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(It's mostly a calculus question and has little to do with optics.)

I'm reading a book on Computer Graphics, Realistic Image Synthesis Using Photon Mapping by Henrik Wann Jensen, and I can't completely understand how to derive the radiance equation given in Chapter 2.

According to the book, the spectral radiant energy $Q_{\lambda}$, in $n_{\lambda}$ photons with wavelength $\lambda$ is

$$Q_{\lambda}=n_{\lambda}\frac{h \; c}{\lambda}\;,$$

where h is Planck's constant. Hence

$$Q=\int_{0}^{\infty}Q_{\lambda}d\lambda\;.$$

Radiant flux $\Phi$ is the time rate of flow of radiant energy:

$$\Phi=\frac{dQ}{dt}\;.$$

And radiant flux area density is

$$E(x)=\frac{d\Phi}{dA}\;.$$

The radiant intensity $I$ is the radiant flux per unit solid angle $d\omega$:

$$I(\omega)=\frac{d\Phi}{d\omega}\;.$$

So the radiance $L$ is the radiant flux per unit solid angle per unit projected area:

$$L(x,\omega)=\frac{d^{2}\Phi}{\cos\theta\;dA\;d\omega}=\int_{0}^{\infty}\frac{d^{4}n_{\lambda}}{\cos\theta\;d\omega\;dA\;dt\;d\lambda}\;\frac{h\;c}{\lambda}d\lambda$$


My question is: Could anyone explain how to derive the last integral, and why it is not

$$\int_{0}^{\infty}\frac{d^{3}n_{\lambda}}{\cos\theta\;d\omega\;dA\;dt}\;\frac{h\;c}{\lambda}d\lambda\;?$$

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1 Answer 1

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This is an error in the book. By the first three equations, the units of

$$ \frac{n_\lambda}{\mathrm dt}\frac{hc}{\lambda}\mathrm d\lambda $$

are those of $\Phi$, and $\cos\theta\,\mathrm dA\,\mathrm d\omega$ is the same on both sides, so there's an unmatched inverse unit of length from $\mathrm d\lambda$ in the denominator on the right-hand side.

Your version of the expression looks OK to me.

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Thanks. Actually I've found other confusing formulas after posting that, which I'm quite sure are mistakes. So is this one, probably. –  Stupident Nov 2 '12 at 15:06

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