Re: Questions about radiation and em waves



FrediFizzx wrote:
"RP" <no_mail_no_spam@xxxxxxxxx> wrote in message
news:Pb6dnbRJF-r2T3PeRVn-rA@xxxxxxxxxxxxxxxxx


The Ghost In The Machine wrote:


In sci.physics.relativity, guskz@xxxxxxxxxxx
<guskz@xxxxxxxxxxx>
wrote
on 11 Feb 2006 17:50:56 -0800
<1139709056.492361.287420@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:


I always thought that a radio wave was a bunch of
ELECTRONS traveling through the air instead of
through an electric wire (tv cables instead of air
transmission) until someone told me NO that it's
only an electromagnetic wave and there is ABSOLUTELY
NO ELECTRONS that travel????


A radio wave is a collection of large photons (large in
the sense of wavelength). Electrons *are* involved,
but not during the transit; in the case of Earth radio,
power drives electrons in an antenna and they radiate
that power away to receivers in various locations (cars,
boom boxes, transistor radios, etc.) whose antennae
intercept the energy and drive an amplifier/discriminator.

(It's a colossal waste of energy, in some respects, but it's
also been very useful and profitable to broadcast news and
events as they happen. But then, we leave the lights on
all the time, too, in such locales as parking lots and
stores that have closed for the night, and that's probably
more energy wastage than the radio transmitters.)



1. I still find it strange since recend atom models
declare even the electrons as more like waves than particles???


Electrons are weird; depending on context they can be either.
They are much like light waves (and any electromagnetic
radiation, for that matter) in that respect.



2. Also the Big Bang produced radiation that latter became
matter(particle)....Is radiation an electromagnetic wave(radio

wave)?

And if so how does radiation and/or EM waves become particles with
TIME???



The creation of matter from electromagnetic radiation requires
rather specific circumstances; the best I can do here is that
first a photon from that radiation has to have enough energy
(a positron-electron pair has 2 * 0.510 MeV = 1.020 MeV; therefore
the photon has to have at least this energy, which corresponds
to a maximum wavelength of hc/E = 1.2155 pm or so). The background
radiation from the Big Bang is in the microwave region, so it's
not quite energetic enough now. (It probably was in the
distant past.)

Also, a charge has to be nearby, usually an atomic nucleus or
perhaps an electron.


I do believe it has to have more mass than an electron in "normal" type
of pair production. A second virtual photon is involved in the
interaction that is connected to the heavy nucleus.


AIUI, given these conditions, the photon gets converted into
a positron and an electron, and any excess energy gets
converted into momentum.


Not "photon". Photons. It takes at least two photons for pair
production.


That's a nice trick, energy turning into momentum that is.


Yes, I think he meant to say kinetic energy of the outgoing electron and
positron is where the excess energy goes. ;-)


It isn't the photon that gets converted, its the partice with which it
interacts that gets converted.


Which "particle" would that be? But correct that there is no conversion
really. The two photons are annihilated and and an electron and
positron are created from the quantum "vacuum".


The energy of a photon is frame dependent. Regardless of its energy

wrt

your frame, it has zero energy wrt some other frame, or arbitrarily
close to zero anyway. From that perspective there will also be other
photons with greater energy, i.e. that have blue shifted through the
transformation between frames, and these would, according to your
theory, not produce a charge pair via interaction with the electron,
while the lower energy photon would. Apparently these higher energy
photons receive instructions not to interact with the electron at all

so

that the theory of pair production can be salvaged. Or perhaps it is
the energy of the photon wrt the electron that matters. Regardless, it
seems apparent that your theory is highly flawed, since a zero energy
photon, which is the energy wrt some frame, cannot possibly be

converted

into two particles wrt that same frame. It must be instead that it is
the particle that is converted, eh?


It is the energy of the photons wrt the center of mass frame of the
interaction that matters.

Pair of photons or no pair, wrt some frame one of those photons essentially doesn't exist, in yet another frame it is the other photon that doesn't exist. Therefore, which photon is it that is converted? A nuclei may well have an electron-positron pair stripped from it via interaction with the zpf, but no particle creation takes place.

I agree more with your version: The shear force created by the converging *waves* (there are always infinite photons superposed over any point in space-time) separates the electron and positron from each other in a preexisting positronium atom. Though these particles are believed to self annihilate upon forming a positronium atom, that isn't what my math predicts. There is a ground state with an orbital radius somewhere around 10^-15 meters. Energy is released upon formation of the atom, but this loss of energy doesn't do away with the particles, only with the mass of the system that was composed by them. E=mc^2, where E is the electromagnetic potential energy. At ground state there is no potential energy left in this system. The resultant atom is sufficiently neutral wrt normal atoms that it goes undetected, even passing through them without inducing transitions. It is essentially a neutrino. With it's mass radiated away in the form of energy, and no further collapse possible (a prediction of Weberian electrodynamics), the electromagnetic PE is essentially but not precisely a constant zero. Disturbances will still upset the orbits, and thus this atom will hover at just above zero mass.

Richard Perry

.

Relevant Pages

  • Re: Weirdos square of negative mass.
    ... less than a free electron + a proton. ... Every example you've vomited up so far has a hydrogen atom ... electron gains energy, then emitted a photon in order to move; ... kg c of photons hit a PERFECT mirror (one that reflects all ...
    (talk.origins)
  • Re: Classical wave theory and tired light
    ... that the exchange of energy between matter and the radiation field is ... electrons on a target and that you observe that an electron is freed ... Because localized least action travel of all elementary EM events is ... "Travels as discrete quanta" is typical of "photons as billiard ...
    (sci.physics)
  • Re: Electromagnetic wave and photon spin
    ... Can someone explain to me the details of how electromagnetic radiation? ... Sun electron shakes, eye electron shakes 8 minutes later. ... Magnetism is the geometric superposition of Coulomb ... Photons are a mathmatical abstraction not a propagation model. ...
    (sci.physics)
  • Re: physical principle responsible for EMwave propagation
    ... relationships of EM waves. ... but as quantized quantities (photons) ... tranquil pond and watching that energy radiate as waves.It seems clear ... energy as is captive in an electron mass, ...
    (sci.physics.electromag)
  • Re: Question about Vacuum Gravity
    ... You seem pretty confident that 'EVERY FORM of Energy exhibits ... There is still a voltage gradient around a free electron. ... Our interactions are ALL by photons, ... >> The meteor itself doesn't slow down AT ALL, ...
    (sci.physics.relativity)