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Propagating planar wave of the Coulomb potential

  • Thread starter Phrak
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1. Homework Statement

There is a propagating planar wave of the Coulomb potential, [tex] \phi = sin(kx - \omega t) [/tex]. What other fields result when it is assume the magnetic potential, [tex]\textbf{A}[/tex] is everywhere constant?

[tex]\phi[/tex], Coulomb potential
[tex]\textbf{B}[/tex], magnetic field strength
[tex]\textbf{E}[/tex], electric field strength
[tex]\textbf{A}[/tex], magnetic potential
[tex]\textbf{J}[/tex], current density
[tex]c[/tex], speed of light in a vacuum
[itex]\rho[/itex], charge density

2. Homework Equations

[tex] \nabla\times\textbf{B} - (1/c) \partial \textbf{E} / \partial t = \textbf{J} [/tex]
[tex] \nabla \cdot \textbf{E} = \rho [/tex]
[tex] \nabla\times\textbf{E} + (1/c) \partial \textbf{B} / \partial t = \textbf{0} [/tex]
[tex] \nabla \cdot \textbf{B} = 0 [/tex] ,
where
[tex] \textbf{E} = - \nabla \phi - (1/c) \partial \textbf{A} / \partial t [/tex]
[tex] \textbf{B} = \nabla \times \textbf{A} [/tex]
and
[tex] \omega/k = c [/tex] .

3. The Attempt at a Solution

Starting with
[tex] \phi = sin(kx - \omega t) [/tex] , and [tex] \textbf{A} = \textbf{0} [/tex] ,
I get
[tex]\textbf{E} = -kx\ cos(kx - \omega t) [/tex]
[tex]\textbf{B} = \textbf{0} [/tex]
[tex]\textbf{J} = (-k \omega /c) sin(kx- \omega t) [/tex]
[tex]\rho = k^2 sin(kx- \omega t) [/tex] .

But in each case the velocity of propagation is [tex]c=\omega/k[/tex]. This includes nonzero charge density [tex]\rho[/tex] traveling at c, so I came up with a nonphysical solution. Where did I go wrong?
 

Answers and Replies

4,222
1
There was no intent to be misleading, but the problem is my own--not a school problem. That's exceptable under the guidelines, right?

I think the problem set-up itself must be nonphysical. That is, [tex] \phi = sin(kx - \omega t) [/tex] with [tex] \textbf{A} = \textbf{0} [/tex] can't exist alone somehow.

It's easy to see how one could generate fairly planar waves in Phi between the plates of two capacitors. But a changing electric field over time, as it oscillates, would produce a corresponding B magnetic field that runs in hoops around the centerline of the capacitor.

I dunno, but perhaps there is no physical way to apply an additional oscillating magnetic field so that [tex]\textbf{B}_{cap} + \textbf{B}_{additional} = \textbf{0}[/tex] over at least some small region.

I don't think one could prove it one way or the other.
 
Last edited:
4,222
1
[tex]\textbf{REGRESSION_{REGRESSION}}[/tex]
 
4,222
1
I see no one cares to solve this, so I'll just clean-up the math and logical errors.

3. The Attempt at a Solution

For a planar wave in [tex]\phi[/tex] (assumed propagating in the x-direction),

[tex] \phi = sin(kx - \omega t) \hat{\textbf{i}}[/tex] .

[tex] \textbf{E} = - \nabla\phi [/tex]
[tex] E_{x} = - \partial \phi / \partial x \ \ \ \ \ \ E_{y} = 0 \ \ \ \ \ \ E_{z} = 0 [/tex]
[tex] \textbf{E} = E_{x} \hat{\textbf{i}} [/tex]
[tex] \textbf{E} = -k \ cos(kx - \omega t)\hat{\textbf{i}} [/tex]

From [tex] \nabla\times\textbf{B} - (1/c) \partial \textbf{E} / \partial t = \textbf{J}[/tex] and [tex] \textbf{B} = \textbf{0} [/tex] :

[tex] \textbf{J} = - (1/c) \partial \textbf{E} / \partial t[/tex]
[tex] \textbf{J} = (k \omega /c) sin(kx- \omega t) [/tex]
[tex] \textbf{J} = sin(kx- \omega t) [/tex]

From [tex] \nabla \cdot \textbf{E} = \rho [/tex]:

[tex] \nabla \cdot \textbf{E} = \partial E_{x} / \partial x [/tex]

[tex] \rho = k^2 sin(kx- \omega t) [/tex]

4. An attempt at analysis

[tex] \nabla\times\textbf{E} = 0[/tex]
so that
[tex] (1/c) \partial \textbf{B} / \partial t = \textbf{0} [/tex]

This means that the requirement that the magnetic field be zero is consistent.

My original analysis was wrong. There are no hoops of magnetic field generated when the capacitor plate is uniformly charged. Obtained is a non-physical result.

I am left to conclude that maxwell's equations of electromagnetism, based upon a 4-vector potential, are incomplete.
 
Last edited:

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