Electromagnetic mass (4/3 problem) revision
David Rutherford
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
Uncle Al
>
> 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.
Stephen Wolfram says it is a consequence of Kuratowski's theorem in a
discrete network reality. He's wrong and you're wronger.
> 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.
Oh brother,
[snip]
> 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.
So sad.
[snip]
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Old Man
news:3D8FB75A...@softcom.net...
> 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
Really David, both Feynman and Jackson say that the energy
density of any electric field is u = (e_0 / 2)*E^2. This is a
general statement, and applies as well to the field of an isolated
electron. So, David has made another error, along with his
audacious claims of competence. Furthermore, contrary to
David's claim of following Feynman, Feynman explicitly states
that classical electrodynamics is inconsistent with the case
wherein all of the energy of an electron is in the electric field.
Electrodynamics does not allow it. This is not the first time
that David has attempted to misrepresent Feynman.
[Old Man]
David Thomson
news:3D8FB75A...@softcom.net...
> 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.
That was an excellent treatment of mass and charge relationship. However,
since the equation can be transposed to:
m_elec
e^2 = -------
a*c^2
What you have found is a mutual relationship between mass and charge. Mass
is not originated in charge any more than charge is originated in mass.
Without going through the long form (unless you request it) I have found the
following relationship of charge to mass:
e^2 = 8 * pi * a * h * Cd
where a is the fine structure constant, h is the angular momentum of the
electron and Cd is a conductance constant derived from Coulomb's constant
equal to 2.112 x 10^-4 siemens.
In the unified charge equation above h * Cd is equal to the strong charge
(electromagnetic charge) of the electron. e^2 is equal to the elementary
charge. When we do electronics calculations for circuits we are actually
using and measuring the electromagnetic charge of the electron. The
elementary charge of the electron is virtually ignored.
Your equation and mine are in agreement. The importance of these equations
shows that a single dimension of mass is directly related to a distributed
dimension of charge. Today's units, such as potential, current, resistance,
magnetic moment, magnetic flux, magnetic field, etc. use only one dimension
of charge per dimension of mass. Were these units corrected to reflect the
proper dimensions of charge we would add more utility to our dimensional
math. We would have expressions like:
potential
----------- = frequency^2
inductance
Inductance, capacitance, conductance, Coulomb's constant, and the
gravitational constant are already in the correct units of mass per
distributed charge.
Great work!
Dave
David Rutherford
David Rutherford wrote:
>
> 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
^^^^^^^^^^^^^^^^^^^^
This should be "field momentum density", not "field energy density",
sorry.
> 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
^^^^^^^^^^^^^^^^^^^
Again, this should be "momentum of the field", not "energy of the
field".
Uncle Al
>
> David Rutherford wrote:
> >
> > 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
> ^^^^^^^^^^^^^^^^^^^^
> This should be "field momentum density", not "field energy density",
> sorry.
You don't know what you are talking about. You are demonstrably
incompetent in process. You are empirically ineducable. That sums to
your being a crackpot.
Go away until you learn something.
Franz Heymann
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
No.
> not
>
> u = (e_0/2)*E^2
This is the correct expression for the energy density in an
electrostatic field.
If you are working under such a gross misapprehension, the rest of your
post is hardly worth reading, so I snipped it.
[...]
Do you realise that the radius of the electron has been measured?
Do you realise that it is one hell of a lot less than that which your
formula might require?
Franz Heymann
David Thomson
news:amqc8g$jjr$3...@knossos.btinternet.com...
> Do you realise that the radius of the electron has been measured?
This ought to be good. Are the radii of electrons orbiting an hydrogen atom
and uranium atom the same?
Dave
FrediFizzx
| electromagnetic in origin on my webpage
Of course the mass of the electron is entirely electromagnetic in origin.
What else could it be? The Higgs boson? AHAHAHAHAHA What a joke. Even if
it is, the Higgs could probably be attributed to EM forces at a higher
energy level. Believe me, there is nothing else in the Universe to make
mass other than EM. *All* "particles" are made from it. In the beginning
all that existed was EM energy. Everything in the Universe comes from that
including space-time and gravity.
FrediFizzx
Old Man
news:f%4k9.948$Nj2....@newsread2.prod.itd.earthlink.net...
Fool. Rutherford is a crackpot, and Thomson doesn't even
qualify. [Old Man]
David Thomson
news:l96k9.2403$V64.1...@newsfeed.slurp.net...
> Fool. Rutherford is a crackpot, and Thomson doesn't even
> qualify. [Old Man]
It's time to put you in my blocked senders list.
David Thomson
Hah! I thought Old Man was referring to Rutherford and Thomson the Nobel
prize winner. I wasn't paying attention to the name of the guy who started
this thread. It's rather interesting that Rutherford and Thomson worked
together to unlock the secrets of the electron and that Rutherford and
Thomson today both independently show the relationship between charge and
mass.
Dave
Franz Heymann
If you think that the radius of the electron has anything whatsoever to
do with the spatial distribution of its wavefunction, you should go and
learn a little physics before committing any further blunders.
Franz Heymann
Franz Heymann
Thus proving Old Man's point to the hilt.
Franz Heymann
>
>
>
>
David Thomson
You're the one talking about the radius of the electron as having a finite
value. The "radius" of a subatomic particle is its fine structure constant
and is a function of its charge. The distance of the particle radius will
change as it needs to in order to maintain the ratio of strong charge to
elementary charge. The subatomic particles don't have a fixed radius. Any
measured values for an electron radius will be specific to a particular set
of circumstances and may change for a different set of circumstances.
Dave
FrediFizzx
I have been suspecting that particles change their effective size also
depending on the situation. However, it is tough to make everything work
out mathematically.
FrediFizzx
David Thomson
news:eJyk9.3100$EX5.73...@newssvr21.news.prodigy.com...
> |The "radius" of a subatomic particle is its fine structure constant
> | and is a function of its charge. The distance of the particle radius
will
> | change as it needs to in order to maintain the ratio of strong charge to
> | elementary charge. The subatomic particles don't have a fixed radius.
Any
> | measured values for an electron radius will be specific to a particular
set
> | of circumstances and may change for a different set of circumstances.
>
> I have been suspecting that particles change their effective size also
> depending on the situation. However, it is tough to make everything work
> out mathematically.
If you start with my unified charge equation and develop from there, the
math works out great. What particular situation do you have in mind?
Dave
Pmb
Howdy Franz
> Do you realise that the radius of the electron has been measured?
What was the result? I'd be grateful if you have a referance handy.
Are you familiar with the paper
"On the self-energy and the electromagnetic field of the electron,"
V.F. Weisskopf, Phys. Rev. 56, 72-85 (1939)
This seems to be a pivitol paper on the self-energy of the electron.
Do you know whether the conclusions still valid. Or has the progress
in physics some how outdated this work?
Note: Weisskopf states in the abstract
"... It is found that, as a result of Dirac's positron theory, the
charge and the magnetic dipole of the electron are extended over a
finite region:.."
If still valid do you know how well it corresponds to measurement?
Thanks in advance
Pmb
Franz Heymann
> "Franz Heymann" <Franz....@btopenworld.com> wrote in message
> news:ams3t5$3fu$1...@paris.btinternet.com...
> >
> > "David Thomson" <ne...@volantis.org> wrote in message
> > news:f%4k9.948$Nj2....@newsread2.prod.itd.earthlink.net...
> > > "Franz Heymann" <Franz....@btopenworld.com> wrote in message
> > > news:amqc8g$jjr$3...@knossos.btinternet.com...
> > >
> > > > Do you realise that the radius of the electron has been
measured?
> > >
> > > This ought to be good. Are the radii of electrons orbiting an
> > hydrogen atom
> > > and uranium atom the same?
> >
> > If you think that the radius of the electron has anything whatsoever
to
> > do with the spatial distribution of its wavefunction, you should go
and
> > learn a little physics before committing any further blunders.
>
> You're the one talking about the radius of the electron as having a
finite
> value. The "radius" of a subatomic particle is its fine structure
constant
> and is a function of its charge.
The latest measurements of the radial distribution of the charge on an
electron showed that it has an RMS radius less than 10^-18 m.
> The distance of the particle radius will
> change as it needs to in order to maintain the ratio of strong charge
to
> elementary charge.
The electron carries only an electric charge. It does not participate
in the strong interaction. Also its electric charge is a strictly
conserved quantity. No theory indicates that it might be otherwise, and
no experiment has ever given contrary results.
And just what do you mean by "The distance of the particle radius "?
> The subatomic particles don't have a fixed radius.
That is an assertion. Please provide evidence of any measurements which
show that you have a basis for that statement
> Any
> measured values for an electron radius will be specific to a
particular set
> of circumstances and may change for a different set of circumstances.
Please provide evidence in favour of that assertion.
Franz Heymann
Franz Heymann
"Pmb" <pm...@hotmail.com> wrote in message
news:8ac61757.02092...@posting.google.com...
No, I am not familiar with Weiskopf's paper, and unfortunately I no
longer have access to a physics library, but it sounds interesting. I
will contact one of my ex-colleagues at UCL to try and get a photocopy.
I have a strong feeling that time will have overtaken many papers from
those days, including this one.
The data on the radius of the charge distribution of the electron comes
from measuring the electric form factor involved in electron-positron
elastic scattering measurements at LEP at CERN. (Essentially determining
possible deviations from the differential cross section to be expected
from point-particle scattering). The result is that the charge
distribution of the electron is compatible with being a point-like
particle, the upper limit to its radius being 10^-18 metres.
This is three orders of magnitude smaller than predictions of the radius
(ca 10^-15 metres.) based on the assumption that the mass arises only
from the energy of the electric and magnetic fields of the electron.
Franz Heymann
David Thomson
> The latest measurements of the radial distribution of the charge on an
> electron showed that it has an RMS radius less than 10^-18 m.
I'm sure it will under certain conditions. Can you cite a reference for the
experiment so I can read up on it?
> > The distance of the particle radius will
> > change as it needs to in order to maintain the ratio of strong charge
> to
> > elementary charge.
>
> The electron carries only an electric charge. It does not participate
> in the strong interaction. Also its electric charge is a strictly
> conserved quantity. No theory indicates that it might be otherwise, and
> no experiment has ever given contrary results.
The charges of the electron are strictly conserved, I agree. But it's
pretty obvious the electron does have a strong charge. As you are well
aware, electrons and photons directly interact with each other electrically.
Together they produce electromagnetic systems. The electron has an
elementary charge, but the photon does not. Yet, the photon is well known
to have a charge associated with it. Not only does my unified charge
equation accurately predict the strong charge associated with the electron,
but my strong charge equation also predicts the strong charge associated
with the photon.
The electron and photon do electrically interact with each other. You're
saying the electron does not have a strong charge and the photon does not
have an elementary charge. Yet you also agree that the photon has a charge
associated with it. If you don't accept the mathematical proof I have given
for the strong charge of the electron, how to you reconcile this with a
photon without an elementary charge but nonetheless has a charge of some
kind?
> And just what do you mean by "The distance of the particle radius "?
Even though the particle radius is not fixed, there obviously is a radius.
Angular momentum has a circumference so it must also have a radius.
> > The subatomic particles don't have a fixed radius.
>
> That is an assertion. Please provide evidence of any measurements which
> show that you have a basis for that statement
There is the Bohr radius and Classical electron radius and now the radius
you referenced. All three are different lengths. What does the new value
of the electron radius do for the Classical distribution of charge
calculation for the electron? How does it affect the electron radius of the
hydrogen atom? Does this mean that the electron is now conclusively proven
to be a ball of something that has a planetary like orbit around an atom?
> > Any measured values for an electron radius will be specific to a
particular set
> > of circumstances and may change for a different set of circumstances.
>
> Please provide evidence in favour of that assertion.
As I just mentioned. The Bohr radius and Classical electron radius have
different values for the electron based on different circumstances. What
were the circumstances of the electron measurements you referenced? I
gather the electrons in this case were measured as free electrons?
Dave
Franz Heymann
> "Franz Heymann" <Franz....@btopenworld.com> wrote in message
> news:amvmk7$gho$3...@knossos.btinternet.com...
> > The latest measurements of the radial distribution of the charge on
an
> > electron showed that it has an RMS radius less than 10^-18 m.
>
> I'm sure it will under certain conditions. Can you cite a reference
for the
> experiment so I can read up on it?
Forget about the "certain conditions" lark. The *only* experiments
which have ever measured the charge distrubution of the electron are
experiments on Bhaba scattering. (The elastic scattering of electrons
and positrons. I disregard earlier e-e- elastic scattering experiments
in which the spatial resolution was worse by factors in the region of
1000)
Here is one of very many references on the topic:
Dmitri Bourilkov, arXiv:hep-ph/0002172v1 16 Feb 2000
This is a combined analysis of the Bhaba scattering data from all three
the the ALEPH, L3 and OPAL collaborations at the LEP accelerator of
CERN.
>
> > > The distance of the particle radius will
> > > change as it needs to in order to maintain the ratio of strong
charge
> > to
> > > elementary charge.
> >
> > The electron carries only an electric charge. It does not
participate
> > in the strong interaction. Also its electric charge is a strictly
> > conserved quantity. No theory indicates that it might be otherwise,
and
> > no experiment has ever given contrary results.
>
> The charges of the electron are strictly conserved, I agree. But it's
> pretty obvious the electron does have a strong charge.
On the contrary, it is obvious that it *does not* carry a strong charge,
since it does not participate in the strong interaction at all.
> As you are well
> aware, electrons and photons directly interact with each other
electrically.
> Together they produce electromagnetic systems. The electron has an
> elementary charge, but the photon does not. Yet, the photon is well
known
> to have a charge associated with it
The photon has no charge of any kind whatsoever associated with it. It
is an elementary neutral vector boson.
>. Not only does my unified charge
> equation accurately predict the strong charge associated with the
electron,
> but my strong charge equation also predicts the strong charge
associated
> with the photon.
Balderdash.
>
> The electron and photon do electrically interact with each other.
You're
> saying the electron does not have a strong charge and the photon does
not
> have an elementary charge.
That is correct. The only EM interaction consists of the emission or
the absorption of a photon by a charged particle. The whole of quantum
electrodynamics and classical electrodynamics depend from this very
simple phenomenon. Electric motors rely on this selfsame interaction
for their operation.
> Yet you also agree that the photon has a charge
> associated with it.
There is no way in which I could be persuaded to utter such nonsense.
If you believe otherwise, please say when and where I said so, and how I
actually expressed myself. Please retain enough context in your
response to allow me to tear your statement to shreds. If indeed I ever
said that there is a charge associated with a photon, I must have benn
blind drunk at the time.
> If you don't accept the mathematical proof I have given
> for the strong charge of the electron, how to you reconcile this with
a
> photon without an elementary charge but nonetheless has a charge of
some
> kind?
You have no clue as to what a "strong charge" might mean. What are its
field bosons? What is the range of this so-called strong charge of the
electron?
>
> > And just what do you mean by "The distance of the particle radius "?
>
> Even though the particle radius is not fixed, there obviously is a
radius.
> Angular momentum has a circumference so it must also have a radius.
There are no experiments in existence which indicate that the electron
has anything but a point charge distribution.
I have no clue about what a statement like "Angular momentum has a
circumference so it must also have a radius" might mean. Frankly, you
don't either.
>
> > > The subatomic particles don't have a fixed radius.
> >
> > That is an assertion. Please provide evidence of any measurements
which
> > show that you have a basis for that statement
>
> There is the Bohr radius and Classical electron radius and now the
radius
> you referenced. All three are different lengths. What does the new
value
> of the electron radius do for the Classical distribution of charge
> calculation for the electron?
There is no classical distribution of charge calculation for the
electron. What there is, is a calculation which determines the radius
of the charge distribution of an electron, on the assumption that its
mass is determined entirely by its classical electrical field energy.
That is clearly a bad assumption.
> How does it affect the electron radius of the
> hydrogen atom?
I know of no reason why the electron radius should depend on where it
finds itself.
> Does this mean that the electron is now conclusively proven
> to be a ball of something that has a planetary like orbit around an
atom?
No, it does not. The dynamics of the electron is determined by the
details of a wave function. QM does not empower us to calculate
trajectories of electrons in atoms.
>
> > > Any measured values for an electron radius will be specific to a
> particular set
> > > of circumstances and may change for a different set of
circumstances.
> >
> > Please provide evidence in favour of that assertion.
>
> As I just mentioned. The Bohr radius and Classical electron radius
have
> different values for the electron based on different circumstances.
Those are but quantities with the dimensions of a length. You might
have added the Compton length to your list if you so desired or indeed
the square root of the total cross section for elastic scattering of
electrons on protons. (The latter would have given you considerable
pause for thought)
> What
> were the circumstances of the electron measurements you referenced? I
> gather the electrons in this case were measured as free electrons?
If you are talking about the form factor measurements, yes they were
free electrons and positrons..
I am afraid you are stuck with that. The radius which you are trying to
probe is so small that the probing particle has to have an exceedingly
small wavelength. This means an energy so large that your target
particle would certainly be free after the collision, even if it was
bound before. The electron binding energies in atoms are measured in a
few eV, whilst the "diffraction" experiments which determine the
electron radius need probe energies in the region of 10^10 times
larger..
Franz Heymann
Pmb
> > Note: Weisskopf states in the abstract
> >
> > "... It is found that, as a result of Dirac's positron theory, the
> > charge and the magnetic dipole of the electron are extended over a
> > finite region:.."
> >
> >
> > If still valid do you know how well it corresponds to measurement?
>
> No, I am not familiar with Weiskopf's paper, and unfortunately I no
> longer have access to a physics library, but it sounds interesting. I
> will contact one of my ex-colleagues at UCL to try and get a photocopy.
Actually you don't need to. There is a copy online from a website I
found. In fact this web site is GREAT!! :-)
I just updated my web site with this link.
You can search it pretty easily. That's how I found it.
Thanks Franz. I'd be interested in your response.
Pmb
Franz Heymann
>
>
> > > Note: Weisskopf states in the abstract
> > >
> > > "... It is found that, as a result of Dirac's positron theory, the
> > > charge and the magnetic dipole of the electron are extended over a
> > > finite region:.."
> > >
> > >
> > > If still valid do you know how well it corresponds to measurement?
> >
> > No, I am not familiar with Weiskopf's paper, and unfortunately I no
> > longer have access to a physics library, but it sounds interesting.
I
> > will contact one of my ex-colleagues at UCL to try and get a
photocopy.
>
> Actually you don't need to. There is a copy online from a website I
> found. In fact this web site is GREAT!! :-)
>
> I just updated my web site with this link.
>
> fangio.magnet.fsu.edu/~vlad/
>
> You can search it pretty easily. That's how I found it.
I found the site. It stinks. I am square. Idon't wish to be amused by
funny little moving pictures when I am trying to read something.
I cannot find anything paper of Weiiskopf in that URL.
I will wait til I get a copy from UCL.
ksh95
attributed to the electrons "self energy". ie, its interaction with
virtual particles in the vacuum, this effect is electromagnetic in
nature and gives rise to the electron form factor and apparent radius.
This is what renormalization theory is all about.
When we neglect all the above effects we get a so called "bare
electron" whose mass is attributed to the Higgs field. But no one has
ever seen a bare electron in nature since one would have to turn of
the vacuum.
Let me give you the quick "dummies guide to to mass renormalization".
Bare electron mass (attributed to the higgs field) + mass due to the
interation with the virtual cloud of electrons that dress the bare
electron (attributed to electromagnetics) = The mass of the electron
that you find in textbooks.
Same thing happens with charge.
You can find out more about this in every introductory quantum field
theory textbook ever written.
pm...@hotmail.com (Pmb) wrote in message news:<8ac61757.02092...@posting.google.com>...
Franz Heymann
"ksh95" <ks...@yahoo.com> wrote in message
news:38a3ceaf.02092...@posting.google.com...
Don't top post. I shifted your post to the bottom for want of knowing
where in the thread you wanted it to be. My comments are also there,
in-line.
> Jesus people........ Part of the electrons mass and charge are
> attributed to the electrons "self energy". ie, its interaction with
> virtual particles in the vacuum, this effect is electromagnetic in
> nature and gives rise to the electron form factor and apparent radius.
> This is what renormalization theory is all about.
I take it that you realise that the electron form factor and charge
radius are compatible with those of a point particle.
> When we neglect all the above effects we get a so called "bare
> electron" whose mass is attributed to the Higgs field. But no one has
> ever seen a bare electron in nature since one would have to turn of
> the vacuum.
>
> Let me give you the quick "dummies guide to to mass renormalization".
>
> Bare electron mass (attributed to the higgs field) + mass due to the
> interation with the virtual cloud of electrons that dress the bare
> electron (attributed to electromagnetics) = The mass of the electron
> that you find in textbooks.
I think you have omitted the energy associated with the virtual photon
cloud which surrounds the electron.
>
> Same thing happens with charge.
>
> You can find out more about this in every introductory quantum field
> theory textbook ever written.
Franz Heymann
Gordon D. Pusch
> "Pmb" <pm...@hotmail.com> wrote in message
> news:8ac61757.02092...@posting.google.com...
> > "Franz Heymann" <Franz....@btopenworld.com> wrote
>>> I will contact one of my ex-colleagues at UCL to try and get a
>>> photocopy.
>>
>> Actually you don't need to. There is a copy online from a website I
>> found. In fact this web site is GREAT!! :-)
>>
>> I just updated my web site with this link.
>>
>> fangio.magnet.fsu.edu/~vlad/
>>
>> You can search it pretty easily. That's how I found it.
>
> I found the site. It stinks. I am square. Idon't wish to be amused by
> funny little moving pictures when I am trying to read something.
> I cannot find anything paper of Weiiskopf in that URL.
A Google search on the keywords "Weisskopf self-energy electromagnetic
field electron" almost imediately found (4th reference):
<http://fangio.magnet.fsu.edu/~vlad/pr100/100yrs/html/chap12_toc.htm>,
from which the paper can easily be accessed.
Popping up a few levels finds <http://fangio.magnet.fsu.edu/~vlad/pr100/>,
which appears to be an online version of the APS's CD-ROM, ``The Physical
Review - The First 100 Years.'' Since this is copyrighted material
(the copyright notice is included on the webapge!), unless the publisher
(Springer-Verlag) has given the website owner permission to post this
material, it is quite likely that the website owner is violating copyright
law by placing this CD-ROM on the web. Moreover, he has been careless enough
to forget to put a copy of 'robots.txt' in the parent directory containing
directives to keep webcrawlers such as Google out, so anyone can find the
evidence of his copyright violation. If the Feds or the copyright owners
catch on, he could be in serious trouble...
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
ksh95
Kinda, but it's a little more complicated than that. The electron is a
point particle dressed in a cloud of other virtual particles that
contribute to the bare electrons apparent mass and charge. The closer
you get to an electron the more it looks like a point particle. The
farther away you are the less it looks like a point particle. It's not
like the virtual cloud is an impassable wall.......BTW this is why
coupling constants run, and this is why high momentum collisions make
electrons look more point like. If the electon always looked point
like, we wouldn't need form factors and coupling constants wouldn't
run...........So in essense, far away the electron looks fuzzy, but if
you have a large enough momentum you can get through the cloud.
>
> > When we neglect all the above effects we get a so called "bare
> > electron" whose mass is attributed to the Higgs field. But no one has
> > ever seen a bare electron in nature since one would have to turn of
> > the vacuum.
> >
> > Let me give you the quick "dummies guide to to mass renormalization".
> >
> > Bare electron mass (attributed to the higgs field) + mass due to the
> > interation with the virtual cloud of electrons that dress the bare
> > electron (attributed to electromagnetics) = The mass of the electron
> > that you find in textbooks.
>
> I think you have omitted the energy associated with the virtual photon
> cloud which surrounds the electron.
No. In the canonical formulation all that stuff is in the second term.
Virtual electrons and virtual photons are inexorably linked, you can't
have interactions between the bare electron and its virtual electron
cloud without considering virtual photons. What do you think all those
squiggly lines and closed photon loops are in the higher order
diagrams.
Franz Heymann
"Gordon D. Pusch" <gdp...@NO.xnet.SPAM.com> wrote in message
news:gi8z1ki...@pusch.xnet.com...
Many thanks. It looks as if I had better get to it quickly.
Franz Heymann
Franz Heymann
You have me worried. I have never heard of any e-e- or e-e+ elastic
scattering experiments at any energy in which the differential cross
sections indicated form factors which were any different from the
predictions of the usual point-charge interactions. Maybe they have not
been done properly at energies substantially below the LEP energies?
> >
> > > When we neglect all the above effects we get a so called "bare
> > > electron" whose mass is attributed to the Higgs field. But no one
has
> > > ever seen a bare electron in nature since one would have to turn
of
> > > the vacuum.
> > >
> > > Let me give you the quick "dummies guide to to mass
renormalization".
> > >
> > > Bare electron mass (attributed to the higgs field) + mass due to
the
> > > interation with the virtual cloud of electrons that dress the bare
> > > electron (attributed to electromagnetics) = The mass of the
electron
> > > that you find in textbooks.
> >
> > I think you have omitted the energy associated with the virtual
photon
> > cloud which surrounds the electron.
>
> No. In the canonical formulation all that stuff is in the second term.
> Virtual electrons and virtual photons are inexorably linked, you can't
> have interactions between the bare electron and its virtual electron
> cloud without considering virtual photons. What do you think all those
> squiggly lines and closed photon loops are in the higher order
> diagrams.
Of course.
Pmb
> > I just updated my web site with this link.
> >
> > fangio.magnet.fsu.edu/~vlad/
> >
> > You can search it pretty easily. That's how I found it.
>
> I found the site. It stinks. I am square. Idon't wish to be amused by
> funny little moving pictures when I am trying to read something.
> I cannot find anything paper of Weiiskopf in that URL.
> I will wait til I get a copy from UCL.
Sorry. My mistake. The link I gave was incomlete. I meant to give
http://fangio.magnet.fsu.edu/~vlad/pr100/
The paper is at
http://fangio.magnet.fsu.edu/~vlad/pr100/100yrs/html/author/fs2au_12004weisskopf.htm
Pmb
Franz Heymann
hope Fangio has clearance for all the copyright stuff in it!.
Franz Heymann
vertner vergon
Just for fun
how does this fit your knowledge?
The electron is composed of *concentric* electric fields that expand and
contract.
The expansion/contraction creates the electric potential.
Each field has mass (Goldhaber & Nieto, etc.)
The sum of the masses of the individual fields creates the
mass of the electron.
The spin creates the magnetic moment.
Just as the mass of astronomical bodies is considered to be at the
center -- so it is with electrons. Thus, they appear to be point particles.
The density of the electron dissipates as the fourth power of the
distance from the center. So at a distance the electron could appear
"fuzzy".
I have what I call the "effective diameter" of the electron which is the
diameter of the volume region of maximum density. 2.4 x 10^-10 cm.
V. Vergon
------------------------------------------------
FREE: For Fun Gambling
Learn what the Pros know.
www.HalfbackCasinoAndSportsbook.com
"Franz Heymann" <Franz....@btopenworld.com> wrote in message
news:an7q0s$1hj$1...@knossos.btinternet.com...
ksh95
There's no discrepency here. The form factor is 1 (point like) to
first order. It's when you add all the higher order diagrams
(interaction with virtual particles etc.) that the form factor
deviates from unity. It's all the radiative corrections and such. It
depends how precise your measurment is.
Uncle Al
[snip]
We thought you were dead. Please don't disappoint us again.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Matthew Nobes
> "Franz Heymann" <Franz....@btopenworld.com> wrote in message news:<an7q0s$1hj$1...@knossos.btinternet.com>...
[snip]
> > You have me worried. I have never heard of any e-e- or e-e+ elastic
> > scattering experiments at any energy in which the differential cross
> > sections indicated form factors which were any different from the
> > predictions of the usual point-charge interactions. Maybe they have not
> > been done properly at energies substantially below the LEP energies?
> >
>
> There's no discrepency here. The form factor is 1 (point like) to
> first order. It's when you add all the higher order diagrams
> (interaction with virtual particles etc.) that the form factor
> deviates from unity. It's all the radiative corrections and such. It
> depends how precise your measurment is.
Ummm. That doesn't mean it's not pointlike. The form factors
are determined entirely on the basis of a point particle theory.
An interesting excerise is to compare the analogous form factors
for the proton to those of the electron. This illustrates the
difference between ``point-like'' and ``non-point-like'' form
factors.
--
``We may feel that at last, unlike all previous generations, we
have found certitude. They thought so too'' -Robert Conquest
Matthew Nobes
c/o Physics Dept. Simon Fraser University, 8888 University
Drive Burnaby, B.C., Canada
http://normland.phys.sfu.ca
Franz Heymann
"vertner vergon" <ver...@hisurfer.com> wrote in message
news:uphhee8...@corp.supernews.com...
[...]
Are you tring to crack a joke?
Franz Heymann
Franz Heymann
It is my understanding that the analysis of the LEP experiments on the
e+e- differential cross sections *did* include appropriate higher order
corrections. I have no references to hand to check up on this
ksh95
> "ksh95" <ks...@yahoo.com> wrote in message
> news:38a3ceaf.0209...@posting.google.com...
> > "Franz Heymann" <Franz....@btopenworld.com> wrote in message
> news:<an7q0s$1hj$1...@knossos.btinternet.com>...
> > > "ksh95" <ks...@yahoo.com> wrote in message
> > > news:38a3ceaf.02092...@posting.google.com...
> > > > "Franz Heymann" <Franz....@btopenworld.com> wrote in message
> news:<an559b$c64$1...@paris.btinternet.com>...
> > > > > "ksh95" <ks...@yahoo.com> wrote in message
> > > > > news:38a3ceaf.02092...@posting.google.com...
> > > > > > Jesus people........ Part of the electrons mass and charge are
I don't know what to tell you. The electon form factor is 1 to first
order. It wouldn't even make sense to have a form factor for a
particle withou form.
If I had to guess, I'd say that the LEP people are refering to the
fact that if the electron was not a point particle the form factor
should look different. This doesn't mean that an electron physically
looks like a singularity, this means that the electron has a structure
consistant with a point particle dressed by a virtual cloud. If a bare
electron had structure it's virtual cloud would be completely
different leading to a completely different form factor. The fact that
it shows the predicted (not-equal-to-unity) form factor is indeed
consistent with a point particle based theory.......but that doesn't
mean an electron looks a point..........
Am I making any sense.
My ge
Franz Heymann
Not really. At least not to me. That may be my lack of understanding.
Do you mean to say that the electron is a point particle, dressed with a
cloud of virtual particles, mainly photons, but a wee admixture of
massive (weak) bosons?
If so, I understand you. If not, I am lost.
Franz Heymann
ksh95
Here's a physical picture. An electron is fundementally a point
particle. However, the vacuum state of the universe consists of a
brewing sea of virtual particles poping in and out of existance.
Photons, electrons, positrons, protons, maybe gravitons.....etc, these
short lived particles are all virtual.
As our point electron moves through space it will interact with the
virtual particles it can interact with i.e. the electron has no color
charge so we wouldn't expect it to be influenced by gluons. It can,
however interact with, other particles having electromagnetic charge.
So during our point electrons journey from the sun to mercury we might
expect the elctron to colloid with a virtual photon created when some
virtual electron positron pair annhilated. Or, our point electron may
attract some virtual positrons and repel some virtual
electrons.....the possibilities are endless. Then net effect of all
this stuff is that the electron carries with it a cloud of virtual
dust. This point electron + its cloud of virtual dust is what we see
when we look at (measure) an electron. This cloud contributes to the
point electrons apparent mass and charge. This cloud also gives the
electron a "make-believe" apparent size, this is the information
contained within the form factor.
We can use quantum field theory to calculate what this "make believe"
electron form factor should look like, BUT we work under the
assumption that the bare electron is a point particle. If we would
have assumed that a bare electron is really a finite sized sphere we
would have calculated a different form factor, but I think I remember
learning something about finite sized particles having fatal
mathematical difficulties???
Anyways, my guess is that the LEP people aren't saying that an
electron doesn't have a virtual cloud contributing to its apparent
mass and charge. They are most likely saying we measured the cloud and
it looks just like it should if the bare electron were a point
particle.
Franz Heymann
One of my ex-colleagues is in fact involved in analysing the data from
these LEP experiments. I propose to raise him and ask him what the
actual situation is.
Franz Heymann
ksh95
Now that I think about it, I'm not so sure about composite particles
like protons. I don't know too much QCD. I would guess it's possible,
but I could be wrong.
ksh95
I'll ask around tomorrow and see if I can find any one who worked on
this specific experiment. What experiment exactly are you refering to.
Franz Heymann
Any of the experiments on either OPAL or ALPHA in which Bhaba scattering
mearurements were made. My colleagues are in the OPAL team.
Are you at CERN?
Franz Heymann
Franz Heymann
[[...]
> > Here's a physical picture. An electron is fundementally a point
> > particle. However, the vacuum state of the universe consists of a
> > brewing sea of virtual particles poping in and out of existance.
> > Photons, electrons, positrons, protons, maybe gravitons.....etc,
these
> > short lived particles are all virtual.
>
> Now that I think about it, I'm not so sure about composite particles
> like protons. I don't know too much QCD. I would guess it's possible,
> but I could be wrong.
Protons have, as a minimum, an associated sea of virtual mesons of all
kinds.
[...]
Franz Heymann