Attached is page 34 of "Advanced Global Illumination", 2nd edition, by Dutre et al. I don't understand how (2.22) is derived from (2.21) and the given incident radiance distribution $L(x\leftarrow\Psi)=L_{in}\delta(\Psi-\Theta)$. Could you please give me a detailed derivation? Thanks a lot!

enter image description here


1 Answer 1


The only thing you really need to know in order to derive (2.22) from (2.21) is that the $\delta$-distribution satisfies

$$ \int \delta(x) f(x)\,\text{d}x = f(0). $$

(The $\delta$-distribution is not a function in the common sense, so the above integral is merely a useful notation and not to be read as e.g. a Lebesgue or Riemann integral.)

Unfortunately, the notation that the authors use for the substitution of the incident radiance distribution is misleading. The $\Theta$ in $L(x\leftarrow\Psi) = L_{in}\delta(\Psi-\Theta)$ is completely unrelated to the integration variable $\omega_\Theta$, so I chose to denote it by $\tilde\Psi$ instead.

Here is the derivation of (2.22) from (2.21):

  1. Substitute $L(x\leftarrow\Psi) = L_{in}\delta(\Psi-\tilde\Psi)$:

$$\frac{\int_{\Omega_x} \int_{\Omega_x} f_r(x, \Psi \rightarrow \Theta) L_{in}\delta(\Psi-\tilde\Psi)\cos(N_x,\Theta)\cos(N_x,\Psi)\,\text{d}\omega_\Psi\,\text{d}\omega_\Theta}{\int_{\Omega_x}L_{in}\delta(\Psi-\tilde\Psi)\cos(N_x,\Psi)\,\text{d}\omega_\Psi}$$

  1. Substitute $\Psi$ with $\Psi+\tilde\Psi$:

$$\frac{\int_{\Omega_x} \int_{\Omega_x-\tilde\Psi} f_r(x, \Psi+\tilde\Psi \rightarrow \Theta) L_{in}\delta(\Psi)\cos(N_x,\Theta)\cos(N_x,\Psi+\tilde\Psi)\,\text{d}\omega_\Psi\,\text{d}\omega_\Theta}{\int_{\Omega_x-\tilde\Psi}L_{in}\delta(\Psi)\cos(N_x,\Psi+\tilde\Psi)\,\text{d}\omega_\Psi}$$

  1. Use the above property of the $\delta$-distribution:

$$\frac{\int_{\Omega_x} f_r(x, \tilde\Psi \rightarrow \Theta) L_{in}\cos(N_x,\Theta)\cos(N_x,\tilde\Psi)\,\text{d}\omega_\Theta}{L_{in}\cos(N_x,\tilde\Psi)}$$

  1. Simplify:

$$\int_{\Omega_x} f_r(x, \tilde\Psi \rightarrow \Theta) \cos(N_x,\Theta)\,\text{d}\omega_\Theta$$

Now you only need to notice that we want equation (2.21) to be true for all such "test functions" $L_{in}\delta(\Psi-\tilde\Psi)$ for the incident radiance, i.e. for all possible values of $\tilde\Psi$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.