Linear extent of an accretion disk

Question

I have a homework assignment question on accretions discs (essentially an estimation of the number of electron scatterings, but this is just for background).

There are a few parameters, one of them being $L$, which is the linear size of the medium (the medium in this case being an accretion disc around a blackhole)

Now, I have been given the mass of the black hole. Other than that, nothing else which could give me the linear size of the accretion disc.

Could I assume that the linear extent of the accretion disc is perhaps of the order of a few Schwarzschild radii? Which could be calculated from the mass, which is given.

If anyone could shed some light on this I would be very appreciative. I need a nudge in the right direction on this.

Answer 1

I think the outer edge of an accretion disk is not well-defined, and observationally the radius will depend on which wavelength you consider, since the farther you get from the BH, the softer the radiation will be. But if you look in the UV, then Morgan et al. (2010) find the following relation between $R_{2500}$ (the radius when observed at $\lambda = 2500$ Å) and the mass $M_\mathrm{BH}$ of the black hole: $$ \log\left( \frac{R_{2500}}{\mathrm{cm}}\right) = 15.78 + 0.80 \log\left( \frac{M_\mathrm{BH}}{10^9M_\odot}\right), $$ (modulo some uncertainties that you can look up in the paper).

That is, if your BH has a mass of $10^8 M_\odot$, its radius will be $R_{2500}\sim64\,\mathrm{AU}$, or roughly 1/3 lightdays.

For comparison, its Schwarzschild radius is $\sim2\,\mathrm{AU}$, so your estimate was actually pretty good.

This result is consistent with Edelson et al. (2015), who find 0.35 lightdays, also in the UV. However, in you look in longer wavelengths, the disk is much, much larger. If you're interested beyond your homework assignment, take a look at the accretion disk theory review by Armijo (2013), who shows that in the radio regime, the disk is thousands of AU, and even up to ~100 pc.

Answer 2

$10^{8} M_{\odot}$ SMBH. Eddington Luminosity (dodgy estimate, since assumes spherical accretion) is $ L/L_{\odot} \simeq 3\times 10^{4} (M/M_{\odot}) = 3\times 10^{12} L_{\odot}$.

Let's assume we are seeing all this emerge from the face of a flat disc of total area (front and back) $\pi R^2$ and temperature $T \simeq 10^{5}$ K.

Assume black body emission, so $L = 2\pi R^2 \sigma T^4$. And thus $$ R \simeq \left[\frac{3\times 10^{12} L_{\odot}}{2 \pi \sigma T^4}\right]^{1/2}$$

Putting in the numbers I get $R \simeq 40$ au. However it is very sensitive to the assumed temperature (much smaller for $T = 10^{6}$ K).