Re: 4/3 problem resolution/Action-reaction paradox in pdf format

[David Rutherford <drutherford@SOFTCOM.NET>]

On Sat, 16 Jun 2001 18:49:30 -0400, Bob Sciamanda <trebor@VELOCITY.NET>
wrote:

> David,
> I have not yet found the time to read your provoking writings, but let me
> raise one question which I have noticed.  You correctly state that the
> Maxwellian forces of interaction between two electrons are not, in the
> general case, "equal and opposite."  You then state that these forces ARE
> equal snd opposite in the CM frame, "because of symmetry".  I doubt the
> truth of this latter statement - please prove.

Hi Bob,

The electrons in the CM frame are always travelling on parallel paths at
the same speed, but opposite directions, or are both at rest, so v = -v'
or v = v' = 0. The Lorentz force on q due to q' is

F = q(E + v x B)

Substituting B = (1/c^2)(v' x E), you get

F = q(E + (1/c^2)(v x (v' x E)))        (*)

and since

v x (v' x E) = v'(v . E) - E(v . v')

(*) becomes

F = q(E + (1/c^2)(v'(v . E) - E(v . v')))

The electric fields are equal in magnitude and opposite in direction, so
E = -E', and since q = q', we can say for the case v = -v' in the CM
frame,

F = -q'(E' + (1/c^2)(v(v' . E') - E'(v' . v)))       (**)

 From the vector identity

v' x (v x E') = v(v' . E') - E'(v' . v)

we can write (**) as

F = -q'(E' + (1/c^2)(v' x (v x E')))       (***)

or, since B' = (1/c^2)(v x E'), we can write (***) as

F = -q'(E' + v' x B')

and, since F' = q'(E' + v' x B'),

F = -F'

Now, for the case v = v' = 0, we have simply

F = qE = -q'E' = -F'

So for all cases in the CM frame, the forces are equal and opposite,
using just the Lorentz force equations.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555