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



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

--
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>