# Oscillating Electric Field Lines

1. Aug 16, 2009

### jchodak2

I'm have some conceptual problems with electric fields and waves. When I think of electromagnetic waves I imagine field lines emanating from a charge that is oscillating, consequently causing the field lines to oscillate. In diagrams showing field lines, the electric field runs parallel with the field lines, yet with diagrams showing electromagnetic waves the electric field is perpendicular to the direction the wave propagates. Why isn't there a component of the electric field that runs parallel with the wave?

2. Aug 16, 2009

### Born2bwire

Because in an unconfined propagating wave, the electric and magnetic fields are always perpendicular to the direction of travel. Only in a few cases like guided waves, surface waves, maybe plasmas, etc. are there exceptions. And often, the exceptions are not exactly true. In many waveguides, the electric and magnetic fields still travel perpendicular to the direction of propagation, they however, are not traveling perpendicular to the direction of guided propagation.

Electromagnetic waves are governed by Maxwell's equations and there are only a few cases where the solution to these equations allows for a wave to have a component in the direction of propagation.

3. Aug 17, 2009

### jchodak2

So that's the way it simply is. Somehow that's not a very satisfying answer. So be it, but I have a quick follow-up: So if I had some oscillating charges, and some distance away I had some other charged particles, would there no longer be an attractive force pulling them together or pushing them apart? They would oscillate up and down due to the perpendicular force, but they wouldn't feel a force parallel to the wave?

4. Aug 17, 2009

### Born2bwire

The charges can experience a force in the direction of propagation due to the magnetic vector component of an electromagnetic wave. The Lorentz force on a charge due to the magnetic field is the velocity cross the B field. The electric field will give the charges a velocity along its direction, which is perpendicular to the magnetic field. Since the velocity vector will be in the same plane as the electic and magnetic fields, then the cross with the magnetic field will be along the direction of propagation. However, when it comes to a good or perfect conductor, the currents are isolated near or on the surface of the scatterer. So there is little or no penetration of the currents in the direction of propagation. This would be different for a low loss material like a dielectric.

5. Aug 17, 2009

### Staff: Mentor

Loosely speaking, the radial "static" component of the electric field still exists, but it decreases as $1/r^2$ from the source, whereas the amplitude of the transversely oscillating electromagnetic wave decreases as 1/r. So when you get far enough from the source, the static component becomes negligible compared to the wave component.

The "near field" is pretty complicated because of the contributions from the two components, so we usually talk only about the "far field" in order to simplify things.

6. Aug 17, 2009

### jchodak2

Thank you.

7. Aug 17, 2009

### jchodak2

Ah, so the radial static electric field still does exist! I should find this out for myself, but can you give me the relevant equations or a link to a site expressing them? I find it very interesting that the magnitude of the force "felt" should be different for a static field as opposed to an oscillating field!

8. Aug 17, 2009

### Born2bwire

There isn't any general equations though. But a static field is different from a wave. To be static means that the field is time-independent, which will never form an electromagnetic wave. jtbell is talking about the near-field that occurs around an electromagnetic wave source. What happens is that the near-field, the fields in the space surrounding the immediate volume of the source, traps energy in and around the electromagnetic source. I would not call it static since it is still a time-harmonic field, it is more or less static only in the sense of spatial propagation. The portions of the near-field that contribute fields in the direction of propagation do not propagate out, they are trapped and stored. The waves that do propagate out and leave the near-field will not have a component in the direction of propagation.

jtbell uses the term "radial" component because with a point source, or any source viewed from a large enough distance away such that it becomes point-like, the propagating waves in the far-field confine their electric and magnetic fields to the theta and phi directions in terms of spherical coordinates. The radial component would be in the direction of propagation.

If you look at the exact electric field equations for an idealized dipole antenna,

$$E_r=\frac{Z}{2\pi}\,I_0\,\delta l\left(\frac{1}{r^2}-i\,\frac{\lambda}{2\pi\,r^3} \right) e^{i(\omega t-k\,r)}\,\cos(\theta)$$

$$E_\theta=i\frac{Z}{2\lambda}\,I_0\,\delta l\left(\frac{1}{r}-i\,\frac{\lambda}{2\pi\,r^2}-\frac{\lambda}{4\pi^2\,r^3} \right) e^{i(\omega t-k\,r)}\,\sin(\theta)$$

You will notice that the radial component will drop off as 1/r^2 and 1/r^3 as opposed to the theta component which drops off as 1/r and more. When we move any appreciable distance away from the source, say where

$$kr >> 1$$

what happens is that when you look at the wave equations as a whole, the contribution from 1/r dominates in comparison to the contributions from 1/r^2 and 1/r^3. So, in the far-field, the theta component dominates dramatically. This is why we state that the near-field is trapped power, the radial component and higher order terms of the transverse components do not contribute to the fields at any distance away from the source.

The reason why we have this radial component as opposed to what I stated earlier is the presence of the source. If you solve for the wave equations with no sources, then my earlier statements hold. But a source can allow for components in the direction of propagation. At the same time though, since these components are non-propagating, can you still define a direction of propagation at all for them? In that sense, these are akin to the confined wave condition that I stated earlier in which you can get components in the direction of propagation.

EDIT:
In response to this, the magnitude of the force acting upon a charge is independent of whether or not the applied field is static or time-harmonic. The Lorentz force,
$$\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})$$
does not differentiate between the two. The only consequence with an oscillating field is that the force will oscillate along with it.

Last edited: Aug 17, 2009