On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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

On Sun, 15 Jul 2007 07:55:06 -0700, s...@microtec.net wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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.

The missing charges are the virtual pairs in pair-space.
John Polasek

On 15 juil, 17:49, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 07:55:06 -0700, s...@microtec.net wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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 ?

Any absorbtion of such a spread out photon would involve
instant regrouping of all of that energy at the single localized
point of absorption (an electron that would jump to an orbital
further away from its nucleus).

This regrouping would instantly violate the speed of light as
a maximum velocity of energy, it seems to me, if the
spreading even slightly exceeds initial localization.

It seems to me that this does set some limit to such
spredding out. that limit being the speed required for
final regrouping, that is the speed of light, no ?

André Michaud

The missing charges are the virtual pairs in pair-space.
John Polasek

On Sun, 15 Jul 2007 15:28:26 -0700, s...@microtec.net wrote:
On 15 juil, 17:49, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 07:55:06 -0700, s...@microtec.net wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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.

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.

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.

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.

The same cannot be said of a photon.

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

On 15 juil, 22:31, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 15:28:26 -0700, s...@microtec.net wrote:
On 15 juil, 17:49, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 07:55:06 -0700, s...@microtec.net wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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

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

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

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

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)]

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.

On Sun, 15 Jul 2007 20:00:39 -0700, s...@microtec.net wrote:
On 15 juil, 22:31, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 15:28:26 -0700, s...@microtec.net wrote:
On 15 juil, 17:49, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 07:55:06 -0700, s...@microtec.net wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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)

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

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 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.

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.

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.

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

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.

You have abandoned SI for cgs.

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

On 16 juil, 13:51, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 20:00:39 -0700, s...@microtec.net wrote:
On 15 juil, 22:31, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 15:28:26 -0700, s...@microtec.net wrote:
On 15 juil, 17:49, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 07:55:06 -0700, s...@microtec.net wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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

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.

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

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)

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

But there remain problems. It turns out that "displacement currents"
actually don't exist. If you put a capacitor into a (black) box and
ramp a current into it you do NOT find a magnetic field about the box
equivalent to if the box just held a straight wire. It seems nobody
has ever been able to measure this. Oops!

On Mon, 16 Jul 2007 13:15:48 -0700, s...@microtec.net wrote:
On 16 juil, 13:51, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 20:00:39 -0700, s...@microtec.net wrote:
On 15 juil, 22:31, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 15:28:26 -0700, s...@microtec.net wrote:
On 15 juil, 17:49, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 07:55:06 -0700, s...@microtec.net wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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.

On Mon, 16 Jul 2007 13:15:48 -0700, s...@microtec.net wrote:
On 16 juil, 13:51, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 20:00:39 -0700, s...@microtec.net wrote:
On 15 juil, 22:31, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 15:28:26 -0700, s...@microtec.net wrote:
On 15 juil, 17:49, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 07:55:06 -0700, s...@microtec.net wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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.

SNIP
But there remain problems. It turns out that "displacement currents"
actually don't exist. If you put a capacitor into a (black) box and
ramp a current into it you do NOT find a magnetic field about the box
equivalent to if the box just held a straight wire. It seems nobody
has ever been able to measure this. Oops!

Quite the contrary. It has been very precisely measured.

I think you simply have not read the right chapter in any
EM undergrad ref.

I suspect Benj was wondering how long it might take for me to jump in on
THIS one!

The unfortunate fact is that the EM undergrad references seem to ALL have it
wrong. So do the post grad references. This is because the modern
interpretation of Maxwell's Equations is that E *causes* H and H *causes* E.
This is wrong.

Maxwell's equations are not causal. They are descriptive.

Enormous amounts of time, effort and money have been spent trying to measure
the magnetic field that is "caused" by Maxwell's Displacement Current. No
One Has Ever Succeeded

Please see D.F. Bartlett, "Conduction current and the magnetic field in a
circular capacitor." Am. J. Phys. 58, 1168-1172 (1990). This chronicles the
futile attempts to observe/measure the magnetic field "caused" by
displacement current. There are many others, including Rosser's "Classical
EM via relativity."

Then please see Oleg D. Jefimenko, "Causality Electromagnetic Induction and
Gravitation" especially Chapter 1. He succinctly shows what *really* is
going on.

Bill Miller

On 16 juil, 20:25, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Mon, 16 Jul 2007 13:15:48 -0700, s...@microtec.net wrote:
On 16 juil, 13:51, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 20:00:39 -0700, s...@microtec.net wrote:
On 15 juil, 22:31, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 15:28:26 -0700, s...@microtec.net wrote:
On 15 juil, 17:49, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Sun, 15 Jul 2007 07:55:06 -0700, s...@microtec.net wrote:
On 12 juil, 08:31, Rudolf Drabek <newsr...@aon.at> 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)

[Don't know what happened to the answer I posted yesterday, so I
answer again]


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.

Since all matter (including lenses) is made up of electromagnetic
particles,
just like photons, this is no surprise to me that photons could
interact
with matter and that their trajectories could be deflected.
They can interact and they can slow up, but their trajectory would

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.

Not at all.

I am talking about the de Broglie photon, whose energy oscillates
between maximum electrostatic extention as two half photons and
maximum magnetostatic extention as a single magnetic quantity,
at both maximum of which the transverse velocity is zero, and
maximum transverse velocity (c) is reached when half the energy
has crossed over in either direction (LC oscillation)

See the equation below.

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.

E = hc/(2 lambda) +[e^2/2C cos^2(wt)+(Li^2)/2 sin^2(wt)]
With lambda = Lc, first term = .5mc^2. Then you have .5Ce^2 and .5Li^2

C = (2 eps_0 lamda alpha)
Cps = eps0*Lp2 = 1.56e-25 farad
L = (mu_0 lambda alpha)/(8 pi^2)
Lps = mu0*Lp2 = 2.22e-20 henry
i = (2 pi e c)/(lambda alpha)
You need both c and C in this equation. I assume c/lambda alpha = w

and of course

w = 1/sqrt(LC)
I have similar rsonant phenomena in my pair space model.

I could have used any other photon value and would have
obtained the very same equations, but with more figure
keying in as I did the actual calculations. Less trouble
going for the set of constants built into my calculator.

These equations are for quanta moving at c, not for
electron at rest.

André Michaud
John Polasek

On Wed, 18 Jul 2007 06:07:05 -0700, s...@microtec.net wrote:
On 16 juil, 20:25, John C. Polasek <jpola...@cfl.rr.com> wrote:
On Mon, 16 Jul 2007 13:15:48 -0700, s...@microtec.net wrote:
On 16 juil, 13:51, John C. Polasek <jpola...@cfl.rr.com> wrote:

[Don't know what happened to the answer I posted yesterday, so I
answer again]

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.

Since all matter (including lenses) is made up of electromagnetic
particles, just like photons, this is no surprise to me that photons
could interact with matter and that their trajectories could be deflected.

They can interact and they can slow up, but their trajectory would
remain unaltered, since, presumably, they have no structure.

(I lost yesterday's reply, so here goes).

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 meant the units of velocity and acceleration don't match.

Not at all.

I am talking about the de Broglie photon, whose energy oscillates
between maximum electrostatic extention as two half photons and
maximum magnetostatic extention as a single magnetic quantity,
at both maximum of which the transverse velocity is zero, and
maximum transverse velocity (c) is reached when half the energy
has crossed over in either direction (LC oscillation)

See the equation below.

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

I have studied the equation a bit. I have to recast the problem in
terms of lambda being the CWL = Lc to gain some traction.

Please post the equation.

E = hc/(2 lambda) +[e^2/2C cos^2(wt)+(Li^2)/2 sin^2(wt)]

With lambda = Lc, first term = .5mc^2.

Then you have .5Ce^2

and .5Li^2

on a phasor rotating at w radians/sec. (except that, squared, you get
double frequency 2w).

I will note the parallel between your findings and those in my
pair-space cell which is Lps = 2 alpha lambda = 3.5e-14 meters and Lp2
is half that.

C = (2 eps_0 lamda alpha)

Cps = eps0*Lp2 = 1.56e-25 farad

L = (mu_0 lambda alpha)/(8 pi^2)

Lps = mu0*Lp2 = 2.22e-20 henry

i = (2 pi e c)/(lambda alpha)

You need both c and C in this equation. I assume c/lambda alpha = w

Then the principle that i = Cde/dt = Cwe

and why 2 pi?

and of course

w = 1/sqrt(LC)

I have similar rsonant phenomena in my pair space model.
I get two different expressions for resonance in my pair cell.
Computing 2 lump parameters on the cell for one version, and the 2d
version is from an entirely different quadrant,K/m but give identical
frequencies.
wps = sqrt(K/me) = 1/sqrt(Lps*Cps) = 1.69e22 rad/s

using the lump constants of Cps and Lps.
From basic analysis of what it takes to make eps0 in space I derived
K = 2.61e14N/m as the spring restraint that makes eps0
eps0 = 2e*e/K*Lps^3 = 8.8e-12F/m