# What are the units of color matching functions?

In some computer vision book I read lately, the color matching function is invoked without clear definition of its units. I suspect the color matching functions are spectral irradiance or spectral illuminance, i.e. with unit $\mathrm W/\mathrm m^3$ or $\mathrm{lm}/\mathrm m^3$, i.e. $\mathrm W$ or $\mathrm{lm}$ per square meter per wavelength. The inconsistency seems a common problem in books written by engineers. Am I correct?

I know this is only partially a physical question, but I guess physicists are more suitable for answering the question.

Edit: never mind, I figured it ought to be dimensionless. But if I am wrong, please still correct me.

The CIE standard colorimetric tables give $\bar x(\lambda)$, $\bar y(\lambda)$ and $\bar z(\lambda)$ as pure numbers, normalized so that $\int\bar x(\lambda)d\lambda$ is the same for the three components.
In the end, it boils down to what units you expect your tristimulus signals $X$, $Y$ and $Z$ to have. For dimensionless $\bar x(\lambda)$, $$X=\int I(\lambda)\bar x(\lambda)d\lambda$$ has the units of intensity. This is actually perfectly natural, because it depends directly on $I(\lambda)$ and has not been normalized, so if you double the brightness of the source then $X$ will double as well. To get rid of this, the actual chromaticity variables used are $$x=\frac{X}{X+Y+Z}$$ and similarly for $y$ and $z=1-x-y$. These are naturally dimensionless, do not change with the light intensity, and are independent of what units you give to the color matching functions. If you're matching XYZ colors to other color spaces, it's these functions that matter.
• Thx. In fact you mean $I(\lambda)$ is the spectral intensity? that is, with an additional $\mathrm m^{-1}$ or $\mathrm{nm}^{-1}$ or whatever that is for the wavelength. – Troy Woo Apr 7 '15 at 19:21
• Exactly. The spectral power distribution $I(\lambda)$ is the energy flux (power per unit area) per unit wavelength. – Emilio Pisanty Apr 7 '15 at 19:26
• Apologies, that was loose language on my part (mostly intended to get the character >15chars). $I(\lambda)$ is either spectral radiance or spectral irradiance, depending on the setting (the difference between those two being integration over solid angle, which is irrelevant for an optical system with fixed aperture). Note that irradiance is the same thing as energy flux, which is distinct from radiant flux. (The flux of a quantity is the amount of it that passes a unit area per unit time.) Confusing language indeed! – Emilio Pisanty Apr 7 '15 at 20:14