I'm reading through a computer graphics paper and author says that the radiant intensity emitted by an infinitely small surface point $p$ with normal $ n $ into direction $ \omega $ is

$$ I_p(\omega) = \phi_p max(0,<n|\omega>) $$

where $\omega_p$ is the radiant flux emitted by the surface, and $<a|b>$ denotes the dot product.

First question: Am I correct in assuming that this is only true for an infinitely small solid angle?

The author then claims that the irradiance due to this surface point $p$ arriving at another surface point is

$$ E_p(x,n) = I_p(x-x_p) \frac{ max(0,<n|x-x_p>) }{||x-x_p||^4} = \phi_p \frac{max(0,<n_p | x-x_p>) max(0,<n|x-x_p>) }{||x-x_p||^4} $$

where $x_p$ is the location of the emitting surface point, $x$ is the location of the surface point where the light arrives, $n$ its normal, and $||x||$ is the norm (length) of $x$. The author doesn't mention it but I'm assuming he's always talking about infinitely small surface points, since no areas seem to be involved.

Second question: Where does $||x - x_p ||^4$ come from? I know that the radiant flux incident on a surface point due to a light source diminishes with the square of the distance from the surface point to the light source, but this in my mind only constitutes a term of $||x - x_p||^2$. Where does the 4 come from?


2 Answers 2


Seems like an awfully complicated way to go about it.

For starters, Radiant Intensity, measured in Watts per steradian, is only defined for "point" sources, so no surface areas, or distances are involved. But even a "point" source would generally have some angular distribution pattern, that seems undefined in your citation. The simplest concept, would be an isotropic point source, but no such thing can physically exist.

Small area plane sources, will often have "Lambertian" radiation patterns, where the off axis intensity is given by: $$ I = I(0)\cos(\theta) . $$ This is only accurate when observed at distances greater than ten times the source diameter, where errors in measured intensity, will be less than 1/2%.

The irradiance on some distant surface normal to the source axis, at some surface point off at an angle theta, is given by $$ I(0)\cos^4(\theta) / d^2 , $$

where $d$ is the axial distance from source to surface.

The $\cos^4(\theta)$ term arises from $I(\theta) = I(0) \cos(\theta)$, then two more $\cos(\theta)$ terms for the oblique distance squared, and the fourth $\cos(\theta)$ from the incidence angle of the receiving surface, which spreads the radiation further.

For a Lambertian source, the total radiated power (Watts) is $\pi I(0)$.

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The answer popped into my head moments after hitting the submit button.

$x-x_p$ is a direction, but not a normalized one. For the equation of $I_p(\omega)$ to be correct, the direction has to be normalized:

$$\omega_{x-x_p} = \frac{x-x_p}{||x-x_p||}$$

Since $x-x_p$ appears twice in the equation, the normalization is done twice, which in combination with the division by the squared distance is the reason for the power of 4.

edit: I can't actually answer my own first question, though.


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