I don't really know where to look in the world wide web for this specific problem. So I hope you can point me into the right direction.

Here is my specific problem:

I have a deformed spherical surface which emits light homogeneously. I want to obtain the intensity distribution on a larger, differently deformed spherical surface.

Here is a picture of what I trying to say:

enter image description here

The light is getting emitted isotropically from Surface A and goes onto Surface B, creating different intensities, here indicated in red (high intensity) to blue (low intensity).

Assuming Surface A and B are well behaved I think there might be a possibility to achieve that intensity by a map from surface A to surface B. Am I right? How exactly would I need to proceed?

Edit: I can assume that there is no self reflection of A, no scattering or reflection effects on any surface. A light source emitted in A gets absorbed in B no matter what.

Is a Gauss map applicable for this problem?


This is a specific case of the general problem of "radiative transfer," well studied in physics, engineering, and (in my area of interest) computer graphics. The key thing describing the process in my field is called "The Rendering Equation", and expresses the light arriving at a point of surface $B$ in terms of an integral of light leaving surface $A$, taken over all of $A$. There are multiple factors involved:

  • The distribution of emitted light from each point of $A$ (your "isotropically" appears to address this, but it's subtle; I suspect you mean that the emitted radiance from $a$ in direction $v$ is a constant function of $v$ (and possibly of $a$ as well).

  • The occlusion of points of $A$: not every point of $A$ is visible from every point of $B$, so there's a "visibility term" in the equation (one that typically cannot be expressed analytically)

  • A change-of-variable from integration over the sphere of arriving-directions-at-$b \in B$ to "integration over the surface $A$", which involves a couple of cosines of incident and exitant angles, and an inverse-squared-distance factor.

  • Self-reflection of $A$, and multiple-bounce reflections for $B$: light emitted from one point of $A$ may hit another, and then be reflected towards a point of $B$. Light hitting $B$ from $A$ may be reflected back towards $A$, and then back yet again towards $B$.

This last fact makes the rendering equation into an integral equation, in which the radiance field $L$ is related to an integral of $L$, just as in a differential equation like $y' = -3y + 2$, the function $y$ is related to its derivative.

Solving integral equations exactly is seldom possible; much of the field of computer graphics consists of approximating solutions to this particular equation in ways where the errors are (somewhat) bounded, and the time involved in computing the solution is (somewhat) bounded as well.

Short answer: this is just plain tough, and to do it well requires reading a lot. If you want an approximate but quick solution, just read any graphics paper from the 1970s. (But don't expect them to mention units like Watts per square meter steradian!)

  • $\begingroup$ To your first point: Yes, I can assume that every point radiates the same amount as any other point on the surface. And it furthermore also radiates isotropically. I also can assume that every point is visible and can be reached. There is no self reflection of A, no scattering or reflection effects on any surface. a light source emmitted in A gets absorbed in B no matter what. $\endgroup$ – Mr Puh Apr 17 '19 at 12:26
  • $\begingroup$ These simplifications are the reason why I think there might be a simple geometric map which solves this for me. I suspect something like a Gauß map (en.wikipedia.org/wiki/Gauss_map) but am unsure if this really captures my problem. $\endgroup$ – Mr Puh Apr 17 '19 at 12:27
  • $\begingroup$ My first point was that the thing that's constant as a function of direction is the radiance (not, for instance, power or energy); that's a subtle point. Assuming that this is what you mean, then the irradiance at a point $b$ of surface $B$ is a cosine-weighted integral of that constant radiance, the integral being taken over the solid angle subtended by $A$ from $b$. This is exactly the integral that's estimated in the basic "radiosity" solutions in computer graphics. Even for the simplest of shapes (one rectangle sending light to another), the integral turns out not to be elementary. $\endgroup$ – John Hughes Apr 17 '19 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.