Electromagnetic nature of light

[Ellispson]

I was recently taught in school about the propagation of light as a combination of changing electric and magnetic fields.
I had read somewhere that a magnetic field was basically an electric field in a moving frame of reference.
So,I wanted to know if in some frame the propagation of light could be explained only by an electric field?
These ideas are new to me and I have not yet grasped them completely,so I apologise if my question is overtly stupid.

 

[Simon Bridge]

The short answer is "no".

 

[Ellispson]

Could you please explain?

 

[Noctisdark]

This is a result of special relativity, magnetic field are electric field seen from a moving frame, this is mainly due to lorentz contraction, neutral field from our frame of reference may not be neutral in other frames, so they exert a force orthogonal to the motion (because contraction is parralel to the motion), but one thing to note here is that the change in the E field will result in a magnetic field(change is due to motion of a charge), and because light is the change in the E field and no one(from any frame) can deny that change in the electric field, then also no one will deny the existence of the B field traveling along with it, so in every frame light is an E and M wave !, I encourage you to google this and learn more ;)

 

[Simon Bridge]

In the model you are trying to come to grips with, light is an electric and a magnetic field together... if you had only an electric field, there would be no light.

The only way that light can propagate is by having both electric and magnetic components... since light propagates (famously: at the same speed) in all frames, it follows that there is no frame an observer could shift to where the light is replaced by just an electric field.

 

[Ellispson]

This is a result of special relativity, magnetic field are electric field seen from a moving frame, this is mainly due to lorentz contraction, neutral field from our frame of reference may not be neutral in other frames, so they exert a force orthogonal to the motion (because contraction is parralel to the motion), but one thing to note here is that the change in the E field will result in a magnetic field(change is due to motion of a charge), and because light is the change in the E field and no one(from any frame) can deny that change in the electric field, then also no one will deny the existence of the B field traveling along with it, so in every frame light is an E and M wave !, I encourage you to google this and learn more ;)

In the model you are trying to come to grips with, light is an electric and a magnetic field together... if you had only an electric field, there would be no light.

The only way that light can propagate is by having both electric and magnetic components... since light propagates (famously: at the same speed) in all frames, it follows that there is no frame an observer could shift to where the light is replaced by just an electric field.

Oh yes.What you guys said does make sense.I'll surely try to learn more about this.Thanks a lot :D

 

[stedwards]

The electric and magnetic fields can be thought of as the result of what is called the 4-vector potential. This is a vector that has a time-like part.

Imagine an isolated region of space populated only with static charges. Only electric fields result. There are no magnetic fields present, and the 4-vector only has a time-like part. It only points in time. Now, with all the charges in motion, the relativistic effect is to change the direction of the 4-vector so that it has almost the same time-like part, but also has very small space-like parts. These space-like parts give rise to magnetic fields.

In the case of light, or electromagnetic radiation, I can only guess that the space-like part is equal to the time-like part so that a change in inertial frame leaves the 4-vector potential unchanged. In relativity-speak, it would point along a null curve.

Does anyone know? What is form of the 4-vector potential of a plane linearly polarized wave?

 

[Simon Bridge]

In relativity, electromagnetism is covered by the faraday tensor.
https://en.wikipedia.org/wiki/Electromagnetic_tensor

You want to know about the 4-potential
https://en.wikipedia.org/wiki/Electromagnetic_four-potential

 

[vanhees71]

First let me answer the question on the labelled Beginner's Level. One cannot answer the question on that level completely since for that one needs a bit of vector calculus.

There are two separate issues. The first is about the discovery by Maxwell that light is in fact an electromagnetic wave. This discovery has been made from theory quite precisely 150 years ago, and that's why there's a worldwide celebration of this groundbreaking discovery in terms of the Internation Year of Light. Maxwell's theory of all electromagnetic phenomena (as far as one can neglect quantum effects) is however solidly based on empirical facts, most importantly by the careful experiments done by Faraday and also Faraday's conceptional breakthrough in discovering the notion of fields. At the time physicists thought that these fields are due to a mechanical substance called aether (or ether), but it has turned out that this is a pretty strange model, and nowadays (for sure since Einstein's famous article on special relativity of 1905) we consider electromagnetic fields as a fundamental thing, not explainable by any mechanical models, i.e., it's an object extended in spacetime which has energy, momentum, and angular momentum and thus dynamics similar to mechanical objects, but also somewhat different.

The precise dynamics of the electromagnetic field, including its generation from electric charges (which are also a fundamental property of matter, which cannot be explained somehow by simpler even more fundamental properties, today) and their motion, i.e., charge-current distributions, is Maxwell's great achievement. It consists (in the non-relativistic formulation) of four equations, which describe basic facts about the fields:

(a) the sources of electromagnetic fields are charge and current distributions (2 socalled inhomogeneous Maxwell equations, containing the charge-current distribution of the matter)
(b) there are no magnetic charges
(c) Faraday's Law of induction: a time varying magnetic field is always accompanied by an electric vortex field

A charge at rest relative to an inertial frame of reference comes with a purely electric field and not magnetic field (one should better talk about electric and magnetic components of the electromagnetic field, but this is not the usual way to express these things). Now according to the special principle of relativity the physics looks the same in all inertial reference frames. So if I look on the situation from the point of view of an observer moving with respect to the charges that are static in the original referencd frame, these charges move and thus I have not only a static charge density but also a stationary electric-current density and thus there's also a magnetic field. This tells us that in the transformation laws to switch from one inertial frame to another the electromagnetic field must transform in a way that makes a purely electrostatic field to an electric and a magnetic field in a frame of reference moving against the former frame. These transformations are necessarily the Lorentz transformations of special relativity and that was in fact the way Einstein (and also other physicists like Lorentz and Poincare) discovered the special theory of relativity. That's why sometimes people claim that magnetic fields can be derived from electrostatics and relativity, which is however true only to a certain extent.

This brings me to the second part of the question, concerning electromagnetic waves. The point is what happens when I consider a charge that is accelerated relative to an inertial frame. Maxwell's equations tell us that then we have not only the stationary fields of a charge moving with constant velocity (which would be the picture if Newtonian spacetime was the right description and not relativistic spacetime) but also electromagnetic wave fields, carrying their own energy, momentum, and angular momentum. These fields are qualitatively different from the stationary fields of charges moving with constant velocity or stay at rest. For these fields you cannot find any inertial reference frame, where you have only an electric or only a magnetic field but you always have both electric and magnetic fields, which propagate like waves. The electric and magnetic fields are vectors pointing always in a direction perpendicular to the direction of motion of the wave, i.e., you have what's called a transverse field and, most importantly from the point of view of relativity, the phase velocity of the wave in free space has always the same magnitude, the speed of light, which is a universal constant appearing in the Maxwell equations (if written in appropriate Gaussian or Heaviside-Lorentz units; in the SI units this physical fact is a bit hidden by introducing the additional base unit Ampere for electric currents).

To answer #7, I must use math and relativity (leaving the B-level of the thread). The point is that there are two invariants you can build from the electromagnetic-field components. In Gaussian units they are given by $\vec{E}^2-\vec{B}^2$ and by $\vec{E} \cdot \vec{B}$. Written in manifestly covariant form in terms of the Faraday tensor they are $F_{\mu \nu} F^{\mu \nu}$ and $\epsilon_{\mu \nu \rho \sigma} F^{\mu \nu} F^{\rho \sigma}$, showing that both field combinations are scalars under proper orthochronous Lorentz transformations and the first a scalar and the second a pseudoscalar under space reflections.

Anyway, for a plane-wave mode of a free electromagnetic field (which is not existing in nature but can be used to build true fields in terms of Fourier series and/or integrals) you have $\vec{E} \cdot \vec{B}=0$ and $\vec{E}^2-\vec{B}^2=0$. The first equation says that electric and magnetic field are always perpendicular to each other and the second that they have the same magnitude. These fields are never static, because if you set $\omega=0$, which means static a static field, in your plane-wave ansatz to solve the free Maxwell you automatically get also $\vec{k}=0$ and finally $\vec{E}=\vec{B}=0$. This shows that the free plane waves are distinct from the stationary fields associated with stationary charge-current distributions (also in the part of space, where no charges are present).