Radius of the accretion disc?

[tiny-tim]

What is the inner radius of an accretion disc?

Is it simply the radius of the smallest possible orbit of a massive particle (the marginally stable radius), 6M ?

 

[stevebd1]

Hi tiny-tim

In most cases, the inner edge of the accretion disk is considered to coincide with the marginally stable which is 6M for a static black hole, reducing to 1M for a maximal Kerr black hole where a/M=1, though there seems to be some debate that the properties of the accretion disk itself can have an effect on this-

A couple of links-

'Where is a Marginally Stable Orbit in Luminous Accretion Disk' by Ken-ya Watarai Watarai, Shin Mineshige
http://quasar.cc.osaka-kyoiku.ac.jp/~watarai/presen/KoreaPoster2004.pdf

'The radiation properties of an accretion disk with a non-zero torque on its inner edge' by Xinwu Cao, Y.D. Xu
http://arxiv.org/abs/astro-ph/0211220

'Black Hole Accretion Disks On The Edge' by Kristen Menou
http://arxiv.org/abs/astro-ph/0304421

 

[Entropia]

The radius of an accretion disc can vary depending on the size and mass of the central object and the amount of material being accreted. However, the inner radius of an accretion disc is typically defined as the marginally stable radius, which is the smallest possible orbit of a massive particle around the central object. This radius is calculated using the Schwarzschild metric and is approximately 6 times the mass of the central object (represented by M). Therefore, the inner radius of an accretion disc can be written as 6M. However, it is important to note that this is just an approximation and the exact inner radius can vary depending on other factors such as the spin of the central object.