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- Thread starter rmberwin
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- #2

jedishrfu

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http://en.wikipedia.org/wiki/Square_wave

which would be multiple frequencies added together.

- #3

Vanadium 50

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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.

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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.

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jtbell

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http://en.wikipedia.org/wiki/Plane_wave

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You must have misunderstood the text.

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Vanadium 50

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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.I saw it in one of the standard texts

- #10

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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.I saw it in one of the standard texts.

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

Khashishi

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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).

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[tex]\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}|[/tex]

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

[tex]\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.[/tex]

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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