# Electric Field Propogation in LIGHT

1. Oct 13, 2013

### san203

Hello. I started studying wave optics a while back and here is my first question

From what i understand, electric field is a region where charged particles feel the coulomb forces.
And light is basically time varying magnetic and electric waves being propagated in some medium or vacuum.

What i don't understand is how can something like an electric field be propagated in the first place?
How can it move. Isnt it supposed to be a mathematical description of Force?
Or is it something like introducing electric field at points in space with varying magnitude?
I hope i have made my question as clear as possible. Thanks.

2. Oct 13, 2013

### Staff: Mentor

The FIELD isn't moving. The CHANGE in the field is.

3. Oct 13, 2013

### mikeph

The fields propagate by a mutual interaction of the time derivative terms in Faraday's law and Ampere-Maxwell's law. If that sentence didn't make sense, wait until you study Maxwell's equations, because they really are a fundamental prerequisite to the understanding of propagating EM waves.

4. Oct 13, 2013

### UltrafastPED

If you were to suddenly unshield an electric charge its field lines would propagate at the speed of light; if the source charge is an electron then the field lines will expand as radial lines from that point.

Now jiggle the electron, and watch what happens to the field lines: the lines will now originate from a new central point, but if you observe them at distance d=ct you will see the new old field lines "kink" as the charge moves to the new position - and then settle into the new configuration.

If you have an oscillating charge (perhaps a current running up and down an antenna at a fixed rate then the oscillating charge will generate oscillating kinks in the electric field lines ... and the changing electric fields generate changing magnetic field lines in accordance with Faraday's law. These oscillating, coupled electric and magnetic field lines constitute electromagnetic radiation - perhaps your favorite AM radio station!

You can find other visualizations by searching for "radiation kink", etc.

5. Oct 14, 2013

### san203

It makes sense when i read it from textbook or from your answers, but what troubles me is imagining electric fields moving.Can i think of it as being constantly generated from the source and from the changing electric and magnetic field which in turn induce each other?

If we have a varying magnetic field in a isolated region, then will a electric field just come out of nowhere, without any other source?

Last edited: Oct 14, 2013
6. Oct 14, 2013

### UltrafastPED

Yes ... this is correct.

7. Oct 14, 2013

### vanhees71

It doesn't make much sense to say the magnetic field induces an electric field and vice versa. There is one and only one electromagnetic field which we describe, after defining a reference frame, in terms of electric and magnetic field components.

8. Oct 14, 2013

### UltrafastPED

I think the field structure of light in vacuo (transverse electric and magnetic fields which are coupled) remains the same for any relativistic observer ... after all, they all see it travelling at c!

9. Oct 14, 2013

### WannabeNewton

^That is certainly true PED because $E\cdot B$ is a relativistic invariant so a radiation field propagating in vacuo will maintain the orthogonality of the $E$ and $B$ fields in every inertial frame.

10. Oct 14, 2013

### mikeph

But why would you make things more complicated than they need to be when explaining the basics of the fields? No need to being electrodynamics into it when the goal is to explain the basics of a propagating field.

11. Oct 15, 2013

### vanhees71

If you have a "propagating field" for me this implies you have electromagnetic waves, and it is way less complicated to learn the correct physical meaning of the electromagnetic field than to wrongly think about it as if electric and magnetic field components were something independent of each other. The great achievement of Faraday and Maxwell in the 19th century has been to combine these fields to the one electromagnetic field, although this becomes clear in full beauty only when discussed as a (paradigmatic!) case of a relativistic classical field theory.

The problem is that in many textbooks the electromagnetic field is introduced for the special cases of the static and stationary field, where the electric and magnetic field components separate (if the constitutive relatitions are approximated in the non-relativistic approximation, which is of course justified almost always for practical situations).

12. Oct 15, 2013

### UltrafastPED

I've not seen light described in terms of the Maxwell tensor -ever! Can you point me to something that does?

My research area was nonlinear optics and ultrafast lasers ... everything I've seen is in terms of E, B.

13. Oct 15, 2013

### WannabeNewton

Well what you'll mostly find if you're after covariant treatments of electromagnetism/radiation are treatments in terms of the 4-potential, from which you can easily get the EM field tensor anyways. For example for a vacuum radiation field in the Lorenz gauge, one has $\partial^{\gamma}\partial_{\gamma}A_{\mu} = 0$ and a simple solution is a wave with constant amplitude: $A_{\mu} = C_{\mu}e^{i\varphi}$; $\varphi$ is the phase of the wave as usual and level sets of $\varphi$ define null hypersurfaces orthogonal to the wave 4-vector i.e. $k^{\mu} = \partial^{\mu}\varphi$. For plane waves in particular we have $\varphi = k_{\mu}x^{\mu}$ where $\{x^{\mu}\}$ is a set of global inertial coordinates.

You could check out section 12.11 of Jackson if you're interested but I personally don't know of any text that does e.g. waveguides, waves in matter, and scattering all in a covariant framework; that would be extremely impractical so it isn't surprising.

14. Oct 15, 2013

### vanhees71

I have not said that you should write everything always manifestly covariant. That indeed would be pretty inconvenient for practical purposes and also not very intuitive. Here the three-dimensional formalism is surely better suited.

However, I'd nevertheless carefully choose the names of objects, e.g., I'd call $(\vec{E},\vec{B})$ the components of the electromagnetic field with the electric components $\vec{E}$ and the magnetic components $\vec{B}$ (wrt. to a given reference frame!), when introducing the fields. Further you can say that $(\vec{E},\vec{B})$ is an abbreviation of the six independent components $F_{\mu \nu}=-F_{\nu \mu}$ of Faraday tensor (not Maxwell tensor, which name is usually reserved for the spatial part of the energy-momentum, i.e., the stress tensor of the em. field) in the reference frame under consideration.

15. Oct 15, 2013

### UltrafastPED

I take it all back: I have seen this before. I even did a bunch of homework problems years ago. I suppose that it is convenient for theoretical physicists who work with particles and other high energy problems.

But it doesn't seem to be used much when people are just talking about light, which is ordinarily described as an electromagnetic wave. If it comes up in the future I will be sure to call it the electromagnetic tensor.

16. Oct 15, 2013

### mikeph

I don't agree that it's a problem. Teaching a student electromagnetism and starting with the field tensor would be a worse approach in my opinion. Students have a hard enough time conceptualising the various cross products in quasistatics to worry about how they'd interpret the full relativistic approach.

For example I have never seen the tensor forms being used in waveguide analysis. So I could say that this approach is equally restrictive (if not more so) to certain people.

17. Oct 15, 2013

### Staff: Mentor

I agree, I would not try to associate a velocity with the fields themselves. As Drakkith said in the first response, it is the change in the field which moves. Similarly, a pressure wave in air travels at mach 1, even though there is no bulk movement of the air at those speeds. So I would think of the velocity of the wave (c) as being different from the velocity of the fields (not even defined).

Yes, and Yes, except that as others point out, the E and B fields are not independent, so it isn't "out of nowhere".

18. Oct 17, 2013

### Superposed_Cat

OR to simplify couldn't we just say that the E field creates a perpendicular B field which recreates the H field ad infinitum until the energy is transferred. And you could explain the propagation in terms of an analogy to sound You would not need that much math to explain it to him or am I completely wrong?

19. Oct 18, 2013

### vanhees71

I don't know, what it should help to "explain" the wave propagation in this way. There is an electromagnetic field, consisting of the electric and magnetic components $\vec{E}$ and $\vec{B}$, obeying the free vacuum (for simplicity) Maxwell equations (in Heaviside-Lorentz units). It doesn't make physical sense to separate them in the sense that one field is the cause of the other.

Starting from the homogeneous Maxwell equations
$$\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0,$$
you find that the electromagnetic field can be described by a scalar and a vector potential
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
These potentials are only determined up to a scalar field, i.e., changing the potentials to
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+\frac{1}{c} \partial_t \chi,$$
leads to the same physical field components $\vec{E}$ and $\vec{B}$. Thus one can give one constraint, the gauge-fixing condition. One example is the Lorenz-gauge condition
$$\frac{1}{c} \partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0.$$
The free fields, i.e., if there are no charges and currents, then obey the wave equation
$$\Box \Phi=0, \quad \Box \vec{A}=0,$$
with the d'Alembert operator
$$\Box=\frac{1}{c^2} \partial_t^2 - \Delta.$$
For the free-field case the Lorenz-gauge condition is not sufficient for a unique solution, because one can still use a gauge transformation with a scalar field $\chi$ that fulfills $\Box \chi=0$. Thus you can give one more constraint. The most simple one is the "radiation gauge",
$$\Phi=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
Then the Lorenz condition is still fulfilled, and you have unique solutions for $\vec{A}$ in terms of, e.g., a plane-wave mode decomposition.

To that end you make the ansatz
$$\vec{A}=\vec{A}_0(\vec{k}) \exp[-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}].$$
The gauge condition tells you
$$\vec{k} \cdot \vec{A}_0=0,$$
and the wave equation for $\vec{A}$ gives the dispersion relation
$$\omega = c |\vec{k}|.$$
$$\vec{A}_{\lambda,\vec{k}}=A_0(\vec{k},\lambda) \vec{\epsilon}(\lambda,\vec{k}) \exp[-\mathrm{i} c t |\vec{k}|+\mathrm{i} \vec{k} \cdot \vec{x}] + \mathrm{c.c.}$$
Here the vectors $\vec{\epsilon}(\lambda,\vec{k})$ for $\lambda \in \{1,2 \}$ are two arbitrary unit vectors perpendicular to $\vec{k}$ (conveniently chosen perpendicular to each other).

Thus electromagnetic waves are always transverse, and there are two polarization states for each wave number $\vec{k}$.

As you see, there is no physical possibility here to say the electric field is caused by a time-changing magnetic field or vice versa.

The "cause" of an electromagnetic field are of course always charge-current distributions. The corresponding solutions are given by the corresponding retarded potentials.

20. Oct 19, 2013

### Claude Bile

Think of a wave on a string.

The atoms in the string aren't propagating forward in the direction of the wave; instead it is the change in displacement is propagating forward.

In the same sense, it is not the electric field propagating forward, it is the change in electric field that is propagating (as originally pointed out by Drakkith).

If you think about it, it makes no sense to say that the electric field propagates, since it is a quantity that has a value as a function of space, rather than having a location in space (as a chunk of matter would).

Claude.