# Violating the principle of relativity

1. Apr 9, 2010

### nonequilibrium

It seems really basic, like a remark which probably everyone asks himself when first coming upon electromagnetism, but I can't see the error in my reasoning, which leads to violating the principle of relativity:

Imagine two protons passing each other by, each with a speed v, opposite to each other, but one is flying just higher so that they don't collide. As the bottom one passes beneath the top one and they're at their closest point, the bottom one is creating a magnetic field that is pushing the top one away (well, the resulting Lorentz force is, anyway). (note: there is also an electric force)

Now imagine the same experiment, but viewed from an inertial reference frame with a speed v flying in the direction of the bottom proton. Now, as they pass at the same point, the bottom proton is not creating a magnetic field because in this reference frame it is standing still. So there is no magnetic force pushing the top one away, only the electric force, which hasn't changed.

Much obliged,
mr. vodka

2. Apr 9, 2010

### D H

Staff Emeritus
No contradiction. All you have discovered is that what looks like magnetism in one frame looks like the electrostatic force in another.

3. Apr 9, 2010

### nonequilibrium

That sounds interesting... but sadly I don't understand? The formula for the electrostatic ("static"?) force is kqQ/r², which is identical in both cases...

4. Apr 9, 2010

### LukeD

And you've hit on the other piece!

That formula is only valid for electrostatics!

In electrodynamics, you need to take into account that changes in the electric field only propagate at the speed of light.

5. Apr 9, 2010

### espen180

This is where you are mistaken. Here is an applet which lets you play with the E-field of a moving charge. The velocity bar is given as a c fraction.

http://www.its.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html

6. Apr 9, 2010

### nonequilibrium

Wow! The formula changes for moving charges? Bombshell. Where does this come from? I just saw Maxwell laws in my first year of university (the integral-forms). Out of which one does this follow? I must say I hadn't expected this. So all those times I've used Coulomb's law for charges with v not zero, I was doing something wrong? It's too hard to believe...

EDIT: but weren't there like at least 100 years between this part of physics and the discovery of the theory of relativity? How did they solve this obvious problem?

7. Apr 9, 2010

### espen180

What do you mean? Relativity theory is founded on the predictions of Maxwell's Equations and the Principle of Relativity. Relativity assumes the predictions of maxwell are correct.

The E-field of the moving charge is given here by Dalespam (The correct expression is the one on the bottom.):
https://www.physicsforums.com/showpost.php?p=2625933&postcount=40

8. Apr 9, 2010

### nonequilibrium

Interesting formula. (and hefty)

And ah, I thought the formula that gave the correct electrodynamic forces was derived from special relativity, my bad. Don't know why I assumed that.

But does this mean EVERY situation where you're in a reference frame that has moving charges, Coulomb's law is simply wrong? I'm flabbergasted. Was the above situation just an exception?

9. Apr 9, 2010

### Pengwuino

It's really not as bad as you think. The electric field can be written as, assuming no magnetic field:

$$E(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ) = \frac{1}{{4\pi \varepsilon _o }}\int {\frac{{\rho dV'\hat r'}}{{|\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} - \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} '|}}}$$

as you would expect. However, what if $${|\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} - \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} '|}$$ is time dependent? All of your methods for solving electrostatics falls apart. If you had a charged particle flying away from you, the magnitude of the distance is changing. To add to that, it's not an instantaneous change either! Since the speed of e/m fields is finite, movement in the source won't translate into an instantaneous change in the field at the observation point.