experimentally measured electron radius
cherring
An interesting equation can be derived in pair production if you set the
energy of the photon needed to create the particle pair equal to the rest
mass energy of the particles produced. It can be shown that for the rest
mass of an electron and the corresponding wavelength of the photon that
produces the lepton pair that 2mowo = h/c where mo is the rest mass of an
electron and wo is the wavelength of the photon (2 particles are produced
hence the 2). These come directly from E=hv, vw=c, and E(total)=2moc^2 (I
don't have a lambda symbol on my keyboard). It seems that as the wavelength
gets shorter the mass get larger. You might even conclude that the particles
are really rotating, vibrating strings, perhaps not point sources at all but
more balloon like static fields and that the size of subatomic particles is
inversely proportional to mass (or energy and for that matter velocity)! If
the
wavelength of the photon that produces the particle somehow carries over
into
a property of the particle you might have what you need. Postulating that a
particle
is formed when a photon's Poynting vector turns in on itself to produce the
particle (see
pair production posting on the 14th this news group), an increase in energy
would
translate to a shorter wavelength and for a particle that would mean a
smaller radius.
This radius may not have any relationship to the classical electron radius.
"Robert D. Morrison" <r...@micron.net> wrote in message
news:3A80161E...@micron.net...
> To s.p.p folks,
>
> Is there any evidence that the experimentally measured electron
> radius varies with its velocity?
>
> The reason I ask is that I have looked at the properties of the
> DeBroglie wave loop model of particles such as the electron, and
> found several properties of the model that seem to compute
> correctly. There is even a logical extension of the Schroedinger
> wave atomic solution that would appear to work for a free electron.
> However, the thing that always kills it is the fact that
> experimental evidence shows that the electron radius is orders of
> magnitude smaller than that predicted by the DeBroglie wave model.
> However, a simplified version of the wave model, which I have called
> a charge loop, says that the velocity of the particle affects the
> radius, with the limiting case being zero as the particle approaches
> the speed of light. It is easy to see this by picturing one
> geometrical view of the charge loop, and two interesting revelations
> pop out when you do that:
> If you orient the charge loop as a circle about an axis, and then
> give the loop velocity along the axis, the loop becomes a helix. The
> wave along the helix is limited by velocity c, thus forcing the
> radius of the helix to the axis to decrease. If you unroll the
> helix, it is easy to see that this radius is the adjacent side of
> a right triangle, so it will change as:
>
> r(helix) = r(e) Sqrt[1 - (v / c)^2]
>
> (r(e) is size of rest electron, v is applied velocity along the axis)
> Thus, this derives a relativistic style behavior just from geometry,
> predicting that the apparent size of the electron will decrease
> the more velocity it has in the lab frame of reference. Obviously,
> this only works for this orientation, but it appears to still be
> valid for other orientations (it's just not as easy to visualize).
> In the limiting case close to the speed of light, the helix
> radius becomes zero.
>
> The other interesting thing that comes out of this is if you
> multiply the wavelength times the momentum of this helix (imagine
> it stretched all the way out like a tortured slinky), you get h
> (Planck's constant), since the frequency in the DeBroglie model is
>
> f = m(e) c^2/h,
>
> so the wavelength is c * (1/f) = c h / m(e) c^2 = h / m(e) c.
> Then multiplying by the momentum m(e) c, you get h--so you could
> argue that this results in:
>
> X * P = h,
>
> the uncertainty relation.
> So you see why I'd be interested in an answer to the question--do
> scattering experiments show any evidence that the electron could
> vary in size depending on its velocity?
>
> Thank you,
>
> Bob Morrison
> r...@nospam.hp.com
>
> replace the nospam with hpbs1326
>
Jim Carr
} radius varies with its velocity?
} However, the thing that always kills it is the fact that
} experimental evidence shows that the electron radius is orders of
} magnitude smaller than that predicted by the DeBroglie wave model.
} However, a simplified version of the wave model, which I have called
} a charge loop, says that the velocity of the particle affects the
} radius, with the limiting case being zero as the particle approaches
} r(helix) = r(e) Sqrt[1 - (v / c)^2]
} (r(e) is size of rest electron, v is applied velocity along the axis)
In article <95pqmk$kjf$1...@lure.pipex.net>
"franz heymann" <franz....@care4free.net> writes:
>
>All experiments done to date are consistent with a point-electron.
>And the upper limit is tiny, tiny.
On the order of 10^{-18} m, or maybe a few orders of magnitude smaller.
>Many orders of magnitude smaller
>than the kind of thing you are considering here.
It is not clear what he is considering here, since he does not
present it in terms of an electron form factor or describe how
it scales. What does seem evident is that he is talking about
a model that ignores QED, so his first challenge should be to
calculate the Lamb shift.
--
James Carr <j...@scri.fsu.edu> http://www.scri.fsu.edu/~jac/
"The half of knowledge is knowing where to find knowledge" - Anon.
Motto over the entrance to Dodd Hall, former library at FSCW.
Bob Morrison
"Jim Carr" <j...@dirac.csit.fsu.edu> wrote in message
news:96pmjr$f4v$1...@news.fsu.edu...
I'll try to clarify a bit--but first let me thank you for taking
the time to respond to my posting, I know it must get tiring to
respond to people who keep flogging dead horses! I will make
this promise--however you choose to respond, I will accept your
expertise and not get contentious, simply because I appreciate
your time and understand that you have done your homework in
this area. To save you time, this response has three parts that you
can choose to read:
a. QED and charge loop
b. How the charge loop scales with velocity
c. Experiments that measure the electron size
Let me first state that if I am ignoring QED, then I should shut up
fast, I am not trying to replace known science. The reason I am
taking the approach I do is that I am hypothesizing
that all of the perturbative components that make up the QED theory
of free electron motion (interaction of an electron with photons, in
particular) are part of the electron. Perhaps another way to state it
is I am hypothesizing that the virtual e-p pairs + virtual photons are
an intrinsic part of what the electron is--and if this is so, then the
perturbative
components make up a structure that is similar to the construction
of a generic function via a Taylor series or a Fourier series. So when
I hypothesis this charge loop, I am thinking that there is another
geometrical representation that will yield insight to the electron
that is not readily seen from the perturbative deconstruction of the
electron shown and proven by experiment in QED. The thing that
makes me think there is something to this is the fact there there are
literally 10 or more unique derivations that come from the
charge loop representations (some of which I list in the
original posting or other postings I have made here or in s.p.r).
I accept that the naked electron is shown to be less than the 10^-18
size. What I was hypothesizing is that the charge loop model
will approach this size if the electron is going
close to the speed of light. In the orientation that is
easiest to visualize, the charge loop becomes a helix that
will decrease in radius significantly (because the loop
stretches out, yet the helix curve components cannot exceed
the speed of light, so to compensate the radius must decrease.
That is how I geometrically got the (rather interesting!)
Sqrt[1 - v^2/c^2] factor).
For example, to reduce the radius 5 orders of magnitude
(from 10^-13 to 10^-18), the electron must have a velocity
of about c * ( 1 - 4 10^-10). This will correspond to
an energy of about 10^5 times the rest mass energy of the electron
if I did my homework right--and it seems like almost any
accelerator scattering measurement would be likely to pass this
energy level, and in fact I wouldn't be at all surprised if
electrons exceed this speed in electronic circuits. Furthermore,
it is so easy to accelerate the electron to relativistic speeds
that I cannot readily come up with a simple experiment that
stops the electron in space for measurement--but I don't
know if there is something obvious I am missing here. This is
why I asked the question in the original posting. I will look
at my QED sources to see if it indicates that the virtual
particle distribution will change as the electron gains
velocity and see if the charge loop hypothesis then matches QED.
If you wish, and are willing to take the time to discuss this a bit,
could you educate me what you are looking for when you ask
for an electron form factor or how it scales?
Thanks again,
Bob Morrison
r...@nospam.boi.hp.com
replace nospam with hpbs1326
Jim Carr
"cherring" <cherr...@home.com> writes:
>
>This might help...
It does not.
>An interesting equation can be derived in pair production if you set the
>energy of the photon needed to create the particle pair equal to the rest
>mass energy of the particles produced.
That equation is not interesting. It only converts mass into
different units (wavelength or frequency or energy).
>It can be shown that for the rest
>mass of an electron and the corresponding wavelength of the photon that
>produces the lepton pair that 2mowo = h/c where mo is the rest mass of an
>electron and wo is the wavelength of the photon (2 particles are produced
>hence the 2). These come directly from E=hv, vw=c, and E(total)=2moc^2 (I
>don't have a lambda symbol on my keyboard). It seems that as the wavelength
>gets shorter the mass get larger.
It seems that way because you assumed it would. Your argument is
a tautology that ignores the fact that higher energy photons can
also create a particular particle pair.
>You might even conclude that the particles
>are really rotating, vibrating strings, perhaps not point sources at all but
>more balloon like static fields and that the size of subatomic particles is
>inversely proportional to mass (or energy and for that matter velocity)!
If you concluded that *and* that they had the size you came up
with above, you get other predictions that disagree with experiment.
--
James Carr <j...@scri.fsu.edu> http://www.scri.fsu.edu/~jac/
Dopeler Effect: The tendency of stupid ideas to seem smarter when they
come at you rapidly. (anon source via e-chain-letter)
cherring
> In article <cf1j6.79983$bb.11...@news1.rdc1.tx.home.com>
> "cherring" <cherr...@home.com> writes:
> >
> >This might help...
>
> It does not.
>
> >An interesting equation can be derived in pair production if you set the
> >energy of the photon needed to create the particle pair equal to the rest
> >mass energy of the particles produced.
>
> That equation is not interesting. It only converts mass into
> different units (wavelength or frequency or energy).
>
> >It can be shown that for the rest
> >mass of an electron and the corresponding wavelength of the photon that
> >produces the lepton pair that 2mowo = h/c where mo is the rest mass of an
> >electron and wo is the wavelength of the photon (2 particles are produced
> >hence the 2). These come directly from E=hv, vw=c, and E(total)=2moc^2 (I
> >don't have a lambda symbol on my keyboard). It seems that as the
wavelength
> >gets shorter the mass get larger.
>
> It seems that way because you assumed it would. Your argument is
> a tautology that ignores the fact that higher energy photons can
> also create a particular particle pair.
I assumed nothing of the kind. If we don't speak of a particle for the
moment then m*w= h/c or mass is inversely proportionally to wavelength (the
other 2 parameters being constants). That is what this equation says; that
is all it says. When a photon does convert into particles this equation
states that shorter wavelengths (higher energy) will produce greater masses
(just another perspective). This is exactly what happens when higher energy
photons produce proton/anti-proton pairs (your other particular particle
pair I presume; yes, there are even others but they are not stable) but I
also stated that this equation does not say anything about stability - why
only electron/positron pairs and proton/anti-proton pairs as the only stable
states in free space? Yes, I believe the proton is stable. Why not all other
states all the way up the spectrum?! Such highly quantized states are
amazing to say the least!
I also stated that perhaps a property related to the photon's wavelength
(probably 1/2 wavelength to be more precise as a rotating vector) gets
carried over into the electron (some form of dimension) but I was clear to
say I don't think it has anything to do with the classical electron radius.
Maybe there is something in the way the energy is stored in space that is
related to the photon's 1/2 wavelength. This of course would have to be the
wo at which mo occurs, be it electron/positron or proton/anti-proton. It
could explain why an electron that is being accelerated changes mass (a
physical reason), that as the energy increases, when it is sufficient, a
change in harmonic occurs as the easiest way to absorb the energy, which
from the equation is an increase in mass. It is only a possible
visualization of what might be happening. It would also, physically, explain
why such particles can never have a velocity that exceeds c. Energy for
acceleration gets continuously converted to harmonic mass. At such small
dimensions any stair-stepping to harmonics would be hard to detect, I would
think, especially if the energy required to initiate a step is more than is
required for the step (climbing over a hill). Quantized motion, isn't that
what we see (probably all but disappearing at very high energies)?
At high enough energy near a strong coulomb field I would expect
pseudo-stable states (the mesons for instance?) which can only remain stable
if space is warped as near a strong coulomb field; electro-statically the
particles "see" space in way that appears as free space to this higher
harmonic state caught in the very strong coulomb field (K capture?). Maybe a
neutron is really something like a hydrogen atom (no, I am not saying it has
the same charateristics). I can then image neutron pairing and a =VERY=
strong sticky surface due to gravity especially if the physically-larger
size of the 2 neutrons block the coulomb field of 2 protons from somehow
affecting each other (they can't penetrate the neutrons' neutral fields).
Here gravity would take over at a very short distance and would appear to be
something other than 1/r^2.
Neutron pairing something like...
p-N-N-p for helium but becoming ...
p
N-N when stacked for higher atomic masses.
p
Something like...
p N
N-N p | p
P N
sitting on top of each other. Then at certain heights other groups would
fill into the notches along the stacks. Spin stability would also have to be
taken into account an would magnetic moment field matching.
I also can imagine electric field nodes of the photon giving rise to charge.
The electron/positron transition is probably a 2-node transition, hence no
smaller parts contained within. A proton/anti-proton is probably a higher
order node transition giving rise to parts of 1/3 charge, 2/3 charge
(quarks?). This last statement about higher order nodes would also explain
why an electron when accelerated to the mass of a proton would not convert
to an anti-proton but would only become more point like. The node-states
don't match.
> >You might even conclude that the particles
> >are really rotating, vibrating strings, perhaps not point sources at all
but
> >more balloon like static fields and that the size of subatomic particles
is
> >inversely proportional to mass (or energy and for that matter velocity)!
>
> If you concluded that *and* that they had the size you came up
> with above, you get other predictions that disagree with experiment.
Experiments that tell us electrons are point like are done with high energy
collisions (which is what I assume you are referring to). The equation
2m0*w0 = h/c implies that high energy electron will appear smaller, more
point like. The higher the energy of the collision the more point like you
get! But I also already stated that this wavelength property in the particle
has nothing to do with the classical electron radius and this implies its
classical physical size as well.
cherring
Compton wavelength, c - speed of light in FS, m_e - electron mass) but this
doesn't acknowledge that 2 particles are produced, therefore,
2*m_o*w_o = h/c (m_o - electron rest mass, w_o - minimum wavelength of
photon required for PP)
is better. Further, if we rewrite and divide by 2*pi, we get...
[(2)*(m_o)*(w_o)*(c)]/(2*pi)=h/(2*pi)
the right side of which is beginning to look like a spin. Mutilple by 1/2
while remembering that w_o divided by 2 is the length of a single node, we
get...
[(m_o)*(w_1/2)*(c)]/(pi) = (1/2)*[h/(2*pi)] (w_1/2 - is 1/2 of w_o)
Since a spin of 1/2 is in units of h/(2*pi), this sure seems to look as
though a linear vector has changed into a rotating vector. Using the above
equation and computing an equivalent radius, r (turning the half wavelength
into a circle), while using...
h = 6.626075x10^(-34) Js
m_o = 9.1093897x10^(-19) C
c = 2.99792458x10^8 m/s
pi = 3.1415927
r_e = 9.6540x10^(-14) m (electron)
r_pion = 3.5363x10^(-15) m (pion)
r_p = 5.2577x10^(-17) m (proton)
which seem to be in the proper dimensional ranges.
"cherring" <cherr...@home.com> wrote in message
news:Qd0p6.139766$bb.14...@news1.rdc1.tx.home.com...
franz heymann
'Fraid not. If you look up the form factor data for the electron and
the proton you will notice that you are out by factors of more than
1000. Moreover, the electron is much smaller than the proton, and may
indeed even be a point particle.
Franz Heymann
cherring
"franz heymann" <franz....@care4free.net> wrote in message
news:988g8o$jee$5...@lure.pipex.net...
As I keep saying, these dimensions are probably not related to the what we
typically associate as size for the various particles. It is simply a
statement that perhaps a dimension property related to the original photon
wavelength somehow gets carried over into the particle. It would be more
like a containment parameter. The radii above are not the classical particle
radii. When I said they appear to be in the correct range I meant that they
are in the range of scale for atomic & subatomic particles.
If we carry the formula a little further we can get...
1/2 = [(2)*(m_o)*(w_1/2)*(c)]/h
The spin is associated with the a single node length from the original wave.
This may be numeric only but then again it might not be. Its an awfully big
coincidence if it is.
Since a singularity occurs between nodes it seems plausible that the wave
could "BREAK" into 2 parts under the right conditions, therefore, under the
right coulomb field strength, at the correct angle, at a given precise time
for a given interaction, probably when 2 nodes are exactly equal in size
then a split could occur trapping the nodes in FS. As the strong coulomb
field attracts one and repels the other then the magnetic field lines that
allow the electric field to exist in a short coupled state would be
stretched, finally given way to stable magnetic moments for each particle
that exactly support their respective static electric fields. Singularities
can describe how water droplet form, so why not particles? We have all but
accepted the singularity in GR, the blackhole, why not allow them in EM?
Again, the equation does not speak to stability so I cannot answer the
question of why only the particle pairs we currently see?
franz heymann
The wavelength of a charged particle has nothing whatsoever to do
with its charge distribution.
> It would be more
> like a containment parameter.
That is devoid of meaning.
The radii above are not the classical particle
> radii. When I said they appear to be in the correct range I meant
that they
> are in the range of scale for atomic & subatomic particles.
So what?
>
> If we carry the formula a little further we can get...
>
> 1/2 = [(2)*(m_o)*(w_1/2)*(c)]/h
>
> The spin is associated with the a single node length from the
original wave.
> This may be numeric only but then again it might not be. Its an
awfully big
> coincidence if it is.
Crap.
>
> Since a singularity occurs between nodes
What on earth makes you think that?
> it seems plausible that the wave
> could "BREAK" into 2 parts under the right conditions,
Bullshit
> therefore, under the
> right coulomb field strength, at the correct angle, at a given
precise time
> for a given interaction, probably when 2 nodes are exactly equal in
size
What is "the size of a node"?
You are burbling
> then a split could occur trapping the nodes in FS. As the strong
coulomb
> field attracts one and repels the other then the magnetic field
lines that
> allow the electric field to exist in a short coupled state would be
> stretched, finally given way to stable magnetic moments for each
particle
> that exactly support their respective static electric fields.
Singularities
> can describe how water droplet form, so why not particles? We have
all but
> accepted the singularity in GR, the blackhole, why not allow them in
EM?
Nonsense.
>
> Again, the equation does not speak to stability so I cannot answer
the
> question of why only the particle pairs we currently see?
I am afraid there are many other questions you cannot answer either.
The truth od the matter is that nothing you have said makes any
contact with physics. This fact is what made it impossible for me to
respond in one or two other places in your post with words like
"bullshit" and "crap".
[Snip much more of the same stuff. It is not worth using bandwidth
transmitting it once again]
Franz Heymann
cherring
This is also not the DeBroglie wavelegth which a linear vector of motion. It
would be a rotating node vector. Why is that hard to visualize?
>
> > It would be more
> > like a containment parameter.
>
> That is devoid of meaning.
>
> The radii above are not the classical particle
> > radii. When I said they appear to be in the correct range I meant
> that they
> > are in the range of scale for atomic & subatomic particles.
>
> So what?
>
> >
> > If we carry the formula a little further we can get...
> >
> > 1/2 = [(2)*(m_o)*(w_1/2)*(c)]/h
> >
> > The spin is associated with the a single node length from the
> original wave.
> > This may be numeric only but then again it might not be. Its an
> awfully big
> > coincidence if it is.
>
> Crap.
Why? Because you say so?!
> >
> > Since a singularity occurs between nodes
>
> What on earth makes you think that?
>
> > it seems plausible that the wave
> > could "BREAK" into 2 parts under the right conditions,
>
> Bullshit
Why? Water drops do it. Oil drops do it. It can't be possible because you
don't think so?!
> > therefore, under the
> > right coulomb field strength, at the correct angle, at a given
> precise time
> > for a given interaction, probably when 2 nodes are exactly equal in
> size
>
> What is "the size of a node"?
> You are burbling
Come on! Photons are waves with nodes in space-time. They are real
disturbances. They are sinusoidal three (four) dimensional electric &
magnetic fields. Their E & M fields are at right angles to each other and
the Poynting vector is a right angle to both of these. Motion is in the
direction of the Poynting vector, the direction of energy transfer through
space-time.Take a cross section right through the Poynting and their shape
is similar to any electronic sinusoidal voltage or current wave form seem on
an oscilloscope. The Poynting vector guarantees a velocity (and momentum).
The field extension, in the direction of the fields not the Poynting vector,
into space-time is limited by c. A maximum and a minimum will exist
(positive-north at right angles and negative-south at right angles). The
leading node is always expanding into space-time. It reaches some maximum
and then decays. The decay creates another leading node that begins to
expand just in front of the previous node. At the point the leading node
equals the decaying node there is a singularity between them where the field
is zero in all dimensions.
> > then a split could occur trapping the nodes in FS. As the strong
> coulomb
> > field attracts one and repels the other then the magnetic field
> lines that
> > allow the electric field to exist in a short coupled state would be
> > stretched, finally given way to stable magnetic moments for each
> particle
> > that exactly support their respective static electric fields.
> Singularities
> > can describe how water droplet form, so why not particles? We have
> all but
> > accepted the singularity in GR, the blackhole, why not allow them in
> EM?
>
> Nonsense.
Why? Because you don't think it is possible?! You do accept black holes,
right? Do you agree that the forming of water and oil droplets at the break
away point is described mathematically by a singularity? Why not a real,
physical, spatial EM field the size of atomic and nuclear fields whose the
coulomb field strengths are enormous?
franz heymann
Because the three words "rotating node vector" strung side by side is
symptomatic of a severe case of verbal diarrhea
>
> >
> > > It would be more
> > > like a containment parameter.
> >
> > That is devoid of meaning.
> >
> > The radii above are not the classical particle
> > > radii. When I said they appear to be in the correct range I
meant
> > that they
> > > are in the range of scale for atomic & subatomic particles.
> >
> > So what?
> >
> > >
> > > If we carry the formula a little further we can get...
> > >
> > > 1/2 = [(2)*(m_o)*(w_1/2)*(c)]/h
> > >
> > > The spin is associated with the a single node length from the
> > original wave.
> > > This may be numeric only but then again it might not be. Its an
> > awfully big
> > > coincidence if it is.
> >
> > Crap.
>
> Why? Because you say so?!
No, because it just is crap.
Just smell this, for example:
"The spin is associated with the a single node length from the
original wave."
>
> > >
> > > Since a singularity occurs between nodes
> >
> > What on earth makes you think that?
> >
> > > it seems plausible that the wave
> > > could "BREAK" into 2 parts under the right conditions,
> >
> > Bullshit
>
> Why? Water drops do it. Oil drops do it. It can't be possible
because you
> don't think so?!
Water drops and oil are fairly remote from "waves".
Oh dear
>
> > > then a split could occur trapping the nodes in FS. As the strong
> > coulomb
> > > field attracts one and repels the other then the magnetic field
> > lines that
> > > allow the electric field to exist in a short coupled state would
be
> > > stretched, finally given way to stable magnetic moments for each
> > particle
> > > that exactly support their respective static electric fields.
> > Singularities
> > > can describe how water droplet form, so why not particles? We
have
> > all but
> > > accepted the singularity in GR, the blackhole, why not allow
them in
> > EM?
> >
> > Nonsense.
>
> Why? Because you don't think it is possible?! You do accept black
holes,
> right? Do you agree that the forming of water and oil droplets at
the break
> away point is described mathematically by a singularity?
No I don't agree. You appear not to know what a singularity might be.
cherring
They are just made of molecules, atoms, electrons, neutrons, protons -- all
of which are built from fields which have real, physical presence in
space-time. We separate these fields all the time. We break the field
bonding of electrons to a nucleus, we break covalent bonds, we break ionic
bonds, we break gravitational bonds, we break nuclear bonds. The photon has
been treated as some sort of "golden calf" in physics. Its the same stuff
that matter is made up of just in a different state. It can be affected the
same way. I say that state is a rotating vector.
FrediFizzx
You are confusing a photon with a classical EM wave. A photon is a quantum
mechanical particle by definition. I don't think your description above is
correct for a single photon according to QM. Though it may be logical to
assume that a photon has similar properties to a classical EM wave, I am not
sure that it does. So you have to be careful when using the term "photon".
In your case, EM particle wave may be a better terminology. It is the same
old story; trying to bridge the gap between classical EM and QM. Someone
might figure it out some day, so keep trying and thinking about it. For
now, QED works pretty good.
Regards,
FrediFizzx
cherring
"FrediFizzx" <FrediFi...@HoHohotmail.com> wrote in message
news:98f5ob$lvq$1...@slb6.atl.mindspring.net...
But the photons at work in QM are real EM waves. Just because we haven't
been able to describe them correctly enough mathematically to make the
computations come out doesn't change reality. The EM fields of radio waves,
microwaves & inferred don't change characteristics when they reach the
frequency of light and higher. They just happen to be on the scale of atomic
particles and interact, in some cases, more like billiard balls. This does
suggest a space localization on the same scale as particles. We can convert
them into particles under certain nuclear scale interactions. The particles
must be built from the parts of the original photon. Photons are linear EM
vectors with momentum (guaranteed by the Poynting vector). Particles are
rotating EM vectors (spin 1/2). They are electric fields with magnetic
moments. Build particles from photons and I say it happens by separating the
positive and negative fields in space by some distance (done by attraction
and repulsion in a strong Coulomb field be it pair production, triplet
production, quasi-atomic collision production or head on photon-photon
collision production) causing them to convert to the only other stable state
available - a static field with a magnetic moment, that the process is not
magic and that it can be visualized.
This equation, 1/2 = [(2)*(m_o)*(w_1/2)*(c)]/h, describes the relationship
of spin in particle created in pair production to w_1/2 of the photon that
creates the particle pairs (of course that would be +/-1/2 as the 2
particles spin in opposite direction. Further the spin of the electron
should be opposite of its cousin electron created in 2 different pair
productions relative to the nuclear magnetic moment as would also be the
case for the spins of the positron cousins.
> Regards,
>
> FrediFizzx
>
>
franz heymann
No. They are quantised waves and also, sometimes they are real and
sometimes they are virtual.
> Just because we haven't
> been able to describe them correctly enough mathematically to make
the
> computations come out doesn't change reality.
Speak for yourself. QED is capable of making extremely precise
quantitative predictions. It has *never* yet been found wanting when
compared with observations.
The rest of your post is simply proof of the fact that you don't
understand the situation at all. Maxwell's equations, used in their
purely classical are *incapable* of explaining effects on the atomic
and sub-atomic scales. Many folk more knowledgeable than yourself
have tried and failed. His equations *have* to be used in the
quantised sense required by QED.
Why on earth else do you suppose people went to all the trouble of
developing QED?
Franz Heymann
cherring
The virtual I dsiagree with. This is a mathematical construct to make the
math work out. Interactions are always real in the real world.
> > Just because we haven't
> > been able to describe them correctly enough mathematically to make
> the
> > computations come out doesn't change reality.
>
> Speak for yourself. QED is capable of making extremely precise
> quantitative predictions. It has *never* yet been found wanting when
> compared with observations.
>
> The rest of your post is simply proof of the fact that you don't
> understand the situation at all. Maxwell's equations, used in their
> purely classical are *incapable* of explaining effects on the atomic
> and sub-atomic scales. Many folk more knowledgeable than yourself
> have tried and failed. His equations *have* to be used in the
> quantised sense required by QED.
> Why on earth else do you suppose people went to all the trouble of
> developing QED?
I didn't say anything about QED but it doesn't talk about the transitions.
It ignores them. And yes, purely classical Maxwell equations don't work
either.
>His equations *have* to be used in the
> quantised sense required by QED.
The picture I am painting is quantized. Photons (here I would talking about
light frequencies and higher) aren't continuous in the same sense radio
waves are. Particle motion is quantized hence as disturbances are created a
short time delay occurs before additional movement of the particle which
creates (or leaves) a long null (in atomic time domain) between packets,
thus we have quantized EM fields, short bursts.
franz heymann
That statement guarantees that you don't actually know anything about
QED. It behoves you to familiarise yourself with it instead of
pontificating. I look forward to post from you after two years or so,
by when you may be able to speak with a little more authority.
>
> > > Just because we haven't
> > > been able to describe them correctly enough mathematically to
make
> > the
> > > computations come out doesn't change reality.
> >
> > Speak for yourself. QED is capable of making extremely precise
> > quantitative predictions. It has *never* yet been found wanting
when
> > compared with observations.
> >
> > The rest of your post is simply proof of the fact that you don't
> > understand the situation at all. Maxwell's equations, used in
their
> > purely classical are *incapable* of explaining effects on the
atomic
> > and sub-atomic scales. Many folk more knowledgeable than yourself
> > have tried and failed. His equations *have* to be used in the
> > quantised sense required by QED.
> > Why on earth else do you suppose people went to all the trouble of
> > developing QED?
>
> I didn't say anything about QED but it doesn't talk about the
transitions.
> It ignores them. And yes, purely classical Maxwell equations don't
work
> either.
Please specify these transitions of which you speak in sufficient
detail for me to be able to comment.
>
> >His equations *have* to be used in the
> > quantised sense required by QED.
>
> The picture I am painting is quantized. Photons (here I would
talking about
> light frequencies and higher) aren't continuous in the same sense
radio
> waves are. Particle motion is quantized hence as disturbances are
created a
> short time delay occurs before additional movement of the particle
which
> creates (or leaves) a long null (in atomic time domain) between
packets,
> thus we have quantized EM fields, short bursts.
There is one hell of a lot more about quantising Maxwell's equations
than you appear to be aware of. I suggest you read Bjorken and Drell,
Relativistic Quantum Mechanics. Practically he whole book is devoted
to a development of the quantisation of electrodynamics. It is most
emphatically not only to do with short bursts of radiation.
FrediFizzx
>
> "franz heymann" <franz....@care4free.net> wrote in message
> news:98fpii$eqr$9...@lure.pipex.net...
> >
> > The rest of your post is simply proof of the fact that you don't
> > understand the situation at all. Maxwell's equations, used in their
> > purely classical are *incapable* of explaining effects on the atomic
> > and sub-atomic scales. Many folk more knowledgeable than yourself
> > have tried and failed. His equations *have* to be used in the
> > quantised sense required by QED.
> > Why on earth else do you suppose people went to all the trouble of
> > developing QED?
>
> I didn't say anything about QED but it doesn't talk about the transitions.
> It ignores them. And yes, purely classical Maxwell equations don't work
> either.
>
> >His equations *have* to be used in the
> > quantised sense required by QED.
>
> The picture I am painting is quantized. Photons (here I would talking
about
> light frequencies and higher) aren't continuous in the same sense radio
> waves are. Particle motion is quantized hence as disturbances are created
a
> short time delay occurs before additional movement of the particle which
> creates (or leaves) a long null (in atomic time domain) between packets,
> thus we have quantized EM fields, short bursts.
A photon is a photon whether it be for light frequencies or radio
frequencies. There is no difference other than the frequency and thus
energy. A radio wave photon is not continuous either. I kind of see what
you are getting at but I think you need to word it more carefully.
Regards,
FrediFizzx
cherring
"FrediFizzx" <FrediFi...@HoHohotmail.com> wrote in message
Yes, but I think the separation is insignificant compared to the wavelength
of wave radio waves, actually, below light frequencies. One or two nm is
not much of an error for a 2 meter wavelength. Only when the frequency is
that of light and above does the separation become significant compared to
the wavelength hence the energy density occurs in quanta that are on the
atomic and subatomic level where the momentum is a closer match. If we take
a full sine wave form through the Poynting vector and rotate it in space 360
degrees it forms 2 geometric solids whose surfaces map out a sine wave. With
real world fields under a =VERY VERY= strong Coulomb field perhaps this 2
field packets could be separated. But as Franz point out this may be too
simple a picture. My contention though is that when we correctly describe
this field packet we will be able to visualize the transition to particles
and develop a complete mathematical model describing this transition or just
the before an after states as though some magic occurred.
> Regards,
>
> FrediFizzx
>
>
cherring
"cherring" <cherr...@home.com> wrote in message
news:iIgr6.162346$bb.15...@news1.rdc1.tx.home.com...
Jim Carr
in message news:97hdos$b2n$1...@news.fsu.edu...
|
| In article <cf1j6.79983$bb.11...@news1.rdc1.tx.home.com>
| "cherring" <cherr...@home.com> writes:
| >
| >This might help...
|
| It does not.
|
| >An interesting equation can be derived in pair production if you set the
| >energy of the photon needed to create the particle pair equal to the rest
| >mass energy of the particles produced.
|
| That equation is not interesting. It only converts mass into
| different units (wavelength or frequency or energy).
|
| >It can be shown that for the rest
| >mass of an electron and the corresponding wavelength of the photon that
| >produces the lepton pair that 2mowo = h/c where mo is the rest mass of an
| >electron and wo is the wavelength of the photon (2 particles are produced
| >hence the 2). These come directly from E=hv, vw=c, and E(total)=2moc^2 (I
| >don't have a lambda symbol on my keyboard). It seems that as the wavelength
| >gets shorter the mass get larger.
|
| It seems that way because you assumed it would. Your argument is
| a tautology that ignores the fact that higher energy photons can
| also create a particular particle pair.
In article <Qd0p6.139766$bb.14...@news1.rdc1.tx.home.com>
"cherring" <cherr...@home.com> writes:
>
>I assumed nothing of the kind.
You sure did. Your "2mowo = h/c" contains a constant on the right
and has another constant (mo) on the left, but wo can by anything
larger than sqrt{s} for the pair. If you don't make that tautological
assumption you get nonsense.
>If we don't speak of a particle for the
>moment then m*w= h/c or mass is inversely proportionally to wavelength (the
>other 2 parameters being constants).
That is just more numerology, signifying nothing other than what
wavelength you would need to make a pair at rest if it were
physically possible to do so with one photon -- which it is not.
You appear to be the sort who believes that physics follows
from randomly selected equations rather than that the equations
follow from the physical principles that apply to a given problem.
<... snip nonsense that follows ...>
Jim Carr
>
>Additionally, the Compton wavelength is given by w_c = h/m_e*c
>(w_c - Compton wavelength, c - speed of light in FS, m_e - electron mass)
It has nothing to do with anything. The "Compton wavelength" is
not a size, it is only a recoil factor that shows up in the
momentum conservation equation for Compton scattering. It would
be the Compton frequency (with corresponding different constant
factors) if the experiments had measured frequency rather than
wavelength.
The rest of your comments look like numerology, not physics
and totally unrelated to the subject of this thread or the
article you include at the bottom (which I snip on the
assumption that you would have added your comments after
it had they been of relevance to it.
cherring
The equationis not random. E= 2m_o*c^2 for the particles.
For the photon E=h*v.
For the minimum energy of conversion E(particles.) = E(photon) ==>
2*m_o*c^2 = h*v = h*c/w_o ==> 2*m_o*w_o = h/c
But in another thread - because w_o/2 = w_1/2 (half wavelength)
then m_o*w_1/2*c = h/4. Divide by pi and we get
m_o*w_1/2*c/pi = (1/2)*(h/(2*pi)). The right side is spin.
Jim Carr
in message <98s9g1$f9h$1...@news.fsu.edu>:
|
| | It seems that way because you assumed it would. Your argument is
| | a tautology that ignores the fact that higher energy photons can
| | also create a particular particle pair.
|
| In article <Qd0p6.139766$bb.14...@news1.rdc1.tx.home.com>
| "cherring" <cherr...@home.com> writes:
| >I assumed nothing of the kind.
|
| You sure did. Your "2mowo = h/c" contains a constant on the right
| and has another constant (mo) on the left, but wo can by anything
| larger than sqrt{s} for the pair. If you don't make that tautological
| assumption you get nonsense.
|
| >If we don't speak of a particle for the
| >moment then m*w= h/c or mass is inversely proportionally to wavelength
| >(the other 2 parameters being constants).
|
| That is just more numerology, signifying nothing other than what
| wavelength you would need to make a pair at rest if it were
| physically possible to do so with one photon -- which it is not.
|
| You appear to be the sort who believes that physics follows
| from randomly selected equations rather than that the equations
| follow from the physical principles that apply to a given problem.
In article <sWAu6.189464$bb.17...@news1.rdc1.tx.home.com>
"cherring" <cherr...@home.com> writes:
>
>The equationis not random.
You may not think so, but it looks that way to someone who
knows what the equations mean and how they must be used.
>E= 2m_o*c^2 for the particles.
Only when produced at rest, so p=0.
>For the photon E=h*v.
And p = E/c, which is not zero.
>For the minimum energy of conversion E(particles.) = E(photon) ==>
>2*m_o*c^2 = h*v = h*c/w_o ==> 2*m_o*w_o = h/c
Except that this says that momentum is not conserved, so your
analysis is invalid due to your assumption that all you have to
do is find two equations with "E" in them rather than conserve
momentum as well.
>But in another thread - because w_o/2 = w_1/2 (half wavelength)
>then m_o*w_1/2*c = h/4. Divide by pi and we get
>m_o*w_1/2*c/pi = (1/2)*(h/(2*pi)). The right side is spin.
Which is, as I pointed out
| <... snip nonsense that follows ...>
nonsense because it follows from a false premise.
cherring
That is what I said - minimum energy to convert (see later below)....
, so p=0.
Not if the linear momentum (of photon) is converted to angular momentum due
to spin! True, the 2 particle's combined linear momentum after conversion is
0 but their spin momentum is not 0. The linear momentum is conversed by
being converted to angular momentum just as an ice skater can convert linear
p to angular p.
> >For the photon E=h*v.
>
> And p = E/c, which is not zero.
>
> >For the minimum energy of conversion E(particles.) = E(photon) ==>
> >2*m_o*c^2 = h*v = h*c/w_o ==> 2*m_o*w_o = h/c
>
> Except that this says that momentum is not conserved, so your
> analysis is invalid due to your assumption that all you have to
> do is find two equations with "E" in them rather than conserve
> momentum as well.
>
> >But in another thread - because w_o/2 = w_1/2 (half wavelength)
> >then m_o*w_1/2*c = h/4. Divide by pi and we get
> >m_o*w_1/2*c/pi = (1/2)*(h/(2*pi)). The right side is spin.
Oh, for crying out load! E(photon of energy 1.022 Mev and above) = E(energy
from photon that produces conversion) + E(remaining) = E(particles masses) +
E(photon energy absorbed for recoil + KE of particles). Because of the way I
defined the 2 enegry components I mentioned above, minimum energy of
conversion, I can state E(energy from photon that produces conversion) =
E(particles) which implies E(remaining) = E(photon energy absorbed by recoil
+ KE of particles). Everything is conserved. I can make the equality and
substitution stated above. E(energy from photon that produces conversion)
can be treated as E(h*v_o) or E = h*v_o.
And in another form this, m_o*w_1/2*c/pi = (1/2)*(h/(2*pi)), becomes this,
E/(a_m_o) = (1/2)*(h/(2*pi))
where a_m_o is the angular frequency corresponding to photon frequency v_o
whose wavelength is w_o, i.e., 2*pi*v_o = a_m_o. Angular frequency implies
an angular vector.
Yes, w_0 is a constant because it is the exact wavelength that is changed
from a linear vector to an angular vector. I didn't say the equation was
contiguous, allows conversion at all energies. I said it didn't speak to
stability, that it doesn't explain why the first stable particle occurs only
at w_o. What it does do is define particle spin in terms of a component of
the photon whose energy is the minimal energy that produces the
paired-particles. However, note that when paired-particles are produced by
any mechanism to date, then m*w(half)*c = (1/2)(1/2)*(h/(2*pi)) where m is
m_o of either particle of the particle pair and w(half) is w_1/2 of a photon
of (h*c)/ w_o such that this is the equivalent energy to the minimum energy
absobed when the particle pair is produced.
cherring
Oops, sorry, I forgot the momentum part but it works out as well.
Jim Carr
In article <TIvy6.238007$bb.19...@news1.rdc1.tx.home.com>
"cherring" <cherr...@home.com> writes:
>
>That is what I said - minimum energy to convert (see later below)....
Which means p=0.
| ... so p=0.
>Not if the linear momentum (of photon) is converted to angular momentum due
>to spin!
Wrong. Linear and angular momentum are conserved separately.
>True, the 2 particle's combined linear momentum after conversion is 0 ...
Then why did you say "not"? Are you a troll? Do you realize that
this is in conflict with p = E/c initially as pointed out next?
| >For the photon E=h*v.
|
| And p = E/c, which is not zero.
|
| >For the minimum energy of conversion E(particles.) = E(photon) ==>
| >2*m_o*c^2 = h*v = h*c/w_o ==> 2*m_o*w_o = h/c
|
| Except that this says that momentum is not conserved, so your
| analysis is invalid due to your assumption that all you have to
| do is find two equations with "E" in them rather than conserve
| momentum as well.
|
| >But in another thread - because w_o/2 = w_1/2 (half wavelength)
| >then m_o*w_1/2*c = h/4. Divide by pi and we get
| >m_o*w_1/2*c/pi = (1/2)*(h/(2*pi)). The right side is spin.
>Oh, for crying out load!
You are objecting to your own statement? Or trying to ignore
the fact that I pointed out that your statement is nonsense
because it follows from a false premise as noted above?
>E(photon of energy 1.022 Mev and above) = E(energy
>from photon that produces conversion) + E(remaining) = E(particles masses) +
>E(photon energy absorbed for recoil + KE of particles).
What does that have to do with your failure to conserve momentum?
All you show here is your ability to grab a random equation and
use it without paying attention to what you are doing.
Hence your nonsense result.
cherring
I guess you don't watch much ice skating. Ice skaters manage to convert
linear mometum into angular momentum all the time.
I will say again...
The derivation is simply but one must assume that linear EM vectors can be
forced to wrap into rotating vectors. With this assumption and starting with
E = hv...
Take E/v=h and multiply by 1/2 then divide by 2*pi and knowing that for the
conversion energy v = v_o and that E/2 equals E_o, the energy mass of either
the electron or positron, one can easily show...
E_o/(2*pi*v_o) = (1/2)[h/(2*pi)] eq 1 (the right side is spin
units)
Further, since 2*pi*v_o is angular frequency w_o, we get...
E_o/w_o = (1/2)[h/(2*pi)] eq2 (simpliest form)
Which means that half the energy from the photon that can create an
electron-positron pair, roughly 1.022 Mev, divide by the angular frequency
equivalent to the linear frequency of the photon is particle spin.
Going back to the left side of eq 1 and knowing that v_o = c/wl_o where wl_o
is the photon wavelength and that wl_o/2 = wl_1/2, half wavelength and that
E_o = m_o*c^2...
E_o/(2*pi*v_o) = (m_o*c^2)/ {2*pi*[c/(2*wl_1/2)]} = (m_o*wl_1/2*c)/pi
hence...
(m_o*c)*(wl_1/2)/pi = (1/2)[h/(2*pi)] eq 3. (the mass and half
wavelength form)
Which means that a form of momentum related to the photon, m_o*c, times a
conversion factor from meters to radians, wl_1/2/pi, equals particle spin,
angular momentum.
Robert D. Morrison
cherring wrote:
> "Jim Carr" <j...@dirac.csit.fsu.edu> wrote in message
>
> > >Not if the linear momentum (of photon) is converted to angular momentum
> due
> > >to spin!
> >
> > Wrong. Linear and angular momentum are conserved separately.
>
> I guess you don't watch much ice skating. Ice skaters manage to convert
> linear mometum into angular momentum all the time.
>
I don't think so, as usual, J Carr is right.... There are many effects in ice
skating
that are wonderful demonstrations of the independence of angular momentum
and linear momentum. The most famous one is when a skater pulls in his arms
to cause an increase in angular velocity--but if you watched as much skating
as I used to, you would see a lot of not so good skaters get the same effect,
but don't quite have their linear momentum under control as they skitter across
the ice. The conservation of angular momentum is independent of any linear
momentum the skater may have. You may be thinking that because the
skater often has linear momentum before entering an angular momentum
spin that he is converting linear momentum to angular momentum, but I
think this is a bad way to think about it--it is actually the offset application
of forces (in other words, "angular forces") on a complex multibody (!)
system that gets the result we see. To do a spin, a skater applies
force (digs in his/her skates) at one point while pushing off with the other
foot
in an offset direction to either side of that point, thus causing a net angular
force
to be applied to his body (he can do this when standing still, too!). To obtain
angular momentum, I think it is always necessary to apply "angular force"
(typically two forces at different locations of a body such that a twist
results).
Linear momentum of a simple object by itself (without a displaced force, such
as offset resistance by the ice skates), I don't think, can ever be converted
directly to angular momentum.
Bob Morrison
PS, just for kicks, since I posted the original query, could we get back to
the idea that the *stationary* electron must be a point particle? Does anyone
have a pointer to experiments that test this? My charge loop hypothesis is
waiting to be skeptically squashed based on such experimental results, thanks!
The hypothesis shows that the electron will become very tiny with almost any
real-lab applied velocity, so it needs to be stationary (see the base
note, if it still exists, or, if it doesn't annoy people, I will repost it since
the
subject seems to have rather severely digressed..), thanks..
FrediFizzx
news:3AE3528E...@micron.net...
>
> Bob Morrison
>
> PS, just for kicks, since I posted the original query, could we get back
to
> the idea that the *stationary* electron must be a point particle? Does
anyone
> have a pointer to experiments that test this? My charge loop hypothesis
is
> waiting to be skeptically squashed based on such experimental results,
thanks!
> The hypothesis shows that the electron will become very tiny with almost
any
> real-lab applied velocity, so it needs to be stationary (see the base
> note, if it still exists, or, if it doesn't annoy people, I will repost it
since
> the
> subject seems to have rather severely digressed..), thanks..
How do you propose to ever have a *stationary* electron? Stationary
relative to what?
Regards,
Fredifizzx
Robert D. Morrison
FrediFizzx wrote:
I didn't realize this was a complicated thing to do. I was thinking of a free
electron
that was held in place by, say an appropriate cavity, then doing an experiment
on it that determines its point nature (bombard it with other electrons or
higher
energy particles?). Your comment implies that this is not practical to do or
that it is pointless (sorry!) in some way, could you elaborate? I realize that
vacuum is almost certainly required for such an experiment. Isn't there
a way to confine a free electron in the appropriate set of magnetic
or electrostatic fields? I thought I read something where a free electron had
been experimented on in this manner. So, to answer your question, stationary
relative to a lab frame of reference...?
Thanks for your reply, Bob Morrison
>
>
> Regards,
>
> Fredifizzx
Ralph E. Frost
>
>
> FrediFizzx wrote:
>
> > "Robert D. Morrison" <r...@micron.net> wrote in message
> > news:3AE3528E...@micron.net...
> > >
> > How do you propose to ever have a *stationary* electron? Stationary
> > relative to what?
>
> I didn't realize this was a complicated thing to do. I was thinking of a
free
> electron
> that was held in place by, say an appropriate cavity, then doing an
experiment
> on it that determines its point nature (bombard it with other electrons or
> higher
> energy particles?). Your comment implies that this is not practical to do
or
> that it is pointless (sorry!) in some way, could you elaborate? I realize
that
> vacuum is almost certainly required for such an experiment. Isn't there
> a way to confine a free electron in the appropriate set of magnetic
> or electrostatic fields? I thought I read something where a free electron
had
> been experimented on in this manner. So, to answer your question,
stationary
> relative to a lab frame of reference...?
Do a search on TOPAZ or go to http://tophp1.kek.jp
Jim Carr
in message news:9bdpm2$rvn$1...@news.fsu.edu...
|
| "Jim Carr" <j...@dirac.csit.fsu.edu> wrote
| in message news:9adr8j$qgi$1...@news.fsu.edu...
| |
| | | and has another constant (mo) on the left, but wo can by anything
| | | larger than sqrt{s} for the pair. If you don't make that tautological
| | | assumption you get nonsense.
| | |
| | | >If we don't speak of a particle for the
| | | >moment then m*w= h/c or mass is inversely proportionally to
| | | >wavelength (the other 2 parameters being constants).
| | |
| | | That is just more numerology, signifying nothing other than what
| | | wavelength you would need to make a pair at rest if it were
| | | physically possible to do so with one photon -- which it is not.
| | |
| | | You appear to be the sort who believes that physics follows
| | | from randomly selected equations rather than that the equations
| | | follow from the physical principles that apply to a given problem.
| |
| | In article <sWAu6.189464$bb.17...@news1.rdc1.tx.home.com>
| | "cherring" <cherr...@home.com> writes:
| | >The equationis not random.
| |
| | You may not think so, but it looks that way to someone who
| | knows what the equations mean and how they must be used.
| |
| | >E= 2m_o*c^2 for the particles.
| |
| | Only when produced at rest, ...
|
| In article <TIvy6.238007$bb.19...@news1.rdc1.tx.home.com>
| "cherring" <cherr...@home.com> writes:
| >That is what I said - minimum energy to convert (see later below)....
|
| Which means p=0.
I assume you agree with this observation now.
| | ... so p=0.
|
| >Not if the linear momentum (of photon) is converted to angular momentum
| >due to spin!
|
| Wrong. Linear and angular momentum are conserved separately.
In article <YxEE6.30983$JI6.1...@news1.rdc1.tx.home.com>
"cherring" <cherr...@home.com> writes:
>
>I guess you don't watch much ice skating. Ice skaters manage to convert
>linear mometum into angular momentum all the time.
You are either very confused or introducing a red herring to
distract the reader from your failure to address the core
issues under discussion.
This issue is really off topic here, so if you wish to followup
we can take the discussion to sci.physics where first-year mechanics
explanations belong. I will just point out that ice skaters do one
of two things: they either apply a force that simultaneously produces
a torque, so that the linear momentum is decreased and the angular
momentum is increased, or they take a motion that has both linear
and angular momentum and amplify the angular momentum by changing
the moment of inertia. In general they do some of both, particularly
during jumps.
The fact is that F = dp/dt and tau = dL/dt are separate rules
related by a common origin, hence p and L are conserved separately.
| >True, the 2 particle's combined linear momentum after conversion is 0 ...
|
| Then why did you say "not"? Are you a troll? Do you realize that
| this is in conflict with p = E/c initially as pointed out next?
I see no reply to the core issue in this discussion.
Thank you for further documenting that your comments are nonsense
even if you were not violating conservation of momentum.
<... snip rest of bogus argument ...>
The criticisms posted earlier (see above) to the first version of
your argument still apply.