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In Chapter 14.5.3 of Physically Based Rendering - From Theory to Implementation (3rd edition), the authors claim that a probability density $p_\omega$ according to solid angle can be converted to a density $p_A$ according to area by $$p_A=p_\omega\frac{\left|\cos\theta_i\right|}{\left|p_i-p_{i+1}\right|^2},\tag1$$ where $\theta_i$ is the angle between the surface normal at $p_{i+1}$ and the vector to $p_i$.

The problem is: $(1)$ doesn't make sense. If $p_\omega$ is a density with respect to the surface measure on the upper unit hemisphere $\Omega_{p_i}$ over $p_i$, then $p_\omega$ is by definition a function $\Omega_{p_i}\to[0,\infty]$. Analogously, if $p_A$ is supposed to be a density with respect to the surface measure on the manifold $M$ build by all surfaces in the scene, then $p_A$ is by definition a function $M\to[0,\infty]$.

So, there's at least an argument transformation missing. I would really appreciate, if anybody could clarify.

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  • $\begingroup$ What's wrong with a probability density having values in $[0,+\infty)$? Perhaps you are confusing the PDF with the probability mass function? $\endgroup$
    – Hubble
    Commented Sep 5, 2018 at 18:21
  • $\begingroup$ @Hubble You've missed the point. The point is that the function on the left-hand side is defined on $\Omega_{p_i}$, while the function on the right-hand side is defined on $M$. And mathematically, the equation $f=g$ for functions $f,g$ only makes sense when both functions have a common domain. $\endgroup$
    – 0xbadf00d
    Commented Sep 5, 2018 at 18:30
  • $\begingroup$ $p_A$ and $p_\omega$ are not defined on the same domain, but the Jacobian term ensures that you can go from projected solid angle to surface area, and vice versa. Read Section 5.5 of PBRT3, especially Equation 5.6 for more info. $\endgroup$
    – Hubble
    Commented Sep 5, 2018 at 20:51
  • $\begingroup$ Isn't there a bijection between points on the upper hemisphere over $p_i$ and points $p_{i+1}$ on the visible surfaces in the scene? $\endgroup$
    – user106
    Commented Sep 6, 2018 at 5:32
  • $\begingroup$ @Rahul There is a bijection: Let $V_{p_i}$ denote the set of all surface points $q$ such that $p_i$ and $q$ are mutually visible and $D_{p_i}$ denote the set of all $\omega\in\Omega_{p_i}$ such that there is a $t>0$ with $p_i+t\omega\in V_{p_i}$. Then there is obviously a bijection between $V_{p_i}$ and $D_{p_i}$. We can define a transformation $$T:V_{p_i}\to D_{p_i}\;,\;\;\;q\mapsto\frac{q-p_i}{\left|q-p_i\right|},$$ but the determinant of the Jacobian of $T$ is $0$ everywhere and hence the transformation formula can't be applied. Am I missing something? $\endgroup$
    – 0xbadf00d
    Commented Sep 6, 2018 at 11:49

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I don't understand what's the problem. We want to make a change of variables so that instead of integrating over the range of directions $d\omega_i$ in the upper hemisphere we integrate over patches of area $dA$ of a region $R$.

This whole thing is given in section 5.5 as Hubble said but let me post it again.

We know by definition that,

$\omega = A * cos(\theta)/r^2$

Now if you differentiate this, you get

$d\omega = dA * cos(\theta) / r^2$

where,

$Jacobian = cos(\theta) / r^2$

as Hubble said. I don't know what are you trying to do by calculating Jacobian through determinant but as far as I know, that Jacobian determinant is for multi-variables and involves partial derivatives. For change of variables involving a single variable the Jacobian can be gotten by simply derivating as I did.

Now this might seem the inverse but wait a moment. What you posted in the question is a relation between the PDFs.

Consider a single sample of the rendering equation integrated over the domain of $\omega_i$

$L_s = BRDF * L_f * (\omega_i . n) / p(\omega) $

After changing the domain by above equations this would become

$L_s = BRDF * L_f * (\omega_i . n) (\omega_i . n_r)\, / \, (p(A) * r^2) $

$L_s = BRDF * L_f * (\omega_i . n) * JN\, / \, p(A) $
where $Jacobian = JN$

This means, $ 1 / p(\omega) = JN/p(A)$

$ p(A) = p(\omega) * JN$

$ p(A) = p(\omega) * cos(\theta)/r^2$

This is what you see above. Simple as ****.

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