What is the waveform of white light?

Question

I don't know very much about the theory of light, but since people talk about light as a wave and talk about the wavelengths of light, I'm going to assume that whenever you see any source of visible light, it's meaningful to talk about the waveform of light, i.e. a real-valued (or possibly vector valued) periodic amplitude function over time, $A(t)$. If two different light waves $A_1(t)$ and $A_2(t)$ pass through the same space, they produce a waveform $A_1(t)+A_2(t)$, just like with sound. When we say "the wavelength of blue is $\ell$ nanometers", what we really mean is that a beam of light whose waveform is a sinusoid of wavelength $\ell$ looks blue. When we say that a certain light source is "a mixture of blue and red", we really just mean that there is a single wave of light passing through the area with a single well-defined amplitude function, but that that function is mathematically what we would get if we added two sinusoids of blue and red wavelength - we do not mean that there are somehow two separate waveforms somehow occupying the same space.

At least to degree to which it's meaningful to talk about light as a wave at all (i.e. setting aside wave/particle duality stuff), is everything I said correct? If so, then what is the waveform of white light? Of course it's "a mixture of all colors of light", but at the end of the day when you add together multiple functions you get one function. What is the "final" waveform of white light? Or are there many different waveforms (corresponding to slightly different mixtures of the colors), all of which our brain perceives as "white"?

Answer 1

When we talk about that sort of waveform, we call it "white noise". It has random amplitude, with variance representing the intensity.

Answer 2

"White" is a very vague description of light. There's a multitude of different spectra that we'd see as white (see metamerism). Additionally, there are different color temperatures that we could consider white depending on lighting in the place we are (see chromatic adaptation). So there's no unambiguous waveform for white light.

Answer 3

You are making an important assumption that is very commonly conveniently swept under the rug when we teach physics to students.

Professors of physics knows that, for any linear system, the complicated behaviour they exhibit can be studied by considering pure waveforms, and then you can think of the actual behaviour as the classical mixture of them. Classical mixtures, classical ignorance, is physically intuitive and easy to understand.

Pure waveforms are easy to mathematically model, and when we want to extract the mathematical solutions from the fundamental equations, these are what we do. And then we also prove that linear superpositions work for them, so we can have a great variety of pure waveforms, with which to explain a great variety of physical behaviour.

The part that is swept under the rug is that white noise is actually not representable that way. It is not obvious that we have one single waveform!

The issue is this: the mathematically nice sinusoidal stuff we use to model physics, is not actually that nice in practice. Any waveform we generate, there is a finite coherence time and coherence length, outside of which the waves cannot be taken to interfere sensibly. The real fields that we are normally not seeing, are actually often jagged Brownian motion messes rather than the smooth waveforms that we tend to think about.

Let us consider the blue and red light that you brought up. If we have a pure waveform made by some fixed combination of blue and red light, then, yes, you can add the amplitude (electric field, say) of both, and get a well-defined waveform. But when we speak of white noise, the problem is that the phase difference between the blue and red light is no longer a constant, but is rather a random phase uniformly distributed over $[0,2\pi)$, and that is just not something so obvious to put into mathematical form.

Only if you are really good with mathematics, will you get taught how to deal with classical mixtures of different pure states. It is actually not that difficult, just ugly and tedious. It brings very little new understanding of physics to students, so there is not much sense for professors to force students to learn about it.

Answer 4

It was Isaac Newton who demonstrated that white light was a composition of multiple colours, by passing it through a prism, which broke it up into components, and reassembled it by passing it though a second prism.

The screen of a colour television, for example, produces white by mixing the primary colours of light: red, green and blue, at high intensity. Hence the "waveform" which you seek could be an amalgamation of all these colours. Unfortunately, the latter is a misconception. Nevertheless, we only see a small portion of the electromagnetic spectrum - which we call visible light - and this is not visible above 700 nm or below 400 nm. Since white light consists of wavelengths of different colours, its overall wavelength is 400 - 700 nm.

Answer 5

Of course it's "a mixture of all colors of light", but at the end of the day when you add together multiple functions you get one function. What is the "final" waveform of white light?

I'm uncertain what description you're expecting here. There is no simpler way to express the waveform than as "the summation of multiple other visible frequencies".

The addition of two discrete frequencies leads to something that looks pretty simple. You might even be able to recognize it directly without analyzing the specific components.

But even that shape isn't completely uniform. Unless the two components are an integer ratio of each other, the phase difference of the two will be constantly shifting, changing the specifics of how the waveform appears.

Now instead of two discrete frequencies you add up a continuous spectrum and there's just no way to describe the result other than the power of the component frequencies.

And that's true just for one particular spectrum. In fact there are many different spectra that could all be described as "white".

Answer 6

Radio engineers call $\omega_0$ a carrier frequency of a time function $s(t)$, i.e., a waveform, if it can be represented as $s(t)=A(t)cos(\omega_0t +\phi(t))$ where the time rate of change (modulation rate) of the amplitude $A(t) \ge 0$ and phase $\phi(t)$ functions is much smaller than $\omega_0$. Formally this means that $$\left | \frac{1}{A(t)}\frac{dA}{dt} \right |\ll \omega_0, \\ \left | \frac{d\phi}{dt}\right |\ll \omega_0 .$$

Not all functions can be represented this way but if yes then we call it "narrowband". When we talk about blue or red light then we talk about a narrowband waveform. In the case of blue light, for example, $\omega_0/(2\pi) \approx 650 \text{THz}$. As long as the modulation rates does not exceed $\approx \pm 20\text{THz}$ the waveform's color is essentially blue. Almost all communications and radar systems use this kind of narrowband waveforms for very practical reasons that are not important to this question.

When you talk about white light you have a waveform that has no carrier frequency because the bandwidth between red to violet is almost an octave and white light has frequency content from the whole spectrum.

You can still write a composite waveform consisting of a narrowband red and a narrowband blue signal as $s_{rb}(t)=A_r(t)cos(\omega_rt +\phi_r(t)) + A_b(t)cos(\omega_bt +\phi_b(t))$ and this may correspond to a plane wave, say, propagating in the $\hat z$ direction and linearly polarized along the $\hat x$ as $\mathbf E = s_{rb}(t)\hat x $ and $\mathbf H = \sqrt{\frac{\epsilon_0}{\mu_0}}s_{rb}(t)\hat y $.

If you add up about a dozen equal amplitude uniformly spaced sinusoids the resulting signal has a nearly 2D normal (Gaussian) distribution, $X\cos(\omega_0t)+Y\sin(\omega_0t)$, where $X,Y \in \mathcal N(0,\sigma^2)$ are i.i.d. normal random variates, and as the number of sinusoids increase the sum converges to that true normal distribution. And this is what happens when you fill the spectrum with roughly equal amplitudes colors.

Answer 7

This isn't really my own answer. I just want to squeeze what naturallyinconsistent said into a nut shell.

In that nutshell: The "waveform" of white light is uninteresting. It's uninteresting because white light is the superposition of many different "waveforms." Usually, it's so many that describing it's shape would be a futile exercise. But, because superposition is a linear phenomenon, it's also completely unnecessary to describe it. Everything we want to know about white-light-as-waves can be understood as the sum of what we know about its individual component waves.