WARNING
Content Under Development
See
release page for
latest official PDF version.
**Chapter 3: MC Integration on the Sphere of Directions**
In our ray tracer we pick random directions, and directions can be represented as points on the
unit-sphere. The same methodology as before applies. But now we need to have a pdf defined over 2D.
Suppose we have this integral over all directions:
$$ \int cos^2(\theta) $$
By MC integration, we should just be able to sample $cos^2(\theta) / p(direction)$. But what is
direction in that context? We could make it based on polar coordinates, so $p$ would be in terms of
$(\theta, \phi)$ . However you do it, remember that a pdf has to integrate to 1 and represent the
relative probability of that direction being sampled. We have a method from the last books to take
uniform random samples in a unit sphere:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++
vec3 random_in_unit_sphere() {
vec3 p;
do {
p = 2.0*vec3(drand48(),drand48(),drand48()) - vec3(1,1,1);
} while (dot(p,p) >= 1.0);
return p;
}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To get points on a unit sphere we can just normalize the vector to those points:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++
vec3 random_on_unit_sphere() {
vec3 p;
do {
p = 2.0*vec3(drand48(),drand48(),drand48()) - vec3(1,1,1);
} while (dot(p,p) >= 1.0);
return unit_vector(p);
}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now what is the pdf of these uniform points? As a density on the unit sphere, it is $1/area$ of the
sphere or $1/(4\pi)$. If the integrand is $cos^2(\theta)$ and $\theta$ is the angle with the z axis:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++
inline float pdf(const vec3& p) {
return 1 / (4*M_PI);
}
int main() {
int N = 1000000;
float sum;
for (int i = 0; i < N; i++) {
vec3 d = random_on_unit_sphere();
float cosine_squared = d.z()*d.z();
sum += cosine_squared / pdf(d);
}
std::cout << "I =" << sum/N << "\n";
}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The analytic answer (if you remember enough advanced calc, check me!) is $\frac{4}{3} \pi$, and the code
above produces that. Next, we are ready to apply that in ray tracing!
The key point here is that all the integrals and probability and all that are over the unit sphere.
The area on the unit sphere is how you measure the directions. Call it direction, solid angle, or
area-- it’s all the same thing. Solid angle is the term usually used. If you are comfortable with
that, great! If not, do what I do and imagine the area on the unit sphere that a set of directions
goes through. The solid angle $\omega$ and the projected area $A$ on the unit sphere are the same thing.
Now let’s go on to the light transport equation we are solving.