B Why doesn't a charged particle moving at a constant speed radiate?

AI Thread Summary
A charged particle moving at a constant speed does not radiate electromagnetic (EM) waves despite changes in the electric field as it approaches or recedes from an observer. The energy associated with the fields remains localized around the charge, and the Poynting vector indicates no energy flow away from it. While the changing fields can be expressed as a sum of sine waves through Fourier transform, they interfere destructively at distances from the source, preventing radiation. Even with constant proper acceleration, a particle does not radiate in its own rest frame. Therefore, the conditions for producing EM waves are not met in this scenario.
alikim
Messages
20
Reaction score
0
If I stand by a flying with constant speed charged particle, at my location the electric field will change as the particle get closer and further from me. So this change in electric field should create magnetic field and so on, producing an EM wave?
 
Physics news on Phys.org
The simplest argument is energy conservation. If the charge radiated, where would the energy carried by the radiation come from? If the charge is accelerating it comes from whatever is doing the acceleration, but if it is moving at a steady speed there is no energy source.

Looking at the fields, the electric field is radial from the charge and the magnetic field is axisymmetric about the direction of motion. If you compute the Poynting vector you will see that the energy flows along with the charge - unsurprisingly given that the field remains localised around the charge. So there's no flow of energy away from the charge. In short, an EM wave is a particular pattern of changing electric and magnetic fields, and although there are changing electric and magnetic fields here they are not the right pattern to be an EM wave.

It's worth noting that you can always use the Fourier transform to write a changing EM field as a sum of inifinitely many sine waves. So it is possible to write this field as a sum of waves, but they interfere destructively away from the source so no energy is radiated even if you view the EM field this way.
 
Last edited:
  • Like
  • Informative
Likes DaveE, alikim and Dale
alikim said:
If I stand by a flying with constant speed charged particle, at my location the electric field will change as the particle get closer and further from me. So this change in electric field should create magnetic field and so on, producing an EM wave?
Note that even if a particle undergoes constant proper acceleration, it doesn't radiate in its own rest frame.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'Gauss' law seems to imply instantaneous electric field (version 2)'
This argument is another version of my previous post. Imagine that we have two long vertical wires connected either side of a charged sphere. We connect the two wires to the charged sphere simultaneously so that it is discharged by equal and opposite currents. Using the Lorenz gauge ##\nabla\cdot\mathbf{A}+(1/c^2)\partial \phi/\partial t=0##, Maxwell's equations have the following retarded wave solutions in the scalar and vector potentials. Starting from Gauss's law $$\nabla \cdot...
Back
Top