Saturday, September 18, 2004

More corrections. Puthoff's argument is still not correct. It's easy to make a mistake of a - sign. Appealing to text books is not good enough. They may all be wrong about these ZPE calculations because the rules of QED conflict with the rules of GR and one cannot really use tricks like normal ordering of the photon creation and destruction operators. Milonni's book, for example, completely ignores GR. The discovery of dark energy shows one cannot ignore GR even at small scales.

On waking next morning at 5AM I realized we better take another look.

Given the asymmetric boundary conditions of parallel plates, we need to be careful about how to do the integration in momentum space. We cannot use spherical coordinates in momentum space. We must use Cartesian rectangular coordinates, i.e. for the spectral density

dpxdpydpz not 4pip^2dp

Let the parallel plates be separated by x along the x-axis.

Between the plates, the limits are from px(min) = h/2x to px(max) = h/a.

However, in the plane of the plates the limits are from 0 to p(max) = h/a.

How do we do the virtual photon ZPE integral?

That integral must split into longitudinal and transverse parts.


ZPE(x) = (1/2)c[px(max)^2 - px(min)^2]py(max)pz(max)xYZ/h^3

= (1/2)c[(h/a)^2 - (h/x)^2](h/a)^2xYZ/h^3

= (1/2)(h/a)^2(h/a)^2xYZ/h^3 - (1/2)c(h/x)^2(h/a)^2xYZ/h^3

= (1/2)(hc/a)(xYZ/a^3) - (1/2)(hc/x)(YZ/a^2)

The ZPF force from this term is

-(d/dx)ZPE(x) = -(1/2)(hc/a^2)(YZ/a^2) - (1/2)(hc/x^2)(YZ/a^2)

= -(1/2)(hc/a^4)YZ - (1/2)(hc/x^2)(YZ/a^2)


ZPE(X-x) = (1/2)c[px(max)^2 - px(min)^2]py(max)pz(max)(X-x)YZ/h^3

= (1/2)c[(h/a)^2 - (h/(X-x))^2](h/a)^2(X-x)xYZ/h^3

= (1/2)(hc/a)((X-x)YZ/a^3) - (1/2)(hc/(X-x))(YZ/a^2)

The ZPF force from this term is from -(d/dx)ZPE(X-x) =

+(1/2)(hc/a^2)(YZ/a^2) + (hc/2)(YZ/a^2)(d/dx)(1/(X-x))

(d/dx)(1/(X-x)) = +(X - x)^-2

Next look at the transverse free virtual photon integral ZPE(xYZ).

ZPE(xYZ) = (c/2)xYZ[(px(max) - px(min)](Integral of 2pip^2dp from 0 to p(max))/h^3

= (hc/2)[(1/a) - (1/x)]xYZ(Integral of 2pip^2dp from 0 to p(max))/h^2

= (hc/2)[(1/a) - (1/x)]xYZ(2pi/3)(1/a)^3

This term gives a net attractive force along x of -(hc/2a^4)YZ.

ZPE((X-x)YZ) = (hc/2)[(1/a) - (1/(X-x))](X-x)YZ(Integral of 2pip^2dp from 0 to p(max))/h^2

= (hc/2)[(1/a) - (1/(X-x))](X-x)YZ(2pi/3)(1/a)^3

This term gives a net repulsive force along x of +(hc/2a^4)YZ.

Therefore, as expected, we need only look at the longitudinal integral for these non-spherically symmetric parallel plate boundary conditions.

Area of plates A = YZ.

Thus the NET ZPF force along x at the plate located at x is

-(1/2)(hc/a^4)A - (1/2)(hc/x^2)(A/a^2) +(1/2)(hc/a^2)(A/a^2) + (hc/2)(A/a^2)(d/dx)(1/(X-x))

- (1/2)(hc/x^2)(A/a^2) + (hc/2)(A/a^2)(1/(X-x)^2)

Assume x << X. As X --> infinity we can ignore the second term from outside the plates.

This is a net attractive force ~ -1/x^2 not - 1/x^4, but it is too weak to contain unbalanced charges in the EVO.

Therefore, this calculation corrects the wrong calculation below.

On Sep 17, 2004, at 6:18 AM, Jack Sarfatti wrote:

On waking this morning, I realized that I ignored the lower infrared limit pmin in the calculations below integrating from 0 to pmax. The boundary conditions determine the nature of this term. For parallel plates perpendicular to x reaching to the end of the BIG BOX we need to integrate from pmin = h/2x to pmax = h/a for the little box.

This gives the additional total ZPE term in 3D case for the small box

-(pi/32)(hc/x)(V/x^3) = -(pi/32)(hc/x^3)yz

for the little box.

Similarly for the BIG BOX pmin = h/2(X - x)

-(pi/32)(hc/(X - x))(V/(X - x)^3) = -(pi/32)(hc/(X - x))((X - x)(Y - Y)(Z - Z)/(X - x)^3) = 0

So this asymmetric boundary condition does give a net ZPF force ~ - 1/x^4.

This is wrong.

Spherical shells will give a different answer.

*Therefore, I retract my previous remark that there is no net free virtual photon ZPF force under any conditions.

Corrected 3rd draft below.

On Sep 16, 2004, at 6:24 PM, Jack Sarfatti wrote:

Review of standard phase space calculations of zero point energy.

Spectral density in 3D space is from a spherical shell in momentum space.

4pip^2dp/h^3 = 3D Spectral Density

Where for the zero point virtual photon the energy is pc/2.

Therefore, integrate from p = 0 to p(max) to get

Total virtual photon zero point energy is in volume V

ZPE(3D) = Integral of (pc/2)4pip^2dp/h^3V = (4pic/8)p^4(max)V/h^3 = (pi/2)(hc/a)(V/a^3)

where p(max) = h/a

a = short-wave cutoff

This assumes Euclidean geometry continuum in 3D momentum space with basically ignoring oddly shaped cavities.

Same kind of calculation in 2D with an annulus in 2D momentum space gives (ignoring factors of 2, pi etc)

2pipdp/h^2 = 2D Spectral Density

ZPE(2D) ~ (hc/a)(A/a^2)

And in 1D along a line

dp/h = 1D Spectral Density

ZPE(1D) ~ (hc/a)(L/a)

In 1D one can do an exact calculation using finite series Sum of n from 1 to N = (1/2)N(N+1), but when L/a >> 1 you get essentially the same result as using the continuum integration.

Next go back to 3D with ZERO VACUUM COHERENCE.

3D Pressure = -d(Internal Energy)/dV

In this case

3D ZPF Pressure = -(pi/2)(hc/a^4) Independent of V! Also the pressure is negative!

w = Pressure/(Energy Density) = -1 not Hal Puthoff's "w = + 1/3" which is true only for real photons not virtual ZPF photons.

If you have a little box of volume v = xyz inside a large BOX of volume V = XYZ, then the total ZPE inside the little box is in 3D

ZPE(v) ~ (hc/a^4)v

The total ZPE inside the BIG outer BOX is

ZPE(V - v) ~ (hc/a^4)(V - v)

Note the LINEAR additive rule

ZPE(v) + ZPE(V - v) = ZPE(V)

Similarly in 2D for ZPE(A) and in 1D for ZPE(L).

Back to 3D what is the total ZPF force perpendicular to the wall x of the little box?

It is obviously

Fx(ZPF) = -dZPE(v)/dx - dZPE(V - v)/dx = -yz(hc/a^4)(1 - 1) = 0

Similarly in 2D and 1D. The number of effective space dimensions in conditions of different symmetries on the boundary conditionns makes no essential difference to this general conclusion.

Typo-corrected 2nd Draft

On Sep 16, 2004, at 7:26 AM, Jack Sarfatti wrote:

My previous calculations below for the free photon field without any j.A coupling to electric charges assumes zero vacuum coherence. Putting in the virtual electron-positron pairs is equivalent to one virtual quantum of negative energy per transverse polarized mode. This is because of the Pauli exclusion principle's anti-commutation rules for the Dirac electron field. The electron has spin 1/2, the photon spin 1, neglecting supersymmetry is OK since there is no evidence for it at all, therefore, the longitudinally polarized virtual photon mode is not compensated for by the virtual electron-positron pairs.

In general with the macro-quantum vacuum coherence, the net virtual photon + virtual electron positron, zero point energy density is positive ~ (hc/2a^4)(1 - a^3|Vacuum Coherence|^2), which is adjusted to ZERO in the non-exotic vacuum without any dark energy and any dark matter. The latter two are simply exotic vacuum phases where

a^3|Vacuum Coherence|^2 < 1

a^3|Vacuum Coherence|^2 > 1


Ken Shoulders EVOs are stabilized by DARK ENERGY core that holds the unbalanced charge -Ne together via

V(r) = +(Ne)^2/mr + c^2/\zpfr^2

Note that the gradients of these two terms are of opposite sign allowing dynamical stability. This also explains the stability of a single spatially-extended electron as a Bohm hidden variable possibly allowing Vigier's tight atomic states as a new form of atomic energy investigated by Maric and Dragic in Beograd. We need to add the rotation term of course.


/\zpf = (8piG*/c^4)(ZPE Density) > 0

[/\zpf] = 1/Area, i.e. curvature

[8piG*/c^4] = [String Tension]^-1 = [Energy/Length]^-1 = Length/Energy

[ZPE Density] = Energy/Volume

G* is the effective gravity constant at the small scale ~ 10^-5 - 10^-3 cm of the EVO, which may, or may not be, Newton's G. That is an empirical question.


N(h/mc)^2 = (Space-Warp Factor)4pi(Observed EVO Radius)^2

Where Space-Warp Factor << 1, i.e. non-Euclidean 3D space geometry for the mesoscopic EVO is my prediction.

On Sep 16, 2004, at 3:31 AM, Jack Sarfatti wrote:

In 3D case for a little box of volume xyz inside a big box of VOLUME XYX

The total free virtual photon ZPE in the little box is, with short-wave cutoff a:

~ (hc/2a)(xyx/a^3)

The total free virtual photon ZPE in the BIG BOX outside the little box is

~ (hc/2a)((X-x)(Y-y)(X-x)/a^3)

Note that in both cases the total 3D free virtual photon positive ZPE density is the same, i.e. ~ hc/2a^4, on both inside and outside of the little box inside the BIG BOX and is - the NEGATIIVE pressure, i.e. w = -1 not Puthoff's wrong value w = +1/3 that he uses in his most recent paper about Ken Shoulders EVO data.

On Sep 15, 2004, at 7:09 PM, Jack Sarfatti wrote:

typo-corrected 2nd draft

If you posit that momentum space is a CONTINUUM metric space and assume FLAT Euclidean geometry with the Pythagorean theorem, then

1. Virtual photon ZPE on a 1D line of length L with short wave cutoff "a", L >> a, with standing wave modes ~ (L)^-1/2sin(kx)

Total ZPE per polarization ~ pi(hc/4a)(L/a) = (pi/2)(ZPE of shortest wave virtual photon)(Number of cells of phase space)

1D ZPE PRESSURE = STRING TENSION = -(d/dL)(Total ZPE) = pi(hc/4a^2) = constant = -ZPE LINE DENSITY, i.e. w = -1

2. Virtual photon ZPE on a 2D lattice of area A with short wave cutoff "a", A >> a^2, with standing wave modes ~(L)^-1sin(kx)sin(k'y)

Total ZPE per polarization ~ pi(hc/4a)(A/a^2) = (pi/2)(ZPE of shortest wave virtual photon)(Number of cells of phase space)

2D ZPE PRESSURE = Membrane Tension = -(d/dA)(Total ZPE) = pi(hc/4a^3) = constant = -ZPE Membrane DENSITY, i.e. w = -1

3. Virtual photon ZPE on a 3D volume V with short wave cutoff "a", V >> a^3, with standing wave modes ~ (L)^-3/2sin(kx)sin(k'y)sin(k"z)

Total ZPE per polarization ~ pi(hc/4a)(V/a^3) = (pi/2)(ZPE of shortest wave virtual photon)(Number of cells of phase space)

3D ZPE PRESSURE = -(d/dV)(Total ZPE) = pi(hc/4a^3) = constant = -ZPE DENSITY, i.e. w = -1

In all three cases, no contribution of virtual photons to any vacuum pressure difference that can be attributed to a VdW "Casimir force".

The effect of EFFECTIVE dimension is in how the "pressure" scales with short-wave cutoff. One could also imagine FRACTALS with a non-integer power of the short-wave cutoff.

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