What is "white light" ? Uniform wavelengths or uniform frequencies ?

Question

Suppose you have a mixture of electromagnetic waves of wavelengths spreaded on the visible spectrum only (from $\lambda_{\text{min}} \sim 400 \, \text{nm}$ to $\lambda_{\text{max}} \sim 700 \, \text{nm}$). At some ideal detector, the light spectral distribution is described by a functional like this : $$\tag{1} I = \int_{\lambda_{\text{min}}}^{\lambda_{\text{max}}} L(\lambda) \, d\lambda. $$ Since $d\omega \propto \lambda^{-2} \, d\lambda$, we could also define the spectral distribution with angular frequencies : $$\tag{2} I = \int_{\omega_{\text{min}}}^{\omega_{\text{max}}} F(\omega) \, d\omega, $$ where $F(\omega) = \lambda^2 \, L(\lambda)$. So the two functions $L(\lambda)$ and $F(\omega)$ are two complementary ways of defining the spectral distribution.

Usually, "white light" is described or defined as an uniform mixture of waves. But on which distribution ? Wavelengths or frequencies ? i.e. $L(\lambda) = \textit{cste}$ or $F(\omega) = \textit{cste}$ ? It cannot be both at the same time ! Why favor one or another function ? A photon's energy depends on frequency ; $E = \hbar \, \omega$, but we could also say that it depends on the wavelength ; $E = h c \, / \lambda$ !

Answer 1

Your assertion that

Usually, "white light" is described or defined as an uniform mixture of waves

is pretty much completely incorrect: this is not how the term "white light" is treated in the literature. The meaning of the term is relatively well captured by this glossary at Plastic Optics:

light, white. Radiation having a spectral energy distribution that produces the same color sensation to the average human eye as average noon sunlight.

However, the term is not normally taken to have a strict technical meaning, a fact which is well reflected by the observation that in the first page of a search for "optics glossary" only a single resource has an entry for "white light".

The meaning of the term is even more complicated because it depends on who is using it:

• If it is a spectroscopist that needs a white-light source to obtain a reflectivity or absorptivity spectrum, they will usually require the light to have a broad bandwidth, with support over the entire visible-light range, to be called "white".

• However, if it's a manufacturer of light bulbs, they will only require that the light be perceived as white, even if it is produced e.g. by three-colour LEDs with narrow-band spectra like this one, and their use of the term will be completely justified.

In terms of its use within the physics literature, it is much more usual to require a broadband source, with a large continuum of wavelengths contributing significantly to the spectrum. However, there isn't a requirement that all the frequencies contribute equally (partly because, as you note, that doesn't even begin to make sense).

Thus, a flat wavelength spectrum (over a broad enough range) will normally be called "white", but so will a flat frequency spectrum over an equivalent range. Moreover, many of the standard models of white light do not have a flat spectrum in either representation, with the most famous model being, of course, blackbody radiation. This has frequency and wavelength spectral distributions of the form $$ P_\nu(\nu,T) = \frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}-1} \quad\text{and}\quad P_\lambda(\lambda,T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_B T}-1} \quad \text{resp.}, $$ and at high enough temperature (i.e. $T\approx 5500\:\mathrm K$) it models white sunlight. At lower temperatures, such as those in incandescent light bulbs, it produces a rather different spectrum, which is still called white light in the literature.

Answer 2

There isn't just one "white light". There's Illuminants A,B,C,D,E,F, there's blackbody white (with a continuum of possible temperatures), etc.

That's why when you buy a lightbulb, they mark the color temperature, but no matter what the color temperature is, they still (properly) label it a "white light".

You say "Usually, "white light" is described or defined as an uniform mixture of waves." This is not true in my experience. It is defined as broadband light, and/or as light that a reasonable non-technical person would describe as "white" if they looked at it, but not as any specific spectrum.

Answer 3

All answers miss a very interesting point, which might be not quite what you asked, but which explains why there is such a concept as "white" though it is not a physical concept but rather a biological issue.

You will see your green plants on your table as green (nearly)regardless of the light in your room; to accomplish this, your brain has to not simply measure the spectral distribution arriving from the plant, but has to compare it with the light source - in order to "compute" what fraction of the amount of light of each frequency is reflected by the object. This is obviously a very useful feature of our neural system, since it "measures" the property of the object and not of accidental external circumstances (illumination).

How is this accomplished?

Well, you have to compare the wavelengths arriving from the object with the average incoming light in your field of view. It would be more precise to compare it with the source, but the biologically evolved mechanism has to be quick and versatile; it would need too much intelligence and delay time to search for the source (or an object that is known to be white) every time you want to check a colour.

Of course this is not done rationally, it's much simpler. Colours are "defined" (by the brain) to be in pairs of complementary colours, which "cancel out" giving... white! White can be a mixture of red and green, or blue and yellow, etc. When light falls on some part of your retina, the sensation of a colour is produced there - and at once the complementary colour is sent to the rest of the picture, and superpones with the real colour there! Obviously, if some frequency should be missing in the illuminating light, this is compensated by this mechanism if the objects are sufficiently randomly coloured (white on average :)). If the light is white, then all these imaginary colours cancel out, and don't have any effect.

This is why biology invented the colour white, though it does not correspond to any frequency. A very important mechanism - think about it, of how little use it would be to detect the incoming frequencies, without taking into account how they depend on the illumination. This allows you to get the sensation of a green plant even if there are no green wavelengths at all in the light present; the green is produced by the surplus of red in the light from all the other objects.

This is why the very concept of white is only useful in this context of human colour-perception. The spectra can be wildly different. Its definition is: white is what you see to be white. This may vary slightly between individuals.

Answer 4

While I was thinking about this, during my bus travel to my job, I had a simple idea that may be a proper answer. I need a confirmation that this is actually making sense.

The notion of white light is dependant on the detector. An ideal detector (call it an "eye-brain", or a digital camera connected to a computer) reacts to frequency or to wavelength, but not both at the same time. In the case of a frequency reacting detector, light is said "white" (i.e. the detector perceives white light) if the frequencies are uniformly distributed ; $F(\omega) = \textit{cste}$. The wavelength reacting detector would then see some blueish light instead, since we have $$\tag{1} d\omega = -\: 2 \pi \lambda^{-2} \; d\lambda, $$ and thus $$\tag{2} L(\lambda) \propto \lambda^{-2} \, F(\omega) \propto \lambda^{-2}. $$

An uniform distribution of wavelengths ($L(\lambda) = \textit{cste}$) implies that small frequencies are favoured ; $$\tag{3} F(\omega) \propto \lambda^2 \, L(\lambda) \propto \omega^{-2}. $$ The frequency detector then "sees" some reddish light, while the wavelength detector sees "white" light.

The conclusion is that "white light" is a relative concept, and doesn't have any physical sense without a detector, which cannot react both to frequencies and wavelengths at the same time ! In other words, the question "What is the color of that light ?" cannot have an answer without specifying the kind of detector used.

Does that makes sense ? Any comments on this ?

Answer 5

If you are able to build your "ideal detector," you will find that if it responds uniformly to "white light" frequencies, it will give you **the same response* to their wavelengths. In other words, the frequency detector and the wavelength detector - are one and the same!