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Let's say I want to model a star of radius $R$ at a distance $r$ from the Earth. I need to show that the apparent luminosity for frequency $\nu$ is equal to

$$\ell(\nu)=\frac{2\pi h}{c^2}\left( \frac{R}{r}\right)^2\frac{\nu^3}{\exp\displaystyle\left(\frac{h\nu}{kT}\right)-1} $$

Which is Planck's law multiplied by $$\pi\left(\frac{R}{r}\right)^2$$

Given the nature of the factor (adimensional function of $R,r$) I would say that the result follows from geometrical considerations, however I fail to see which ones.

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First, calling $\ell(\nu)$ a luminosity is a bit misleading since the quantity you've shown has dimensions of power/area (usually called flux, at least in astronomy), while a luminosity has dimensions of power.

To answer your actual question, consider first the total power emitted by the star (should involve $R$) and the spatial distribution of that power once it reaches the distance of the detector (might help to consider a detector with area $1m^2$ for argument's sake). This should involve $r$. You're right, it's really just a bit of geometry.

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  • $\begingroup$ The total power is given by Steffan-Boltzmann's law, but I will loose the $\nu$ dependence integrating with respect to $\nu$. For the spatial distribution may I invoke a $r^{-2}$ law? $\endgroup$
    – Jorge Lavín
    Jun 13, 2013 at 16:33
  • $\begingroup$ Ah, sorry, by total power I meant $L_\nu d\nu$, not its integral. And yes, a $r^{-2}$ law would be thinking along the right lines :) $\endgroup$
    – Kyle Oman
    Jun 13, 2013 at 16:51
  • $\begingroup$ So the factor $r^{-2}$ comes from the inverse square law. Now for the $\pi R^2$ factor is what troubles me, shouldn't be $4\pi R^2$ ? $\endgroup$
    – Jorge Lavín
    Jun 13, 2013 at 17:37
  • $\begingroup$ The origin of the $R^2$ should be somewhat obvious - $B_\nu$ (Planck's law) gives power per unit area, so to get the total power emitted you need to multiply by an area, intuitively that of the emitter, which will involve $R^2$. You've already figured out the $r^2$. The factor of $\pi$ is a little puzzling, though. It could be you're looking for a subtly different quantity that the one I've discussed in my answer. Can you give a formal definition of what you're calling "apparent luminosity" in your question? In other words, in general $\ell(\nu) = ?$ $\endgroup$
    – Kyle Oman
    Jun 13, 2013 at 18:15
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    $\begingroup$ Right, so it sounds to me like that $\pi$ might actually be an error... might want to check with the source of the problem (teacher/professor?). $\endgroup$
    – Kyle Oman
    Jun 13, 2013 at 18:29
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The specific flux (defined as power per unit area per unit frequency) per coming from the surface of the star is given by $\pi I_{\nu}$ where in the case of a blackbody, the specific intensity $I_{\nu}$ is given by the Planck function.

The reason for this relationship is shown here. To summarise, the factor $\pi$ arises from integrating the specific intensity multiplied by the cosine of the zenith angle over a hemispherical solid angle. i.e. $$ F_\nu = I_{\nu} \int \cos(\theta)\ d\Omega = I_\nu \int^{2\pi}_{\phi=0} d\phi\ \int_{0}^{\theta=\pi/2} \cos(\theta) \sin (\theta)\ d\theta = \pi I_\nu\ . $$

The total luminosity of the star is therefore $\pi I_{\nu} 4\pi R^2$.

At the Earth, at a distance $r$, the received flux is therefore the total luminosity spread over a sphere of area $4\pi r^2$ $$\pi I_{\nu} 4 \pi R^2/(4\pi r^2) = \pi I_\nu \left(\frac{R}{r}\right)^2\ , $$ which is what you were looking for.

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