2
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

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.

So, the real answer is "it doesn't matter that much".

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – Troy Woo
    Commented Apr 7, 2015 at 19:21
  • $\begingroup$ Exactly. The spectral power distribution $I(\lambda)$ is the energy flux (power per unit area) per unit wavelength. $\endgroup$
    – Emilio Pisanty
    Commented Apr 7, 2015 at 19:26
  • $\begingroup$ Thanks again. In fact since you mentioned flux, I am really confused since flux should just be power instead of power per unit area according to wiki. However, my computer vision teacher also use the same one as yours...it may be relevant to consider electric flux and magnetic flux which both being integral over a surface, rather than an "intensity"...could you come up with an explanation? $\endgroup$
    – Troy Woo
    Commented Apr 7, 2015 at 20:06
  • $\begingroup$ 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! $\endgroup$
    – Emilio Pisanty
    Commented Apr 7, 2015 at 20:14
  • $\begingroup$ ..........................................well explained, I think I could figure out. Many thanks from a physics illiterate. Also, if you may, please kindly answer my another question. It is being edited now. $\endgroup$
    – Troy Woo
    Commented Apr 7, 2015 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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