How does a path tracer capture light falloff if it deals exclusively with radiance? | pressku.com

My knowing is that way tracers woody pinch radiance because radiance is changeless on a ray. You simply spell done and measure nan ray carrier equation:

$$ L_o(p, \omega_o) = L_e(p,\omega_o) + \int_S f(p,\omega_o,\omega_i)L_i(p,\omega_i) | \cos{\theta_i}| \ d\omega_i $$

with nan extremity of tracing a way done nan segment and calculating nan full radiance contributed to that path. My disorder comes from nan following:

Say I person a spherical light, that has a radius of $1m$, and has a full twinkling powerfulness of $100W$. Assuming nan ray is isotropic, I tin cipher nan twinkling exitance from this ray by dividing nan full twinkling power, by its aboveground area, giving:

$$ M_e = \frac{100 W}{4\pi(1m)^2} = 7.9577 \frac{W}{m^2} $$

Using nan pursuing relations:

\begin{align} M_e &= \frac{\partial \Phi_e}{\partial A} \\ L_{e,\Omega} &= \frac{\partial^2 \Phi_e}{\partial\Omega \partial (A\cos{\theta})}\\ \end{align}

I tin cipher nan emitted radiance from its aboveground by nan following:

\begin{align} L_{e,\Omega} &= \frac{\partial M_e}{\partial \Omega} \frac{1}{\cos{\theta}}\\ \int_SL_{e,\Omega} \cos{\theta} \ d\Omega &= M_e\\ \int_0^{2\pi}\int_0^{\pi/2} L_{e,\Omega}\cos{\theta}\sin{\theta} \ d\theta \ d\phi &= M_e\\ L_{e,\Omega} \pi &= M_e\\ L_{e,\Omega} &= \frac{M_e}{\pi}\\ \boxed{L_{e,\Omega} = 2.533 \frac{W}{sr \cdot m^2}} \end{align}

Now if I formed 2 rays into nan scene, 1 which lands 100m from nan light, and nan different which lands 10m from nan light, they should beryllium illuminated very differently. Yet, erstwhile I sample nan ray source, I'll get a ray which carries a radiance of $2.533 \frac{W}{sr \cdot m^2}$. How is nan truth that little full power (and frankincense little "illumination") makes it to nan further distant spot, accounted for successful this process?

My first thought would beryllium to relationship for nan angular size of nan light. That is to say, for a constituent $100m$ distant ($d$) I tin cipher nan coagulated perspective subtended by nan ray pinch respect to nan constituent I'm evaluating using:

\begin{align} \Omega &= 2\pi\left(1 - \frac{\sqrt{d^2 - R^2}}{d}\right)\\ &= 3.1417\times10^{-4} \ sr \end{align}

I tin usage this to get nan irradiance by multiplying this coagulated perspective by nan antecedently calculated radiance, I get:

\begin{aligned} E_e &= \Omega\cdot L_{e,\Omega}\\ &= (3.1417\times10^-4 \ sr) \cdot (2.533 \frac{W}{sr \cdot m^2})\\ &= 7.9579\times 10^{-4} \frac{W}{m^2} \end{aligned}

This matches pinch nan irradiance you'd cipher from a constituent ray pinch a twinkling powerfulness of $100W$, truthful it appears arsenic if each has been calculated correctly.

But that amount is irradiance, pinch units of $\frac{W}{m^2}$. And without multiplying by this coagulated angle, nan radiance remains changeless sloppy of nan region to nan ray source. So really precisely does a way tracer accurately simulate ray falloff, if it only tracks nan radiance on nan full path?

Looking astatine PBR's conception of Sampling Light Sources still leaves maine confused. Part of it states:

The Light’s Pdf_Li() method returns nan probability density pinch respect to coagulated perspective for nan light’s Sample_Li() method to sample nan guidance wi from nan reference constituent ref.

This feels akin successful tone to what I person done. As nan ray is further away, it subtends a overmuch smaller coagulated angle, and truthful conceivably you'd weight its radiance publication little (as nan further distant nan ray gets, nan little apt it is you'd intersect nan light). But I'm struggling to spot really this weighting process is different than simply multiplying nan radiance by nan coagulated perspective subtended by nan light, which would output irradiance, frankincense breaking nan full way tracing conception of summing radiance on a path.