<p>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 <a href="http://arxiv.org/abs/0707.0305" rel="nofollow">Morgan et al. (2010)</a> 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).</p>

<p>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.</p>

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

<p>This result is consistent with <a href="http://arxiv.org/pdf/1501.05951v2.pdf" rel="nofollow">Edelson et al. (2015)</a>, 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 <a href="http://arxiv.org/abs/1203.6851" rel="nofollow">Armijo (2013)</a>, who shows that in the radio regime, the disk is thousands of AU, and even up to ~100 pc.</p>

CC BY-SA 3.0
2015-11-04T13:29:22.597Z
2015-11-06T08:38:47.600Z
2015-11-06T08:38:47.600Z
pela
<p>$10^{8} M_{\odot}$ SMBH.  Eddington Luminosity (dodgy estimate, since assumes <em>spherical</em> accretion) is $ L/L_{\odot} \simeq 3\times 10^{4} (M/M_{\odot}) = 3\times 10^{12} L_{\odot}$.</p>

<p>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.</p>

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

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

CC BY-SA 3.0
2015-11-05T21:35:23.447Z
2015-11-05T21:35:23.447Z
ProfRob
<p>I have a homework assignment question on accretions discs (essentially an estimation of the number of electron scatterings, but this is just for background). </p>

<p>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)</p>

<p>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. </p>

<p>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. </p>

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

CC BY-SA 3.0
2015-11-03T20:00:27.807Z
couple of links http://www.scholarpedia.org/article/Accretion_discs and http://arxiv.org/abs/1002.4160. and finally http://relativity.livingreviews.org/Articles/lrr-2013-1/fulltext.html
James K
CC BY-SA 3.0
2015-11-03T21:42:19.597Z
I don't think accretion disks have a well-defined outer radius, but according to [this](http://astro.physics.uiowa.edu/~kaaret/s09/L19_accretion.pdf) presentation, the inner radius is i $3 R_\mathrm{S}$, with a rather sharp edge because the material is slowed due to conservation of angular momentum.
pela
CC BY-SA 3.0
2015-11-03T22:57:07.553Z
@pela Hi Pela, I should add: I'm looking for rough estimates...so maybe for an SMBH are we looking at perhaps the order of 10-50$r_s$?? ($r_s$ being the Schwarschild radius) Just by visual inspection of accretion disc diagrams?
MichaelJRoberts
CC BY-SA 3.0
2015-11-04T00:12:32.757Z
There really is no way to answer this, because it depends on a whole load of stuff. Giving *us* the mass of the black hole might be a start!
ProfRob
CC BY-SA 3.0
2015-11-04T00:14:43.750Z
Why does the linear extent matter? And what are "electron scatterings"?
ProfRob
CC BY-SA 3.0
2015-11-04T09:42:44.963Z
It would just be a rough back of the envelope type calculation, I'm not looking for anything more. Mass is $10^{8}\,M_{\odot}$, temperature is given as $10^5-10^6$K. So we are talking SMBH size.
MichaelJRoberts
CC BY-SA 3.0
2015-11-04T20:18:01.530Z
And are you also given the luminosity of the accretion disc to go with this temperature...? Or do you want to assume the Eddington luminosity?
ProfRob
CC BY-SA 3.0
2015-11-04T21:45:51.857Z
Assume Eddington luminosity, as no luminosity is given.
MichaelJRoberts
CC BY-SA 3.0
2015-11-05T21:25:13.117Z
In that case, don't your luminosity and disc temperature give you a size for the disc... ?
ProfRob
<black-hole><accretion-discs><accretion>
Linear extent of an accretion disk
MichaelJRoberts
# Linear extent of an accretion disk

(Asked by: username_0 on 2015-11-03)

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.

**Tags:** black-hole, accretion-discs, accretion
> **Comments:**
> **username_1**: couple of links http://www.scholarpedia.org/article/Accretion_discs and http://arxiv.org/abs/1002.4160. and finally http://relativity.livingreviews.org/Articles/lrr-2013-1/fulltext.html  
> **username_2**: I don't think accretion disks have a well-defined outer radius, but according to [this](http://astro.physics.uiowa.edu/~kaaret/s09/L19_accretion.pdf) presentation, the inner radius is i $3 R_\mathrm{S}$, with a rather sharp edge because the material is slowed due to conservation of angular momentum.  
> **username_0**: @pela Hi Pela, I should add: I'm looking for rough estimates...so maybe for an SMBH are we looking at perhaps the order of 10-50$r_s$?? ($r_s$ being the Schwarschild radius) Just by visual inspection of accretion disc diagrams?  
> **username_3**: There really is no way to answer this, because it depends on a whole load of stuff. Giving *us* the mass of the black hole might be a start!  
> **username_3**: Why does the linear extent matter? And what are "electron scatterings"?  
> **username_0**: It would just be a rough back of the envelope type calculation, I'm not looking for anything more. Mass is $10^{8}\,M_{\odot}$, temperature is given as $10^5-10^6$K. So we are talking SMBH size.  
> **username_3**: And are you also given the luminosity of the accretion disc to go with this temperature...? Or do you want to assume the Eddington luminosity?  
> **username_0**: Assume Eddington luminosity, as no luminosity is given.  
> **username_3**: In that case, don't your luminosity and disc temperature give you a size for the disc... ?  

---

## 2 answers

### Answer by username_2 on 2015-11-04. Score: 1 | Accepted answer

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)](http://arxiv.org/abs/0707.0305) 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)](http://arxiv.org/pdf/1501.05951v2.pdf), 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)](http://arxiv.org/abs/1203.6851), who shows that in the radio regime, the disk is thousands of AU, and even up to ~100 pc.


---

### Answer by username_3 on 2015-11-05. Score: 1

$10^{8} M\_{\odot}$ 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).


---

