- #1

zb23

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Why is the divergence of an amplitude of an electric field of a monochromatic plane wave zero?

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- Thread starter zb23
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- #1

zb23

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Why is the divergence of an amplitude of an electric field of a monochromatic plane wave zero?

- #2

DaveE

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Divergence is a vector function, not just amplitude, which is why we can use it on E-fields, which are vectors.

One of Maxwell's equations (Gauss' Law) says ∇⋅E=ρ/ε

- #3

zb23

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ampitude can be written as a vector compex function in most generalised way.

- #4

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What does ## \nabla \cdot \bf D = 0 ## say?

(no charge density assumed)

(no charge density assumed)

- #5

PeterDonis

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Divergence is a vector function, not just amplitude, which is why we can use it on E-fields, which are vectors.

The way you are saying this is a bit confusing. Divergence is an

- #6

DaveE

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yes. I was also kind of sloppy in mixing the derivative and integral forms when I said "measured over a region...".The way you are saying this is a bit confusing. Divergence is anoperatorthat can be applied to vectors; the result of applying this operator to a vector is a scalar.

Last edited:

- #7

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Right, so if I can say ## \nabla \cdot \bf E = 0 ## for any electric field E in the absence of charge, am I home or not?The way you are saying this is a bit confusing. Divergence is anoperatorthat can be applied to vectors; the result of applying this operator to a vector is a scalar.

So,question: can you say that? Did someone say it about 140 years ago?

- #8

PeterDonis

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if I can say ## \nabla \cdot \bf E = 0 ## for any electric field E in the absence of charge, am I home or not?

Yes.

So,question: can you say that? Did someone say it about 140 years ago?

Yes, it's the charge-free case of the first Maxwell equation.

- #9

zb23

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- #10

PeterDonis

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I don't know where "complex vector amplitude" is coming from. In classical EM, the electric and magnetic fields are vectors with real components, and the divergence of the electric field vector when no charges are present, which will be true for an EM wave in vacuum, is zero by the first Maxwell equation.

If you ask "why" this is true, there is no answer beyond the fact that Maxwell's Equations have plenty of experimental confirmation within their domain of validity.

- #11

zb23

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- #12

zb23

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- #13

PeterDonis

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if I write my solution as E*e^i(wt-k*r), where my E is my amplitude written as complex vector

Doing this is a convenience for calculation, but to get the actual physical answer at the end you're going to take the real part, since that's what has physical meaning. The divergence of the real part of ##E## is what is zero.

- #14

zb23

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- #15

zb23

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E is not a vector field that represent electric field*

- #16

PeterDonis

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E is not a vector field that represent electric field*

If these statements are true, then your question about why the divergence of this ##E## is zero makes no sense. The only ##E## whose divergence has to be zero in charge-free space is the ##E## that appears in Maxwell's Equations.

In short, your question appears to be based on a misconception, so it is unanswerable. Thread closed.

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