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Chapter 3 Accretion Disks

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Chapter 3 Accretion Disks Chapter 3 Accretion disks 3.1 Accretion Accretion of matter onto compact objects is perhaps the only mechanism that gives sufficient energy to feed the most luminous objects in the Universe. Examples can be found in different classes of narrow binary systems and nuclei of active galaxies and quasars. How much energy can be produces by accretion, is shown by the following. The amount of energy ∆E gained by accretion of a mass m at a star of mass M and with radius R is ∆E = GMm/R. For a typical neutron star we have M = 1 M⊙ and R = 10 km and thus ∆E is about 1.3 1016 J per swallowed kg of matter! Most of this energy is released again as electromagnetic× radiation. For comparison: burning wood gives 1.6 107 J/kg, and nuclear fusion (the hydrogen cycle) gives ∆E 0.007mc2 or 6 ×1014 J/kg. It is clear that the more compact a source is (for larger≃ M/R), the× more efficient is the energy conversion process. Therefore in particular accretion on neutron stars and black holes is very important. For white 4 dwarfs (M = 1 M⊙, R = 10 km) nuclear fusion is 25–50 times more efficient than accretion. But even in white dwarfs accretion plays a dominant role: the time scale for nuclear fusion is so short (nova outburst) that during the main part of the life of the white dwarf accretion is the most important energy source. Binaries with accretion onto a white dwarf are called cataclysmic variables. They are found frequently in our Galaxy. In general we can distinguish accretion onto two types of objects: accretion onto Galactic objects (black holes, neutron stars, white dwarfs) of 1 M⊙, and accretion ∼ 6 9 onto extragalactic supermassive black holes (nuclei of galaxies) of 10 –10 M⊙. The luminosity that is released around an accreting source cannot be unlimited: with increasing luminosity the radiation pressure increases, until this is so large that all matter that wants to fall onto the compact object is blown away again. The radiation pressure works primarily on electrons (Thomson scattering): for protons scattering is a factor of me/mp smaller. Because of the electrostatic interaction between electrons and protons however, the protons must follow the electrons. The radiation force on a pair (electron + proton) is therefore given by σTS/c, where S is the incoming flux (W m−2). The total luminosity is L =4πr2S, and therefore the net force F onto an electron-proton pair is given by (GMmp LσT/4πc) F = − (3.1) r2 If that quantity becomes zero, the maximum luminosity is reached. This defines 90 the so-called Eddington limit: 4πGMmpc 31 LEdd = = 1.3 10 (M/M⊙) W (3.2) σT × The Eddington luminosity therefore increases with increasing mass. For M = 9 40 13 10 M⊙ one gets the typical luminosity of a bright quasar: 10 W (or 3 10 L⊙). The actual luminosity L of an accreting source can be expressed in the× amount of matter that is being accreted per unit of time. For a neutron star or white dwarf we have L = GMM/R˙ (3.3) This can be written also as 26 L =1.3 10 M˙ 13M0/R7 W (3.4) × or 29 L =1.3 10 M˙ 13M0/R4 W (3.5) × 13 where M˙ 13 is the amount of accretion in units of 10 kg/s, M0 the mass of the 0 7 star in units of 10 M⊙ and R7 the radius in 10 m. (3.4) is characteristic for accretion onto white dwarfs, (3.5) is more relevant for neutron stars. The left hand side of both equations gives a reasonable impression of the typical luminosity of both types of objects. For black holes, on the other hand, we can write 2 L =2ηGMM/R˙ S = ηMc˙ (3.6) 2 where the Schwarzschild radius (Rs = 2GM/c ). Further, η is the efficiency of the energy release. Later we will show that η is usually of the order of 0.1 (compare to η =0.15 for accretion onto a neutron star and η =0.007 for hydrogen fusion). For 8 a black hole of 10 M⊙ radiating at the Eddington limit, one finds a Schwarzschild radius of 3 1011 m (2 astronomical units – an Earth orbit) and an amount of × 23 accretion of 1.4 10 kg/s or 2.3 M⊙/year. A bright quasar therefore must swallow × the equivalent of a few stars per year in order to maintain its luminosity! It is also possible to say something about the characteristic spectrum of the source. If all radiation is emitted as black body radiation, Stefan-Boltzmann’s law is applicable: 2 4 L =4πR σTb (3.7) −8 −2 −4 where σ is the constant of Stefan–Boltzmann (σ 5.67 10 W m K ). Tb is the temperature of the black body. Therefore we have≃ × 1 2 4 Tb =(L/4πR σ) (3.8) We can also define the temperature Tth that the matter would have if all potential energy would be converted into thermal energy. Per electron–proton pair the gained potential energy is GMmp/R (note that me mp) and the thermal energy is 2 3 kT , yielding ≪ × 2 91 Tth = GMmp/3kR (3.9) If the accretion flow is optically thick, the radiation will reach thermal equilib- rium with the matter and we have Trad = Tb. However, if the accretion flow is optically thin, then all radiation can escape and we have Trad = Tth. In general we therefore have: Tb < Trad < Tth (3.10) ∼ ∼ We now approximate for the emitted photons E kTrad. For a white dwarf, 8 ≃ neutron star and 10 M⊙ black hole at the Eddington limit we therefore find using (3.4), (3.5), and (3.2), respectively: white dwarf: 3 eV < E < 50 keV neutron star: 0.6 keV ∼< E ∼< 50 MeV AGN: 20 eV ∼< E ∼< 50 MeV ∼ ∼ 3.2 Gas dynamics Matter falling onto a compact object must obey a number of boundary conditions. We discuss these conservation laws in some detail. First, there is mass conservation: ∂ρ + ρv = 0 (3.11) ∂t ∇· Further we have conservation of momentum: ∂v ρ + ρ(v )v = p + f (3.12) ∂t ·∇ −∇ The term with ρ(v )v represents the convection of matter. This terms allows for stationary solutions·∇ with v = 0 for those situations where ∂/∂t = 0. The force term with f can be composed6 of different components: a gravity term f = ρg, − magnetic forces j B or viscosity. As the third conservation law we have the energy equation: × ∂ 1 2 1 2 ( ρv + ρǫ)+ [( ρv + ρǫ + p)v]= f v Frad q (3.13) ∂t 2 ∇· 2 · −∇· −∇· Here ǫ is the internal energy per unit mass, and is given by 3 ǫ = kT/µm (3.14) 2 H where µ is the so-called mean molecular weight. It should be noted that the designation ”mean molecular weight” is misleading, but that word is used widely in the community, unfortunately. In a plasma, we have in general few molecules. More accurately, µ is the mean mass per particle expressed in atomic mass units, so that one can write the mass density ρ (in kg s−1) as ρ = µmHntot, (3.15) 92 with ntot the total number of particles (ions and electrons). For a neutral plasma with cosmic abundances, we have ntot = 1.10nH and µ = 1.30; for a fully ionised plasma with cosmic abundances, ntot =2.30nH and µ =0.62. The left hand side of (3.13) is very similar to the continuity equation (3.11), with 1 2 as a difference that the conserved quantity now is not given by ρ but by 2 ρv + ρǫ. The last term within the square brackets is the force exerted by the pressure. The term with Frad is the gradient of the radiation pressure. This term generally depends in a complicated way on the specific intensity of the radiation Iν. In some special cases, however, it can be determined quite easily: for optically thin matter we have ∞ τ 1 : Frad = ǫνdν (3.16) ≪ ∇· Z 0 −3 −1 where ǫν is the emissivity (W m Hz ). For this one can take for example a cosmic plasma as treated before in these lectures. In the other extreme case the optical depth is very large (like in the interior of a star). In that case Frad approaches the black body flux and we can use the Rosseland approximation: 16σ 3 τ 1 : Frad = T T (3.17) ≫ −3κRρ ∇ Here σ is again the universal radiation constant of Stefan–Boltzmann and κR is the Rosseland opacity. For Bremsstrahlung we have (Kramers law): 21 −7/2 2 −1 κR =6.6 10 ρT m kg (3.18) × For pure Thomson scattering we have 2 −1 κR = κes =0.040 m kg (3.19) Note that in (3.18) ρ is expressed in kg m−3 and T in K. Finally the term with q in (3.13) is the heat conduction coefficient: for classical, unsaturated heat flow one∇· can approximate this well with: q 10−17T 5/2 T W m−3 (3.20) ≃ ∇ Finally we have of course the ideal gas law: p = ρkT/µmH (3.21) An important illustrative example is given by stationary flows. In that case all terms with ∂/∂t in the conservation laws can be ignored. We ignore here also the radiation pressure and heat conduction. It then follows that: ρv = 0 (3.22) ∇· ρ(v )v = p + f (3.23) ·∇ −∇ 1 [( v2 + ǫ + p/ρ) ρv]= f v (3.24) ∇· 2 · We now substitute (3.22) into (3.24) and find 93 1 f v = ρv ( v2 + ǫ + p/ρ) (3.25) · ·∇ 2 On the other hand, it follows from (3.23) that (check this): 1 f v = ρv ( v2)+ v p (3.26) · ·∇ 2 ·∇ Combination of (3.25) with (3.26) immediately gives: ρv (ǫ + p/ρ)= v p (3.27) ·∇ ·∇ We can write this as 1 v [ ǫ + p ] = 0 (3.28) · ∇ ∇ρ It follows that along a stream line dǫ + pd(1/ρ) = 0.
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