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Re: "4/3 Problem" Resolution (fwd), comment on



Dear David-
The point is that I did NOT have your permission to post our
private conversation on the web, nor do you have mine. I have stated
my position to the group. If you wish to explain your position, you
are free to do so.
It is clear to me that we have two totally disparate opinions
about almost everything. Since we seem to have almost no means of
communication with each other, we have no alternative but to communicate
through the group. That, by the way, is what peer review is all about.
Regards,
Jack



On Mon, 14 May 2001, David Rutherford wrote:

On Mon, 14 May 2001 13:35:44 -0500, Jack Uretsky <jlu@HEP.ANL.GOV> wrote:

Hi all-
David's calculation makes no sense to me because it makes use
of a special gauge for the EM field. After discussion with David I
see no way to resolve our disagreement. I would firmly reject any
calculation that forswears gauge invariance. Gauge invariance merely
says that it is the fields that are physical, not the potentials.
Regards,
Jack

I really wish you had posted your questions _here_ so that our conversation
could have been public. That way, other people could have seen it and given
their views on the matter. But instead, all they see is _your_ conclusions
on the matter, Jack. Please post the whole thing, you have my permission. I
think I made some excellent points which you seem to have completely
ignored.

Regards,
Dave


--
Franz Kafka's novels and novella's are so Kafkaesque that one has to
wonder at the enormity of coincidence required to have produced a writer
named Kafka to write them.
Greg Nagan from "The Metamorphosis" in
<The 5-MINUTE ILIAD and Other Classics>

On Thu, 10 May 2001, David Rutherford wrote:

I would like to offer a resolution to the famous "4/3 problem" of
electrodynamics. The theory of relativity implies that the momentum of
the field of an electron must be the same as the rest energy of the
field times v/c^2, where v is the velocity of the electron. However, the
momentum of the field, calculated from the Poynting vector, is 4/3 times
the energy of the field times v/c^2. Until now, this "4/3 problem" has
not been satisfactorily resolved.

I will show, here, that the mass of an electron is _entirely_
'electromagnetic' in origin. I'll be using a derivation which closely
parallels the one that Richard Feynman uses in "Lectures on Physics",
vol. 2, sections 28-1 through 28-3. I've included, however, the
necessary additions that Feynman didn't consider.

The value that the mass, m_elec, must have to be considered entirely
'electromagnetic' in origin is the energy of the field, U_elec, divided
by c^2, or

m_elec = U_elec/c^2

The value for U_elec, which Feynman calculated in section 28-1, eq.
(28.2), is

U_elec = (1/2)e^2/a

where

e^2 = q^2/4*pi*e_0

and q and a are the charge and radius of the electron, respectively. So
our m_elec needs to be

m_elec = (1/2)e^2/ac^2 (*)

Suppose that the electron is in uniform motion with velocity v<<c. The
Poynting vector, which Feynman refers to as the momentum density g=ExB
(with e_0 set equal to one), is directed obliquely to the line of motion
for an arbitrary point P at a distance r from the center of the charge
(refer to fig. 28-1). The magnetic field is B=(vxE)/c^2, which has the
magnitude (v/c^2)*E*sin(b), where b is the angle between E and v. The
momentum density g, then, has the magnitude

g = (v/c^2)*E^2*sin(b) (**)

The fields are symmetric about the line of motion, so when we integrate
over space, the transverse components will sum to zero, giving a
resultant momentum parallel to v. The component of g in this direction
is g*sin(b) or, from (**)

g*sin(b) = (v/c^2)*E^2*sin^2(b) (***)

However, the momentum due to g*sin(b) alone, when integrated over all
space, as Feynman points out later, does not give the correct value for
m_elec.

I would like to consider, now, a contribution to the momentum density
from h=E*(-div(A)), where A is the vector potential. We can also write
-div(A) as

-div(A) = v.(-grad(phi))/c^2 = v.E/c^2

where phi is the static electric potential. The magnitude of -div(A)
is v*E*cos(b)/c^2, so the magnitude of h is

h = (v/c^2)*E^2*cos(b) (****)

The momentum density term h might seem, at first, to be an ad hoc
addition, but is just as physical and valid a contribution to the
momentum density of the field as g. In fact, they are inseparable.

The component of h in the direction of v is h*cos(b) or, from (****)

h*cos(b) = (v/c^2)*E^2*cos^2(b) (*****)

So the total momentum density is given by g*sin(b)+h*cos(b). We now have
to integrate the total momentum density over all space to find the total
field momentum p (refer to fig. 28-2). Feynman takes the volume element
as 2*pi*r^2*sin(b)dbdr, so that the total momentum is

p = (integral){(g*sin(b) + h*cos(b))2*pi*r^2*sin(b)dbdr}

or, using (***) and (*****),

p = (integral){(v/c^2)*E^2(sin^2(b) + cos^2(b))2*pi*r^2*sin(b)dbdr}

Since sin^2(b) + cos^2(b) = 1, this reduces to

p = (integral){(v/c^2)*E^2*2*pi*r^2*sin(b)dbdr}

The result of integrating this over all space is

p = (1/2)(e^2/ac^2)v

The mass of a particle is equal to the coefficient of the velocity v, so
we can call the coefficient (1/2)(e^2/ac^2) the mass of the electron
m_elec, or

m_elec = (1/2)(e^2/ac^2)

As you can see, this is _exactly_ the same value as we got in (*). The
value for the mass of the field derived from the energy of the field,
and the value for the 'electromagnetic' mass are _identical_, meaning
that the mass of the electron is _entirely_ 'electromagnetic' in origin.

This 3-dimensional, non-relativistic derivation is based on my
4-dimensional, relativistic equations at the URL below.

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


--
Franz Kafka's novels and novella's are so Kafkaesque that one has to
wonder at the enormity of coincidence required to have produced a writer
named Kafka to write them.
Greg Nagan from "The Metamorphosis" in
<The 5-MINUTE ILIAD and Other Classics>


--
Franz Kafka's novels and novella's are so Kafkaesque that one has to
wonder at the enormity of coincidence required to have produced a writer
named Kafka to write them.
Greg Nagan from "The Metamorphosis" in
<The 5-MINUTE ILIAD and Other Classics>