# Magnetism and Relativity

1. Jul 28, 2009

### nassboy

I've seen in feynman and on the internet a derivation that the force on a negatively charged particle from neutral current carrying wire can be shown to be purely magnetostatic in one frame of reference and purely electrostatic in the other frame of reference using the length contraction. The contracted wire has a net positive charge, and therefore attracts the negatively charged particle.

I've tried to extend this idea to two current carrying wires(the direction of the current the same in both wires)...but it seems that they both would have positive charge and repel when they should attract.

What am I doing wrong?

2. Jul 28, 2009

### tiny-tim

Hi nassboy!

That doesn't look right …

E2 - B2 is invariant (independent of the frame of reference), so you can't have the same field with just E or just B in different frames.

3. Jul 28, 2009

### nassboy

4. Jul 28, 2009

### tiny-tim

5. Jul 28, 2009

### Creator

Nassboy....
You are failing to take into account the Magnetic fields, in particular the DIRECTION of each B field around each wire.This is the same mistake 'meemoe_uk' was making in a previous thread, (except he would never admit it).

Remember the purpose of that exercise is to derive the MAGNETIC FIELD around the wire....

The DIRECTION of that field around each wire DETERMINES the direction of the force (in the lab frame) on a test charge or upon ANOTHER CURRENT A CARRYING WIRE according to the Lorentz force equation (below).....

F = qv X B (for point charges) (B = mag.field; q = test charge with velocity v outside the wire)

F = Integral ( IL X B) .....( for wires) (where L = length of wire, I = current)
Remember these are the forces YOU as the observer sees in the lab frame.

(Actually, the full Lorentz equaton is F = qE + (qv X B)...but the electric field E is zero in the lab frame since the each wire is electrically neutral in labe frame).

And remember its the 'right hand rule' that determine the direction of the force.
See here for a little more: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html

Creator

P.S. the link you cited is not a very great explanation since it arrived at the "magnetic force" (eqn. # 3) in somewhat non traditional form...and without the Lorentz force eqn. you cannot see in which direction that force is acting.

However, you will see that the mag. force eqn. (equation # 3 in your link) is the same as my equation above because the term in the parenthesis is simply equal to B.....IOW, simply substitute B = uI / 2(pi)R (Ampere's law) for the term in parenthesis and you will recover Lorentz force equation I gave above.)

see:http://hyperphysics.phy-astr.gsu.edu/HBASE/magnetic/magcur.html

Last edited: Jul 28, 2009
6. Jul 28, 2009

### marcusl

Let's put the same current in each wire so the mean electron drift velocities are equal. Each moving electron in wire 2 sees a net positive charge in wire 1--this is the case that you say Feynman considers--so the electrons are attracted towards the other wire. The positive copper ions in wire 2 are at rest, however. They see an electrically neutral wire 1, and feel no force. The system is symmetric, so everything we said about wire 2 applies also to wire 1. The mutual force is attractive.

Creator: you've missed the point of this thread. The OP's link demonstrates the relativistic unity of electric and magnetic forces.

7. Jul 28, 2009

### nassboy

I still don't see why both wires aren't length contracted and therefore have a positive charge...

Only one wire is moving and not both?

8. Jul 28, 2009

### jostpuur

It is possible that $\boldsymbol{E}=0$ and $\boldsymbol{B}\neq 0$ in some frame, so that a particle feels only a magnetic force, and $\boldsymbol{E}\neq 0$ and $\boldsymbol{B}\neq 0$ and $\boldsymbol{v}=0$ in some other frame, so that a particle feels only an electric force.

9. Jul 29, 2009

### marcusl

The wires aren't moving, only the electrons in the wires.

10. Jun 18, 2011

### lovetruth

If they both have positive charge they should have repulsive electrostatic force. But the attractive magnetic force due to moving proton will be more. Thus the total force on wire will be attractive. U failed to see the magnetic force.

Last edited: Jun 18, 2011
11. Jun 18, 2011

### Staff: Mentor

Nonsense. Length contraction is most definitely considered.