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I am having trouble grasping how the equation

$\bar{E} \left( \bar{r}, t \right) = \bar{E}_{0} \exp \left[ i \left( \bar{k} \cdot \bar{r} - \omega t \right) \right]$

fully describes a plane wave.

My current understanding is if we consider only the $\exp (i \omega t)$ part, that describes a point on the complex plane. This point moves around a circle as time increases. We can consider only the real part (x axis projection) of this point to essentially trace out our wave. I have tried to illustrate this below:

plane wave

My problem now is I cant see where the $\exp (i \bar{k} \cdot \bar{r})$ comes into play. I have read up about $k$ being the wave number, but just can't grasp its purpose or what it means physically, since my current plot can give a full illustration of an E-M wave.

Is there a way to visualize the purpose of the $i \bar{k} \cdot \bar{r}$ bit, like I have done with the $i \omega t$?

Thanks

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    $\begingroup$ The $\mathrm{e}^{\mathrm{i}\omega t}$ traces out the time behaviour of an oscillator at a point, not that of a wave. $\endgroup$
    – ACuriousMind ♦
    Commented Nov 13, 2014 at 14:41

2 Answers 2

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A picture is worth a thousand words. Here's how it looks as a function of space, evolving in time:

enter image description here

Here blue is real part, and purple is imaginary part of the complex exponent $\exp(i(kx-\omega t))$.

If you instead just look at $\exp(-i\omega t)$, you'll get this:

enter image description here

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  • $\begingroup$ I like the animations, but would suggest a couple of small improvements. First, change the text just above your second plot - "If instead you look at just $e^{-i\omega t}$ at all points in space (without the $ikr$ part in the exponential) then all points would move at the same time:" - and then below, a takeaway like "As you can see, you need the $ikr$ part to describe the spatial variation of the wave". Because even pictures need some words, sometimes. $\endgroup$
    – Floris
    Commented Nov 13, 2014 at 15:32
  • $\begingroup$ Thanks, I never figured that about the just $\exp(-i \omega t)$. So is there a way to visualize just the $\exp(i k r)$, like you've done the $\exp(-i \omega t)$? $\endgroup$
    – Steve Hatcher
    Commented Nov 13, 2014 at 15:51
  • $\begingroup$ @SteveHatcher - yes, just stop the animation of the top picture and you have it... $\endgroup$
    – Floris
    Commented Nov 13, 2014 at 15:53
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If you look at a wave at a moment in time, you can see how it varies spatially by plugging in different values of r: $e^{ikr}$. If you look at a point in space, you can see how it varies in time by fixing r and varying t: $e^{-i\omega t}$.

If you want the behavior in both space and time, you end up with the expression you have - and you can see how the speed of propagation relates to $k$ and $\omega$.

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  • $\begingroup$ Thanks for your answer, just to clear some things up: so the $\exp(-i \omega t)$ represents the temporal evolution of the wave, and the $\exp(i k r)$ represents the spatial disturbance based on that temporal evolution? I don't fully understand what you mean by "how it varies spatially by plugging in different values of r". $\endgroup$
    – Steve Hatcher
    Commented Nov 13, 2014 at 15:54
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    $\begingroup$ Yes. If you set $t=0$ for example, you will compute a different amplitude / phase for different points on the wave. Just consider the 1D case (r = x) - $e^{ikx}$ which describes a cosine centered on x=0. When you work in 3 dimensions, the same thing still holds (although you need the dot product to figure out how many wave cycles you have traversed in a given direction). Is that clearer? $\endgroup$
    – Floris
    Commented Nov 13, 2014 at 15:59

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