On Sat, 21 Jul 2007 10:38:20 -0700, s...@microtec.net wrote:
On 19 juil, 21:33, John C. Polasek <jpola...@cfl.rr.com> wrote:
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:

[mega snip]

[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.

Really :-)

What about micro-bubble capture? You must have read about
these experiments supposedly showing light "slowing down".

When you read the proper articles, you see that individual
photons are simply captured for a while inside micro bubbles
whose size is in sync with the photon's longitudinal wavelength.

To me this is only an added confirmation of photon trajectory
deflection on top of gravitational deflection and lens deflection.

jp: Both lens deflection and gravity well deflection depend on one end
of a "wave-front" advancing slower than the other and thus tilting the
wavefront and redirecting its vector normal. In gravity the part
farther out is faster than the inner part, but they are joined,
stick-like, so a definite turning moment exists. I don't see such
possiblity for photons. I guess you would have to assert that it
matters that now photon 1 arrives after photon 2, when we know they
started together. Does that bend their path?

(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.
Could you clarify ? I don't see what you mean here.

jp: You say: "acceleration (LT^-2) can't possibly exceed c" (= LT^-1).

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.

No problem with this. CWL is as legitimate a photon wavelength
as any other. Very practical in fact.

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.

In the case of a photon, this part is plain kinetic energy keeping
the electromagnetic half in motion at c.

Then you have .5Ce^2

No doubt you mean here .5e^2/C

jp: No I mean the correct expression for capacitive energy is .5Ce^2.

and .5Li^2

Yes. That's my equation says.

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

Why square anything here. Could you clarify ?

jp: It's in your equation, sin^2(wt).
sin^2 A = 1 - cos^2 A = .5(1 - cos 2A). It's a trig identity, may not
mean anything. Power wave varies between +0 and +1 at 2w frequency.

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

jp: or Cps = eps0*Lps = 3.13e-25 farad like yours .

Using CWL I get

C = (2 eps_0 lamda alpha) = 3.135381415E-25 Farad

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

jp: Why 8pi^2 which is 78.8?

Lps = mu0*Lp2 = 2.22e-20 henry

jp: My Lps = mu0 lambda alpha = 2.22e-20H; when divided by 8pi^2 is
same 2.81e-22H as yours.

Using CWL, I get

L = (mu_0 lambda alpha)/(8 pi^2) = 2.817940285E-22 henry

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

For i, I get

i = (2 pi e c)/(lambda alpha) = 17045.08865 Coulomb/sec

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

Almost. Again, a matter of using full transverse integrated
wavelength
in your model and me using transverse integrated amplitude. See below.

Then the principle that i = Cde/dt = Cwe

In my model energy is not induced as a function of time but as a
function of transverse amplitude, this is why it took me so long to
figure out the correct math, since I had to generally refrain from
deriving on a time basis. So I did not use this approach.
See below.

and why 2 pi?

Why! Its the divisor of the integrated wavelength to obtain the
transverse amplitude, what else

A = (lambda alpha) / (2 pi)

jp: Notice A is the classical radius of the electron 2.8e-15m if
lambda = CWL and lambda alpha is my half-cell Lp2 1.77e-14m which is
2pi times Rclasselec.

this is why i = (2 pi e c)/(lambda alpha)

which is the same as i = ec/A

jp: or in other words i = e*c/Rclass

and of course

w = 1/sqrt(LC)

Here is how w is derived for the de Broglie photon

w = 1 / sqrt(LC)

w = 1 / sqrt[(mu0 lambda alpha)/*8 pi^2) (2 eps0 lambda alpha)]

jp: I can't see the 8pi^2, nor the 2 in 2 eps0

w = 1 / [(mu0 eps0 lambda^2 alpha^2) / (4 pi^2)

jp: or in other words 1/sqrt(mu0 eps0 A^2) A = class. radius of
electron (needs sqrt sign)

Mine is similar without the /4pi^2: w^2 = 1/(mu0 eps0 Lp2^2)

w = 1 / [(lambda^2 alpha^2) / (4 pi^2 c^2)]

w = (2 pi c) / (lambda alpha)

With the CWL

w = (2 pi c) / (lambda alpha) = 1.063870869E23 rad/s

jp: or w = c/Rclasselec>> 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

Not far from my own calculation for w. See above.

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-12 Coul^2/m^2*N=F/m

jp: is electron charge x charge density divided by spring K. Lps > 3.5e-14m. From the simple consideration of cell size Lps and spring K
we make eps0 happen.

I am confused as to your notation here so I could not do the calc
with my own parameters.

The recall constant is calculated in this manner for the de Broglie
photon

E = -KA^2

K = - 2E/A^2

K = - 4 pi E/(lambda alpha) here E= .5 m0 c^2

jp: K = E/Lp2 (w/o 4pi) figures to be 2.3N, the same as the average
force to remove an electron from my cell (Eq. 8 my permittivity paper
on my web).

K = 1.031019177E-16 N/m

K is not newton/meter, it's Newton.

So, F = KA = 29.05350473 Newton

Now how did I verify that this figure was ok ?

Here is the logic.

Since F is proportional to KA, if you multiply
the equation by alpha, you will get the corresponding
force for the longitudinal wavelength (This F is for
transverse amplitude/force relation) then

F alpha = 0.212013666 Newton

Now, this relates to the electron CWL which
is the usual longitudinal wavelength associated
with this quantum of energy.

We know also that the energy induced at the
Bohr rest orbital is equal to the electron rest mass
energy multiplied by alpha^2.

Since force is proportional to energy, we can
further find the force associated with a photon
of same energy as Bohr rest orbital energy

F alpha alpha^2 = 1.12900148E-5 newton

Now, this is the force for a photon, obviously
moving at c, but we know that force is proportional
to velocity. Andr we know besides that the velocity
at the Bohr rest orbital is equal to c multiplied
by alpha.

So, a final multiplication by alpha should
restitute the well known force associated
with the Bohr rest state

F alpha alpha^2 alpha = 8.238721808E-8 Newton

QED

This proves in my view that the Force / Energy / K / C / L / i / w
parameters of the de Broglie photon LC equation are self consistant.

in my model in a cube of Lps with a harmonic oscillator frequency
wps.

Also it happens that wps*Lp2 = c. The tip of Lp2 sweeps out c from
wps.
Of course my model is all about electrons, not photons. Nevertheless
the values I came up with Lps and K are all self-consistent. (It could
be modeled by a slinky and a pair of rubiks cubes).

I see that also.

André Michaud

I cannot afford more time here right now, as I have a lecture to do
Monday, but I am seeing more and more from your use of lambda alpha
that you are into my pair-space.

Pair space is a model of our world reduced in size by alpha, so that
alpha x CWL = my Lp2 or half-cell.
The idea is that physical laws are acted out in pairspace, not in the
vacuum.

I'll say it flat out: alpha, the mystery coefficient, is the scale
factor between our ordinary space and pairspace, it's not the fuzzy
electromagnetic coupling coefficient. Your photons fly in my
pair-space where the transmission velocity is c.
(You cannot be serious when you talk of using a hand calculator?)

"John C. Polasek" <jpola...@cfl.rr.com> wrote in messagenews:7kj4a3dn8gaq16qpuupp4akb5pbbqhqb0g@4ax.com...

I cannot afford more time here right now, as I have a lecture to do
Monday, but I am seeing more and more from your use of lambda alpha
that you are into my pair-space.

Well, I think André's model is different where he is trying to
incorporate "your" "pair-space" mechanism into the actual photon.

But
it doesn't work because photons have no charge and no dipole moment.
Tough to model a photon with those restrictions like he is trying to do.

The easy solution is to model photons as phonons in a relativistic
medium of bound charge. Of course, that leads to other problems and
requires the concept of "less than virtual". ;-)

Pair space is a model of our world reduced in size by alpha, so that
alpha x CWL = my Lp2 or half-cell.
The idea is that physical laws are acted out in pairspace, not in the
vacuum.
I'll say it flat out: alpha, the mystery coefficient, is the scale
factor between our ordinary space and pairspace, it's not the fuzzy
electromagnetic coupling coefficient. Your photons fly in my
pair-space where the transmission velocity is c.

IMV, the sqrt(alpha) is simply the ratio of electronic charge to the
bound quantum "vacuum" charge (your pair-space mechanism) that can be
associated with photons. If this is true, then bound quantum "vacuum"
charge should be the basic bound charge associated with all gauge
bosons. In a loose way, alpha is a geometric scale factor but I would
think the real scale factor could be as high as 10^20 or more (see L.
Randall's warped geometries).

Best,

Fred Diether
Moderator sci.physics.foundations

On 21 juil, 17:44, "FrediFizzx" <fredifi...@hotmail.com> wrote:

"John C. Polasek" <jpola...@cfl.rr.com> wrote in
messagenews:7kj4a3dn8gaq16qpuupp4akb5pbbqhqb0g@4ax.com...

I cannot afford more time here right now, as I have a lecture to
do
Monday, but I am seeing more and more from your use of lambda
alpha
that you are into my pair-space.

Well, I think André's model is different where he is trying to
incorporate "your" "pair-space" mechanism into the actual photon.

I think you misunderstood. I am not trying to incorporate
his pair space mechanism. I was explaining the LC oscillating
de Broglie photon to him.

On Jul 21, 7:15 pm, s...@microtec.net wrote:



On 21 juil, 17:44, "FrediFizzx" <fredifi...@hotmail.com> wrote:

"John C. Polasek" <jpola...@cfl.rr.com> wrote in
messagenews:7kj4a3dn8gaq16qpuupp4akb5pbbqhqb0g@4ax.com...

I cannot afford more time here right now, as I have a lecture to
do
Monday, but I am seeing more and more from your use of lambda
alpha
that you are into my pair-space.

Well, I think André's model is different where he is trying to
incorporate "your" "pair-space" mechanism into the actual photon.

I think you misunderstood. I am not trying to incorporate
his pair space mechanism. I was explaining the LC oscillating
de Broglie photon to him.

Sorry, what I said didn't come out quite right from your perspective.
Well, you can't have LC without charge. So where is it? If a photon is
to be a distinct entity, which I think you are claiming in your model,
then the charge for the LC of a photon can't be from real electrons,

I forgot to mention K = 2.4e14N/m and the cell size L = 3.514x10^-14m
is twice the FSC alpha (= 1/137) x the Compton wavelength,
L/2 = alpha*CWL.
which shows it is interestingly related to physicists' yardstick, the
CWL.

But 2 electrons per cell L means mass density 4.1e10kg/m^3 which is
2.5 million times the density of iron (FSC^3 = 137^3) so this cell
structure cannot be in our vacuum, but only appears so, electrically.

This proves in my view that the Force / Energy / K / C / L / i / w
parameters of the de Broglie photon LC equation are self consistant.

On Jul 14, 9:14 am, John C. Polasek <jpola...@cfl.rr.com> wrote:
I forgot to mention K = 2.4e14N/m and the cell size L = 3.514x10^-14m
is twice the FSC alpha (= 1/137) x the Compton wavelength,
L/2 = alpha*CWL.
which shows it is interestingly related to physicists' yardstick, the
CWL.

But 2 electrons per cell L means mass density 4.1e10kg/m^3 which is
2.5 million times the density of iron (FSC^3 = 137^3) so this cell
structure cannot be in our vacuum, but only appears so, electrically.

Wrong, the elèctròn's givven mass is for all breadth, or spase.
http://google.com/groups?q=Autymn+electron+size+infinitely
http://google.com/groups?q=Autymn+sun+bird

Autymn D. C. wrote:
On Jul 14, 9:14 am, John C. Polasek <jpola...@cfl.rr.com> wrote:
I forgot to mention K = 2.4e14N/m and the cell size L = 3.514x10^-14m
is twice the FSC alpha (= 1/137) x the Compton wavelength,
L/2 = alpha*CWL.
which shows it is interestingly related to physicists' yardstick, the
CWL.

But 2 electrons per cell L means mass density 4.1e10kg/m^3 which is
2.5 million times the density of iron (FSC^3 = 137^3) so this cell
structure cannot be in our vacuum, but only appears so, electrically.

Wrong, the elèctròn's givven mass is for all breadth, or spase.
http://google.com/groups?q=Autymn+electron+size+infinitely
http://google.com/groups?q=Autymn+sun+bird

Wrong, Looneybin. Elektron haz no givven masss. Masss is only aparent.
Elektronz are LESS THAN nothing.
Quit trying to pretend you know something about scienz!

On Jul 25, 8:19 am, Benj <bjac...@iwaynet.net> wrote:
Autymn D. C. wrote:
On Jul 14, 9:14 am, John C. Polasek <jpola...@cfl.rr.com> wrote:
I forgot to mention K = 2.4e14N/m and the cell size L = 3.514x10^-14m
is twice the FSC alpha (= 1/137) x the Compton wavelength,
L/2 = alpha*CWL.
which shows it is interestingly related to physicists' yardstick, the
CWL.

But 2 electrons per cell L means mass density 4.1e10kg/m^3 which is
2.5 million times the density of iron (FSC^3 = 137^3) so this cell
structure cannot be in our vacuum, but only appears so, electrically.

Wrong, the elèctròn's givven mass is for all breadth, or spase.
http://google.com/groups?q=Autymn+electron+size+infinitely
http://google.com/groups?q=Autymn+sun+bird

Wrong, Looneybin. Elektron haz no givven masss. Masss is only aparent.
Elektronz are LESS THAN nothing.
Quit trying to pretend you know something about scienz!

apparent and givven; your knowledh is lear than nothing.
What ever happened to the F7 key?