Does J go to inf when v goes to c?

[GiftOfPlasma]

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?

Thanks for your responses.

 

[tiny-tim]

Hi GiftOfPlasma!

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

Yes.

 

[GiftOfPlasma]

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

Thanks,

 

[quZz]

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

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.

 

[GiftOfPlasma]

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"

Thanks for the effort.