Wednesday, September 15, 2004

On Sep 15, 2004, at 12:38 PM, main_engineering wrote:

I understand your math and the physics perfectly well. I don't understand
why you are using the wrong math in this example and limiting yourself to a
1 dimensional problem when in fact 3 are required to model reality.

Because, my model is true any dimension. Also the piston in the cylinder is the simplest situation and it is effectively 1D by symmetry since the important parameter is the movement of the piston in the cylinder.

The same calculation works in all dimensions because the space-dimensions are LINEARLY INDEPENDENT.

The key conclusion that, contrary to Hal Puthoff's frequent claims to science journalists, FREE virtual zero point photons without any j.A coupling to electrons and protons make exactly ZERO contribution to the measured Casimir force, which is not at all a direct zero point force, but is, rather, an electrostatic Van Der Waals force between real charges placed outside the vacuum in spatially separated charge neutral configurations. The free virtual photon pressure is CONSTANT in the case of zero vacuum coherence and it obeys w = -1 with a positive ZPE density and an equal in magnitude but opposite in sign NEGATIVE PRESSURE. I prove this explicitly below using the standard continuum spherically symmetric Lorentz-invariant spectral density of field oscillators, which is an APPROXIMATION to the more exact discrete sum in a finite cavity problem. Contrary to QED, in GR you cannot subtract off even these constant free virtual photon ZPE densities, which have STRONG direct space-time warping power when not suppressed by MACRO-QUANTUM VACUUM COHERENCE out of which Einstein's gravity and the inertia of ordinary matter emerges. Note Hal Puthoff's category confusion here. I am NOT claiming that the Casimir force is in anyway related to the origin of gravity and inertia as Hal has suggested in numerous pop interviews with the media.

Look the idea is very simple.

Suppose you have a finite 3D box cavity, independent x,y,z axes with a little box of variable size X, Y, Z inside a big box of fixed size Lx, Ly, Lz. The counting of the LINEARLY INDEPENDENT standing wave modes along each direction works the same way.

Count the number of independent modes along each space direction exactly as I did. Assume a common short-wave cutoff a for simplicity. The EXACT total number of modes is then inside the little box

N = NxNyNz = (1/8((X/a)(X/a + 1)(Y/a)(Y/a + 1)(Z/a)(Z/a + 1)

But there is only 1 virtual photon per independent mode, and their virtual zero point energies add linearly even though the modes in different independent directions multiply.

That is the Hamiltonian for a sum of independent field oscillators is

H = Sum over all independent modes of hck(a*kak + 1/2)

where we sandwich the LINEAR ENERGY OPERATOR between the VACUUM STATE

|0> that is a product of the vacuum states |0>k one for each independent mode. Standing wave cavity modes in different directions are independent as well as those along the same direction with different nx, ny, nz and

k<0|a*kak|0>k = 0

using 2nd quantization in Fock occupation number space in the traditional textbook way.

a* creates a real photon, a destroys a real photon.

The wave number for each independent standing wave cavity mode field oscillator, that in the vacuum has NO real photons is

knx = nxpi/X

where nx = 1 to X/a, similarly for y, z

DO NOT CONFUSE THESE MODE INTEGERS WITH THE EIGENVALUES OF a*kak for the number of REAL PHOTONS in each independent cavity mode field oscillator k that is a standing wave obeying the cavity boundary conditions.

Note that

Nx = SUM of ALL the MODE INTEGERS nx from 1 to X/a (in sense of nearest least integer to X/a).

Similarly for ny & nz.

Each independent mode of wave number knx = nxpi/X contributes a zero point energy hckni/2 with a fixed coefficient hcpi/2X, so that the key quantity is sum of nx from 1 to (X/a) which is (1/2)(X/a)(X/a + 1). Similarly for y and z and you simply ADD the separate linearly independent zero point virtual photon energies for EACH independent oscillator! So my EXACT calculation obviously works in 1D, 2D, 3D ... ND.

This is simple and obvious.

you're not just saying that Hal is wrong, you're trying to tell me that
Planck, Casimir, Milonni and everything else QED has to say about the ZPF is
wrong without any justification other than your desire to prove you're right
and everyone else is wrong. I know better than to believe such nonsense.

I never said Planck was wrong. Show me where Planck did such a calculation?

Show what is wrong with the above simple exact DISCRETE counting?


Density of oscillator states in 3D is 4pik^2dk per polarization.

k^2 = kx^2 + ky^2 + kz^2

= (nxpi/X)^2 + (nypi/Y)^2 + (nzpi/Z)^2


But this discrete sum is approximated by a continuous integral below and spherical symmetry is assumed. These two approximations introduce mathematical artifacts LIKE LORENTZ INVARIANCE!

The total zero point energy density is then

Integral (hc/2)4pik^3dk = 4pi(hc/8)k(MAX)^4

k(MAX) = 2pi/a

a = short wave cut off

Total zero point energy ZPE is then

ZPE = 4pi(hc/8)(2pi/a)^4V > 0 in a spatial volume V.


Note that even in this calculation, the PRESSURE is - d(ZPE)/dV = - 4pi(hc/8)(2pi/a)^4 is CONSTANT independent of changes in volume V. AND IT IS NEGATIVE and it is - the ZPE DENSITY! That is w = -1!

Therefore even in this continuum spherical symmetric model one gets the same effective result I get with my more exact purely discretum model with NO SPHERICAL SYMMETRY IMPOSED ADHOC! There is NO net ZPE pressure differential on any wall. You have exactly the same negative free virtual photon pressures on each side of the wall NO MATTER HOW YOU CALCULATE IT!

If the formula used to derive the ZPF mode pressure presented by Milonni is
wrong, then so is the spectral energy density of the ZPF and the Lorentz
invariance of that density.

No it's not as I just showed. To the extent that V --> infinity and a --> 0 which is what you need for globally flat continuum of special relativity with EFFECTIVE SPATIAL SPHERICAL SYMMETRY (no finite cavity walls of arbitrary shape), you get same essential result! That same essential result is that there is NO contribution to the Casimir force from purely free virtual zero point photons! The Casimir force requires the j.A coupling to properly compute as a Van Der Waals force between real charges. This is NOT a pure vacuum problem!

That cannot be so. It has been verified by more
than 1 kind of experiment. QED is the best tested theory there ever was.
IMO, test after test it has proven itself correct. It is even more battle
tested and proven in day to day life than GR ever was.

Not true! The domain of validity of QED is much more limited than anyone realized even though the accuracy of the QED measurements is very high and even though renormalization methods work very well - too well, although not even Richard Feynman understood why! Feynman told me that directly BTW!


-----Original Message-----
From: Jack:

I count the number of modes exactly. There is one FREE virtual photon
per mode (AKA field oscillator). Therefore, the calculation is exact
and simple.

No comments: