# Does J go to inf when v goes to c?

1. Jul 10, 2009

In SR, does the current become infinite when the velocity of the charge approaches c?

I'm thinking it doesn't because of the SR Lagrangian for a charged particle:
$$L = -\frac{m c^{2}}{\gamma} - q \phi + q \vec{v} \cdot \vec{A}$$
doesn't have a gamma in the qv term.

Another way of putting this is: Does the four current equal the charge times the four velocity?

2. Jul 11, 2009

### tiny-tim

Yes.

3. Jul 11, 2009

So the Lagrangian should be:
$$L = -\frac{m c^{2}}{\gamma} - q \phi + q {\gamma} \vec{v} \cdot \vec{A}$$

Thanks,

4. Jul 11, 2009

### quZz

No, it shouldn't. The original one is ok, but you've probably lost $$c$$ somewhere. I do not know what you mean by 4-current... 4-current density? If so recall that the volume is also contracted is SR.

5. Jul 11, 2009

Ok, I think I see what is going on, the original Lagrangian is for a charged particle.
The classical particle is always a point, so it doesn't contract.
Furthermore, the number of particles in a volume is the same in all reference frames.
The volume is contracted with respect to an observer moving relative to the volume.
So the relativistic observer sees a higher charge density and current density than a non-relativistic observer.

I found a good explanation here:
http://farside.ph.utexas.edu/teaching/jk1/lectures/node16.html" [Broken]

Thanks for the effort.

Last edited by a moderator: May 4, 2017