OK I will say this most politely, or, rather, "collegially". ;-)

Read sections 7.1 and 7.2 in Milonni, quote..

"In attempting to better understand this result, Casimir found that the van

der Waals interaction could be attributed to the change in the zero point

energy of the field due to the presence of the two atoms, much as in

Feynman's later argument for the Lamb shift. He then considered the simpler

example of two parallel conducting plates... (pg. 218)

Jack: Yes, that is precisely my point. The key words are "THE CHANGE in the zero point energy due to the presence of the two atoms. There are NO ATOMS in my EXACT CALCULATION using ONLY THE FREE EM FIELD with boundary conditions. It is not however any effect of FREE EM FIELDS WITHOUT THAT INTERACTION! That's the point you keep missing!

On Sep 12, 2004, at 1:34 PM, main_engineering wrote:

Todd: Sorry, I did not catch that distinction before. So you disagree on what the

ZPF spectral energy density is?

Jack: No. You are thinking sloppy with too loose associations. Your statements are ill-posed. ZPF spectral energy density in what kind of situation?

Your method of calculating ZPEnergy by finite series will not give you the

well known spectral energy density,

rho(w) = (hbar*w^3)/(2*pi^2c^3) = (hbar*k^3)/(2*pi^2)

will it? As on pg. 49 in Milonni?

No, because there two distinct physical situations here. I did a cylindrical symmetry cavity problem that essentially is a good approximation to the flat plates. You could also try spherical symmetry.

The usual mode density for isotropic 3D space takes a shell in momentum space of radius p and thickness dp. The volume of this Euclidean shell is obviously 4pip^2dp. The total phase space volume is then

4pip^2dpV. The total number of cells of phase space in this shell is then (3 polarizations for virtual photons not 2)

dN = 12pip^2dpV/h^3

Therefore

dN/V = 12pip^2dp/h^3

Each virtual photon has energy pc/2

Virtual photon energy density is then (pc/2)dN/V - 6pip^3dp/h^3

But you are making an unjustified CONTINUUM APPROXIMATION if you do a Riemann integral to get

Virtual photon energy density ~ (3/2)cp(Max)^4/h^3 (Energy/Volume)

Where you integrate from p(Min) = 0 to p(Max) = h/a

Your p(Min) already assumes an INFINITE LONG WAVE CUTOFF! So it does not work exactly for a FINITE CAVITY!

Imagine a large sphere in 3D of radius R whose surface is an electrically neutral metal conductor.

What is the volume of the phase space for electromagnetic field modes?

Put in the adhoc short-wave cutoff a.

The longest wave that will fit is 4R.

The shortest wave that will fit is ~ a.

Ignore different polarizations for now to keep the math simple.

The largest spherical wave number that will fit is

k(Max) ~ (R/a)2pi/R = 2pi/a

I may be off by a factor of 2 or 4 here. Will look at it more carefully later.

Notice that R cancels out of the problem for k(Max).

The total volume of phase space per polarization state for these modes in flat space is therefore

(4pi/3)R^3(4pi/3)(1/a)^3 = (16pi^2/9)(R/a)^3

Therefore, the total number of allowed spherical modes here is

N = (48pi^2/9)(R/a)^3 = (48pi^2/9)(R/a)^3

because there are 3 independent virtual photon polarizations for spin 1 imposed by local U(1) gauge symmetry on the phase of the charged sources.

R = long-wave cutoff

a = short-wave cutoff

The total number of modes for FREE virtual photons.

N(virtual photons) = (48pi^2/9)(Long-Wave Cutoff/Short-Wave Cutoff)^3

in sense of nearest integer with a small round-off error.

However, I have made the same careless error AS ABOVE of including infinitely long waves here by using Euclid's continuum geometry for the volume of a sphere in continuous space! I have inconsistently neglected the DISCRETE boundary conditions on the allowed spherical waves! Physicists do this sloppy all the time! They are Onanites when they do this! :-)

Now you can do the problem for two concentric spheres and see if it is topologically equivalent to the cylinder problem with the movable piston inside.

Back to a thin shell

The allowed spherical wave numbers are

k{n} = npi/R

n = 1 to N(max) = 4R/a (nearest integer)

Therefore, the total ZPF virtual photon energy is

Sum from n = 1 to ~ R/a of (hc/2)(pi/2R)n

= (hc/4)(pi/2R)(R/a)(R/a + 1) --> (hc/4a)(R/a)

where R >> a.

This is essentially same problem as before. It also works for spherical cavities!

With 2 concentric cavities you have inner radius r and outer radius R.

What you want is -d/dr to get the CONSTANT free virtual photon ZPF force that IS NOT, I repeat, IS NOT, the Casimir VDW force! There are no real charges here in the problem! The VDW problem is a different more complicated problem.

"ZPF FORCE" =/= "Casimir Force" = "VDW Force"

The FREE VIRTUAL PHOTON ZPF Force is ~ hc/4a^2 = hc/4(Short-Wave Cutoff)^2

But it is same in magnitude and opposite in direction on both sides of each sphere in a set of concentric spheres.

That is on each spherical surface the total ZPF Force = 0 independent of Short-Wave Cutoff!

What about for EXPANDING SPACE OF UNIVERSE?

There the analogy breaks down because there is NO OUTSIDE of the UNIVERSE and there is work done by the guv metric field such that

w(virtual quanta of all types) = (Vacuum Pressure/Vacuum Energy Density) = -1

in order to reconcile Einstein's general relativity of gravity with Heisenberg's uncertainty principle as shown by John Peacock in "Cosmological Physics" ~ p. 26

The introduction of VACUUM COHERENCE missing from above will make some differences i.e. new phenomena not previously anticipated. However, I have never said that the observed VDW Casimir force between flat plates has anything at all to do with this NEW VACUUM COHERENCE EFFECT. It does not. Puthoff is not even in the right ball park to claim that Ken Shoulders charge clusters are explained by the VDW Casimir force. That is obvious nonsense since the unbalanced charge -Ne of the EVO cluster is not like the VDW force between two NEUTRAL molecules! The EVO stability is a genuinely new mesoscopic exotic vacuum coherence phenomenon completely under Puthoff's radar. Tweaking that stability will lead to micro-fusion energy sources including weapons like new kinds of powerful bombs unfortunately - in my opinion at this time. I could be wrong of course, but don't bet on that.

On Sep 12, 2004, at 12:34 PM, Jack Sarfatti wrote to Main Engineering:

Todd,

I AM SAYING THE CASIMIR FORCE IS THE VAN DER WAALS FORCE!

IT IS NOT THE ZPF FORCE!

On Sep 12, 2004, at 12:18 PM, main_engineering wrote:

Read sections 7.1 and 7.2 in Milonni, quote..

"In attempting to better understand this result, Casimir found that the van

der Waals interaction could be attributed to the change in the zero point

energy of the field due to the presence of the two atoms, much as in

Feynman's later argument for the Lamb shift. He then considered the simpler

example of two parallel conducting plates... (pg. 218)

Yes, that is precisely my point. The key words are "THE CHANGE in the zero point energy due to the presence of the two atoms. There are NO ATOMS in my EXACT CALCULATION using ONLY THE FREE EM FIELD with boundary conditions.

It is not however any effect of FREE EM FIELDS WITHOUT THAT INTERACTION! That's the point you keep missing!

and on pg. 220-221

"...motivated Lifshitz (1956) to develop a macroscopic theory of the forces

between dielectrics. His results reduce to those of the microscopic theory

with additive intermolecular forces when the dielectric constants are close

to unity, but are substantially different otherwise. In the limiting case of

perfect conductors, the predicted force between two plates reduces to the

Casimir force (7.1). Except for a few details discussed in the following

chapter, the Lifshitz theory is now generally accepted and, as we shall see,

has been supported by some careful experiments."

Nothing I do contradicts that.

It seems to me that Milonni is saying that the van der Waals force is in

fact responsible for the Casimir effect AND vice-versa, and that Lifshitz's

macroscopic theory of intermolecular (van der Waals) forces and the Casimir

effect are one in the same.

THAT'S WHAT I AM SAYING! You keep missing the significant distinction! What you cite is not the free-field completely random virtual photon ZPF force!

Now, in your "exact calculation" you are deriving the force on each side of

the piston by finding the gradient. The gradients are equal on both sides,

no doubt. However, the energy density on each side of the piston is not

identical. The longer side of the partitioned cylinder has more energy

density because it permits more modes to exist in that volume. The energy

density exerts a ZPF pressure on the piston that is not equal on both sides

and leads to the Casimir force. Your force is proportional to 1/a^2 where

"a" is your high end cut-off. Where as, the Casimir force is proportional to

1/d^4 where "d" is the separation between the plates. To my knowledge, this

has been confirmed empirically.

What you say here is completely false as my complete mathematics showed. Let's see your math of back up this false hand waving. The longer side does have more total EXTENSIVE FREE virtual photon energy because it can fit more modes, but it has exactly the same INTENSIVE FREE virtual photon energy density since you need to divide by the bigger volume and the two effects exactly cancel when you do the sum over the modes correctly. Also note, there is not a molecule in sight here. When you stick them in and do the full QED you get the correct VDW answer.

On Sep 12, 2004, at 12:18 PM, Jack Sarfatti wrote:

PS Hal Puthoff also wrote:

In summary, we see how Casimir's positive-radiation-pressure

"blunder" model leads to agreement with experiment; we don't see how

your opposite-sign dark-energy w = -1 repulsion force model would

lead to the observed attraction. Please show this. If you can show

this for the standard parallel plate Casimir Effect, then

consideration of its use for more exotic geometries (e.g., Shoulders'

charge cluster phenomenon) would be warranted. If not .....

As I just showed in detail, the bogus mode-counting argument rests on a strange fluke using that ad hoc exponential fudge factor for short-wave cutoff that allows the exact sum of an infinite series of ZPF FREE FIELD virtual photon modes that SYNCHRONISTICALLY gives the correct Van Der Waals force answer, but for entirely the wrong physical reasons! I admit this is very curious and it has thrown ALL OF US way off base for more than 50 years. Even I fell for this BLUNDER!

In fact, the real problem for the ZPF FORCE is completely FINITE as I showed in detail today.

Again lest there be no confusion in this very confused situation.

1. The observed Casimir force is NOT a free virtual photon ZPF force at all. That ZPF force (not to be confused with the Van der Waals (VDW) Casimir force) always cancels out exactly! That it is zero, for vanishing vacuum coherence, which you assume, is CUTOFF INDEPENDENT.

2. The VDW Casimir force plays no important role in Ken Shoulders "charge cluster" EVOs.

3. The EVOs are a micro-fusion dark energy zero point energy effect from variation in the vacuum coherence completely missing in ALL your models of the phenomena.

4. I never claimed that the observed Casimir force, e.g. in the parallel plates experiment, had anything to do with dark energy. That is your misunderstanding of what I have been trying to teach you.

5. Your most recent Casimir force explanation of EVOs is wrong.

6. What you say in Nick Cook's book "The Hunt for Zero Point" and in Aviation Week's "To The Stars" is seriously wrong and throws everyone off the track.

## Sunday, September 12, 2004

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