Annex: The 4 in PBR - Patrick Bell

This annex to Derivation of Physically Based Rendering derives how $\omega_r$ changes relative to $\omega_m$ for a specular microfacet.

The delta function enforces $\pmb{h} = \pmb{m}$ , so ${d\omega_m = d\omega_h}$ .

To make the problem simpler we can swap what’s changing, that is, for some area on the unit sphere $d\omega_r$ at $\pmb{r}$ , what is the corresponding area $d\omega_h$ at $\pmb{h}$ , keeping $\pmb{i}$ as a constant vector?

Illustration of geometric argument to explain the 4 in PBR

We need to transforms the area $d\omega_r$ onto the corresponding area $d\omega_h$ following the definition of the halfway vector:

So our area at $d\omega_r$ needs to be shifted from $\pmb{r}$ to $\pmb{i} + \pmb{r}$ , then rescaled onto the unit sphere to become $d\omega_h$ .

Shifting $d\omega_r$ to $\pmb{i}+\pmb{r}$ doesn’t change the area, but the area no longer lies perpendicular to a sphere, so it needs to be projected onto the sphere with radius $|\pmb{i}+\pmb{r}|$ . $d\omega_r$ ’s normal is $\pmb{r}$ and the outer sphere’s normal is $\pmb{h}$ at $\pmb{i}+\pmb{r}$ , so the projected area is $dA_{\text{proj}} = (\pmb{r} \cdot \pmb{h}) d\omega_r$ .

We now need to rescale our area from the outer sphere onto the unit sphere. The proportion of to the outer sphere’s surface area is equivalent to the proportion of $d\omega_h$ to the unit sphere’s surface area,

So substituting for and rearranging

We need to find $|\pmb{i}+\pmb{r}|$ , one method is to note that $\pmb{i}+\pmb{r}$ is parallel to $\pmb{h}$ so

Or equivalently, if $\theta$ is the angle between $\pmb{h}$ and $\pmb{r}$ , then $2\theta$ is the angle between $\pmb{i}$ and $\pmb{r}$ so

Substituting $|\pmb{i}+\pmb{r}|^2$ ,