# Black hole accretion disk

Does matter in a black hole's accretion disk eventually fall into the black hole? If no new matter is added to disk, how long does it take or the black hole to consume the entire accretion disk?

Matter does fall in, but it's not trivial for this to happen, and the rate varies a lot between different cases.

The thing is black holes don't suck matter in, just as the Sun isn't sucking in the Earth. A test particle in vacuum placed in orbit around a black hole will orbit forever. In an accretion disk, interactions between particles allow for angular momentum to be transferred outward, allowing for the material to fall inward (closer orbits have lower angular momentum). This is generally accomplished by something reminiscent of viscosity, though in black hole accretion the actual mechanism is turbulence driven by electromagnetic fields -- a complicated process with lots of variability.

Because something has to carry angular momentum outward for some matter to fall inward, in general an accretion disk is never fully depleted. However, you are right it will generally deplete over time if not replenished.

Accretion is generally scaled (for less than compelling historical reasons) to the Eddington luminosity of the accreting object, which we'll take to be $$L_\mathrm{edd} = 3\times10^4\ L_\odot \frac{M}{M_\odot}.$$ Infalling matter will accrete at a rate $\dot{M}$ and be converted to power $L$ with efficiency $\epsilon$. If the luminosity is a fraction $f$ of the Eddington value, and if the disk mass $m$ is a fraction $g$ of the black hole mass, some rearranging tells us the timescale for disk depletion is $$\tau = \frac{gc^2M_\odot}{\epsilon f(3\times10^4\ L_\odot)} = \frac{g}{\epsilon f}\ 5\times10^8\ \mathrm{yr}.$$

Values for $\epsilon$ range from percent-level to tens of percent. $f$ can be anywhere from $10^{-8}$ (something like the supermassive black hole at the center of our galaxy, which doesn't really have a disk to speak of) to $10^3$. $g$ could easily be anywhere from $10^{-8}$ (a star shredded into a disk as it falls into a supermassive black hole) to $0.1$ or so. As you can see, there is plenty of variability, and indeed black hole accretion systems fall into a number of extremely dissimilar regimes describing very disparate systems.

Orbiting matter will eventually fall in because viscosity in the disk extracts angular momentum allowing gas to move inwards. As it does so it will release gravitational potential energy, about half of which is radiated and half goes into kinetic energy. Here's another back of the envelope calculation. I'll go with Chris White's notation that the disk mass is $gM$ for consistency.

The jettisoning of angular momentum problem is a problem until you get to the last stable orbit at 3 Schwarzschild radii (for a non-rotating black hole). At this point the material will rapidly fall into the BH. This means we could approximate the whole process as material falling to $3r_s = 6GM/c^2$ and use a conservation of energy argument.

The potential energy lost in moving from infinity to this inner orbit goes into the orbital kinetic energy plus radiated energy. Thus

$$L = \left[\frac{GM}{3r_s} - \frac{GM}{6r_s}\right] \frac{dM}{dt}$$ $$L = \frac{GM}{6r_s} \frac{dM}{dt} = \frac{1}{12}c^2 \frac{dM}{dt} .$$

From there we can say the timescale for its depletion is $$\tau = \frac{gM}{\dot{M}} = \frac{gMc^2}{12 L} \simeq 10^{13} g \left(\frac{M}{M_{\odot}}\right) \left(\frac{L}{L_{\odot}}\right)^{-1}\ \ yr$$

This explicitly shows how the timescale depends on the mass of the disk divided by the accretion disk luminosity (the rate at which it is consumed). It has the added advantage that both of these quantities can potentially be measured.

A big quasar might have a central black hole of $10^9 M_{\odot}$, but emits $10^{13} L_{\odot}$ solar luminosities. Unless $g > 1$ this demonstrates that quasars must be refuelled if they are to live for more than a small fraction of the lifetime of the universe.

A stellar mass black hole (e.g. a low-mass X-ray binary) might have a luminosity of $10^{4} L_{\odot}$ and a mass of $10M_{\odot}$ and $g \ll 1$, so the timescales in this case could be similar.

Note that this really is order of magnitude stuff. The assumption of a non-rotating black hole is poor, not least because it is accreting material with angular momentum. This can allow the $1/12$ factor to be bigger, accretion disk luminosities to be bigger and lifetimes shorter. There is also the issue of radiation pressure preventing further accretion. However the Eddington luminosity is derived for spherical accretion rather than through a disk, and super-Eddington accretion appears to be possible. I also haven't allowed for the fact that unless $g < 1$ then the mass of the black hole increases significantly as the accretion proceeds.

I believe that the accretion disk would just orbit forever. It might go in the same way that the moon orbits the earth, or the earth orbits the sun. In the case of the moon orbiting the earth, it is actually just falling towards the earth, but the curvature of the earth takes over, so the moon just keeps orbiting. I think this is what would would happen, but I don't know for sure.

## protected by Qmechanic♦Feb 23 '17 at 23:36

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).