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A version of QED without ultraviolet divergences

A version of QED without ultraviolet divergences - Physics Forum

A version of QED without ultraviolet divergences - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


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  #41  
Old 08-09-2004, 08:33 AM
Eugene
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Default A version of QED without ultraviolet divergences



FrediFizzx wrote:


This experiment shows that the strength of electromagnetic interaction
increases with energy. Basically, at large energies we cannot use the
usual Coulomb law e^2/(4 \pi r). We should use e^2 f(E) /(4 \pi r),
where f(E) is a function of the energy of two particles, which is equal
to 1 at E=0 and slowly grows with increasing E. Where do you see
virtual particles? This observation says nothing about the existence of
virtual particles. I agree that existing theory can explain this effect
using the virtual particle idea. But I am pretty sure that my theory
can explain this effect as well without using the concept of virtual
particles. My confidence is based on the fact that in my theory the
S-matrix is forced to be the same as in the standard approach, so all
effects associated with the S-matrix are exactly the same. The growth
of \alpha, or electron charge with energy is one such effect.


Could you give a reference to the experiment where such "kicks"
are observed?



It predicts the time evolution. It also predicts
wave functions of bound states.

An example is hydrogen atom with Lamb-shifted states. Current
theory can predict the energies of such states in good agreement
with experiment. This is possible, because these energies are
reflected in the S-matrix as its poles, and everything related to
the S-matrix is perfectly reproduced by the traditional QED.
However QED cannot predict radiative corrections to the wave functions
of the low-lying hydrogen states. (As far as I know, all calculations
of Lamb's shifts use either non-relativistic wave functions or
solutions of the Dirac's equation, please correct me if I am wrong).

In my theory, the electromagnetic
interaction between electron and proton is described by various terms
(potentials) in each order of the perturbation theory. The largest
contribution is given by the pure two-particle potential.
In the 2nd order of the perturbation theory I derived this potential
and obtained usual Breit's form: Coulomb potential + Darwin (magnetic)
potential + contact + spin-orbit + spin-spin. I have a full set of
rules how to extend these calculations to 4th and higher orders.
Then the eigenvalue problem for the hydrogen atom can be solved
just as in ordinary non-relativistic quantum mechanics. Radiative
corrections for both eigenvalues and eigenfunctions can be obtained.

Now, you may say that eigenvalues are measured, but what is the
use of eigenfunctions? Without eigenfunctions you cannot predict time
evolution. Recently, there was lot of interest in preparing wave
packets on atomic states and measuring their time evolution.
Currently, they do it with Rydberg states, so there are no meaasurable
Lamb's shifts. In the future, these wave packets may include
states affected by radiative corrections, or the accuracy of
measurements will improve so that even small radiative corrections
will become noticeable. Neither QED, nor Dirac's equation, nor
non-relativistic Schroedinger equation will be valid in this
case. My theory will be valid.

Of course, the 2-particle potentials are only part of the story,
there are also potentials (starting from the 3rd order) which
are responsible for the coupling with photons. They are also obtained in
my theory. The presence of
these potentials make excited levels of the hydrogen atom
unstable. Note that spontaneous emission in my theory does not
appear as a result of some mysterious "zero-point vibrations of
the electromagnetic field". It appears, because the full Hamiltonian
contains bremsstrahlung terms describing the coupling with photons.
These terms can
be treated as perturbation to derive the probabilities of the
spontaneous emission of photons, and the lifetimes of different atomic
levels. There are many other terms in the Hamiltonian,
e.g., those responsible for the process (electron+proton) ->
{2 electrons + positron + proton), but they can be safely neglected
in the energy range characteristic for the hydrogen atom.



See above. The same Hamiltonian that I use for studying the
hydrogen atom,
I can also use for studying the time dynamics of low-energy collisions
of the electron and proton. If I like, I can take the classical limit
and obtain trajectories of two interacting particles. The
bremsstrahlung terms (coupling with photons) in my Hamiltonian will
describe the effect of "radiation reaction": When particles accelerate,
they emit bremstrahlung photons, so the energy is spent on the photon
emission, and the particles slow down.



You expressed two parameters m_e and e through two other
parameters alpha and lambda_C. What makes you believe that these new
parameters are more fundamental? How do you get their values? Try
to convince me that "alpha is a geometric factor involving volumes of space"



As I said, there are no virtual particles in my approach, just real
observable particles (electrons, photons, protons, etc.) and
instantaneous potentials (Coulomb, magnetic, spin-orbit,
bremsstrahlung, pair creation, etc.) acting between them. I obtain
these potentials by taking the Hamiltonian of QED (with infinite
counterterms, to ensure the correct S-matrix) in the Coulomb gauge
and applying a unitary "clothing transformation". This gives me
a finite Hamiltonian with all those interaction potentials, which I can
now use to study the time evolution, bound states, etc.

I do not try to derive the QED Hamiltonian from the gauge principle.
This principle does not play any role in my approach, just as
fields do not play any role in my approach. Why should I care about
fields, virtual particles, gauges, etc? They are not observable,
anyway.

Eugene.


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  #42  
Old 08-09-2004, 09:02 AM
Eugene
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Default A version of QED without ultraviolet divergences



Bilge wrote:

I am pretty sure that this decay can be described without invoking
the concept of virtual particles. This concept might be a useful
heuristic tool for making some guesses and predictions, though.
That's entirely different matter. When Kekule invented the model
of the benzene molecule he saw in a dream six monkeys holding
each other by the tail.


Dipole moments, etc. are just theoretical constructs which make it
easier for us to grasp complex phenomena, like charge distributions. I
would agree with you if you say that virtual particles are just
theoretical constructs which make it easier for us to grasp complex
phenomena, like interactions of charged particles. My point is
that interactions of charged particles looked so complex simply
because we used a wrong Hamiltonian all this time. If we change
the Hamiltonian, as I suggested, everything becomes so simple,
that virtual particles are not needed.


Why there sould be something holding electron charge together?
Are you suggesting that electron is not an elementary particle?
In my view electron charge is just a factor in front of the interaction
terms in the Hamiltonian. I don't know why this factor has this
particular value. I don't know why the interaction has this particular
form. I accept this as given. My theory is about different things.


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  #43  
Old 08-10-2004, 04:30 AM
Bilge
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Default A version of QED without ultraviolet divergences

Eugene:


But the question is why would it make more sense to do so? The decay
above is a decay producing propagating particles, so it seems rather
natural to use the same description for the beta decay in a nucleus. It's
the same interaction. Let's use something even more ordinary. How do you
describe the (forbidden) transition 2s -> 1s in hydrogen? I describe it as
proceeding through a virtual p-state. 2s -> (2p) -> 1s. The complete decay
would of course involve all of the virtual states through which the
process could occur, with th 2p being the major contribution. That seems a
lot more physical than just writing down the perturbation expansion for an
assumed ``small perturbation'' to what is otherwise an eigenstate.



I see it as more than that. We _define_ an electron or a photon by
the physical properties we give it. A real photon is no less a mathematical
object than a virtual photon. They solve the same differential equations,
which is why virtual photons are called photons instead of something
else. The only difference is the way in which the continuity equation
enters. For a real photon, the longitudinal polarization is identically
zero, so that p^u e_u = 0 (no sum) while for a virtual photon, the
the longitudinal and scalar components cancel, such that p^u e_u = 0
as a sum.

[...]

I don't see dipole moments as ``just theoretical constructs''.
A dipole moment describes a specific physical attribute (at least
in the case where the dipole and by extension, other multipole
moments cannot be made to vanish by a change of coordinates).


I think that misses the point. One could always just use any
scheme to solve a differential equation without trying to give
some meaning to the mathematics.






I'm suggesting that whether the electron is elementary or not,
really isn't relevant. What is relevant is that its radius is
known to be at least 3 orders of magnitude smaller than can be
explained by equating the classical coulomb energy to the electron
mass. If you don't have any objection to ignoring the self-energy
required to ``assemble'' the charge, then I don't see why you
have a problem with renormalization. Ignoring the self-energy
of the electron is equivalent to ignoring renormalizability.


Well, that's fine, call the number 1 electron charge, if you like,
but regardless of what you call it, energy is required to assemble
the charge. If you are going to ignore the self energy, then any
argument you make regarding renormalization vanishes.


Precisely. I have no problem with the idea of a hamiltonian formalism.
The issue I have is that I think you've overstated what is is capable
of doing.

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  #44  
Old 08-10-2004, 05:20 AM
FrediFizzx
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Default A version of QED without ultraviolet divergences

"Eugene" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
|
|
| Bilge wrote:
| > Eugene:

| > >In my approach, vacuum is just a state without particles. Electron
| > >is a single particle with mass and charge taken directly from
| > >experiment. There is no interaction between single electron and the
| > >vacuum. Interaction is present when there are two or more particles.
| >
| > What holds the electron charge together?
| >
| Why there sould be something holding electron charge together?
| Are you suggesting that electron is not an elementary particle?
| In my view electron charge is just a factor in front of the interaction
| terms in the Hamiltonian. I don't know why this factor has this
| particular value. I don't know why the interaction has this particular
| form. I accept this as given. My theory is about different things.

The electron is an elementary particle. But if you took a "bare" electron
and a "bare" any other fermion, you could not tell them apart from each
other. So what gives them different properties when they aren't bare? It
can only be the quantum vacuum that gives them their particular properties.
The big mystery is why are protons stable? The answer has to lie in the
quantum vacuum. But besides all that, yes, your theory is about something
different. No problem.

FrediFizzx

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  #45  
Old 08-10-2004, 06:35 AM
FrediFizzx
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Default A version of QED without ultraviolet divergences

[Note: Please be careful not to bugger the quote attributes]
"Eugene" wrote in message:
|
| FrediFizzx wrote:
|
| > |"Eugene" wrote in message: [you actually wrote what immediately
follows:]

| > | I am pretty sure that 99.99% of particle physicists believe that
| > | electron is screened by a cloud of virtual particles, and that vacuum
| > | is a "boiling soup" of virtual pairs, and that interaction occurs when
| > | "real" particles "throw virtual particles" to each other. My point is
| > | nobody have seen these virtual particles in experiment. All these
| > | virtual particles, "soups" and "clouds" are just artefacts of
| > | incorrectly written Hamiltonian. I suggested a way how all this
nonsense
| > | can be eliminated from the theory, and, still, all observable
| > | predictions can be preserved (S-matrix) and new predictions
| > | (time evolution) can be added. Could you please give me a reference to
| > | the "Topaz experiment", I'll try to explain it without invoking the
| > | concept of virtual particles.
| >
| > [Only registered users see links. ]
| >
| > There is a postscript article that you can download at the bottom of the
| > page.
|
| This experiment shows that the strength of electromagnetic interaction
| increases with energy. Basically, at large energies we cannot use the
| usual Coulomb law e^2/(4 \pi r). We should use e^2 f(E) /(4 \pi r),
| where f(E) is a function of the energy of two particles, which is equal
| to 1 at E=0 and slowly grows with increasing E. Where do you see
| virtual particles? This observation says nothing about the existence of
| virtual particles. I agree that existing theory can explain this effect
| using the virtual particle idea. But I am pretty sure that my theory
| can explain this effect as well without using the concept of virtual
| particles. My confidence is based on the fact that in my theory the
| S-matrix is forced to be the same as in the standard approach, so all
| effects associated with the S-matrix are exactly the same. The growth
| of \alpha, or electron charge with energy is one such effect.

Actually, I think alpha is a function of momentum transfer squared. Did you
somehow get a different web page than me? It talks about virtual particles
right on the page and also in the article. Did you download and read the
article?

| > I hardly think they are artifacts since real particles are seen to
| > get a momentum "kick" from invisible vacuum virtual particles in actual
| > experiments. So you are wrong about that.
|
| Could you give a reference to the experiment where such "kicks"
| are observed?

[Only registered users see links. ]

The electron (shown in red) from the electron beam is detected to go off at
a different angle from the "kick" that it gets from the gamma photon which
is virtual. Believe me, there are plenty of other examples of real
particles seen to get a momentum kick from unseen virtual particles. You
should be able to find them without too much trouble. I think the old
bubble and cloud chamber photos are full of them.

| > |
| > | Of course, it is impossible to predict the time evolution of the wave
| > | function in the complex picture with virtual particles. That's my
| > | whle point. Of course, Feynman was right, and it has to be simplified.
| > | That's what I did: I simplified the theory without losing a bit of
| > | its predictive power, and even added to its predictive power.
| >
| > What does it predict that is "additional"?
|
| It predicts the time evolution. It also predicts
| wave functions of bound states.
|
| An example is hydrogen atom with Lamb-shifted states. Current
| theory can predict the energies of such states in good agreement
| with experiment. This is possible, because these energies are
| reflected in the S-matrix as its poles, and everything related to
| the S-matrix is perfectly reproduced by the traditional QED.
| However QED cannot predict radiative corrections to the wave functions
| of the low-lying hydrogen states. (As far as I know, all calculations
| of Lamb's shifts use either non-relativistic wave functions or
| solutions of the Dirac's equation, please correct me if I am wrong).

Huh? I don't think they are from solutions of Dirac's equation. See "The
Quantum Vacuum" by Milonni. He gives a few different physicist's
interpretations of the Lamb shift.

| In my theory, the electromagnetic
| interaction between electron and proton is described by various terms
| (potentials) in each order of the perturbation theory. The largest
| contribution is given by the pure two-particle potential.
| In the 2nd order of the perturbation theory I derived this potential
| and obtained usual Breit's form: Coulomb potential + Darwin (magnetic)
| potential + contact + spin-orbit + spin-spin. I have a full set of
| rules how to extend these calculations to 4th and higher orders.
| Then the eigenvalue problem for the hydrogen atom can be solved
| just as in ordinary non-relativistic quantum mechanics. Radiative
| corrections for both eigenvalues and eigenfunctions can be obtained.
|
| Now, you may say that eigenvalues are measured, but what is the
| use of eigenfunctions? Without eigenfunctions you cannot predict time
| evolution. Recently, there was lot of interest in preparing wave
| packets on atomic states and measuring their time evolution.
| Currently, they do it with Rydberg states, so there are no meaasurable
| Lamb's shifts. In the future, these wave packets may include
| states affected by radiative corrections, or the accuracy of
| measurements will improve so that even small radiative corrections
| will become noticeable. Neither QED, nor Dirac's equation, nor
| non-relativistic Schroedinger equation will be valid in this
| case. My theory will be valid.

So this is your prediction. When do you think an experiment might be done
to validate it?

| Of course, the 2-particle potentials are only part of the story,
| there are also potentials (starting from the 3rd order) which
| are responsible for the coupling with photons. They are also obtained in
| my theory. The presence of
| these potentials make excited levels of the hydrogen atom
| unstable. Note that spontaneous emission in my theory does not
| appear as a result of some mysterious "zero-point vibrations of
| the electromagnetic field". It appears, because the full Hamiltonian
| contains bremsstrahlung terms describing the coupling with photons.
| These terms can
| be treated as perturbation to derive the probabilities of the
| spontaneous emission of photons, and the lifetimes of different atomic
| levels. There are many other terms in the Hamiltonian,
| e.g., those responsible for the process (electron+proton) ->
| {2 electrons + positron + proton), but they can be safely neglected
| in the energy range characteristic for the hydrogen atom.
|
|
| >
| > | > The interaction "space" is really a blob and all we care about
| > | > are initial states and final states. But it is not wrong for us to
| > | > eventually want to know exactly what happens in the "blob". I think
we
| > just
| > | > don't know what all the quantum vacuum objects are yet is what is
| > preventing
| > | > us from figuring out the exact complexity.
| > |
| > | You cannot figure out what's inside the "blob" because you are using
| > | wrong QED Hamiltonian. I can predict exactly what's inside the "blob".
| > | Note also that in the macroscopic low-energy limit the "blob" becomes
| > | very big: centimeters, meters, etc. in size. So, the interacting
| > | time evolution becomes directly observable. QED cannot predict this
time
| > | evolution, as you already pointed out. In order to make such
| > | prediction you need to abandon QED and switch to the classical
| > | Maxwell's theory with point particles and electromagnetic
| > | potentials. My theory works in the full range of energies.
| >
| > I am interested in this. Show me a simple example.
|
| See above. The same Hamiltonian that I use for studying the
| hydrogen atom,
| I can also use for studying the time dynamics of low-energy collisions
| of the electron and proton. If I like, I can take the classical limit
| and obtain trajectories of two interacting particles. The
| bremsstrahlung terms (coupling with photons) in my Hamiltonian will
| describe the effect of "radiation reaction": When particles accelerate,
| they emit bremstrahlung photons, so the energy is spent on the photon
| emission, and the particles slow down.
|
|
| > |
| > | It would be nice to see how somebody can derive electron mass and
charge
| > | from something more fundamental. So far, I haven't seen that.
| > | In my approach, vacuum is just a state without particles. Electron
| > | is a single particle with mass and charge taken directly from
| > | experiment. There is no interaction between single electron and the
| > | vacuum. Interaction is present when there are two or more particles.
| >
| > Here is a semi-classical heuristic for electron mass in CGS units.
| >
| > m_e = (sqrt(hbar*c)e/w_C^2*sqrt(alpha))(2pi/lambda_C)^3
| >
| > With w = angular frequency, lambda_C = electron compton wavelength and
alpha
| > = fine structure constant. So this tells use that electron mass is an
| > interaction between vacuum charge and electronic charge divided by
frequency
| > squared per a volume of space. We figure that sqrt(alpha) is a
geometric
| > factor and goes with the volume of space. The above expression reduces
to
| > the familiar electron compton wavelength expression which is known to be
| > true experimentally.
| >
| > m_e = 2pi*hbar/lambda_C*c
| >
| > Now electronic charge is a much tougher one because you have to know the
| > exact geometrical configuration of the quantum vacuum. But the simple
| > heuristic is e = sqrt(alpha*hbar*c). The square root of alpha is simply
the
| > ratio between electronic charge and vacuum charge. I believe that alpha
is
| > a geometric factor involving volumes of space.
|
| You expressed two parameters m_e and e through two other
| parameters alpha and lambda_C. What makes you believe that these new
| parameters are more fundamental? How do you get their values? Try
| to convince me that "alpha is a geometric factor involving volumes of
space"

I think you are missing the point of the heuristics. What is fundamental
can be completely relative. The point is that the mass of an electron can
be described purely in terms of items that mass doesn't appear in. Mass is
emergent from an interaction in a volume of space possibly from quantum
objects that are massless. Read the following article on "Spin foams..."
starting on page 19 to get an idea about alpha, the fine structure constant,
coming from relations of volumes.

[Only registered users see links. ]

| > | It gives a lot of new insight.
| > | First, now you can directly calculate the
| > | time evolution of the wave function during interaction.
| > | Second, you can now forget about the mind boggling picture with
| > | virtual particles and work with real particles and potentials
| > | between them, just as in ordinary quantum mechanics.
| > | Third, I predict that the potentials acting between particles
| > | (Coulomb, magnetic, spin-orbit etc.) are instantaneous. This can be
| > | verified by experiment.
| > | Fourth, the theory predicts small (but fundamentally important)
| > | deviations from Einstein's predictions, for example, in the case of
| > | the decay of fast moving particles.
| > | Finally, this approach demonstrates that Lorentz transformations
| > | are not exact and universal, and that Minkowski space-time is an
| > | approximation.
| >
| > Hmmm... So what is the interpretation of gauge bosons in your idea?
| >
| > FrediFizzx
|
| As I said, there are no virtual particles in my approach, just real
| observable particles (electrons, photons, protons, etc.) and
| instantaneous potentials (Coulomb, magnetic, spin-orbit,
| bremsstrahlung, pair creation, etc.) acting between them. I obtain
| these potentials by taking the Hamiltonian of QED (with infinite
| counterterms, to ensure the correct S-matrix) in the Coulomb gauge
| and applying a unitary "clothing transformation". This gives me
| a finite Hamiltonian with all those interaction potentials, which I can
| now use to study the time evolution, bound states, etc.
|
| I do not try to derive the QED Hamiltonian from the gauge principle.
| This principle does not play any role in my approach, just as
| fields do not play any role in my approach. Why should I care about
| fields, virtual particles, gauges, etc? They are not observable,
| anyway.

Ok, that gives me a better idea of your approach. You are basically just
trying to simplify the situation even more. I have a feeling that you are
going to run into problems when you try to extend this. I have to disagree
that fields and virtual particles are not observable. They certainly are
observable indirectly. Even *real* photons are not directly observable.
You can't ever be in the "frame" of a photon.

FrediFizzx

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  #46  
Old 08-10-2004, 08:38 AM
Eugene
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Default A version of QED without ultraviolet divergences



Bilge wrote:

Because the theory is much simpler and does not use the concepts
which are not observable. Isn't it the goal of theoretical physics to
find the description of nature in the simplest terms, preferably
in terms of observable quantities?


I missed your point here.


Let me tell you how this process is described in my theory. If we keep
just electron-proton interactions in the Hamiltonian we can solve the
eigenvalue problem and obtain the energy spectrum and wavefunctions of
the stationary states (1S^{1/2}, 2P^{1/2}, 2S^{1/2}, etc.),
including fine structure, hyperfine structure,
Lamb's shifts, etc. This solution is approximate, because we need to
take into account other terms in the Hamiltonian. The largest
perturbation comes from bremsstrahlung terms which are written in terms of
creation and annihilation operators as
d^{\dag}a^{\dag}c^{\dag}da and
d^{\dag}a^{\dag}dac, where d, a, c are operators of the proton,
electron, and photon, respectively. With the addition of these terms
the 2s state of hydrogen becomes unstable. This is reflected in two
observable effects. First, the 2s state obtains some width
(the energy is not well defined). Second, if the hydrogen atom is
prepared in the 2s state,
its wave function will evolve with time. Such evolution can be described
in terms of time-dependent decomposition in the basis involving all
non-perturbed atomic states and (real) photon states. Of course, as
time goes to infinity, the final asymptotic state will
include the hydrogen atom in the stable 1S state and a number of freely
propagating photons
with most probable energies E(2S^{1/2}) - E(1S^{1/2}), E(2S^{1/2}) -
E(2P^{1/2}), and E(2P^{1/2}) - E(1S^{1/2}).

My theory can predict this time evolution, e.g., the probability of
population of the 2p level as a function of time. Such
time-dependent evolution is, in principle, observable in experiment.
(However, it would
require much finer time resolution and accuracy than today's
experiments with wave packets on Rydberg states.) Can QED predict
that? I bet it can't.

My description does not use the concept of virtual particles.
I don't see how virtual particles can help in understanding of the
light emission from hydrogen.




You forgot one little difference: real photons are really observable.
Nobody have observed a virtual photon, only theoreticians in their
imagination.



In my view, dipole moments are more useful objects than virtual
particles. Dipole moments are approximations to really existing
things, like charge distributions. Virtual particles are just artefacts
of our usage of a wrong Hamiltonian. Choose the correct Hamiltonian,
and virtual particles are not needed.


I think, there are just too many thing in theoretical physics which
do not have any meaning. That's what bothers me. All these virtual
particles, ghosts, gauges, fields,.. (the list is very long).
They have no relationship at all to anything observable. For me,
this is a sign that the theory is sick. I am trying to rewrite
the theory to get rid of all this garbage. In experiment, we observe
particles, and probabilities for measurements of particle
observables, like position, momentum, spin, polarization, etc.
A healthy theory should be constructed around these quantities.
That's what I am trying to do.


The self-energy of the electron and the necessity of the renormalization
is just a result of using a wrong Hamiltonian in the traditional
theory. They do not have any physical meaning.
The self-energy is a symptom that the theory is sick, and the
renormalization is a half-successful attempt to cure the theory.



Assemble the charge from what? Are there some droplets of charge
which need to be assembled and held together? What are they?

Eugene.


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  #47  
Old 08-10-2004, 08:53 AM
Eugene
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Default A version of QED without ultraviolet divergences



FrediFizzx wrote:

In my approach, there is no distinction between "bare" and "clothed"
particles. Because the finite Hamiltonian I am talking about is obtained
by applying so-called "clothing transformation" to the usual Hamiltonian
of QED. So, you may say that all particles are properly "clothed". I
highly recommend you to read excellent paper, where this approach was
introduced 46 years ago

O.W.Greenberg, S.S. Schweber, Nuovo Cim. 8 (1958), 378.

The difference between two kinds of fermions, like electron and photon
is 1) in their mass and 2) in the form of interaction terms
which depend on creation/annihilation operators of these
particles. For example, you can find proton operators in the
interactions responsible for strong forces, and you cannot find
electron operators there. That's the difference. It has nothing to
do with vacuum.


I don't have a solution for this mystery, but I suspect that you
are wasting your time looking in vacuum. There is nothing in vacuum.
Vacuum is just empty space without particles.

Eugene.


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  #48  
Old 08-10-2004, 09:41 AM
Eugene
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Default A version of QED without ultraviolet divergences



FrediFizzx wrote:

Agreed. I just allowed myself some slack to call it (transferred) energy
for brevity.



I got the same article all right. I read the stuff about virtual
particles. It is the same mumble I can read in every textbook. The
experiment does not confirm a word from this mumble. It says that
the strength of electromagnetic interaction grows with increasing
square of the transferred momentum. The same effect can be explained
without invoking the concept of virtual particles.


Thanks for the reference. I am pretty sure my theory can explain this
experiment without much trouble and without virtual particles.
If gamma photon you are talking about is seen in the detector,
then it is
a real particle. If it is not seen in the detector, then it is not
needed in the theory. Virtual photons have some scary physical
properties, like imaginary mass. So, I am pretty sure nobody have seen
them except theoreticians.


All lines seen in bubble chambers (and even unseen lines corresponding
to neutral particles) are created by real particles with positive
energy and positive (or zero) mass. Virtual particles do not have
these good properties, and have never been
seen directly in experiment. There are some effects which people
attribute to virtual particles, but all these effects can be explained
equally well (and even better) without virtual particles.



Thanks for the reference. I can also suggest you a good reading. In
Weinberg's "The quantum theory of fields" vol. 1 there is pretty
detailed
calculation of the Lamb's shift. On page 593 he says that the main
parameters in this calculation were obtained by using non-relativistic
wave functions. My point is that QED is not capable of predicting
radiative corrections to wave functions. It must borrow wave functions
from non-relativistic theory.


I agree that this experiment is difficult to perform. However, I have a
better experiment to suggest. I think this one should be doable with
minimum efforts in any laboratory. At least it does not cost billions
of dollars. One of predictions of my theory is that the Coulomb and
magnetic potentials acting between charged particles are instantaneous.
The idea of the experiment is simple: take two charges at a distance
R. Wiggle one charge, then I predict that the second charge will
immediately start to move in response. The second charge will get
another kick after time R/c when photons generated by the acceleration
of the charge 1 (or, electromagnetic wave, if you wish) will reach the
charge 2.



Why you are trying to complicate simple things? Why you need
"spin foams" to explain such simple thing as mass of the particle,
which is just a parameter?


I take a photographic plate and see the place where real photons hit it.
Or I can just open my eyes and see real photons. What can be more
direct?


God forbid! I have no intention to be in the "frame" of a photon.

Eugene.


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  #49  
Old 08-10-2004, 09:54 AM
tadchem
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Default A version of QED without ultraviolet divergences


"FrediFizzx" <[Only registered users see links. ]> wrote in message
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<snip>


I'll bet Grandpa's old cloud chamber could tell an electron from a positron.


Tom Davidson
Richmond, VA


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  #50  
Old 08-10-2004, 04:47 PM
FrediFizzx
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Default A version of QED without ultraviolet divergences

"tadchem" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
|
| "FrediFizzx" <[Only registered users see links. ]> wrote in message
| news:[Only registered users see links. ]...
|
| <snip>
|
| > The electron is an elementary particle. But if you took a "bare"
electron
| > and a "bare" any other fermion, you could not tell them apart from each
| > other.
|
| I'll bet Grandpa's old cloud chamber could tell an electron from a
positron.

Sure, as long as it is not removed from the quantum vacuum. I said "bare".

FrediFizzx

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