Let's consider a plane wave of the form $e^{i(\vec{k}.\vec{r}-\omega t)}$

Is the graphical representation the following

reference : https://fr.wikipedia.org/wiki/Onde_plane

or the following ?

enter image description here

I'm wondering what is now the graphical representation for the case of a plane wave of the form $e^{i(k.x-\omega t)}$ ?

Is it just that ?

enter image description here

or is it that :

enter image description here

  • 1
    $\begingroup$ those are all plane waves. A plane wave is one that takes the same value on planes $\endgroup$ Commented Feb 12, 2021 at 20:25
  • $\begingroup$ sure, but I put a formula : for each formula, are the two representations compatible with the formula, or not ? $\endgroup$ Commented Feb 12, 2021 at 20:26
  • $\begingroup$ two of those images don't show the wave vector. Your second picture is the most generic, the fourth picture is the same as the second but it's only showing the vertical component $\endgroup$ Commented Feb 12, 2021 at 20:30
  • 2
    $\begingroup$ I don't understand the difference in the formulas. Are you saying that one has $\vec k \cdot \vec r$ while the other has $kx$? In that case, the latter just has a wavevector $\vec k$ which is aligned with the $x$-axis. $\endgroup$
    – J. Murray
    Commented Feb 12, 2021 at 21:35
  • 1
    $\begingroup$ The first diagram in the post may help? What is the difference in the uses of ωt and kx in wave equation? $\endgroup$
    – Farcher
    Commented Feb 12, 2021 at 22:42

1 Answer 1


Any visualization is going to be incomplete, because $e^{i(\vec k \cdot \vec r - \omega t)}$ defines a complex-valued field at every point in a four-dimensional spacetime. We have to simplify.

Two simplifications we can make are to choose a specific time, such as $t=0$ (which we can relax later by animating our visualization), and to choose a coordinate system in which $\vec k = |k| \hat z$ is parallel to one of our major coordinate axes. Now we have the simpler task of visualizing the complex field $e^{ikz}$ in one dimension only.

Let’s visualize this complex field in one dimension by attaching a visualization of the complex plane to every point along the $z$-axis. At $z=0$ the wave $e^0=1$ is purely real. At all values $z$ the wave $e^{ikz}$ has unit magnitude. The value $1+0i$ recurs whenever the argument $kz$ is an integer multiple of the circle constant $2\pi$ (“one turn”). Purely real or purely imaginary values of the field occur when $kz$ crosses each quarter-turn, in the pattern $1, i, -1, -i, \cdots$. The complex value varies continuously between these specific values as $kz$ varies.

An appealing way to visualize this would be use the ordinary $x,y,z$ axes, interpreted in the following way: the $z$-axis tells us the location in space where we are sampling the field, the $x$-axis tells us the real part of the value of the field, and the $y$-axis tells us the imaginary part of the value of the field. In this visualization it would be natural to draw a corkscrew around the $z$-axis, passing through the points

$z$ $x,y$
$0$ $1,0$
$\frac18\frac{2\pi}{k}$ $\frac1{\sqrt2},\frac1{\sqrt2}$
$\frac14\frac{2\pi}k$ $0,1$
$\vdots$ $\vdots$

But that corkscrew is a misleading visualization, because we have the $x$- and $y$-axes doing double duty. The wave $e^{i(\vec k \cdot \vec r - \omega t)}$ is defined everywhere in spacetime: we can compute the real part and imaginary part of the field for every value of $(x,y,z,t)$. So we actually need the much larger table

$x,y$ $z$ $\Re\ e^{i\vec k \cdot \vec r}$ $\Im\ e^{i\vec k \cdot \vec r}$
$0,0$ $0$ $1$ $0$
$0,1$ $0$ $1$ $0$
$-1,1$ $0$ $1$ $0$
(anything else) $0$ $1$ $0$
$\vdots$ $\vdots$

That’s what’s happening in the second visualization in your question:

many corkscrews

Here we seem to have the $\vec k$-direction coming out of the screen and to the right, and the complex value of the field represented in a plane perpendicular to $\vec k$ with the real axis pointing upward. But we also have a volume-filling array of little arrows in little complex planes emphasizing that this complex field is defined at every point in space.

As $t$ increases, the shape of your visualization will translate in the $\vec k$-direction with speed $v = \omega/k$.

It’s worth pointing out that your fourth illustration,

real-only field

if we use the same convention, represents a field like $e^{i\vec k \cdot \vec r} + e^{-i\vec k \cdot \vec r} = 2\cos\vec k \cdot \vec r$ which is defined at every point in space but whose value can be represented by a real number rather than a complex number.

These visualizations get hairier when you use them in electromagnetism, where the value of the field is two three-dimensional vectors $\vec E, \vec B$ everywhere in space, rather than just a complex number. Further complicating the issue, there are symmetries required between $\vec E$, $\vec B$, and $\vec E\times\vec B$ in discussions of electromagnetic waves and their polarization states, which just so happen to invite a geometrical interpretation similar to the corkscrew-everywhere illustration. You have to read carefully to know what you are getting.

  • $\begingroup$ Couldn't the second figure also just be the real part of a circular polarized wave? $\endgroup$
    – Gilgamesh
    Commented Feb 13, 2021 at 8:30
  • $\begingroup$ Most people who talk about a "circular polarized wave" are talking about the electromagnetic field: a real antisymmetric tensor with six interdependent components (related). It is the case that the electric field in a circularly polarized plane wave has a representation very similar to the complex wave under consideration here, so a complex wave is often used as a representation for circularly polarized light. But the fields of a light wave do not have real parts and imaginary parts; the components of the field tensor are real. $\endgroup$
    – rob
    Commented Feb 13, 2021 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.