Electric Field Divergence of Monochromatic Plane Wave: Why is it Zero?

[zb23]

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

 

[DaveE]

Humm... I don't think I understand your question.

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/ε₀; the divergence of any E-field is 0 unless it is measured over a region that encloses an electric charge.

 

[zb23]

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

 

[rude man]

What does $\nabla \cdot \bf D = 0$ say?
(no charge density assumed)

 

[PeterDonis]

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 operator that can be applied to vectors; the result of applying this operator to a vector is a scalar.

 

[DaveE]

The way you are saying this is a bit confusing. Divergence is an operator that can be applied to vectors; the result of applying this operator to a vector is a scalar.

yes. I was also kind of sloppy in mixing the derivative and integral forms when I said "measured over a region...".

 

[rude man]

The way you are saying this is a bit confusing. Divergence is an operator that can be applied to vectors; the result of applying this operator to a vector is a scalar.

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?

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

 

[PeterDonis]

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.

 

[zb23]

So is divergence of a complex vector amplitude of electric field of a monochromatic plane wave always zero?

 

[PeterDonis]

So is divergence of a complex vector amplitude of electric field of a monochromatic plane wave always zero?

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.

 

[zb23]

I think you didnt understand me. My question is more mathematical than physical. When you derive solution for wave equation from maxwell equations, sometimes it is more general for you to write your solution for wave equation in the form of complex vectors, and then you take just real component. So, if I write my solution as E*e^i(wt-k*r), where my E is my amplitude written as complex vector and k,r are my wave vectors and radii vector, than if you put this solution for wave equation for monochromatic plane wave in gauss law in vacuum you ll get two parts, one of them if div (E), where E is complex amplitude.

 

[zb23]

Griffits and Jackson, that is standard literature for classical electrodynamics, write general solution for wave equation for monochromatic wave in complex notation. Sorry if I am a little bit confusing. I hope I explained my troubles better this time.

 

[PeterDonis]

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.

 

[zb23]

I understand but E is not longer a vector field it is just an amplitude vector that doesn't have to satisfy maxwell equation.

 

[zb23]

E is not a vector field that represent electric field*

 

[PeterDonis]

I understand but E is not longer a vector field it is just an amplitude vector that doesn't have to satisfy maxwell equation.

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.