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

You are using relations which Feynman developed for a charge moving at a
constant velocity, as viewed from an inertial frame (presumably your lab
frame). The CM frame of your two interacting electrons is not inertial.
Feynman explicitly states that his development is invalid if the charge is
accelerating (as viewed from an inertial frame).
Bob Sciamanda
Physics, Edinboro Univ of PA (em)
----- Original Message -----
From: "David Rutherford" <drutherford@SOFTCOM.NET>
To: <>
Sent: Saturday, June 16, 2001 09:30 PM
Subject: Re: 4/3 problem resolution/Action-reaction paradox in pdf format

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

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

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"