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The wave nature of light comes from Maxwell's equations. More precisely, the two wave equations that come from them:

$$\Delta\vec{E}=\mu\varepsilon \frac{\partial^2\vec{E}}{\partial t^2}\\ \Delta\vec{H}=\mu\varepsilon \frac{\partial^2\vec{H}}{\partial t^2}$$

Looking only at the electric field part, the solution of that equation is any function:

$$\vec{E}=\vec{F}(\vec{s}\cdot\vec{r}-vt)$$

That is what we call a wave - we make a disturbance in one point and we can measure the same disturbance elsewhere after a short period of time.

Now my question is why do we always talk about light in terms of wavelengths? It's not necessarily periodic - the wave equation doesn't say that it has to be. One possible explanation that I thought about is the possibility to dissolve any disturbance in Fourier series, but that would mean we would have to add light with infinitely small wavelengths. Since the series is infinite, that would definitely make problems thinking about QM. Why do we say light has wavelength and phase when it's not necessarily a sinusoidal wave or not even periodic at all?

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    $\begingroup$ it's not necesserily a sinusoidal wave or not even periodic at all It doesn’t follow from this that “light waves can’t have a wavelength”. $\endgroup$
    – Ghoster
    Commented May 5 at 21:52
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    $\begingroup$ Well the wavelength I would assume is defined as the distance between two crests which corresponds to the length of one period. Moreover I can't imagine the wavelength would make sense if the periodic wave is not sinusoidal since if it was just any shape of periodic wave there would be differences of energy flow in two waves with different shapes but the same period/wavelength $\endgroup$
    – Krum Kutsarov
    Commented May 5 at 22:00
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    $\begingroup$ $\Delta\vec{E}=\mu\varepsilon \frac{\partial\vec{E}}{\partial t}$ This isn’t a wave equation. A wave equation has a second-order time derivative. $\endgroup$
    – Ghoster
    Commented May 5 at 22:05
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    $\begingroup$ yeah thanks for the correction but still, wavelength and phase implies periodicity $\endgroup$
    – Krum Kutsarov
    Commented May 5 at 22:09
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    $\begingroup$ Wavelength serves as a practical measure to convey the energy of light, but really, it has no direct connection to waves. Light, composed of countless individual photons, behaves as discrete particles rather than continuous waves. Each photon pulsates at a specific frequency as it travels at the speed of light. For instance, a photon designated with a 500nm wavelength exhibits a frequency exceeding 600 trillion cycles per second. It completes a cycles every 500 nanometers as it travels along. While we commonly refer to it as wavelength, the term doesn't accurately reflect its wave-like nature. $\endgroup$
    – Bill Alsept
    Commented May 7 at 21:14

5 Answers 5

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You're right in saying that you can't always attribute a unique wavelength to an electromagnetic field. However, it's a simple consequence of Fourier analysis that you can write any field as a superposition of different frequencies, or equivalently wavelengths. Basically, you can look at sinusoidal fields as a basis in which you can decompose any given field. Since a lot of electromagnetic systems are linear, it's usually convenient to work with the basis functions (sinusoids) instead of a general functional form.

To see this explicitly, just take a (temporal) Fourier transform on both sides of the wave equation $\Delta\mathbf{E}(\mathbf{x},t) = \mu \epsilon ~\partial^2 \mathbf{E}(\mathbf{x},t)/\partial t^2$:

$$\Delta \tilde{\mathbf{E}}(\mathbf{x},f) = - \mu \epsilon (2\pi f)^2\tilde{\mathbf{E}}(\mathbf{x},f), $$

where my convention for the Fourier transform is $\tilde{\mathbf{E}}(\mathbf{x},f) = \int_{-\infty}^\infty dt~\mathbf{E}(\mathbf{x},t)e^{-i2\pi f t}$. The above is typically called the Helmholtz equation, which is just the frequency-domain representation of the wave equation. Whatever the solution is, you can convert it to the original (time-domain) field via an inverse Fourier transform

$$\mathbf{E}(\mathbf{x},t) = \int_{-\infty}^\infty df~\tilde{\mathbf{E}}(\mathbf{x},f)~e^{i 2\pi f t},$$ which explicitly tells you that the field is superposition of sinusoids with varying frequencies ($f$). If you prefer, you can also write this as a superposition of wavelengths with a simple change of variables $\lambda = c/f$: $$\mathbf{E}(\mathbf{x},t) = -c\int_{-\infty}^\infty d\lambda~\frac{\tilde{\mathbf{E}}(\mathbf{x},c/\lambda)}{\lambda^2}~e^{i 2\pi c t/\lambda}.$$

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    $\begingroup$ That is an amazing answer man. One last thing about wavelengths not being able to be smaller than the Planck length. How do we explain that? Is the field not as "smooth" as the mathematical representation is? $\endgroup$
    – Krum Kutsarov
    Commented May 5 at 22:54
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    $\begingroup$ @KrumKutsarov I'm glad it was helpful! The thing about the planck length is pretty complicated. You need to work in the framework of quantum field theory (specifically quantum electrodynamics) to account for quantum effects. There you don't really have this kind of classical field anymore. Instead you have operator-valued fields, with their own uncertainty principle etc. $\endgroup$
    – Sahand Tabatabaei
    Commented May 5 at 23:01
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    $\begingroup$ This post basically discusses your question: physics.stackexchange.com/q/273888 $\endgroup$
    – Sahand Tabatabaei
    Commented May 5 at 23:02
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    $\begingroup$ @KrumKutsarov That is simply not true. There is no known lower limit on the wavelength of light. $\endgroup$
    – Vincent Thacker
    Commented May 6 at 0:09
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    $\begingroup$ @KrumKutsarov The argument about the Planck length is that if you tried to create a photon with enough energy in a small enough volume that it fit inside its Schwarzschild radius, then the photon would collapse into a black hole. This is very speculative and requires a theory of quantum gravity to resolve, for which we have theoretical proposals but no definitive experimental evidence. In practice, this bound is extremely weak because it involves wavelengths much smaller than are ever relevant in practice, so you can safely ignore this subtlety while studying Maxwell's equations. $\endgroup$
    – Andrew
    Commented May 6 at 2:12
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  1. Something does not need to be sinusoidal in order to have a wavelength. It just needs to be periodic or repeat after a certain distance. More generally, if $\lambda$ is the smallest non-negative real number such that $f(x+\lambda) = f(x)$ is true for all $x$ then we would say that the function $f$ has a wavelength $\lambda.$
  2. Working with sinusoids is mathematically convenient for a few different reason. One is that they are just a simple example of a periodic function. But more importantly the mathematical convenience comes from two facts: sinusoidal functions form a basis, and they are eigenvectors of the differentiation operator. The wave function in Fourier space then just takes the simple form of multiplication by the eigenvalue. See @Sahand's answer for more details. Note that many simple periodic functions can also be approximated with a single sinusoid function.
  3. Finally, a lot of physical processes acutally do happen near fixed sinusoidal functions with definite wavelengths. For example, emissions spectrums are a useful scientific tool because atoms/molecules radiate/absorb light, and this has a dependence with the frequency/wavelength of the light. Fundamentally, this is because a particle of light has energy related to its wavelength. A photon at a fixed energy $E$ in free space would actually be a sinusoidal function with corresponding wavelength $E=h\nu$. (Setting aside the non-normalizability of such a function).
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Sahand gave a really great answer but I'd like to offer a different perspective.

Edit:

It was pointed out in the comments that energy conservation isn't enough to give periodic solutions. Which is correct. I went back to a couple of my E&M books and the answer is actually a bit unsatisfying. As far as I can tell everyone decided to assume that the solution should be separable into time and space components. If you do that you get (I believe) only oscillatory solutions and a defined wavelength. To be honest the assumption was probably motivated by the experimental evidence that light was mostly an oscillatory wave and previous experience solving differential equations of that form in ways that described oscillatory waves.

So we guessed a functional form (for EM waves in free space) and that resulted in only oscillatory solutions which explained experimental results.

Old Answer:

If I recall correctly those wave equations are valid only in free space with no charges and no currents. In terms of pure mathematics the most common solutions would be exponentials and sinusoidal functions (which are really exponentials in disguise). In the absence of any other conditions on the nature of the allowed solutions both families of solutions are allowed.

However, there is another condition. In the absence of charges and currents the amount of energy in the disturbance described by these equations must be constant in time i.e. conservation of energy. That definitely rules out pure exponentials and I can't really think of any nonperiodic function that would satisfy conservation of energy. Except maybe the trivial solution of constant electric and/or magnetic fields. Basically, we must choose oscillatory solutions because of conservation of energy. I believe similar reasoning holds for an undamped mass/spring system.

Of course once you put charges and currents back in it is possible to have solutions that are not oscillatory (or at least not oscillatory forever) but still conserve energy. Negative exponentials show up in processes that involve absorption and positive exponentials show up when you're driving an electromagnetic oscillation especially near or at resonance.

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  • $\begingroup$ -1: The question itself contains an example of a non-oscillatory solution (admittedly one with infinitely large energy – but still – locally energy is conserved). Literally, for any (sufficiently non-singular) function $f \colon \mathbb{R} \to \mathbb{R}$ $\vec E(\vec r, t) = \vec E_0 f(\vec k \cdot \vec r - \omega t)$ (such that $\omega = c\left| \vec k \right|$) solves the wave equation and you can construct the corresponding magnetic field. Now you can cut out some finite part of this wavefront of this perpendicular to $\vec k$ – this will give some "diffraction front", but still ... $\endgroup$
    – Sebastian Riese
    Commented May 7 at 14:12
  • $\begingroup$ the central part will remain the bump for some time (simply due to the restricted signalling speed). (And of course $\vec E_0 \perp \vec k$). $\endgroup$
    – Sebastian Riese
    Commented May 7 at 14:13
  • $\begingroup$ Your point is well made. Also, I don't think that solution necessarily has infinite energy which further improves things. Energy conservation is necessary but not sufficient. $\endgroup$
    – ShawnS
    Commented May 7 at 17:06
  • $\begingroup$ The cut out solution does not have infinite energy, the original construction does (as long as it is not the vacuum, that is $\vec E_0 \ne 0$ and $f \not\equiv 0$, since it will be constant along 2D planes, and the fields are non-zero). $\endgroup$
    – Sebastian Riese
    Commented May 7 at 18:38
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Time domain Maxwell's equation never implies periodicity. If time harmonic representation of Maxwell's equation is done than by default idea of periodicity automatically arise in equations.

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    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented May 6 at 11:40
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Panagopoulos, D. J. (2018). Man-made electromagnetic radiation is not quantized. Horizons in world physics, 296, 1-57.

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    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
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