1
$\begingroup$

Consider a proton in harmonic motion along vertical direction.

enter image description here http://physics.weber.edu/schroeder/mrr/MRRtalk.html

Near a point source, direction of electric field is along the curve.

enter image description here https://en.wikipedia.org/wiki/Plane_wave

But at greater distance from point source, in plane wave approximation, electric field is not along the sinusoid but it is perpendicular to axis.

How the direction of electric field is determined in plane wave approximation?

$\endgroup$

1 Answer 1

3
$\begingroup$

The electric field of an oscillating point charge consists of a Coulombic component and a radiative component:

  • the Coulombic part is generally directed away from the charge, and it goes down as $1/r^2$ with the distance from the center of the oscillations.
  • the radiative part is generally transversely polarized, and it goes down as $1/r$ with the distance from the charge.

When we're considering radiation, we keep the $1/r$ component, because it dominates completely over the $1/r^2$ Coulombic near-field when you're far away from the charge. This explains the discrepancy you observe.

As to how the direction of the electric field is determined - that obviously depends on the situation. Plane waves are a model, and nothing more, and they are generally a terrible model for the field radiated by a point charge. (Instead, you normally use spherical EM waves.) Different situations call for different models, and different characteristics of the radiation within those models.

$\endgroup$
2
  • $\begingroup$ How to calculate the radiative part? And how did you determine the radiative part of electric field goes down as 1/r? $\endgroup$
    – shul
    Commented Jun 13, 2018 at 10:28
  • 1
    $\begingroup$ In a general setting, the only calculation possible is the Liénard-Wiechert potentials, which are nontrivial to calculate, but they show an explicit split into radiative and near-field parts. $\endgroup$ Commented Jun 13, 2018 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.