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Some time back, I put a proof that the mass of the electron is entirely^^^^^^^^^^^^^^^^^^^^
electromagnetic in origin on my webpage and posted it to various
newsgroups. I followed the method that Richard Feynman used in chapter
28, volume 2 of "Lectures on Physics". Since then I've found two errors
in my proof, but, fortunately, they cancel each other out. The first
error I found was that my integration of the field energy density
of a^^^^^^^^^^^^^^^^^^^
moving electron over all space is actually twice the value that I
initially calculated, due to the fact that the integral of sin(b)db from
0 to pi is 2, not 1, as I had originally calculated. This would make the
value for the energy of the field
exactly twice what it
should be in order for the electron to be entirely electromagnetic in
origin.
However, I then noticed that Feynman's value for the electromagnetic
energy density of the field of an isolated stationary electron is
incorrect. It should be
u = e_0*E^2
not
u = (e_0/2)*E^2
as Feynman has stated. Since we are not considering particle _pairs_,
we don't need to divide by 2. Therefore, the energy of the field of a
stationary electron is exactly twice what I had originally thought. That
means that everything comes out right; the mass of the electron is
entirely electromagnetic in origin.
Below is a copy of the revised proof with some changes in notation due
to having to use ascii rather than tex. Hopefully, I didn't make any
errors in the translation. You can see the pdf version at
http://www.softcom.net/users/der555/elecmass.pdf
Comments welcome (especially about my correction of Feynman's stationary
electron field energy density). Here's the proof:
"4/3 Problem" resolution
------------------------
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 magnitude of 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 gone unresolved (this non-relativistic,
three-dimensional treatment is derived from my relativistic,
four-dimensional equations at
http://www.softcom.net/users/der555/newtransform.pdf).
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. In order to conform
with Feynman's derivation, I've used SI units, here, in contrast to the
Gaussian units used in the paper, above.
The value that the mass m_elec, derived from the momentum of the field,
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 (1)
The value for U_elec, which Feynman calculated in Section 28-1,
Eq. (28.2), is
U_elec = (1/2)*e^2/a (2)
where a is the lower limit of integration of the field energy density,
and
e^2 = q^2/(4*pi*e_0)
where q is the charge of the electron and e_0 is the permittivity
constant. However, in (2) the factor 1/2 is a correction term arising
from the summation of the energies of the fields of pairs of charges.
But here, we are calculating the energy of the field of a single
electron. Therefore, the factor 1/2 should not appear and the value for
U_elec should be
U_elec = e^2/a
So in order to be considered entirely electromagnetic, in origin, our
m_elec needs to be
m_elec = e^2/(a*c^2) (3)
Suppose that the electron is in uniform motion with velocity v<<c. The
momentum density of the field g = e_0*ExB, where E and B are the
conventional electric and magnetic field three-vectors, 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, page II-28-2 in
Feynman's "Lectures on Physics"). The magnetic field is B = vxE/c^2,
which has the magnitude (v/c^2)*E*sin(b), where b is the angle between
v and E. The momentum density g, then, has the magnitude
g = (e_0*v/c^2)*E^2*sin(b) (4)
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 (4)
g*sin(b) = (e_0*v/c^2)*E^2*sin^2(b) (5)
However, the momentum due to g*sin(b) alone, when integrated over all
space and divided by v, as Feynman points out later, gives a value for
m_elec of
m_elec = (4/3)*U_elec/c^2
which is, clearly, not the same as the value from (1) required in order
for the mass of the electron to be entirely electromagnetic in origin.
This is the "4/3 problem" (actually, since the factor 1/2 in (2) has
been removed, it should be called the "2/3 problem").
I would like to consider, now, a contribution to the momentum density
from h = -e_0*E*div(A), where A is the vector potential (this is not an
ad hoc addition. It is derived from the time component of my electric
field four-vector, and is part of the momentum density components of my
energy-momentum tensor in the paper at the URL above). Since
A = v*phi/c^2, where phi is the static electric potential, we can also
write div(A) as
div(A) = div(v*phi/c^2) = (1/c^2)*v.grad(phi) = -(1/c^2)*v.E
The magnitude of (1/c^2)*v.E is (v/c^2)*E*cos(b), so the magnitude of h
is
h = (e_0*v/c^2)*E^2*cos(b) (6)
The component of h in the direction of v is h*cos(b) or, from (6)
h*cos(b) = (e_0*v/c^2)*E^2*cos^2(b) (7)
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, page II-28-2). Feynman takes
the volume element as 2*pi*r^2*sin(b)dbdr, so that the total momentum is
p = \int{(g*sin(b) + h*cos(b))*2*pi*r^2*sin(b)dbdr}
or, using (5) and (7),
p = \int{(e_0*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 = \int{(e_0*v/c^2)*E^2*2*pi*r^2*sin(b)dbdr}
The result of integrating this over all space, with the limits of b
being 0 and pi, and the limits of r being a and infinity, is
p = (e^2/(a*c^2))*v
Since p = mv, the mass of the electron m_elec is
m_elec = e^2/(a*c^2)
As you can see, this is exactly the same value as we got in (3). The
value for the mass of the field derived from the energy of the field
divided by c^2, and the value for the 'electromagnetic' mass derived
from the momentum of the field are identical, meaning that the mass of
the electron is entirely 'electromagnetic' in origin.
--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf