Accretion DisksAngular momentumNow let’s go back to black holes. Black holes being what they are, something that fallsinto one disappears without a peep. It might therefore seem that accretion onto a blackhole would release no energy. It isn’t the case, however, and the reason has to do withangular momentum.Thinking in terms of Newtonian gravity, suppose you have a point source of mass. Askclass: taking gravity into effect, is it possible to hit the mass with another point particle?No, it isn’t. Your aim would have to be perfectly good. In reality, however, there willalways be some lateral component of the motion of the projectile (i.e., there will always benonzero angular momentum). This will cause the projectile to deviate more and more froma radial trajectory as it gets closer; this is a consequence of the `2/r3term in the radialequation of motion. In practice, for compact objects and often even for normal stars, theangular momentum of matter is sufficient to ensure that there is not direct radial accretion.Of course, along the axis of rotation there is less centrifugal support, so this tends to forma disklike structure (the accretion disk).Now, imagine that you have lots of such projectiles moving around the central mass. Ifthey don’t interact, their orbits will be unaffected by the presence of other projectiles. Butif in reality the “projectiles” are streams of gas, they will collide with each other. This willtend to circularize the motion. But will anything happen once the motion is circularized?Disks usually rotate such that each fluid element is moving almost (but not exactly!)in a circular orbit. If there were no interactions between fluid elements, Ask class: whatwould the angular velocity be as a function of radius? Ω ∝ R−3/2, so there is a shearingflow. This means that coupling between adjacent radii exerts a force. Ask class: giventhat the outer parts rotate more slowly, in which direction will the force be and what willbe the effect on the angular momentum and on the movement of mass? Inner tries to speedup outer, giving it a higher velocity. This increases the angular momentum of the outer,decreases the angular momentum of the inner, so net result is that angular momentum istransferred outwards and mass flows inwards (some subtleties, of course). The disk spreadsas a result. Mention: this has similarities to the effect of “shepherd moons” except therethe coupling is purely gravitational.So, gas moving towards a massive object has a tendency to circularize, form a disk,and spread inward and outward. This is an “accretion disk”. If the massive object has asurface, then often the matter spirals in until it hits the surface or interacts with the stellarmagnetic field, whichever comes first. But a black hole has neither a surface nor a magneticfield. However, strong-gravity effects of general relativity mean that the gas can’t spiral allthe way to the horizon, either. This is because of the ISCO, which we have discussed before.Particles spiraling inwards will release little energy inside the ISCO, so the efficiency is justthe binding energy in nearly circular orbits there. This is 6% of mc2for a nonrotating blackhole, and up to 42% for a maximally rotating black hole.Therefore, the accretion efficiency for black holes can in principle be the highestaccretion efficiencies in astrophysics. Unlike stars, for black holes all the emitted energymust come from the accretion disk. We will therefore take a closer look at accretion disks.Now we’ll think more carefully about accretion disks themselves. One model of disks,which has many advantages (e.g., it is robust and does not depend on too many parameters)is one in which the disks are geometrically thin but optically thick. Let’s think about theconditions for this to occur.First, we make the assumption (standard for all models of accretion disks) that thedisk itself has negligible mass compared to the central object. One aspect of that is thatthe gravity in the disk is dominated by the gravity of the central object, not the gas in thedisk itself. On the other hand, pressure forces within the disk are not necessarily negligible.Quantifying this, Ask class: what is the equation of hydrostatic equilibrium? In generalit is ∇P = −ρg, where g is the local acceleration of gravity. We make another standardassumption, which is that the gas is orbiting in almost Keplerian orbits. That means thatwe can focus on the z component (normal to the disk plane) of the hydrostatic equation.If the central object has mass M and the fluid element of interest is an angle θ out of theplane, then (draw diagram) the equation becomesdPdz= −ρGM sin θr2. (1)Let’s call H the half-thickness of the disk; H ¿ r for a thin disk. To rough accuracy,dP/dz = −P/H and sin θ = H/r. ThenPH≈ ρH/r2µGMr¶⇒Pρ= c2s≈µHr¶2(GM/r) . (2)Since v2K= GM/r, this means that the sound speed is cs≈ (H/r)vK. Therefore, thethin disk condition H/r ¿ 1 implies (and is implied by) the condition that the soundspeed is much less than the orbital speed. The sound speed increases with temperature,which increases with luminosity, which increases with accretion rate, so here we have anearly warning that at high enough accretion rates the thin disk approximation is likely tobreak down. “High enough” turns out to mean near the Eddington luminosity. This isnot a surprise, because near the Eddington luminosity radiation forces are strong enoughto significantly modify the behavior of matter, and so having them puff up the disk isreasonable.Temperature distribution of a thin diskNow let’s take a first stab at what the temperature distribution of a thin disk shouldbe. In a moment we’ll do things more carefully, and find a surprising factor of 3.Ask class: suppose that as each fluid element moves inward that it releases its energylocally, and that its energy is all gravitational. How much energy would an element of massm release in going from a circular orbit at radius r + dr to one at radius r? Gravitationalpotential energy is Eg= −GMm/2r, so the energy released is GMmdr/2r2. Warning:presaging here is where the mysterious factor of 3 comes in. It turns out that in reality,far from the inner edge of a disk, the local energy released is a factor of 3 greater than that,due to viscous stresses associated with the transport of angular momentum. However, letus now focus on just the radial dependence, writing dEg∼ GMmdr/r2. That means