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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 ; $E = h c \, / \lambda$ !

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    $\begingroup$ I don't know if it applies to light, but white noise is defined to be equal power content per hertz (i.e., uniform frequency unit). Pink noise is equal power per octave, so there is less power in high frequency ranges than in lower. So, for sound waves, it's generally a frequency distribution. $\endgroup$ – Bill N Oct 26 '16 at 16:28
  • $\begingroup$ @BillN, this is probably because our ears respond to frequencies, not to wavelengths, since it's an oscillating membrane. So frequencies are favored because of our body's workings. However, the sounds we produce depend on the length of our vibrating string in the throat. So wavelength favored ? $\endgroup$ – Cham Oct 26 '16 at 16:30
  • $\begingroup$ I interpret uniform to apply to intensity at different wavelengths or frequencies. It does not require a continuous spectrum to make "white" light. Ask any TV set! I would suggest the perception of white is based more on biology than physics. $\endgroup$ – bpedit Oct 26 '16 at 16:34
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    $\begingroup$ I don't know if there is an official definition, but I would also lean towards frequency, because it's what you get when you do a Fourier transform. $\endgroup$ – Javier Oct 26 '16 at 16:38
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    $\begingroup$ As I understand it, the technical term "white" applied to sound or light means the energy follows a continuous distribution of wavelengths or frequencies, take your pick. Whether the distribution is uniform, gaussian, lognormal, or something else, is up to choice. (Video displays typically have a discrete distribution of RBG, to match your eyes, so technically it's not white.) $\endgroup$ – Mike Dunlavey Oct 26 '16 at 19:07
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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.

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  • $\begingroup$ While I generally agree with the description of this answer, it doesn't say what is "white light" on an ideal theoretical point of view. The solar spectrum appears to be white because it is almost uniform in the visible spectrum. What I'm looking is the ideal theoretical definition of "white light". I don't see what it could be, if it isn't an uniform distribution of frequencies (or wavelengths, but not both at the same time). $\endgroup$ – Cham Oct 31 '16 at 17:44
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    $\begingroup$ There is no "ideal theoretical definition" of white light: it is an umbrella term that catches a number of different possible spectra. If you're a theorist and you want to describe white light as an abstract entity, you're free to pick any suitable spectrum; if the physics you want to describe turns out to depend on how you model the white light, then what you're interested in isn't actually white light - it's something more specific. $\endgroup$ – Emilio Pisanty Oct 31 '16 at 18:33
  • $\begingroup$ (And, in case you're wondering: yes, there's plenty of model-independent white-light physics. Absorption and transmission spectra, for example, require a broadband source to get all the data, but you normalize w.r.t. the incoming energy, so the initial distribution matters very little. Similarly, the assertion "white light has shorter temporal coherence than monochromatic light" is in general true regardless of the model - and you don't make any further assertions about that temporal coherence without specifying further assumptions about the source. And on it goes.) $\endgroup$ – Emilio Pisanty Oct 31 '16 at 18:36
  • $\begingroup$ Ok, you mostly confirmed what I believed, that there is no general theoretical definition of "white light". I'll mark your answer after a while, in case someone else give another interesting answer. By the way, in your answer, could you give the formula for the light intensity $I$ emitted by a black body, in watts/m^2 ? $\endgroup$ – Cham Oct 31 '16 at 18:52
  • $\begingroup$ @Cham It's given in Wikipedia, though you should note that the "intensity" emitted by a black body is the wrong question. Thermodynamics gives result for the (spectral) radiance: the power emitted (per unit spectrum) per unit area at the source and per unit solid angle. The intensity at a point away from the body depends on the relative geometry. For more details, see this Wikipedia page. $\endgroup$ – Emilio Pisanty Oct 31 '16 at 19:02
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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.

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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.

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  • $\begingroup$ Does this tend to confirm that a complex detector could be reacting to frequencies, and not to wavelengths, and that there are also detectors that reacts to wavelength instead of frequencies ? And that "white" is thus defined relative to a particular detector ? $\endgroup$ – Cham Nov 2 '16 at 15:18
  • $\begingroup$ White is meningfully defined only relative to a very specific detector: the human eye+brain :) the question whether a detector reacts to $\lambda$ or to $\omega$ is also misleading. $\endgroup$ – Ilja Nov 2 '16 at 17:52
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    $\begingroup$ In practice there are three types of detectors for colours, three chemical substances in the eye. To compute, what signal they measure, take the incoming spectrum, multiply it with the corresponding sensitivity: upload.wikimedia.org/wikipedia/commons/thumb/f/f9/… and take the integral. The curve would be different in respect to the frequency, and you would have to know the incoming distribution of frequencies; but the choice - frequency or wavelength - is not a property of the detector, only of your method of computation. $\endgroup$ – Ilja Nov 2 '16 at 17:54
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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 ?

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  • $\begingroup$ I agree that my answer isn't totally satisfying. The white light defined above isn't a very usefull terminology, since it's equivalent as saying "the distribution is uniform" ! $\endgroup$ – Cham Oct 29 '16 at 17:40
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    $\begingroup$ This is a terrible answer. A detector doesn't "react to frequency or to wavelength" - it reacts to light, period, and if it produces a frequency power spectrum it is equivalent to an output as a wavelength power spectrum. You're right that it boils down to what one means by "white" (i.e. that the term does not have a unique meaning) but everything else here is hogwash. $\endgroup$ – Emilio Pisanty Oct 31 '16 at 14:09
  • $\begingroup$ @EmilioPisanty, I don't agree that this answer is "hogwash". Your detector cannot just "react to light, period", like a magic process. Depending on the way the detector is designed, frequency (or wavelength), or another aspect of light (amplitude ?, number of photons ?) would trigger a physical process and activate a registration. AFAIK, the cones in your eyes don't react to wavelengths (they're not "rulers"), they react to frequency (molecules oscillations or wathever). Or maybe it's the contrary ? The lenght and shapes of the cones may be reacting to wavelenghts ? $\endgroup$ – Cham Oct 31 '16 at 14:24
  • $\begingroup$ @Cham Frequency and wavelength are in a direct one-to-one correspondence. The mechanism of interaction is irrelevant - the detector will produce both a frequency and a wavelength spectrum, because the two are the same thing to within a trivial transformation. $\endgroup$ – Emilio Pisanty Oct 31 '16 at 14:30
  • $\begingroup$ There's no detector with uniform sensitivity in frequency or wavelength. Real detectors will always have some bell-shaped spectral sensitivity curve if not even more complex one. So your talk about "designed to detect wavelengths" or "designed to detect frequencies" is not about real detectors, but rather about two extremes of ideal detectors, and neither of these extremes seems really useful. $\endgroup$ – Ruslan Oct 31 '16 at 15:30
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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!

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