Re: Particle Visualization

----- Original Message -----
From: "PD" <TheDraperFamily@xxxxxxxxx>
Newsgroups: sci.physics.particle
Sent: Wednesday, June 01, 2005 2:53 PM
Subject: Re: Particle Visualization


>
>
> Monitek wrote:
>> "PD" <TheDraperFamily@xxxxxxxxx> wrote in message
>> news:1117573328.198051.145710@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>> >
>> >
>> > Monitek wrote:
>> >> "PD" <pdraper@xxxxxxxxx> wrote in message
>> >> news:1116642764.925698.47790@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>> >> >
>> >> > Nick wrote:
>> >> >> PD wrote:
>> >> snip-----------------------
>> >> >> > Small detail. Electrons and positrons have electric charge.
>> >> >> > Photons
>> >> >> and
>> >> >> > neutrinos do not.
>> >> >>
>> >> >> Photons are not electromagnetic?
>> >> >>
>> >> >
>> >> > yes but they are not charged. which means there is no trilinear
>> >> > photon
>> >> > coupling.
>> >> >
>> >> > PD
>> >> >
>> >> > [snip]
>> >> >
>> >>
>> >> If photons are not electromagnetic:
>> >
>> > Reading comprehension problem. Photons are electromagnetic, but they
>> > are not charged.
>> >
>> >>
>> >> a) What are photons?
>> >
>> > The carriers of the electromagnetic interaction. The carrier does not
>> > need to carry charge itself. The charge identifies the *strength* of
>> > the coupling between a (fermion) particle and the appropriate (boson)
>> > carrier of the interaction.
>>
>> Are ther 2 kinds of photon one for positive charge and one for negative
>> charge?
>
> No, there is one kind of photon, the same for positive and negative
> charge.
>
>> Do the photons travel at the speed of light?
>
> Yes.
>
>>
>> > Think of it as how often the vertex
>> > happens.
>> >
>>
>> I do not understand the relevance of this statement.
>
> I wasn't sure you would. It's a reference to Feynman diagrams. A
> Feynman diagram is a little schematic of an interaction, where the
> different parts of the schematic represent terms in an algebraic
> expression which allows you to calculate the rate at which that
> interaction occurs. For example, a diagram like a "Y" might have for
> the three legs an incoming electron, an outgoing electron, and an
> emitted (or absorbed) photon, and the point where the emission (or
> absorption) occurs is at the junction of the "Y" -- the vertex. The
> term corresponding to the vertex is proportional to the charge of the
> electron. The stronger the charge is, essentially, the more frequently
> the vertex will occur during an electron's path through spacetime.
>
Sorry I switched of the idea behind Feynmans aide memoirs the minute he
introduced an imaginary exchange particle into the proceedings.

> Note that the charge is carried by the electron in this case, which is
> codified in the rule that vertices happen on electron lines but never
> on photon lines. You can't have a "Y" with just photons on the three
> legs. On the other hand, the statement that the electron has charge
> means *nothing more* than the fact that it does have vertices where
> photons are emitted or absorbed. That is, the answer to the question,
> "What is charge, anyway?" is simply that it is a flag that the particle
> that bears it participates in such interactions.
>

The vertices are imaginary. When the rate of change of velocity is large
enough then a charged particle causes an electromagnetic disturbance to
radiate.

>>
>> >>
>> >> b) From what are the electromagnetic properties of EMR produced?
>> >
>> > Too vague. I don't know what you mean by "electromagnetic properties".
>> >
>> >
>>
>> One can measure the electrostatic field associated with EMR.
>> One can measure the magnetic field assoiated with EMR.
>> "electromagnetic properties" is a phrase which encompasses both
>> properties.
>> My question is : Where do the electromagnetic properties associated with
>> EMR
>> arise from?
>> If the photon has no charge (I note the word 'need' as used above leaving
>> open the option for a charged photon), then the "electromagnetic
>> properties"
>> must arise from something else. If so what could that something else be?
>> On
>> the other hand if the photon is charged then one does not have to look
>> for
>> something else to derive the electromagnetic properties of EMR.
>>
>
> No, and this is simple to explain. You are under the impression, it
> appears, that the only sources of electric and magnetic fields are
> charges. This is not so. If you open a textbook to the chapter on
> Faraday's law and Maxwell's equations, you will see that there are
> other sources for electric and magnetic fields; namely, a changing
> magnetic field is a source of an electric field, and a changing
> electric field is a source of a magnetic field. This, in fact, is the
> key thing that lets electric/magnetic fields propagate *together*
> through space, without carrying charge with them. The fact that the
> fields are periodic (oscillating) is central to this, because
> oscillating fields are constantly changing, which is required for them
> to continue to feed each other.
>

Electromagnetic induction is what I would call it. Even that will not work
for EMR because a magnetic field can only induce a current in a conductor.
Where are the conductors in EMR in vacuo? A conductor is a collection of
charges which are free to move. To use electromagnetic induction to
propagate EMR you need charged particles.


Sorry, your starting assumption in your inductive cycle is totally wrong.
You can not start the electromagnetic induction cycle with a magnetic field.
The reason being that moving charges are required to create a magnetic field
so the changing electric field comes first. Electric fields require charged
particles-period. Therefore to create a magnetic field you require charged
particles. Where are your charged particles in EMR?


> The fact that there are non-charge sources for electric and magnetic
> fields came as quite a surprise to Faraday, but experiments don't lie.
> Likewise, the fact that the mathematical solution to propagating,
> chargeless fields happened to propagate at a speed that was numerically
> close to the measured speed of light came as quite a surprise to
> Maxwell, but it too was verified later in experiment.
>

If you have an electrostatic field you have a charged particle responsible
for its creation. There are no electric fields which do not have a charged
particle as a source. Yes there is no doubt that Maxwells equations were a
flash of pure brilliance. The subsequent sweeping of the field source under
the carpet was a mistake.

> This sequence of stunning revelations completely changed everything in
> physics. Indeed, of the four fundamental interactions, the
> electromagnetic interaction is the *only* one that we freely exploit at
> will. (Some will include gravity in that category, but if we didn't
> happen to have a fairly huge source of a gravitational interaction
> nearby, it'd be useless to us.)
>
> PD
>

The weak force has been resolved as electromagnetic. I have modelled the
strong force as electromagnetic. The electromagnetic force is
electromagnetic. So dont be surprised that there may only be two fundamental
forces.

Regards,
Monitek (Arden Barker)


.

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