Can a sinusoidal EM wave exist in nature?

AI Thread Summary
A sinusoidal electromagnetic (EM) wave cannot exist in nature due to the infinite energy requirement for a perfect plane wave solution of Maxwell's equations. Constant electric (E) and magnetic (B) fields do not constitute a wave, as they lack the necessary oscillation. While square waves can be composed of multiple frequencies, they still do not represent a true wave in the context of electromagnetic theory. The discussion emphasizes the importance of citing sources accurately when referencing scientific texts. Ultimately, waves can take various forms, but true sinusoidal waves are theoretical constructs rather than physical realities.
rmberwin
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I saw an example of a hypothetical EM wave that had constant E and B fields. Is that possible? How would it be produced? And wouldn't such a wave have an infinite wavelength?
 
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If the E and B fields are constant, how is it waving? It doesn't sound like a wave of any sort.
 
rmberwin said:
I saw an example of a hypothetical EM wave that had constant E and B fields. Is that possible? How would it be produced? And wouldn't such a wave have an infinite wavelength?

Please describe the source, i.e. where exactly did you see this? We are trying to have members to get into the habit of citing their sources. And has V50 has mentioned, a constant E and B field is not an "EM wave".

I've had someone told me before of a square pulse having a constant E and B field, but this is nothing more than a severe error in understanding what a square pulse is.

Zz.
 
A EM wave like any other wave derived from LINEAR partial diff. eqns. can have any shape whatsoever except for constant because a constant shape isn't waving at all.
 
I saw it in one of the standard texts. I agree that technically there is no wave. But it is an electromagnetic disturbance, traveling at the speed of light. But if there's no wavelength, what is the color of the light?
 
rmberwin said:
I saw it in one of the standard texts. I agree that technically there is no wave. But it is an electromagnetic disturbance, traveling at the speed of light. But if there's no wavelength, what is the color of the light?

You must have misunderstood the text.
 
rmberwin said:
I saw it in one of the standard texts

Please tell us exactly what you read and where. It sounds like you're misunderstanding something, but without knowing what you read, it's hard to help.
 
  • #10
rmberwin said:
I saw it in one of the standard texts.

This is not a valid reference citation in PF. You need to cite: (i) author (ii) title of the text (iii) publication year (iv) page number.

You will have to use such similar formats when you write your term papers etc. So you might as well learn to adopt that style in this forum. It is one of the more valuable lessons you can learn by being here.

Zz.
 
  • #11
Waves don't need to be sinusoidal. Sinusoidal waves are merely a convenient mathematical decomposition. I'm not sure if there is a universal definition of what counts as a wave, but I would go with, "something that solves the wave equation". Several examples are shown in
http://en.wikipedia.org/wiki/Wave_equation

This definition includes propagating waves and evanescent waves (which certainly aren't sinusoidal), and even constant waves (E = B = constant).
 
  • #12
Ironically, a sinusoidal wave, i.e., the plane-wave solution of Maxwell's equations for a free em. field,
\vec{E}(t,\vec{x})=\vec{E}_0 \cos(\omega t-\vec{k} \cdot \vec{x}), \quad \vec{k} \cdot \vec{E}_0=0, \quad \omega=c |\vec{k}|
cannot be realized in nature. That becomes immediately clear when you try to calculate the total energy of the electric field, which is infinity, and since we don't have an infinite amount of energy available, we can never create such a plane wave in the strict sense.

Of course, according to Fourier's theorem you can write any free-field solution in the form of a Fourier integral
\vec{E}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \tilde{\vec{E}}(\vec{k}) \exp[-\mathrm{i} |\vec{k}| c t+\mathrm{i} \vec{k} \cdot \vec{x}], \quad \vec{k} \cdot \tilde{\vec{E}}(\vec{k})=0.
I've used the (complex) exponential form of the Fourier integral, because it's more convenient than the cos-sin form, but is of course equivalent.
 

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