White Light equals Brown Noise

Here is a recent show­er thought:

It is usu­al­ly said that the col­ors of noise are inspired by the spec­tral dis­tri­b­u­tions of cor­re­spond­ing col­ors of light. For exam­ple, with ‘white’ in white noise is an allu­sion to white light which is thought to have a (most­ly) flat spec­trum. But is this so? How does white light actu­al­ly look like, as an elec­tro­mag­net­ic wave?

So I made fol­low­ing fig­ure which shows some pow­er-law dis­tri­b­u­tions in the vis­i­ble range of wave­lengths col­ored by their the­o­ret­i­cal appearance:

Fig 1: Spec­tral dis­tri­b­u­tions with expo­nents from 0 to −4.

The col­or of the flat line is also known as Stan­dard Illu­mi­nant E‍‍ – or equal-ener­gy white light. Com­pared to the D65 white back­ground it has a rather pink­ish appear­ance with a cor­re­lat­ed col­or tem­per­a­ture of about 5500 K.

The oth­er extreme is a dis­tri­b­u­tion where the spec­tral inten­si­ty is inverse­ly pro­por­tion­al to the wave­length raised to the fourth pow­er, or {d}I \sim \lambda^{-4} {d}\lambda‍. This was both the col­or of pure Rayleigh scat­ter­ing (sky blue) and also ther­mal radi­a­tion in the lim­it of infi­nite temperature.

Now wait a minute. White noise is defined as a flat fre­quen­cy spec­trum, but white light is defined as a flat wave­length spec­trum. This is gen­er­al­ly not the same: To con­vert a “per wave­length” dis­tri­b­u­tion to a “per fre­quen­cy” dis­tri­b­u­tion one must do a prop­er change of vari­able.

So let’s con­vert the dis­tri­b­u­tions from above from wave­length \lambda to wavenum­ber \nu (that’s the greek let­ter ν). In doing so, the dis­tri­b­u­tion acquires an addi­tion­al fac­tor of 1 \over \nu^2‍, since \lambda and \nu are rec­i­p­ro­cals of one another:

    \[\int I_\lambda(\lambda) \operatorname{d}\lambda \Leftrightarrow \int I_\nu(\nu) {\operatorname{d}\lambda \over \operatorname{d}\nu} \operatorname{d}\nu \Leftrightarrow \int {I_\nu(\nu) \over \nu^2} \operatorname{d}\nu .\]

Fig 2: The same dis­tri­b­u­tions as in fig. 1, but now as a wavenum­ber spec­trum. The cus­tom­ary unit for wavenum­ber is cm-1.

So who is the flat one now?

It turns out that the mid­dle guy, the dis­tri­b­u­tion that fol­lows an expo­nent of −2 in the wave­length pic­ture (col­or tem­per­a­ture about 10000 K) is the flat one in the wavenum­ber pic­ture. An elec­tro­mag­net­ic wave in the shape of white noise – in the visu­al band at least – should there­fore appear as this color.

On the oth­er hand, what is usu­al­ly called white light now fol­lows an expo­nent of −2 in the wavenum­ber pic­ture. White light equals brown noise!
In fact, for any pow­er-law dis­tri­b­u­tion we have

    \[\int \lambda^{-\alpha} \operatorname{d}\lambda \Leftrightarrow \int \nu^{\alpha-2} \operatorname{d}\nu ,\]

which leads to the fact that the dis­tri­b­u­tion with an expo­nent of −1 is the same in both pic­tures. This is the pale col­ored one with a col­or tem­per­a­ture of about 7000 K.

So to sum­ma­rize, the white in “white noise” is, in fact, a bla­tant misunderstanding :)

pow­er laws noise col­or
I_\lambda \operatorname{const.} I_\nu \sim \nu^{-2} brown 5456 K          Illu­mi­nant E
I_\lambda \sim \lambda^{-1} I_\nu \sim \nu^{-1} pink 7029 K          invari­ant exponent
I_\lambda \sim \lambda^{-2} I_\nu \operatorname{const.} white 9926 K          EM white noise
I_\lambda \sim \lambda^{-4} I_\nu \sim \nu^{2} blue ∞ K          Rayleigh scat­ter­ing


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