Harvard Astronomy 201b

Accretion Luminosity

In Uncategorized on April 27, 2011 at 5:05 pm

Let $m = \dot{M} t$ be the orbiting mass entering and leaving the ring.  $dE = \frac{dE}{dr} dr = \frac{d}{dr} \left(-G \frac{M_* m}{2 r}\right) dr = G \frac{M_* \dot{M} t }{2 r^2} dr$.  In the steady state, conservation of energy requires that the energy radiated by an annulus in the the accretion disk be equal to the difference in energy flux between its inner and outer edges.  Therefore $dE = dL_{\rm ring} t$.  Equating the two sides gives: $dL_{\rm ring} = G \frac{M_*\dot{M}}{2 r^2} dr$. The Stefan-Boltzmann relation gives $dL_\text{ring} = 4 \pi r \sigma T^4 dr$.

Therefore we find that $\sigma T^4 = \frac{G M_* \dot{M}}{8 \pi r^3}$.  We did not take into account the fact that the star has a finite size which introduces a boundary layer into the disk.  A more thorough analysis (e.g. in Lynden-Bell & Pringle 1974) then adds an additional factor, giving: $\sigma T^4 = \frac{G M_* \dot{M}}{8 \pi r^3} \left(1- \sqrt{\frac{R_*}{r}} \right)$