Will I have to eat my hat on my remark

There is no way to bend spacetime by brute force because spacetime is too stiff at the scale of the thickness of the fuselage of the alien birds?

Good thing I have not let Super Cosmos into print yet. Is this what Tricky Hal has up his sleeve? :-)

OK what's wrong with the following numbers that seem to suggest that the Townsend-Brown "lifter" effect is not as crackpot as I first thought? Huh? This shows that once concrete calculation plugging in the numbers is worth "billions and billions" of Zielinski's excess verbal baggage. :-)

These parameters for geodesic gliding (simple warp drive) of operational USAF Black "Anti-Gravity" Birds like the Black Triangles and what Nick Cook is hunting for are well within current technology. Is it really as simple as this to make a George Trimble G-Engine or have I made a colossal arithmetic blunder below?

PS I would like to thank Gordon Novel for stimulating this revolting new development! Without his urging I would not have done this little exercise. Maybe it's wrong? Maybe it's not? Intelligent corrections welcomed. It can't be this easy? Is it? I do not believe what I show below - at least not yet. I have astonished myself. I am very skeptical myself of what I say here. As Richard Feynman admonished me more than once "Jack, always try to prove yourself wrong."

Working with electromagnetic units is a bitch and counter-intuitive.

Here goes:

PS The cgs emu electrical field E energy density is E^2/8pi ergs/cc

where MKS 1 volt per meter = 10^6 abvolts per cm

Therefore

10^21 ergs/cc ~ E^2/8pi

E ~ (25 10^21)^1/2 ~ 10^10 abvolts/cm ~ 10^4 volts/meter

This is NOT a lot! I mean this is easy.

Suppose you have a capacitor with a separation between the plates of say 1 cm.

That's a voltage difference of only 100 volts to get the above E. So how come charged capacitors are not floating all over the place. What's wrong with this picture? Where is my error below? Back to the drawing boards boys and girls.

On Nov 30, 2004, at 9:22 AM, Jack Sarfatti wrote:

Clarifications.

Note that the PV theory of Hal Puthoff promoted by Eric Davis in the USAF report says that black holes do not exist. This is one reason why Lawrence Kraus labeled the USAF report bogus and crackpot. It is also why Cliff Will, Matt Visser (who Davis cites significantly in the USAF report), Bill Unruh and others I talked to at GR 17 reject the PV theory completely. Sir Martin Rees, British Astronomer Royal and Master of Trinity College, Cambridge gave ample evidence for the factual status of black holes as did others.

1. How much energy does it take to make a simple black hole?

Let R be the radius of the total mass M.

The critical energy density T00* to make the black hole of radius r* is determined from

r* = 2GM/c^2 = 8pi(G/c^4)T00*r*^3

This is an induced curvature of

1/r*^2 = 8pi(G/c^4)T00*

The numbers below give the energy density needed to make a black hole of size 10^4 cm.

Let r' be the actual radius of the gravitating source (assumed static, uniform and isotropic for simplicity here in this toy model. Then the dimensionless black hole parameter is

b = (r'/r)^2 = 8pi(G/c^4)T00*r'^2

The event horizon of infinite redshift where time literally stops is when b = 1. When b < 1 we have a black hole.

2. The geodesic glider in Earth's atmosphere.

Let M be the mass of the Earth.

M = 6x10^27 grams

GM/c^2 ~ 0.5 cm

That is if the Earth were a black hole its surface area would be only ~ 3.14 cm^2 that would fit in the palm of your hand. Of course you would be crushed out of existence.

When you stand on the surface of rotating Earth at a radius r' = 6x10^8 cm you are a non-inertial observer pushed off the inertial motion geodesic path in curved spacetime by the electrical reaction forces and the Pauli exclusion quantum correlations in the material rock beneath your feet. You feel the inertial g-force proportional to your "weight" only because of these non-gravity electrical forces and quantum pressures from Fermi-Dirac statistics of identical particles.

The universal g-acceleration at Earth's surface is

g = GM/r'^2 ~ 10^3 cm/sec^2

Note that

b = ~ 10^18 for Earth.

The radius of curvature at the surface of the Earth induced by the Earth is r@ where

1/r@^2 ~ GM/c^2r'^3 = g/c^2r'

Note that the Levi-Civita connection that is not a tensor, and that has zero tensor component, in this LNIF at the surface of the Earth is

(LC)' ~ g/c^2

Paul Zielinski falling off his ladder next to you has (LC) = 0. :-)

Therefore,

r@ = c(r'/g)^1/2 ~ 310^10(6 10^10/10^3)^1/2 ~ 10^13 cm ~ 1 Au

The applied brute force energy density T00@ we need to neutralize the Earth's (LC)' in the LNIF for the geodesic glider is of order

1/r@^2 = 8pi(G/c^4)T00@ = g/c^2r'

i.e.

T00@ ~ (c^4/8piG)(g/c^2r') = (c^2/8piG)(g/r') ~ (1/8pi)10^28x10^3/6 10^8 ergs/cc

~ 10^21 ergs/cc ~ 10^14 Joules/cc

This corrects my late last night off the top of my lazy dog head instant estimate below that was too high by a factor of 100.

On Nov 29, 2004, at 9:47 PM, Jack Sarfatti wrote:

There is no way to bend spacetime by brute force because spacetime is too stiff at the scale of the thickness of the fuselage of the alien birds.

The spacetime stiffness or "string tension" is c^4/G ~ 10^19 Gev per 10^-33 cm.

1 Gev is a billion electron volts of energy.

10^19 is 10^7x10^12 = ten million trillion electron volts of energy

10^19 Gev is ten million billion trillion electron volts of energy

The space warp induced curvature factor is

1/(Area of Induced Spacetime Warp) ~ (G/c^4) (Applied Energy Density)

To get your weightless warp drive as allegedly seen in real flying saucers

suppose the saucer is say 100 meters i.e. 10^4 cm

The warp factor around the saucer must be of order

10^-8 cm^-2 = (10^-33 cm/10^19Gev)(Applied Energy Density)

The required energy density for this Hal Puthoff/Eric Davis PV warp drive metric engineering approach is therefore

(Applied Energy Density) ~ 10^19 Gev x 10^33 x 10^-8 ~ 10^44 Gev/(cubic centimeter)

That's 10^53 electron volts of energy per cubic centimeter, that ~ 10^34 Joules of energy per cubic centimeter.

That's simply too much for any high voltage circuit to make.

Note that the energy density of a proton is

10^9 ev/(10-13)^3 = 10^9 10^39 = 10^48 ev/cc = 10^30 Joules per cc.

This is the energy density of a neutron star that has the mass of the entire Sun in a sphere of ~ 10km diameter. The flying saucer needs ten thousand times more than that following the Puthoff-Davis approach.

It don't work that way quite obviously. The alleged fact that saucers fly easily through our skies without burning us all to a crisp in Darth Vader's "Death Star WMD" means there is a kinder gentler way - there is.

Actually the above is really more for a primitive star gate - to neutralize Earth's 1 g field, i.e. weightless geodesic gliding we only need 10^-18 of the above energy density in the brute force method. I will double check this number later as it is late. Manana.

So that's only 10^14 Joules per cc - still quite a lot, but not as bad as above.

In terms of Poynting EM flux

That's 10^24 Joules per cm^2 per second or 10^24 watts per square centimeter. That's still a hell of a lot.

## Tuesday, November 30, 2004

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