Anisotropy: Bug in the derivation of the or wrong assumption about the resulting $a_t$ and $a_b$

[sherholz]

There seems to be a bug in the mapping from roughness and anisotropy to the GGX parameters $a_t$ and $a_b$, or a misassumption about the characteristics of the resulting values.

The spec text says the following:

The NDF terms $a_t$ and $a_b$ are more conveniently parametrized as the total roughness $r$
and an anisotropy $a∈[0,1]$. We specify the following mapping from $r$ and $a$ to $α_t$, $α_b$:
$a_t=r^2\sqrt{\frac{2}{1+(1-a)^2}}$ and $a_b=(1-a)a_t$.
This formulation satisfies $α^2_t+α^2_b=2α^2$, to preserve the average roughness regardless of the anisotropy.

First, there seems to be a typo in the formula that should be satisfied. The text says that the average roughness should be preserved, but the formula uses $a$ instead of $r$: $2a^2$ -> $2r^2$.
Second, when solving the left side of the formula by inserting the definitions of $a_t$ and $a_b$ one gets $a_t^2+a_b^2 = 2r^4$.
The result does not seem to be related to an average roughness.

To get something that fulfills the goal of a mapping that matches the average roughness, the mapping needs to be:
$a_t=r^{1/2}\sqrt{\frac{2}{1+(1-a)^2}}$ and $a_b=(1-a)a_t$.
and then the following would hold:
$a_t^2+a_b^2 = 2r$.

[portsmouth]

Thanks for the careful read — but I think the formula is actually correct, and the confusion comes down to a symbol collision between $\alpha$ (the GGX roughness) and $a$ (the anisotropy). Let me try to untangle it, and then answer why is $\alpha_t^2 + \alpha_b^2 = 2\alpha^2$ the right thing to require.

First, the algebra checks out

There are two distinct "roughness" quantities in play:

• $r \in [0,1]$ — the user-facing perceptual roughness (specular_roughness)
• $\alpha$ — the GGX NDF parameter, with the standard squaring remap $\alpha = r^2$

Substituting the definitions $\alpha_t = r^2\sqrt{\tfrac{2}{1+(1-a)^2}}$ and $\alpha_b = (1-a)\alpha_t$:

$$\alpha_t^2 + \alpha_b^2 = \alpha_t^2\big(1 + (1-a)^2\big) = r^4 \cdot \frac{2}{1+(1-a)^2}\cdot\big(1+(1-a)^2\big) = 2r^4 = 2\alpha^2$$

So $\alpha_t^2 + \alpha_b^2 = 2\alpha^2$ holds exactly. The RHS is $2\alpha^2$ (alpha, the NDF roughness), not $2a^2$ (anisotropy) — I think that's where the "should be $2r^2$" reading comes from. And it reduces correctly in the isotropic limit $a=0$: $\alpha_t = \alpha_b = r^2 = \alpha$, recovering the base relation $\alpha = r^2$. (A mapping satisfying $\alpha_t^2+\alpha_b^2 = 2r$ instead would give $\alpha = \sqrt{r}$ at $a=0$, contradicting $\alpha = r^2$.)

Why $\alpha_t^2 + \alpha_b^2 = 2\alpha^2$ is the right invariant

This is the better question. The key fact is that $\alpha$ is the scale of the microfacet slope distribution — $\alpha^2$ is a slope variance, and $\alpha$ is its standard deviation. Seen most cleanly in the Beckmann form (same $\alpha$ convention as GGX, but with finite moments):

$$P(x_m,y_m) = \frac{1}{\pi\alpha_t\alpha_b}\exp!\left(-\frac{x_m^2}{\alpha_t^2}-\frac{y_m^2}{\alpha_b^2}\right)$$

This is a 2D Gaussian in slope space with per-axis variance $\alpha_t^2/2$ and $\alpha_b^2/2$, so the total mean-square slope of the microsurface is

$$\big\langle x_m^2 + y_m^2 \big\rangle = \frac{\alpha_t^2 + \alpha_b^2}{2},$$

which equals $\alpha^2$ in the isotropic case. So the requirement $\alpha_t^2 + \alpha_b^2 = 2\alpha^2$ is precisely: hold the total microsurface slope variance fixed. Anisotropy just redistributes a fixed budget of slope variance between the tangent and bitangent axes.

Equivalently, in tensor terms the NDF is governed by the roughness tensor $A = \mathrm{diag}(\alpha_t^2, \alpha_b^2)$, and we're preserving its trace $\mathrm{Tr}A = \alpha_t^2 + \alpha_b^2$ — the natural rotation-invariant, additive scalar summary.

Why quadratic (sum of squares) rather than linear?

Because roughness composes in quadrature, not linearly — variances add, standard deviations don't. This is the same principle behind $\alpha = r^2$ and behind normal-map / LEAN-LEADR roughness filtering, where extra sub-pixel slope variance combines as $\alpha_\text{eff}^2 = \alpha^2 + \alpha_\text{extra}^2$. The physically additive "roughness energy" is $\alpha^2$, so the total over both axes is $\alpha_t^2 + \alpha_b^2$. The alternatives don't hold up:

Invariant	Preserves	Issue
$\alpha_t^2 + \alpha_b^2$ (chosen)	total slope variance / $\mathrm{Tr} A$	natural & additive
$\alpha_t + \alpha_b$	arithmetic mean of $\alpha$	not a physical moment; not how roughness combines
$\alpha_t,\alpha_b (=\det A)$	"area" of the slope lobe	$\to 0$ as one axis sharpens — perceptually wrong

The practical payoff

With this invariant, the no-anisotropy fallback value is exactly the RMS of the two axes:

$$\alpha = \sqrt{\frac{\alpha_t^2 + \alpha_b^2}{2}}$$

i.e. the variance-preserving average. So a renderer that drops anisotropy (LOD, performance, format conversion) keeps the same total slope energy and lands on an isotropic highlight of roughly the same overall size — just circular instead of elliptical. That perceptual match is the whole motivation, and it only works because the conserved quantity is quadratic.

Suggested doc tweak

The math is correct, but the $\alpha$ vs $a$ glyphs are genuinely easy to misread. It might be worth spelling out $\alpha = r^2$ inline right next to the $\alpha_t^2 + \alpha_b^2 = 2\alpha^2$ statement, and noting explicitly that the preserved quantity is the total slope variance (trace of the roughness tensor). Happy to open a small PR for that wording if useful.