# Question on Radiance equation

The radiance equation is $$L = \frac{d}{dA} \frac{2(\phi)}{dW cos(\theta)} (watt/srm^2)$$ where $\phi$ is the flux.

I am thinking, should not be the cosine term on the numerator instead of the denominator? Having the cosine in the denominator will make L goes to infinity if $\theta = 90$ , which does not make sense to me. My understanding is that if $\theta = 90$ (i.e. flux direction is perpendicular to the surface normal), then $L$ should equal to zero (and not infinity).

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Hi Khaled. Physics.SE allows MathJax, so your equations turn out nice and pretty. Have a look at the faq for more info on how to use it. I'll fix your equation for you, but I'm not sure exactly what you're trying to say, so you may have to correct it. – Kitchi Jan 31 '13 at 12:50
@Kitchi, for some reasons I did not see your comment, I guess I did not configure my account to send notifications when I get a reply! Now, it's been a year now and I just want to thank you for editing my equation, and apologies for not replying... – Khaled Feb 28 '14 at 18:30

Generally, the radiative transfer equation is given by

$$\frac{1}{c}\frac{\partial I_{\nu}(\vec{r},\vec{n},t)}{\partial t} + \vec{n}\cdot\frac{I_{\nu}(\vec{r},\vec{n},t)}{\partial \vec{r}} = \rho(\vec{r},t) \kappa_{\nu} (\vec{r},t)\left\{-I_{\nu}(\vec{r},\vec{n},t)+S_{\nu}(\vec{r},\vec{n},t)\right\}.$$

To monochromatic intensity is defined by noting that $I_{\nu}(\vec{r},\vec{n},t)\cos\!\Theta \,d\nu\,df d\Omega\, dt$ at the position $\vec{r}$ gives the radiative energy in the frequency interval $\nu,\nu+d\nu$ transported through the surface element $df$ into the solid angle element $d\Omega$ enclosing the direction $\vec{n}$ during a time increment $dt$, see the appended figure.

Your definition looks a bit strange, there is probably a time differential $dt$ missing in the denominator and the $2\phi$ should be the energy differential $dE$. However, $\cos\theta$ in your definition of the radiance correctly appears in the denominator. Looking at my figure, if $\Theta = 90^{\circ}$ the energy propagates along the area element $df$ instead of though it, so the radiance is indeed ill defined in this case.

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Sorry for this extra late reply :) I thought that I would get a notification when I get a reply, but I did not configure account to do so. So belated thanks for your reply :) – Khaled Feb 28 '14 at 18:33

The formula correctly reflects the definition of radiance, so it's certainly correct in this sense. Is such a definition useful? I guess so - the value defined in this way gives some idea of "brightness" of the source, and "brightness" does tend to infinity when the angle tends to 90 deg - the same flux comes from a very small "perceived" area.

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Sorry for this extra late reply :) I thought that I would get a notification when I get a reply, but I did not configure account to do so. So belated thanks for your reply :) – Khaled Feb 28 '14 at 18:36

I believe the confusion comes from the fact that we speak of radiant flux per PROJECTED area per unit solid angle. We are not looking really at the radiant flux of a small surface element around P or x (depending on which convention you use) but how much flux goes through a projection of that surface in the direction of incidence. And if you consider 1 unit of projected area, as the angle $\theta$ increases, the area of the surface element corresponding to the projection of $dA^\perp$ on the horizontal plane increases (and goes to infinity in the limit of $\theta$ approaching 90 degrees. Thus what you measure here in way is that amount of "light" that is incident or emerging from this very large area back into a very small "projected" area. Looked that way, you can hopefully better understand why it goes to infinity?