Let's imagine an electromagnetic wave that points every direction (i.e., from $\theta = 0$ to $\theta = 2\pi$). For simplicity let's consider only the electric field vectors. The wave goes through a polariser. Setting $\theta$ to be the angle from the vertical line (parallel to the polarising direction) the magnitude of the component of the electric field vector will be $E_0 \cos \theta$. Let's think about the superposition of these infinitely many waves with same amplitude. Then the sum will be

$$\sum E_0 \cos \theta \Rightarrow \lim_{N \rightarrow \infty} \sum_{k=1}^N E_0 \cos \theta_k, \ (\theta_k = \frac{k \times 2\pi}{N})$$

which from the graph of $\cos \theta$ from $0$ to $2\pi$ (symmetry) we can deduce that the value will be equal to $0$.

So the resultant component of the electromagnetic wave is zero, and how can we even talk about things like intensity (which becomes $I_0/2$ after moving through the polariser).

This actually leads me to a more fundamental question which is: if the electromagnetic wave directs towards every direction throughout $0\leq \theta<2\pi$ and their amplitudes are all same, then shouldn't the electromagnetic wave always undergo destructive interference with each other, resulting in no light at all?

This logic contradicts our observations of natural phenomena, so maybe I have a misconception. Could anybody clarify?

  • 1
    $\begingroup$ at some place you have not your sum, but $$\sum E_0cos(\theta_k)*sin(\omega*t+\phi_k)$$ $\endgroup$
    – trula
    Sep 2, 2020 at 16:53
  • $\begingroup$ @trula Oh. So the different $\phi_k$ of individual waves prevents the sum from being always zero? $\endgroup$ Sep 2, 2020 at 16:55
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    $\begingroup$ yes, it is almost never 0 $\endgroup$
    – trula
    Sep 2, 2020 at 17:09
  • $\begingroup$ @trula Aha... I see. It helped really a lot. Thanks! $\endgroup$ Sep 2, 2020 at 17:10

1 Answer 1


You are thinking in terms of waves, which is fine.

Your wave is actually the sum of a very large but finite number of waves, each coming from a single accelerated electric charge. No charge, no radiation. No acceleration on the charge, no radiation.

No two of those charges are in the same position. They are perhaps accelerating in random directions, but Fourier analysis says we can split each of those into a bunch of sine waves, so it's just like they're sending waves at discrete wavelengths. Let's pretend they're all accelerating in circles and it's just one wavelength.

It isn't an infinite sum, it's a finite sum. And they will cancel in some places and add together in others.

So they don't cancel out everywhere all the time.

Here's an analogy that I had to reach for, that might not be all that good -- if an infinite number of people play poker, on average there will be no winners. The losers and the winners will average out completely. But in practice some people do win, and some do lose.

OK, poker has skill involved. Pick a game which is all chance and it still works.


As far as I've ever heard,the ONLY way to create EM radiation is with an accelerated electric charge. Most of the radiation comes from accelerated electrons because they are easier to accerate.

Imagine the simplest case. An electron is traveling up and down in a radio tower, at say 1 time a second, in a sine wave. It makes radiation while it accelerates, and the faster it accelerates the more radiation it makes. The radiation is easiest to detect perpendicular to the direction of acceleration.

I drew some animations that I hope will help. When I showed them to my daughters they got kind of confused by the first simplest one, so if that doesn't make enough sense then skip it.


  • $\begingroup$ Thanks a lot for your answer. I think I don't have a good understanding of EM waves. I didn't quite notice the obvious fact that EM waves occur due to existence of charges. To be honest, I am not grasping the idea of wave well. So do you mind adding a simple diagram illustrating the situation you described? Then it will be much better for me to intuitively understand. $\endgroup$ Sep 2, 2020 at 17:05
  • $\begingroup$ I added a simple animation that shows some things. I didn't think of a good way to show a large but finite number of sources that tend to cancel some places but not others. $\endgroup$
    – J Thomas
    Sep 4, 2020 at 7:38

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