# Why doesn't electric field change in the direction of its source's motion

1. Jan 26, 2016

### Clueless

I understand from my classes that, when looking to calculate the electric field of a charge we can use the electric field measured in its rest frame to find its electric field in the lab frame (where the charge is moving). The gamma factor changes the value for the component electric field if it is perpendicular to the direction of the charge's motion.

I don't understand why it is not also for the parallel part: if length contraction and charge conservation are at play, then in the lab's frame, there would be "more" charges per metre along the axis where the charges is moving?

And so the electric field increases by superposition of "extra" electric fields? And how does is discriminate between the direction of motion and the direction perpendicular?

2. Jan 26, 2016

### Orodruin

Staff Emeritus
You cannot just look at the electric field and make conclusions based on that. The full electromagnetic field behaves as a rank two antisymmetric tensor under Lorentz transformations and its source is a 4-vector and not a scalar. In order to properly understand how it transforms, you must take these things into account.

3. Jan 26, 2016

### Clueless

I see. Thanks for pointing me in the right direction to understanding this!

4. Jan 26, 2016

### Mister T

In the first paragraph you look at a single charged particle, but then in the second paragraph you switch to a collection of charged particles.

It depends on the charge distribution. If you are modelling a long straight wire you have an essentially infinite number of charged particles moving in the direction of motion, not so in the transverse direction..

5. Jan 26, 2016

### pervect

Staff Emeritus
http://web.mit.edu/sahughes/www/8.022/lec12.pdf has the argument I usually see, but it's not as detailed as I'd like. I suspect a missing piece of the argument was covered in a previous section.

The first idea is that you can find out how the electric field transforms by considering a parallel plate capacitor in motion - i.e. figuring out how the electric field in this example shows you how it transforms in general.

The second idea is that you use Gauss's law, and find that the integral of the electric field normal to a boundary is equal to the enclosed charge. Making the boundary in question a cube around one of the plates, and knowing that the electric field is nonzero only between the plates and is perpendicular to the plates (ignoring edge effects) then gives you the needed diagram. For a parallel plate capacitor, this analysis gives you that the electric field is proportional to the charge density.

The third idea is that the charge density doesn't change when you boost the capacitor in the direction normal to the plates. Then, since the charge density doesn't change, the electric field doesn't change either.

An alternative idea that may be more advanced is to use the idea of the retarded potential to compute the electric field in a moving frame (such as Jeffminko's equations, https://en.wikipedia.org/w/index.php?title=Jefimenko's_equations&oldid=695260811). But I usually see the parallel plate capacitor approach in introductory E&M texts.

The advantage of the retarded potential approach (Jeffminko's equations) is that (carried out properly) it demonstrates that you CAN find the electric field in any frame knowing only the electric and magnetic field in one frame. This is not particularly obvious, but is vital to the tensor approach. It shows that you only need to solve the problem once in any frame you like, and gives you a recipie to transform this solution to any other frame you desire.

Last edited: Jan 26, 2016