Rant

People have used up all the English words that mean “light” in photometry, light, lumen, lux, luminance, illuminance. Wikipedia on this subject and many other online sourses are fuzzy with definitions, fuzzy in the sense that I cannot understand enough to be able to use them. Thus I make an attempt to understand in this snippet.

Let \Omega denote the whole space. It is clear at this point that I am no physicist. Otherwise I would use \Omega for the solid angle. I cannot appreciate the assignment of a fixed symbol to a physical object. There are so many objects and yet so few easily accessible symbols. Thus one should assign the symbols adaptively: assign nice short symbols to objects on which one is going to talk most often. Anyway I try to respect the convention if possible, for example, \rho is some density.

Power flux

Let \Omega denote the whole space. A light source is a surface in \Omega that consumes energy and emits light. All surfaces are required to be orientable in this snippet. Let S_{L} denote the surface. Let \mu_{L} denote the measure on S_{L}. All the measures in this snippet are derived from the standard euclidean metric on \Omega. Let \rho_{L}(X) denote the density of the emitted light power at X\in S_{L}. Note that power means energy per unit time. Then the total emitted light power of S_{L} is given by \begin{aligned} \label{eq:total-emitted-light-power}\tag{eq:total-emitted-light-power} \int_{S_{L}} \rho_{L}(X) d\mu_{L}(X). \end{aligned}

The light source S_{L} produces a vector field \mathbf{F} on \Omega which is the light power flux density. For a surface U\subseteq \Omega with measure \mu_{U}, the flux of light power through U is given by \begin{aligned} \label{eq:flux-light-power}\tag{eq:flux-light-power} \int_{U} \mathbf{F}(Y) \cdot \mathbf{n}_{U}(Y) d\mu_{U}(Y), \end{aligned} where \mathbf{n}_{U} denotes the normal vector field on U.

Assume that the contribution to \mathbf{F}(\cdot) from each point X of S_{L} is given by the vector field \mathbf{f}(X,\cdot) such that \begin{aligned} \int_{S_{L}} \mathbf{f}(X,Y) d\mu_{L}(X) = \mathbf{F}(Y). \end{aligned} In general, the contributions may not be additive and the light source can self-obstruct or interfere with itself.

We need to make some further assumptions to link \rho_{L}(X) to \mathbf{f}(X,\cdot). By the principle of conservation of energy, we have \begin{aligned} \int_{S_{L}} \rho_{L}(X) d\mu_{L}(X) = \int_{U} \int_{S_{L}} \mathbf{f}(X,Y) \cdot \mathbf{n}_{U}(Y) d\mu_{L}(X) d\mu_{U}(Y), \end{aligned} where now U is a surface enclosing the light source. We thus assume \begin{aligned} \label{eq:power-density-identity}\tag{eq:power-density-identity} \rho_{L}(X) = \int_{U} \mathbf{f}(X,Y) \cdot \mathbf{n}_{U}(Y) d\mu_{U}(Y), \end{aligned} By the square inverse law, we expect that \mathbf{f}(X,Y) = \frac{\alpha(X)}{d^{3}(X,Y)} \mathbf{v}(X,Y) where \alpha(X) is a constant of proportionality to be determined, d(X,Y) is the distance function and \mathbf{v}(X,Y) denotes the vector \overrightarrow{XY}. Assume now that U is a sphere centred at X with radius R. Then the RHS of [eq:power-density-identity] simplifies to \begin{aligned} \int_{U} \frac{\alpha(X)}{ R^{2}} d\mu_{U}(Y) = 4\pi \alpha(X). \end{aligned} Thus we take \alpha(X) = \frac{\rho_{L}(X)}{4\pi}. If we assumed that \mathbf{f} depends on the distance only with direction given by \mathbf{v}(X,Y), the above computation would show that the only way for \alpha to be independent of Y is for \mathbf{f} to take the form \mathbf{f}(X,Y) = \frac{\alpha(X)}{d^{3}(X,Y)} \mathbf{v}(X,Y).

An attempt to match terminology

Next we try to match with the terminology in Handprint¹.

The luminous flux is the total emitted light power, so it is given by [eq:total-emitted-light-power].

The luminous intensity is the luminous flux per solid angle, so it is given by \begin{aligned} \frac{1}{4\pi} \int_{S_{L}}\rho_{L}(X) d\mu_{L}(X) . \end{aligned}

Now U is again an arbitrary surface in \Omega. The illuminance is the light power received by U per unit area, so, at the point Y\in U, it is given by \begin{aligned} \mathbf{F}(Y) \cdot \mathbf{n}_{U}(Y) &= \int_{S_{L}} \mathbf{f}(X,Y) \cdot \mathbf{n}_{U}(Y) d\mu_{L}(X)\\ &= \int_{S_{L}} \frac{\rho_{L}(X)}{4\pi d^{3}(X,Y)} \mathbf{v}(X,Y) \cdot \mathbf{n}_{U}(Y) d\mu_{L}(X) . \end{aligned} Assume the light source is small relative to the illuminated surface U. Assume \rho_{L}(X) is independent of X\in S_{L} and we write \rho_{L} for this constant value. Then the illuminance at the point Y\in U can be approximated by \begin{aligned} \mu_{L}(S_{L}) \frac{\rho_{L}}{4\pi R^{2}} \cos(\theta), \end{aligned} where R is the approximate distance from the light source to Y and \theta is the angle between \mathbf{n}_{U}(Y) and the approximate unit vector pointing from the light source to Y.

The luminance is the contribution from a point X\in S_{L} to the illuminance at Y\in U adjusted to account for the distance and the surface normal of S_{L} at X, i.e., the illuminance divided by the ‘visual size of the light source’. It is given by \begin{aligned} \lim_{S\rightarrow \{ X \}} \frac{ \int_{S} \mathbf{f}(X',Y) \cdot \mathbf{n}_{U}(Y) d\mu_{L}(X') }{ \frac{\mu_{L}(S)}{ d^{3}(X,Y)} \mathbf{v}(X,Y) \cdot \mathbf{n}_{L}(X) } = \frac{d^{3}(X,Y) \mathbf{f}(X,Y) \cdot \mathbf{n}_{U}(Y) }{ \mathbf{v}(X,Y) \cdot \mathbf{n}_{L}(X) }. \end{aligned} where S runs over open subsets of S_{L} that contain X. It simplifies to \begin{aligned} \frac{\frac{\rho_{L}(X)}{4\pi d^{3}(X,Y)} \mathbf{v}(X,Y) \cdot \mathbf{n}_{U}(Y)}{ \frac{\mathbf{v}(X,Y) \cdot \mathbf{n}_{L}(X)}{ d^{3}(X,Y)} } = \frac{\rho_{L}(X)}{ 4\pi} \frac{ \mathbf{v}(X,Y) \cdot \mathbf{n}_{U}(Y) }{ \mathbf{v}(X,Y) \cdot \mathbf{n}_{L}(X) }. \end{aligned}

I guess the relation between Irradiance and Radiance is the same as Illuminance and Luminance.