Re: physical principle responsible for EMwave propagation

On Mon, 16 Jul 2007 13:15:48 -0700, srp@xxxxxxxxxxxx wrote:

On 16 juil, 13:51, John C. Polasek <jpola...@xxxxxxxxxx> wrote:
On Sun, 15 Jul 2007 20:00:39 -0700, s...@xxxxxxxxxxxx wrote:
On 15 juil, 22:31, John C. Polasek <jpola...@xxxxxxxxxx> wrote:
On Sun, 15 Jul 2007 15:28:26 -0700, s...@xxxxxxxxxxxx wrote:
On 15 juil, 17:49, John C. Polasek <jpola...@xxxxxxxxxx> wrote:
On Sun, 15 Jul 2007 07:55:06 -0700, s...@xxxxxxxxxxxx wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@xxxxxx> wrote:
This is no joke, I ask for help.

Maxwell's theory is wellknown and describes all the math.
relationships of EM waves.
To date I never had the idea to ask for the physical principle behind.

I know and accept the induction principles that charges are the cause
for EM fields.
The vacuum has c, e_o and my_o, but where are the propagating charges?

Would be nice to get really a serious answer of my request.

By definition e_0 (eps_0) is a unit of capacitance per meter of
vacuum and mu_0 is a unit of inductance per meter of vacuum

Both of these are associated when full treatment of em
waves is done. The Poynting vector is a reflection of this
association.

Capacitance implies displacement current, which in turn
implies charges. But it is true that in wave treatment,
there seems to be no traces of charges being associated.

On the other hand, it is known since Planck that light
does not really propagate as a wave despite the usefulness
of wave treatment, but as quantized quantities (photons)

Presumably, when the internal dynamic structure
of individual photons is better understood, we may
finally discover the apparently missing charges.

André Michaud

I have the impression that all light comes from quantized transitions
in the atom, so that each such transition is possessed of some exact
energy E = h*nu = h*c/L? It thereupon transmits as a wave in
pair-space or equivalent with wavelength L.

A close analog is dropping a pebble that has energy E = mgh into a
tranquil pond and watching that energy radiate as waves.It seems clear
the waves must possess energy E, but idt would be a stretch to claim
that E travels as a photonlike particle, just because it's quantized.

Would it ?

If the energy of a single photon emitted by one atom really did
radiated as waves, it would of necessity stop to be localized
and its energy would obviously spread out somehow, no ?

Since you subscribe to the SI system (from your book) you must endorse
eps0 and mu0 and therefore a transmission medium with speed c.

Yes.

I htink> we can both agree that it is not possible to suddenly introduce
a quantum of energy either into the water or the medium without some
natural reaction, which is the formation of radial waves in either the
2D or 3D case.

I would subscribe to this, if I subscribed to vacuum as being or
having
a medium, but I do not. My model makes sense only if there is not
underlying medium that interferes or interacts with the departing
photon.

Electromagnetic properties do not belong to any medium in my
model, but to each individual quanta of energy.

The energy is merely a bookkeeping entry whose value was last known
when the electron changed levels, but must now be accounted in the
departing wave, measurable only with difficulty.

But easy to calculate.

The initial event is a minute flash of light, the wave front spreading
at c with a coherence such that at any point, with a suitable lens it
can be recovered and reproduced as a point image.

I am trying to imagine the size of the lens required to refocus
the spread out energy of even one photon

No, not recover the energy, but produce an accurate image of the event
which is as close as you can ever get. It is routine to capture a
small fraction of the wavefront and refocus the rays (wavefront
normals) back to their originating true image.
In fact, just looking at a blackboard at 1 meter range, your eye deals
with only one part in 4 million of the light that's there, but you
can't argue that it can't work; it does. (4sq meters/10^-6 m^2 area of
pupil)

So you don't think that individual photons are hitting your retina
then ?
It's hard to reconcile photons with the fact that my eyeglasses help
me to see a given point far better and it's hard to connect photons
with eyeglasses.
But, there is no magic in lenses, which operate solely because the
index of refraction is 1.5 vs 1.0 for space. So if photons are
similarly slowed by 1.5, that's one for your side.
But the next step is impossible: the photon does not have a wavefront
surface with vector normals that are turned when one end of the
wavefront strikes the lens and is slowed before the other end does,
due to lens curvature. You need to 'flesh out' a means by which a lens
or prism can do its job. The photons are numerous and unconnected
bee-bees and some arrive later than others, soI don't see how they can
be turned by a lens.
from one hydrogen
atom de-energyzing to rest state coming in merely from Alpha
Centory. How spread out would it be arriving here, and at what
speed would all that spread out energy have to move from
that lens to regroup at the point of detection if all of it
was already moving at c before refocussing ?

Just do the vector operation and you will see that it would
have to regroup faster than c.

The same cannot be said of a photon.

I did some calculation on transverse acceleration of energy
for even one photon. (There was very little math in your old
copy of my book. This has been completed since).

The farthest sidewise that energy of a single quantum can
oscillate transversally for the energy not to internally exceed
c is the integrated transverse amplitude. For all photons,
the absolute transverse acceleration is given by

(2 pi c^2) / (lamda alpha)

wow 3.188e31 m/ss

Yes, if you apply the absolute wavelength of a photon of same
energy as is captive in an electron mass, yes.

But this equation is not meant for massive particles (no more
than E=h nu), but for free energy (photons) moving at c, while
transversally LC oscillating with a transverse acceleration that
can't possibly exceed c.

You probably mean accumulates to c.

I gave the corresponding LC equation more than once here,
even very recently. I can post it again here if you wish.

Please post the equation.

This is valid for photons of any wavelength.

where lamda is the wavelength and (lamda/alpha)/(2 pi) is
the integrated amplitude

This came out of converting the photon energy equation to
the form of something being accelerated.

E = hc = e^2/(2 eps0 lambda alpha)

(see equation (11) in this paper :

http://www.wbabin.net/science/michaud.pdf

I have studied the paper in some detail. I notice you repeatedly make
use of the expression 2 alpha*lambda (lambda =CWL for an electron) but
you have not reduced it to a single letter variable. I have done so.

I think you are missing the point I explained at length at the
beginning
of the paper. The purpose was to identify the lower limit of
integration
of the energy of localized photons. That universal lower limit turned
out to be (lambda alpha)/(2 pi), which also turns out to be the
maximum
transverse amplitude of the photon's LC oscillation (electromagnetic
oscillation). so the expression I repeatedly make use of is not
2 alpha lambda but (lambda alpha)/(2 pi).

I draw your attention to the fact that I have already deduced this
variable in Dual Space theory. It is the size of cell containing the
electron-positron pair in pair-space:
Lds = 2alpha*CWL = 3.514e-14m.
You can see it derived in Eq. 7 in my permittivity paper on my
websitehttp://www.dualspace.net.

Yes. I see how this connects with your model.

Then your Eq. (10) can be rewritten and simplified and expanded:
En = e^2/2eps0*alf*lam = e^2/eps0*Lds = e^2/C_cell = m_ec^2
My cell is a box
Lds = 3.514e-14m having a farad capacity of
C_cell = eps0*Lds = 3.13e-25 Farads.
The energy is m_ec^2.

Again, I must highlight the fact that my equation was not meant
to describe electrons, but photons of any wavelengths.

I have suggested in my book putting a lower limit L2 = L/2 to suit
Heisenbergs search for a lower limit, which equals half a box rather
than your putative radius of the electron. Half a box is the range of
either the electron or positron inside.

I saw that. Yes.

If you resolve eps0 to its other form

eps0 = 1/ (4 pi c^2 10^-7)

Units are missing; not SI valid>and substitute, you get

Well, I had assumed that it was clear that c was the
speed of light, which means that the units are s^2/m^2

Since mu0 = 4 pi 10^-7 and that this is what allows
c= 1/sqrt(eps0 mu0) (m/s)

Yes, I see you are using eps0*mu0*c^2 =1, but this is a mathematical
verity devoid of any physical content such as farads/meter,
nevertheless it can come in handy.

Unless you also consider that this standard equation
is not valid and not SI.

e = hc = (e^2 4 pi c^2 10^-7) / (2 lambda alpha)

or, more simply, divided by Lds my primal cell.

If you regroup so as to have "something" being
accelerated (the form X v^2/r) you get

e = (e^2 10^-7) [(2 pi c^2)/(lambda alpha)]

Notice above I equated the energy En = mc^2. I don't think you can get
that from this last equation.

My equation was not meant at all to deal with
mass, but with energy.
I think it's a significant finding that your energy = mc^2.
You have abandoned SI for cgs.

I have absolutely not. I still work with joules, meters, seconds, and
with the standard value for e (1.602176462E-19 C)

Any further spread involves internally exceeding the speed of light.

So what I'm saying is the the event is a quantum event, and its
continuation becomes a continuum, but with a clearly quantized
value.
John Polasek

This makes sense in your model, but in mine, with no underlying
medium, the only option is constant localization with the
maximum spread I mentionned, which amounts to constant
localization.

André Michaud

You should have a copy of my book Dual Space Theory.
John Polasek

I do.

André Michaud
.

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