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With reference to Black holes in particular, how can you approximate the luminosity of an accretion disk? It is possible to quantify the temperature at a given point, but as the disk is not a black body, and this temperature is at a specific point, I am unsure how to equate this to luminosity - surely you could not do so using the Stefan-Boltzmann constant?

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The simplest answer is that you integrate the emission from annuli radiating at various temperatures. Explicitly, the luminosity per wavelength in this approximation is

$$L_\lambda = 2 \int_{r_{\rm in}}^{r_{\rm out}} 2 \pi r [\pi B_\lambda(r)] dr$$

where the overall factor of 2 is for the two sides of the disk, $2 \pi r dr$ is the area of each annulus, $B_{\lambda}$ is the Planck function, which depends on the temperature at the given radius, and $\pi B_\lambda$ is the flux that arises from integrating the thermal emission over solid angle. According to the model by Shakura and Sunyaev 1973, where this is all explained in much greater detail, the temperature roughly goes as $r^{-3/4}$. To get the bolometric luminosity, integrate $L_\lambda$ over wavelength.

This works well for radiatively efficient, thin disks. The situation gets trickier for thicker disks that can't cool as efficiently.

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  • $\begingroup$ Hi @kleingordon ,thanks for the answer! However, this is a bit above my current abilities in terms of maths - I am studying AS-Levels in the UK at the moment. Could you provide a simpler explanation? I understand integration, but would have no idea how to approach the above example. $\endgroup$ – Noah P May 7 '16 at 4:42
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    $\begingroup$ I would not be surprised if there is no analytic solution. $\endgroup$ – Demi May 7 '16 at 5:09
  • $\begingroup$ Hi @NoahP, the most important point is that a single temperature doesn't characterize the entire disk, as you noted in your question. So the total emission will be a sum over different temperatures along the radius of the disk. So instead of temperature, it often helps to think of another physical quantity that sets the luminosity of the disk. Traditionally this is the rate at which mass is being fed in, or $\dot{M}$. For thin, radiatively efficient disks, the luminosity will be around $10\%$ of $\dot{M} c^2$. Depending on the exact accretion flow, that efficiency can vary. $\endgroup$ – kleingordon May 9 '16 at 22:59

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