# Electric field of an approaching or receding charge

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1. Jan 27, 2016

### jartsa

First I stand next to a point charge, then I start moving away or towards the charge, which causes the charge to be further away from me according to me (my ruler contracts according to an inertial observer), and as electric field of a point charge decreases with distance, the electric field caused by the charge decreases at my position according to me.

Is that above thing right or wrong?

2. Jan 27, 2016

### A.T.

So which one?

How would length contraction increase the distance to the charge?

3. Jan 27, 2016

### jartsa

Let's study both cases, first the easier one, whichever that is.

Simple case: Bell's spaceships. Distance increases according to accelerating spaceships.
Complicated case: Bell's spaceships, but one spaceship is out of fuel, but that does not prevent the other spaceship from observing the Bell's spaceship type of change of distance.

4. Jan 27, 2016

### A.T.

If one ship doesn't accelerate (corresponding to the charge in your question?), the one behind it will crash into it. That is hardly an increase in distance.

5. Jan 27, 2016

### Ibix

You're forgetting the relativity of simultaneity. Put yourself at the origin and the charge at x=X. At time t=0 you accelerate instantaneously to velocity v. Plug the numbers into the Lorentz transforms. Yes, at t=0 the charge is at $x'=\gamma X$. However, at t'=0, which is your current simultaneity convention, it is at $x'=X/\gamma$ - closer to you. For non-instantaneous acceleration you'd have to specify an acceleration profile to figure out how far you and the charge move during your acceleration phase.

You need to take a look at the electromagnetic tensor (see for example equation 1.58 in Carroll's GR notes - apologies for lack of link, but my phone is not cooperating with linking today) to determine what field you will see.