Accelerated charge

1. Aug 2, 2006

quasar987

Here's a question I asked in the Introductory Physics forum that went unanswered. I'm trying it again in a different forum because I'm really curious about this.

Why do we say that electrons emit an electromagnetic wave when they accelerate but not when they travel at constant speed? In other words, what's different btw the "emited radiation" caused by acceleration, and the simple propagating deformation of the electric field caused by a moving charge?

2. Aug 2, 2006

lalbatros

quasar987,

A very interresting question. I found an answer by reading Landau-Lifchitz, field Theory.
Go back to the formula for the retarded potentials for a particle in motion, the so-called Liénard-Wiechert potentials.

$$A^i = e \frac{u^i}{R_k u^k}$$​

where

$$R^k = [c(t-t'),\mathbf{r}-\mathbf{r}']$$​

and where the prime coordinates are related by the light-cone condition:

$$R_k R^k = 0$$​

t and r are the time and position of the observer and
t' and r' are those of the particle at the retarded time from the light-cone condition​
From these expressions, the difference between constant velocity and variable velocity is rather obvious. Indeed, the fields are obtained by derivatives of the potential.

The calculation may be a little tricky, but the result to be expected must contain two terms:

one term where only coordinates and velocity will appear
a second term proportional to the time derivative of the velocity ​

It should be expected that the first term corresponds to static charges and currents and the second to waves.

The second term corresponds to the wave emission. It shows two clear wave-like features.
At large distance from the charge:
this electric field is perpenticular to the "line-of-sight" of the observer on the charge
this electric field varies as 1/R of 1/R² for a static potential
the Poynting vector is non-zero​
These three features are markedly different from static fields and typical of electromagnetic waves in the vacuum. (sorry that the expression are a little bit too complicated for me to try writing them in Tex, this simple post took me already 30 minutes! See eq 10.65 in these lecture notes on the web.)

Michel

Postscriptum:

This shows why we call the second term, proportional to the acceleration, a wave or radiative term.
However, the physical meaning still needs further clarification.
I would translate the missing link as such:

why does this electric field become perpendicular to the line-of-sight ?
why does the Poynting vector become non-zero ?
can these features be simply understood ? (picturaly, without algebra, ...)​

Personally, thinking to ultra-sonic planes and how their bang occurs will help me a little bit to understand that. Indeed, fields "are emitted" with the speed of light, but the places where they are emitted from are not regularly spaced because of the acceleration. Therefore, the "emitted static fields" are deformed, sometimes compressed sometimes expanded depending on the actual motion. These distortions propagate as waves and carry their own energy.

Last edited: Aug 3, 2006
3. Aug 2, 2006

Andrew Mason

It is not entirely clear that charges emit radiation because of acceleration. This is still an unsettled question in physics. After all, charges do not radiate when being accelerated in a gravitational field. Feynman thought that radiation depended on non-uniform acceleration (he refers to the third time derivative of position).

It may be that charges do not radiate because of the acceleration at all. They may radiate because they experience an electro-magnetic field - e.g. a photon and it is the time dependent interaction of the external field with the field of the charge that creates a 'disturbance' which propagates as an electromagnetic wave. Of course, when this occurs, there is a force on the charged body so it accelerates.

AM

4. Aug 2, 2006

pallidin

How interesting. Perhaps more comments will follow.

5. Aug 3, 2006

lalbatros

quasar987,
mason,

This is quite right. The back-reaction of the field on the particle itself is proportional -in first approximation- to the third derivative of the position, also called sometimes the "suracceleration". This means that the radiation emitted by the particle deccelerate the particle with a force proportional to the derivative of the acceleration.

This is indeed an unsolved problem in classical physics. As far as I understood it, it will remain unsolved forever because it knocks on the door of quantum physics. The details of the fields near or even inside the particle must be considered to solve this problem fully, and a realistic solution comes in the realm of quantum physics.

Now, for the question discussed here, we still need to think a little bit.
Why is the emission proportional to the acceleration?
What does it mean and what is the link with the third derivative law for the back-reaction.

The reason for the third derivative is simple. When developping the approximate equation for the motion two terms arise in the radiation reaction: one is proportional to the acceleration and the next is proportional to the surracceleration. The acceleration term can be grouped with the "inertial" term and gives rise to the "effective mass" of the particle. This "effective mass" is the sum of the so-called "bare mass" and the "electromagnetic mass". The hope in this theory (Sommerfeld) was to eventually explain the inertia of electron simply by their electromagnetic energy. As explained, this is apparently too much for classical theory.

Michel

Last edited: Aug 3, 2006
6. Aug 3, 2006

take a radio, a battery, and a conductor.
put the battery near the antena, and shorcut it, over and over.
if radio will sound "white noises", every time u shortcut the battery, and when u switch the currnet off. though if u just leave the electron sream to go on and on, there would be no noises.
also such thing can be observed in your room, when speakers are on, and u turn on or off the light in ur room, it whould make white noise...

i know nothing of equations, though the connection to elecromagnetic waves and accelerating charges is simple:
when u shortcut the battery, electrons change their average velocity from zero to a certain velocity, such proces ofcours is a result of accelerating charges. so, while they accelerate, they emit electromagnetic waves.
and in the end of the process, the average velocity of the electrons is constant, therefor, no electromagnetic waves will be emited.
same thing with opening the circle with the battery, just that the electrons slow down, instead of speeding.

also, this has a connection to why they all whine about the electric poles which are built nearby houses, that it causes cancer, since AC current flosws within the conductors, which is electrons accelerating all the time and emits electromagnetic waves.
(well, they do talk about it a whole lot in israel, donno about other countries...)

7. Aug 20, 2006

leright

um, your post only explains that accelerating charge is connected to EM wave propogation, but it says nothing about WHY they are connected which is the question the OP presented.

8. Aug 20, 2006