical tunneling. The Sun's structure is adjusted so that the temperature near the center is sufficient for the pp fusion to work. At the center of the Sun, approximately T = 1.6x10^7oK and =1.6x10⁵kg/m³.

The solution: The mass loss per second dM/dt = Lo/c² = 5x10⁹kg/s. The related time is Moc²/Lo = 1.5x10¹³years. If only 0.7% of mass is converted, the same amount of mass lasts only 10¹¹years. The expected life of the Sun is 10% of that, namely about10¹⁰years.

Interpretation: The solar mass loss is enormous on a human scale, 5 million tons per second. Yet the life of the Sun is very long compared to a human lifetime.

   Why do stars use only 10% of all hydrogen? Once the hydrogen is exhausted near the center and fusion occurs only slightly further out, the gas at the center becomes isothermal. The pressure gradient is reduced. If the mass of the isothermal core exceeds 0.1 of the star's mass, the core cannot be supported against gravity. This limit is named after Schoenberg and Chandrasekhar.

   A correction is necessary here : The Sun is not exactly constant. After some hydrogen is used up at the center, the hydrogen fusion must proceed more vigorously to provide enough pressure, and the Sun becomes slightly more luminous with time . When the Sun was young 4x10⁹years ago, the Sun was less luminous than now by 30%. This creates a puzzle : It seems that the Earth 4x10⁹years ago should have been too cold for liquid water! Yet we know that primitive (one-celled) life occurred soon thereafter. Presumably this primitive life needed liquid water. What extra heat kept our atmosphere warm enough for liquid water and life at that time? We do not know.

Didactics: The pp reaction is an extremely improbable reaction. The electrostatic repulsion presents a 1-MeV barrier through which the typically 1-KeV protons must tunnel. Just how improbable the pp reaction is can be seen from the end result in the Sun: Since most protons near the center have fused after 10¹⁰ years, every proton near the center must move about and collide with other protons for 5x10⁹years before it has even a 50% chance of fusing with another proton.

   This problem is placed here to complete the story of the Sun's life. The physics here are probably the conceptually most complicated physics of the section. The next two problems return to the (radiation) physics of ordinary thermal gases. If the student has not done problem 5.5, then the one-step integration used here may need some elaboration.

Physics Introduction: Pressure is defined as momentum flux across a surface, P = p_xv_xf(p)d³p, where f(p) is the distribution function, f(p)d³p = n. For an isotropic distribution, p_xv_xf(p) 1/3 pvf(p) and d³p 4p²dp on integration over angles. In most situations, the distribution is thermal, P = 1/3 m < v² > = nkT, where n is the total particle density. However, when electrons are very densely packed, their momentum is influenced by the Heisenberg uncertainty principle xp_xh/2, where x n^-1/3, and now n = electron density. The higher n, the smaller the space available and the higher must be the momentum of most electrons. The distribution f(p) = (3/4)n/p_F³ for 0 < p < p_F = (3²n) ^1/3 (h/2) implies that all quantum states are filled by electrons up to momentum p_F, and higher momentum states are empty.

The problem:

1. Assume non-relativistic electrons, v = p/m. Evaluate the pressure as a function of n. Write n (the electron density) in terms of mass density , assuming the star is made of carbon (one proton and one neutron) per electron. Use this pressure (including all its physical constants) in the equation for hydrostatic equilibrium and do a one-step integration, by replacing r R, etc., still keeping all the physical constants. This should yield a relation between M and R. Evaluate R for M = 0.5Mo and compare with the Earth's radius (6x10³km).

2. Repeat the procedure for relativistic electrons, v = c. Show that the R cancel and you get only one possible value of M.

The astrophysical setting: The Sun in middle age, as a main sequence star, maintains T(center) high enough for fusion of H He plus energy. When the H at the center becomes exhausted, the central gases gradually lose support and must shrink, heat up, and thus regenerate support. The gases heat up until they reach T 2x10^8oK. At that temperature, He nuclei (two positive electric charges) overcome their electrostatic repulsion so that a new version of fusion occurs, 3He C plus energy. This fusion provides a new source of energy to support the center. Meanwhile, the outer layers of the Sun have expanded so that the Sun is a red giant . Its huge luminosity implies the exhaustion of the He fuel in mere millions of years.

   What happens when the He is exhausted at the center? Now the gases at the center are so dense that a new pressure is available to support the gas against gravity, the so-called electron degeneracy pressure. This pressure is strictly a quantum effect: the less physical space each electron occupies, the more momentum space it must occupy, according to Heisenberg's uncertainty principle. When electrons are squeezed into higher density, energy must be provided to give higher momenta to the electrons. This has nothing to do with temperature. Because of this new electron pressure, the solar center can adjust (actually rather violently) to support the rest of the star even when there is no more nuclear fusion at the center.

   Once the Sun has been a red giant for a few million years, the Sun sheds its outer layers. The expelled gases become observable as a "planetary nebula". (Such nebulae appeared like planets in early telescopes, but they have nothing to do with planets physically.) While the nebula expands, the stellar core becomes visible as a very hot star. In a mere few thousand years, its surface temperature decreases until the star becomes what we call a "white dwarf". (The "white" is also an historical misnomer). The Sun at age 10¹⁰ years will become a white dwarf, cooling off ever more slowly, forever.

   Probably the most famous white dwarf is Sirius B, the companion to the bright star Sirius. It was the first white dwarf to be detected visually in 1862, 18 years after Sirius was known to be a binary system. (Sirius B has M = 1Mo, surface temperature about T = 2.7x10^4oK.)

The solution

1. Plugging in v = p/m_e and integrating over p up to p_F yields P = n p_F²/5m_e, where n = /2m_p, so that p is proportional to ^5/3. Used in P/RGM/R²with M = (4/3) R³, one gets R = M^-1/3G^-1 (h/2)²m_e^-1m_p^-5/3 (9/8) ^2/310^-1 = 1.5x10³km, about 1/4 R(Earth).

2. Plugging in v = c and integrating over p up to p_F yields P = 1/4 n c p_F, so that P is proportional to ^4/3. In p/RGM/R² the R cancel, and M = (3/64) ^1/2 (hc/2G) ^3/2m_p^-2 = 0.15Mo.

Interpretation: The radius of a nonrelativistic white dwarf decreases as the mass increases, quite unlike ordinary objects or even ordinary stars. Evidently, it is difficult to support an increasing mass. Moreover, as the density increases, the electrons with momenta near p_F become relativistic. Subsequently, further mass addition causes less pressure increase, and at some point no equilibrium is possible. The "Chandrasekhar limit" for the maximum mass of a white dwarf is 1.44Mo (for negligible rotation and for 2 nucleons per electron, as for carbon). Our estimate was too small by a factor ten.

   The appearance of h in the estimates for the radius and for the maximum mass of a white dwarf makes the white dwarf a macroscopic quantum object. The quantum nature of white dwarfs and their maximum mass due to relativity were first recognized and worked out by S. Chandrasekhar.

   Isolated white dwarfs cannot contain hydrogen because any such hydrogen would have fused long ago (see didactic 6). If new hydrogen falls onto a white dwarf from a binary companion, the fusion is explosive. A supernova or a (recurrent) nova occurs, depending on the rate of accretion. Supernovae of this type are all very similar. They are visible far out in the Universe. Current research suggests that they provide the most accurate distance scale for distant clusters of galaxies. And that scale, in turn, is needed, together with other data, to determine the age of our Universe.

Didactics:

1. The highest energy of non-relativistic degenerate electrons, 1/2 p_F²/m_e, is called the Fermi energy, hence the subscript F.

2. In a mixture of electrons and nuclei, such as in white dwarfs, the degenerate electrons provide the main pressure, while the non-degenerate nuclei maintain a Boltzmann velocity distribution. The ions are supported against gravity by an electric field (due to a very tiny charge separation) coupling the ions to the electrons, which are supported by their pressure.

3. For a non-relativistic white dwarf, the one-step integration is adequate. The factor 1/5 in P comes from p⁴dp. The actual radius for a 0.5Mo white dwarf is about 10⁴km..

4. The factor (hc/2G) ^1/2 = 2.2x10^-8kg appearing in the maximum mass is called the Planck mass.

5. The cancellation of R for a relativistic white dwarfs means that there is no equilibrium radius. A slight expansion leads to permanent expansion, a slight contraction leads to collapse (until the physics change). A fundamental explanation why a white dwarf cannot be supported by relativistic degenerate electrons, based on the Virial Theorem, can be found in didactic 3) of problem 4.6.

6. Within a white dwarf, heat conduction is high, and the temperature of the nuclei is nearly uniform. The thermal energy of the nuclei gradually leaks out through a non-degenerate surface layer. The interior temperature can be estimated from L = (4r²)(c/3n)adT⁴/dr (see problem 5.7). A one-step integration with dr 10^-2R yields T 10^{7 o}K. At this interior temperature, there can be no hydrogen left, as it would have fused long ago.

Required: Students must do problem 5.5 first.

Physics introduction: The diffusion of some quantity Q is given by Q/t=/x[DQ/x] , where the diffusion coefficient is D =1/3 v. Here, v = root-mean-square particle velocity and the mean free path =1/(n), with a collision cross section and n the density of particles that are targets for collisions. The flux of the quantity Q is F= DQ/x . F is independent of x when Q/t = 0.

The problem: In a spherically symmetric star, radiation steadily diffuses outwards through each spherical layer of the star without accumulating anywhere (except near the center, where nuclear fusion adds heat which is turned into radiation). The diffusing quantity Q now represents the radiative energy density, Q = aT⁴, and F = L/(4r²) = DQ/r. Write down an equation for L in terms of n, T, dT/dr, and the constants c, a, and .

   Now do a one-step integration with r R/2, kT kT(center) = p(center)/<n> and use p(center) as derived in problem 5.5. Show that the dependence on R disappears and derive how L depends on M .

   Given the solar lifetime of 10¹⁰years and that stars use only 10% of their H during their lifetimes, estimate the lifetime of a star with M = 20Mo.

The setting: Most stars, including the Sun, are found to be "main-sequence" stars. All main-sequence stars derive their energy from fusion of hydrogen into helium. The hotter and more luminous main sequence stars turn out to be more massive stars. We do not know why they obtained more mass during star formation, but there they are. Among stars whose masses can be measured (see problem 1.4), L is roughly proportional to M³. Stars with the largest masses M 10²Mo have luminosities L 10⁶Lo. If stars during their lifetimes all consume 10% of their hydrogen (see problem 5.4), the luminous stars must consume their nuclear fuel extremely rapidly compared to the Sun.

Nucleosynthesis: Massive stars are of special interest because they end their life by exploding as a supernova. Nuclear reactions during the explosion generate most of the elements. The elements that compose most of Earth and us humans, mainly carbon, nitrogen, oxygen, silicon, and iron, have all been made in such exploding stars. The oxygen we breathe was made in an exploding star. The oldest stars in our Galaxy, about 1.3x10⁹years old, were formed before any stars had exploded, and they may be predicted to contain none of these elements. Indeed, they consist (almost) entirely of hydrogen and helium. The hydrogen and helium are left over from the Big Bang origin of the Universe.

The solution: In the diffusion equation dealing with radiation, v c. Therefore, L = (4r²)(c/3n)a(dT⁴/dr). Since the problem asks only for proportionality, we ignore not only all factors such as 2 or , but also factors such as c, a, , or G. With r R and (from problem 5.5) T p(center)/<n> (M²/R⁴)(R³/M) = M/R, we obtain L R² (R³/M)(M/R) ⁴/R = M³.

   If we calibrate the relation using the Sun, then L/Lo=(M/Mo) ³. The lifetime of a star with structure similar to the Sun is proportional to the amount of nuclear fuel divided by the rate the fuel is used, thus to M/L and therefore to M^-2. A star with M = 20Mo has a lifetime about 2.5x10^-3 of the Sun's lifetime, or about 25 million years.

Interpretation: The derived relation is actually quite good. The observations fit reasonably to L proportional to M^3.3. The one-step integration works well because all the stars on the main sequence have a similar degree of concentration.

   The main simplification in our solution is the assumption that is the same constant for all the stars. The atomic absorption of radiation is a strong function of both temperature and density. It differs significantly between stars of different mass. Within the Sun, the mean free path of an x-ray photon near the center is on the order of cm. As the radiation diffuses outward into less hot gases, the mean free path increases until it is of the order of 100 km for a visible photon near the surface .

   Another structural factor ignored in the problem is convection. Where is too high, dT/dr becomes so high that convection starts and carries the heat upward. Massive stars are convective near the center; the Sun is convective for r > 0.7Ro; and low-mass stars are entirely convective. The transport of heat by conduction is negligible in main sequence stars.

   Finally, the manner of hydrogen fusion changes for the more massive stars. Carbon, nitrogen and oxygen nuclei are used as catalysts to speed up the fusion of hydrogen into helium. Because the electrostatic repulsion of these nuclei is larger, this "CNO cycle" requires higher central temperatures than exist in the Sun. Indeed, with T(center) proportional to M/R and, observationally, R proportional to M^0.6, the more massive stars are hotter at the center. Because the tunneling probabilities increase very rapidly with temperature, even a small increase in central temperature, above that of the Sun, allows a much greater luminosity.

Didactics:

1. The physics introduction presented diffusion in a coordinate x. Strictly, we should write the diffusion equation appropriate to spherical symmetry, Q/t = div(Flux). In the problem, the spherical symmetry was introduced by defining F = L/(4r²).

2. Problem 5.3, didactic 6) deals with the time needed for protostars of various masses to reach the main sequence. This stage is much shorter for massive stars than for the Sun. In fact, every stage of a massive star's life is much shorter than the equivalent stage in the Sun's life. At the end, the massive stars explode violently and leave behind a neutron star or a black hole.

3. How long does radiation take to diffuse out of the Sun? We obtain this answer directly from problem 5.3: the protostars contract at a rate given by how fast radiation can leave the star. According to Problem 5.3, the time is roughly 0 = GMo²/RoLo. The radiation now leaving the Sun started out near the center some 30 million years ago.

Required: Students must do problems 5.5 and 5.7 first.

The problem: In problem 5.7 you derived the equation L = (4r²)(c/3n)adT⁴/dr. The right side, derived in terms of radiation energy density aT⁴, can also be expressed in terms of radiation pressure P(rad) = 1/3 aT⁴. Now suppose that gas pressure is much smaller than radiation pressure so that dP(rad)/dr dominates the equation for hydrostatic equilibrium. Eliminate dP(rad)/dr between the equation for L and the equation of hydrostatic equilibrium. . Eliminate in the equation of hydrostatic equilibrium by expressing it in terms of electron density n_eassuming pure ionized hydrogen, and assume that electrons scatter the radiation, so that n = n_ein the equation for L. Set M(r) = M for the outer parts of a star, and obtain an equation for L in terms of only M, , and constants. Evaluate this L using the Thomson scattering cross section _T = 6.6x10^-29m². Derive L/Lo in terms of M/Mo.

The setting: If one makes models of stars on the main sequence, one finds that the hotter, more massive stars involve an increasing ratio of radiation pressure to gas pressure, throughout most of the star. The gradient in radiation pressure acts outward, as does the gas pressure. If the gradient in radiation pressure becomes sufficient, it overcomes gravity and the star can no longer be in hydrostatic equilibrium. This stage is reached when L equals the "Eddington luminosity", evaluated in this problem. Stars with higher luminosity (depending on M/Mo) cannot be stable.

The solution: The equation of hydrostatic equilibrium is dP(rad)/dr = -GM/r². If radiation scattered by electrons supports the gas, then dP(rad)/dr = (L/4r²)(n_e/c). Also, = n_em_p. On canceling similar terms, L/Lo = 4GMm_pc/Lo = 4.5x10⁴M/Mo.

Interpretation: The Eddington luminosity is the maximum luminosity any static star (of pure hydrogen) can have. For stars on the main sequence with L/Lo = (M/Mo) ^3.3, this limit is reached for M/Mo about 100. Indeed, no steady stars have been measured to have larger M.

   Many stars with L near the Eddington luminosity are observed to have strong winds. This confirms that gravity binds the gas only weakly. For example, the star Eta Carinae has M = 150Mo, L = 6x10⁶Lo and has a huge mass loss rate. In the 1830's it brightened to be the second brightest star in the sky. Altogether it seems to be at the edge of stability. The "Pistol" star, recently observed by Hubble Space Telescope, has an unexplained L = 10⁷Lo. A surrounding nebula 4 light years across contains about 10Mo of gas, which was probably ejected in two eruptions 4x10³ and 6x10³years ago.

   The Eddington luminosity applies to any radiating object. For quasars with a black hole of mass M = 10⁹Mo, the Eddington luminosity is around 4x10¹³Lo. Indeed many quasars have such luminosities, far more than the luminosity of the galaxies whose centers they occupy. Quasars that exceed the Eddington luminosity cannot be static. Typically, gases from equatorial accretion disks are funneled into the powerful outflowing polar jets that help make quasars so spectacular.

Didactics:

1. The Eddington luminosity was derived for pure ionized hydrogen. For stars with 75% H and 25% He by mass (so that n_e /m_p), L/Lo = 3.8x10⁴M/Mo.

2. The coefficient 1/3 in the diffusion coefficient D of problem 5.7 was chosen to match diffusion of nearly isotropic radiation (mean free path short compared to the pressure or temperature scale height). It stands for the hemispheric average of cos². The same factor 1/3 appears in the radiation pressure. Indeed throughout most of the star (all except the photosphere) the radiation is very nearly isotropic.

3. There is a good physical reason that stars cannot involve radiation pressure much larger than gas pressure. In equilibrium, according to the Virial Theorem, = -2K (Problem 5.3, didactic 4). Imagine the whole star expanding or contracting. changes in proportion to 1/R. If internally P changes in proportion to ^a R^-3a, where a = 5/3 for an ideal monatomic gas, then K, an integral of pressure over the volume of the star, changes in proportion to R^3-3a. If a = 5/3, or more generally if a > 4/3, making R smaller makes K increase faster than . Physically, that means the enhanced pressure will cause the star to expand again. As a result, the radius of the star will oscillate. The predicted period of oscillation is of the order of the time needed for a sound wave to cross the star. Indeed, many stars are observed to oscillate according to this prediction. However, if a = 4/3, then and K change with R in the same manner. Therefore, = -2K remains satisfied even as R changes. In principle, the star can be in equilibrium at any R. But if equilibrium is violated even slightly, the star will either continue to expand forever or continue to contract forever. Therefore, no stars with a = 4/3 can exist. Since radiation has a = 4/3, no star can exist supported purely by radiation. Similarly, in problem 5.6, a massive white dwarf with relativistic electrons has a = 4/3 and cannot exist.