1 / 19

5.7 COMPOSITION OF THE GAS

5.7 COMPOSITION OF THE GAS. We need to relate the number of free particles to the density r . If we have a mixture of ideal gases each with a partial pressure P i and number density n i then. The total pressure. Where n is the number of particles per unit volume. Generally we write.

francesca-key
Download Presentation

5.7 COMPOSITION OF THE GAS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 5.7 COMPOSITION OF THE GAS We need to relate the number of free particles to the density r. If we have a mixture of ideal gases each with a partial pressure Pi and number density ni then The total pressure Where n is the number of particles per unit volume Generally we write If we use m as the mean molecular weight, then So that Giving If the gas is fully ionised and the mass fraction for the isotope (z,A) is Fi, then including electrons we have If we crudely divide the composition into three groups : } Hydrogen with a mass fraction of X Helium with a mass fraction of Y Metals (the rest), mass fraction of z Then The number density of particles is NOTE the total number of particles is the sum of all types : i.e. X+Y+z = 1 Typical values are PHYS3010 - STELLAR EVOLUTION

  2. 5.8 SIMPLIFIED STELLAR MODELS We have derived a series of differential equations which describe the major physical processes which take place within stars. In principle it should be able to derive solutions to these equations, provided appropriate boundary conditions can be obtained. Unfortunately exact solutions are not readily available due the fact that there are more variables that equations. Some compromises, however, can be made so that a good overall understanding of the interiors of stars can be obtained. The solutions as a consequence are not exact. THE LINEAR MODEL r We assume the density is a linear function of stellar radius Actual Assumed A ‘reasonable’ approximation. r Now Is the mass of the star As the mean density throughout the star Giving PHYS3010 - STELLAR EVOLUTION

  3. PRESSURE THROUGHOUT THE STAR Using the equation of hydrostatic equilibrium At the surface P = 0, and r = R Thus the central pressure PHYS3010 - STELLAR EVOLUTION

  4. TEMPERATURE RELATIONSHIP Ignoring radiation pressure Using the gas law Now at r = 0 we have T =TC and since For the Sun Thus NOTE107 K gives 3/2 kT ~ 2 10-16 J ~ keV X-rays PHYS3010 - STELLAR EVOLUTION

  5. LUMINOSITY RELATIONSHIP Fortunately effectively all the energy generation e, within a main sequence star takes place near to the centre. This simplifies mathematical modelling since for r¹ 0, e= 0 so that L(r) is a constant. The radiative transfer equation is ……….(1) We assume Kramer’s opacity law …….(2) Another value of dT/dr may be obtained by differentiating the expression for T(r) obtained from the gas equation …..(3) If we adopt the rather crude expedient of equating the two expressions for dT/dr at r = R/2 At R/2 equation (3) gives Whereas at R/2 PHYS3010 - STELLAR EVOLUTION

  6. From equation (2) Now from the m(r) integration So that Approximately PHYS3010 - STELLAR EVOLUTION

  7. L has a strong dependence on the composition For pure hydrogen m= 1/2. If the star were to be almost entirely higher z materials the value becomes m~2 • The luminosity has a very strong dependence on the mass of the star. This is why the lifetime of a massive star is considerably shorter that the lifetime of a lower mass star. Since L µ M5.5and the available mass is µ M, the reason is obvious. • The luminosity expression is indicative rather than a precise quantitative relationship. Since a) We have neglected radiation pressure b) The elimination of T by equating dT/dr at one point only in the star implies the system is not in thermal equilibrium NOTE • If we have 2/3H and 1/3 Hethen m=12/19 and m7.5 = 0.0326 Giving This value compares well with the solar value of L0 = 4 1026 • The strong dependence of luminosity on the gravitational constant G (i.e. G7.5) is of cosmological interest. Geological evidence indicates that the sun is at least 4.5 x 1010 years old. If G had been higher in the past all the Hydrogen would have been exhausted long ago. • Strong dependence of luminosity on mass means that massive stars are convective • Repeating the analysis for the opacity being dominated by electron scattering as found in massive stars, and substituting the opacity value k= 0.02(1+x), see later, one obtains i.e. having a lower mass sensitivity. PHYS3010 - STELLAR EVOLUTION

  8. 1/r Coulomb Potential E EC r ro (~10-15 m) Nuclear 6. NUCLEAR SOURCES OF STELLAR ENERGY We have seen from the Kelvin-Helmholtz timescale that gravitational energy is not capable of sustaining the luminosity of main sequence stars for the period of their lifetimes. Chemical reactions fall even shorter. The only know source of energy has to be nuclear fusion. We now explore the physics of this scenario. Our modelling has shown that the central temperatures of main sequence stars is upwards from about 107 K for the lightest and greater for more massive stars. The corresponding energies of thermal particles will be The potential energy between two like charged nucleons as a function of their separation r is shown in the figure. For distances greater than about 10-15 m the repulsive Coulomb 1/r2 force dominates. At distances less than 10-15 m the attractive nuclear force binds the particles together. The height of the Coulomb barrier is It is clear that, in order for thermonuclear fusion to take place, barrier penetration has to assist the process. NOTE : Since the height of the barrier increases as Z1Z2e2 where Z1e and Z2e are the charges of the two reacting nuclei, higher temperatures are required if heavier elements are involved in the fusion process. Furthermore since higher Z elements have higher masses, they are more difficult to accelerate to the required velocities. Low mass stars can only burn nuclei of low atomic weight PHYS3010 - STELLAR EVOLUTION

  9. 6.1 REACTION RATES AND ENERGY GENERATION RATES The energy released by nuclear reactions depends on the physics of the nuclei involved. Nuclear energy is obtained if the mass of the final products of the reaction is less than the initial masses. For example if the reaction starts with particles a and X and finishes with particles b and Y Binding Energy per Nucleon Fusion Fission 100 200 Then the energy release is A value The figure shows schematically the binding energies of nuclei as a function of their atomic number A. The binding energy per nucleon in the figure can also be interpreted as the mass defect of each atom, which can be defined as the difference between the actual mass of the atom M and the product Am1, where A is the atomic weight and m1 the conventional mass unit of atomic weight. Since the most tightly bound nuclei (i.e. lowest mass per nucleon) exist close to the iron group at A = ~ 50 we can obtain energy from fusion up to this point. For A ³ 50 Q = -ve No energy can be supplied by further fusion! This fact leads to spectacular results for the most massive of main sequence stars. PHYS3010 - STELLAR EVOLUTION

  10. The Reaction Rate This is the number of reactions per unit volume per unit time The probability that a reaction will occur if we have a stationary nucleus X bombarded by projectile particles a approaching with velocity v. The probability per unit path length that the reaction will take place is nXs , where nX is the number density of the targets andsthe cross-section for interaction. If there are many incident particles of type a, all at velocity v, then the number of reactions that take place is Inside a star the particles will have a spectral distribution, f(v) in velocities so that Where na is the number density of the projectile particles The Energy Generation Rate Following the nuclear reactions energy is liberated in the form of g-rays, energetic particles and neutrinos. The neutrinos escape but the rest remains within the star to provide heating. The energy liberated is The Mean Reaction Time t is obtained by dividing the the number of projectile particles per m3 by the reaction rate (reactions per m3 per second). The Lifetime of the Nuclear Fuel. The reactions destroy the X type nuclei with a characteristic time scale The above takes only the destruction of X into account from bombardment by a. X can also be simultaneously produced by other processing, and to further complicate matters many reactions involve the collisions by particles of the same type. PHYS3010 - STELLAR EVOLUTION

  11. 6.2 TEMPERATURE DEPENDENCE OF THE REACTIONS The reaction rate is strongly dependent upon the velocities of the particles, and hence the temperature within the stellar gas. To a reasonable approximation the reaction rate versus temperature can be described by the expression Where Gi is the rate at temperature Ti The energy production rate near to Ti is where For the case of hydrogen burning by the p-p chain reaction at Ti ~ 107 Kn ~ 4. However the value of n can vary up to typically 30 or more for carbon burning. The Barrier Penetration The reaction rates are dependent upon barrier penetration. The height of the barrier is The quantum mechanical barrier penetration probability is proportional to the ratio of the Coulomb to kinetic energies kT ~ 2 keV means the non-relativistic approximation. i.e. Using the Uncertainty Principal we obtain In terms of the energy E = 1/2 mv2of the particles we obtain PHYS3010 - STELLAR EVOLUTION

  12. The constant b is given by the expression It is easy to see why thermonuclear fusion in a star requires considerably higher temperatures for higher atomic number isotopes. Apart from the barrier penetration the cross-section for a nuclear reaction will depend upon the ‘area’ within which the penetration can take place. As a first approximation this range can be considered to be the de Broglie wavelength l. Thus the ‘size’ of the particle is So that the cross-section may be written Where S(E) has the units J m2 and is slowly varying function of energy. Using the Maxwell distribution in energy The reaction rate Becomes PHYS3010 - STELLAR EVOLUTION

  13. 6.3 THE GAMOW PEAK G Gamow peak N(E)dE 0 to 1.0 Rate of barrier penetration e -b/ÖE Maxwell Eo E The Gamow peak arises because the barrier penetration increases as the particle energies increase whilst the Maxwell distribution falls off with increasing energy. The result is an intermediate maximum. A nominal value for E0 may be obtained as follows If S(E) ~ constant Thus The energy value of the Gamow peak depends on the product Z1Z2 and the temperature. It is thus the most effective energy for fusion given isotopes at a given temperature. PHYS3010 - STELLAR EVOLUTION

  14. 7. MAIN SEQUENCE STARS 7.1 HYDROGEN BURNING ON THE MAIN SEQUENCE Hydrogen is the most abundant of the materials from which stars are formed, since it is the element with the lowest z value it is the easiest to burn by thermonuclear fusion. In fact it is the process of converting hydrogen to helium that sustains the main sequence phase of stellar evolution. There are a number of ways in which the hydrogen can be converted to helium. 7.2 THE LOWER MAIN SEQUENCE - THE PROTON-PROTON (PPI) CHAIN. The PP chain converts hydrogen to helium as follows Reaction Energy Release Reaction Time MeV 1.44 14 109 y 5.49 6 s 12.86 106 y Total Energy release = 26.2 MeV The rate of energy release is governed by the first (slowest) reaction, and it is this which effectively dictates that the main sequence phase is a long-lived one. (Fortunately for us!) This first link occurs rarely since two conditions must be satisfied : 1) The KE of the protons must be ~20 kT (10-8 of the protons) 2) During the collision (2ro/v ~ 10-21s) period one of the protons must be converted to a neutron However the large numbers of available protons makes the energy production reasonable Note that the deuterons are rapidly turned into He3 PHYS3010 - STELLAR EVOLUTION

  15. Subsidiary Proton Chains As the number densities of isotopes of helium build up in the central regions of the star other options for thermonuclear fusion become possible. PPII PPIII The latter two chains depend upon reasonable quantities of He4 being available. PPI is the major sequence. NOTE • The neutrinos have a very low cross-section s ~ 10-48 m2 so their energy is completely lost from the star. It does not provide any heating. • Neutrino astronomy enables us to study nuclear processing deep within the Sun, although it should be remembered that the neutrinos from b-decay are not in the form of discrete line emissions. PHYS3010 - STELLAR EVOLUTION

  16. Reaction Energy Release Reaction Time MeV 1.45 1.3 107 y 2.22 7 min 7.55 2.7 106 y 7.29 3.2 108 y 2.76 82 s 4.96 1.1 105 y 7.3 THE UPPER MAIN SEQUENCE - THE CARBON CYCLE (CNO) Since the PP chain suffers from the fact that the initial reaction has a very small cross-section. However if sufficient carbon exists within the star and if T ³ 2 107 K then hydrogen burning may proceed as follows : Total energy release ~ 25 MeV NOTE • The carbon and nitrogen are neither produced or destroyed, they act purely as catalysts • The neutrinos take away about 5% of the energy produced • The reaction times are much less than the 1010 y required for PP reaction. Hence if there is a combination of a suitably high carbon concentration and high temperature then the CNO cycle will dominate PHYS3010 - STELLAR EVOLUTION

  17. Temperature Dependence of Reactions L The reaction rates are temperature dependent CNO in the case of PP reactions n = 4 PP Chains So that (for T6 representing units of 106) T For the CNO cycle Since n = 15 For the case of the Sun only about 2 % of the energy is thought to be derived from the CNO cycle. However since thus and we would expect the CNO cycle to dominate for more massive stars. 7.4 LIFETIME ON THE MAIN SEQUENCE Only the core of a star becomes hot enough to ignite the hydrogen so that only about 10% is typically available as main sequence fuel. The hydrogen to helium conversion liberates about The total energy released by a main sequence star will be ~ 0.007 Mc2 PHYS3010 - STELLAR EVOLUTION

  18. The rate of energy loss is given by the luminosity of the star. Thus if the luminosity remains approximately constant throughout the period on the main sequence, the main sequence lifetime will be The linear model gave So that Note the strong inverse mass dependence For the case of the Sun m ~ 12/19 and we have TMS ~ 2 1010 years 7.5 MINIMUM MASS FOR THERMONUCLEAR FUSION In order to become a star a ball of gas has to able to generate a sufficiently high temperature at the centre so that thermonuclear fusion can be self-sustaining. Now we know that the central temperature depends on the mass of the star i.e. Consequently a minimum mass must exist for main sequence thermonuclear fusion. This mass will depend on the composition and the precise fuel mix for the fusion process. If we take Tmin£ 107 K and with m ~1/2 we obtain Mmin ~ 0.01 Mo PHYS3010 - STELLAR EVOLUTION

  19. 7.6 THE MAXIMUM MASS LIMIT As for the case of estimating the mass for the onset of thermonuclear fusion, an estimate of the maximum mass a main sequence star can have is difficult to estimate in a simple way. If we assume that the maximum mass a main sequence star can have is when gravity can just compete with radiation pressure, then a rough working upper mass limit can be evaluated. Earlier we defined the point at which radiation pressure overcomes gravitation as the Eddington Luminosity Limit. Taking the value of luminosity given by the linear model Then form ~ 1/2, and assuming the radii of a large star is R ~ 100 Ro Mmax ~ 34 Mo This value is very approximate, stars are found to exist up to about 50 - 100 Mo PHYS3010 - STELLAR EVOLUTION

More Related