# B Charge at constant velocity emitting EM waves?

1. Aug 6, 2016

### Dexter Neutron

So far I have came to know that when a charge is accelerated the electric field magnitude around the charge changes and the effect is not felt instantaneously. The change in magnitudes of electric and magnetic field travels outwards at speed of light creating the so called EM wave.

So the EM wave originates only due to "the effect not being felt instantaneously". This case would also arise when a charge is not accelerating but moving at a constant speed. For example if there is a charge moving at a constant speed $v = c/2$ in space. At time $t = 0$ the charge comes into existance at origin and we place an observer at a distance $r = 12 * 10^8$metres from the charge. Now it would take electric field a time $t = 4 seconds$ to travel to the observer i.e. the electric field is not felt instantaneously. This would mean that an em wave has travelled from the charge to the observer. In this 4 seconds of time the charge moves by a distance of $6*10^8$ metres which means that the electric field needs to be changed again because $E \alpha \frac{1}{r^2}$and since r is changing E must change so again the change in field would take some time to reach the observer which means EM wave is produced travelling from charge to the observer.
The charge is not accelerating but still the EM wave is produced. HOW?

2. Aug 7, 2016

### Staff: Mentor

There are two effects here. The EM wave itself, and the change in the EM field as the first charge approaches the observer. I believe the EM wave arrives first, and behind it the gradual change in the field as the particle approaches. The 2nd effect is not an EM wave.

3. Aug 7, 2016

### Dexter Neutron

EM wave itself is the change in the EM field propagating radially outwards from the charge. How are you saying that both are different?

4. Aug 7, 2016

### Staff: Mentor

Because they are different. An EM wave is a propagating disturbance that obeys a wave equation and moves at a velocity of c with respect to all inertial observers. The change in the field associated with the non-accelerating, linear motion of the charged particle is not a propagating disturbance and can be done away with just by changing reference frames. For example, an observer who is stationary with respect to the charge sees no change in the EM field except for the EM wave.

5. Aug 7, 2016

### houlahound

Could the op add a sketch please. A word description is hard to visualize.

6. Aug 7, 2016

### Staff: Mentor

The fields at the observer do change over time, but they are not considered to be waves because the energy does not radiate off to infinity, but rather stays localized near the charge.

I will point you again to the Lienard Wiechert potentials. https://en.m.wikipedia.org/wiki/Liénard–Wiechert_potential

Last edited: Aug 7, 2016
7. Aug 7, 2016

### Sturk200

This post made me wonder, suppose there is a stationary charge held at the origin and an observer standing, say, at x=10 meters. The observer begins to rock catatonically back and forth. This constitutes an acceleration of his frame with respect to the charge at the origin. Does this produce a propagating EM wave in the observer's frame?

8. Aug 7, 2016

Staff Emeritus
Of course not.

9. Aug 7, 2016

### Sturk200

Why? What's the difference between the charge standing still and me accelerating versus the charge accelerating and me standing still?

Sorry if the answer is obvious, but I'm not seeing it.

10. Aug 7, 2016

### Staff: Mentor

Unlike velocity, (proper) acceleration is not relative so there's no reason why the two situations should be equivalent.

Suppose you're watching a spider in its web. If the spider bounces up and down, it will move back and forth in your field of vision and there will be ripples propagating through the web. If you bounce up and down, the spider will move back and forth in your field of vision, but there will be no ripples in the web.

(This analogy is somewhat oversimplified because of the way that electrical and magnetic fields are defined by their effects on the motion of a test particle, but it's good enough to answer your question).

Last edited: Aug 7, 2016
11. Aug 7, 2016

### Staff: Mentor

An accelerating frame of reference is non inertial, so the standard Maxwell's equations won't hold. In those coordinates waves would be difficult to even define since energy would not be conserved so it may not even be possible to show that it is transported.

12. Aug 12, 2017

### FrankMak

Maxwell's equations did not accommodate a moving charge and he stated so. Maxwell did not know the electron existed, but Maxwell suspected that particles might be involved, as his 1864 paper mentioned the possibility of particles acting at a distance. “The mechanical difficulties, however, which are involved in the assumption of particles acting at a distance with forces which depend on their velocities are such as to prevent me from considering this theory as an ultimate one, though it may have been, and may yet be useful in leading to the coordination of phenomena.

13. Aug 12, 2017

### Staff: Mentor

This is simply false. A stationary charge in one inertial frame is moving in another, and Maxwell's equations accommodate all inertial frames.