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Re: Park City Paradox ?



On Mon, 7 Jan 2002, Robert Cohen wrote:

I guess I had the same misconception as Ludwik...

Isn't the magnetic force proportional to (q_1v_1)(q_2v_2)/r^2?
If both charges are moving with speed c in our reference frame, wouldn't
this give rise to a magnetic attraction?

The electrons don't need to move (and, indeed, *can't* move) with
speed c in our reference frame for there to be a magnetic
attraction. But all I am saying is that, since the two electrons
*do* move apart in their own frame, they will do the same in every
frame and that fact will be properly accounted for by a net
repulsive force.


Here is a sketch showing the fields in both frames at the position
of one of the electrons as obtained from the Lorentz
transformation:

v
e : e--->
:
: ^
^ : |
| E : |E_lab = (gamma)E
| : |
e : e--->
* B_lab = (gamma*v/c^2)E, outward

electron frame lab frame

You can see that the electric force is larger (more repulsive) in
the lab frame than in the electron frame, but that the attractive
force, due to the magnetic field that arises in the lab frame,
more than compensates for the increase in the electric force.
The bottom line is that the Lorentz force on either electron in
the lab frame will be reduced by a factor of gamma from what it
was in the electron frame.

So it is true that as gamma goes to infinity the force goes to
zero and *if* gamma could *equal* infinity the force would *be*
zero. However, I think of this as mostly being an effect of time
dilation. The separation that takes place in the electron frame
takes place by *their* clocks. Since we see their clocks running
slow by a factor of gamma, we observe the separation process
running slow by the same factor. We interpret that observation in
terms of a reduced lateral force. Since "time stops" for v = c,
no separation would be observed.

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm