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Home Jack Sarfatti's Blog Do we need Deser's negative bare mass electron as the Bohm hidden variable?

*On Dec 16, 2011, at 5:19 AM, This email address is being protected from spambots. You need JavaScript enabled to view it. wrote:P.S. Positive bare mass electron models are constrained to radii of the Compton wavelength to avoid superluminal surface velocities in generating the magnetic moment and angular momentum. . . . Electrons, though, are known to be R < 10^-16 cm. Negative bare mass solves this problem.*

JS: That ain't necessarily so. You must include quantum theory. The shell of charge as a Bohm hidden variable in a quantum potential Q has zero point vacuum fluctuations. The Compton radius is the rough boundary of the virtual plasma dressing cloud of off-mass-shell electron-positron pairs and off-mass-shell photons (+ everything else with smaller effect). So the shell of charge can be smaller than the Compton radius - no problem. Indeed, Puthoff's mechanism would be one factor outside the shell of charge. That is the increased density of virtual photons relative to virtual electron-positron pairs outside the shell of charge works cooperatively with the opposite inside the shell of charge. Remember I am talking of the w = -1 induced ZPF gravity not the QED Casimir force. I think Puthoff only includes the latter. Both are there in reality.

Therefore, I say that Deser's negative bare mass model is not needed.

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From: JACK SARFATTI <This email address is being protected from spambots. You need JavaScript enabled to view it.>

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Subject: Re: I don't yet see how Price's paper explains how Deser's negative ba re mass electron is stable & 100yr starship

Date: Thu, 15 Dec 2011 15:18:02 -0800

OK, negative mass and the ADM electron.

The quote you clip from Price below is intended to illustrate the sometimes very counterintuitive behavior of negative mass matter when the central point of Price's comments is followed: if any of the masses -- active, passive, or inertial -- is known to be negative, then the Equivalence Principle requires that all of the masses are negative.

Agreed.

??? Examples?

My general points against mass modification are

1) F = DP/ds = MDV/ds + VDM/ds

the 2nd term on RHS is a g-force rocket term even if no propellant is ejected i.e. it's not a zero g-force warp drive.

2) Even if you arrange M ---> M - &M

you will not gain very much advantage.

M drops out altogether in Warp Drive WP since the whole idea of WP is that the ship controls its own timelike geodesic with small amounts of power using perhaps my superconducting metamaterial idea where the electric permittivity is huge and negative.

From Barrow & Tipler & Martin Rees "Just Six Numbers" you cannot tinker very much with e/m ratios without causing havoc and if you only tinker with chemical binding energies they are too small and you are in danger of killing your crew and having equipment malfunction.

3) No practical way to do it - unless Jim's Mach generator works and can be scaled up. Cannot fool with the Higgs field as shown by Lenny Susskind in his Landscape book I seem to recall.

What about negative bare mass ADM electrons? First, note that positive bare mass electrons do not have stable finite radii solutions. But their masses at zero radius are finite (and equal to [e^2/G]^1/2).

False as I showed here:

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From: JACK SARFATTI <This email address is being protected from spambots. You need JavaScript enabled to view it.>

Subject: Re: I don't yet see how Price's paper explains how Deser's negative ba re mas...

Date: December 15, 2011 1:42:29 PM PST

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In my model (attached) it's the virtual ZPE pressure outside the shell that stabilizes it.

Hal

In a message dated 12/15/2011 11:06:33 A.M. Central Standard Time, This email address is being protected from spambots. You need JavaScript enabled to view it. writes:

False in my model the virtual electron positron pair plasma inside the shell of charge stabilize it.

On Dec 15, 2011, at 8:40 AM, "This email address is being protected from spambots. You need JavaScript enabled to view it.:

*What about negative bare mass ADM electrons? First, note that positive bare mass electrons do not have stable finite radii solutions. But their masses at zero radius are finite (and equal to [e^2/G]^1/2).*

In my model, I am not clear about Puthoff's, it's the gravity generated by the virtual particle plasma cloud that stabilizes the positive rest mass electron with finite radius.

simple toy model neglect rotation for now, a static, spherically symmetric metric (static LNIF)

shell of charge of radius r

g00 = 1 + V/c^2

Vcharge = e^2/mr

gcharge = Fcharge/m = - dVcharge/dr = + e^2/mr^2 repulsive unstable

The interior of the thin shell of charge has vacuum zero point fluctuation virtual particle plasma with an effective potential

VZPF = -/\r^2

gZPF = - dVZPF/dr = +2/\r

the AdS case /\ < 0 gives the compensating stabilizing attraction. We get this when the density of virtual electron-positron pairs inside the shell exceeds the density of virtual photons. The excess of virtual photons outside the charged shell will work in an implosive manner similar to what Puthoff proposes - but again this is purely gravity w = -1 for ZPF not electromagnetic Casimir at all. That will be there too of course as a secondary effect.

V = VZPF + Vcharge = - /\r^2 + e^2/mr

Equilibrium is when

dV/dr = 0

-2/\r - e^2/mr^2 = 0

-2/\r*^3 = e^2/m

/\ < 0 (AdS)

r* = (e^2/2m|/\|)^1/3

stability of the equilibrium is a positive second derivative, i.e.

-2/\ - 2e^2/mr*^3 > 0

-2/\ - 2/\e^2/m/2e^2/m = -3/\ > 0

-------------------------------------------------

JW: The negative bare mass version differs from Price's container example because the electrically charged dust of the ADM negative bare mass model has both active and passive masses negative. This means that as the electrical action of one part of the dust on another produces a force of repulsion (like charges and all that), the result, because of the negativity of the inertial mass of all of the parts, is for the dust to accelerate toward, not away from, the center of the dust. That is, the electrical self-energy, instead of being positive as usual, becomes negative like normal positive mass gravity. Note that this interaction is linear in the sense that the electric charge is unaffected by the interaction.

JS: OK, but I need to see the algebra.

e^2/r = -md^2r/dt^2

Yes, OK so now with two like charges, and at least one of them is negative inertial mass, you get attractive electric force OK

If the other equal sign charge is positive inertial mass we have Bondi self-acceleration - I think?

In the case of the gravitational self-energy of the dust, the negativity of all of the dust gives a force of attraction (minus active times minus passive) as for positive masses. But that force acts on material with negative inertial mass. So the effect of the attracive force is to make the dust expand, rather than contract. That is, the self-energy is effectively positive rather than negative because of the negativity of the inertial mass. Note that the gravitational self-energy remains non-linear in the sense that the mass in the quadratic term of the ADM solution is the total mass, not just the bare mass.

Since the effective self-energy terms in the ADM solution are still of opposite sign and the gravitational mass is non-linear, configurations where the two terms cancel for the dust distribution with non-zero radius are possible. If such a distribution is perturbed toward collapse, the effectively repulsive non-linear gravitational force will begin to exceed the strength of the electrical force which is not non-linear. If the dust is perturbed outward, the effectively attractive electrical force will become larger than the non-linearly decreasing effective gravitational force, and the dust will experience a net force that tends to restore the original configuration. So, finite radii dust distributions where the forces balance are stable.

Plausible I need to think about it more, would like to see the algebra, but it sounds right.

Now, for the negative bare mass ADM model to return a realistic value for the electron mass, the dust must lie very close its gravitational radius (so that the gravitational potential of the dust bare mass almost exactly cancels the gravitational potential of the universe -- as explained in MUSH and the Stargates papers. Since electrons have small finite positive masses, somehow this configuration must occur and be maintained. How this might happen was not explained in MUSH -- because though I thought I knew what must be going on, an algebraic error diverted me into frustrating blind alleys.

The problem is: what constitutes the criterion for minimum energy? With the positive mass assumption, the answer is simple: zero. Or the quantum zero point value (not without problems of its own I note). Allowing negative mass means allowing negative energy, and that complicated minimum energy (a problem Dirac struggled with). The obvious conjecture is that allowing negative energy seriously louses things up because energy then is not bounded from below. You can try to finese this by claiming that minimum energy should still be taken as zero and that something like the absolute value of energy is the important consideration in minimum energy. I wouldn't try selling that to anyone though.

The situation for the negative mass ADM model is analogous to the nuclear atom problem. The missing piece is quantum mechanics.

Bohm's quantum potential Q should save the day?

You need the analog of the Bohr conjecture that the minimum energy is that where the electric charge has one quantum of angular momentum. Otherwise there is no reason for the dust not to coalesce at R = 0 and display a classic ADM mass (21 orders of magnitude larger than the observed electron mass). That's what got worked out a year and a half ago and put into the Stargates paper.

If this is right, then hidden in a modest amount of normal matter is a Jupiter mass of exotic matter. And it can be exposed by screening that matter from the gravitational action of the distant matter in the universe. . . .

Not if G gets huge on the small scale of a fermi as in Abdus Salam's f-gravity and in some modern models of extra space dimensions. You can cut Jupiter mass down by about 40 powers of ten from 10^27 kg to 10^-13 kg = 10^-9 grams.

Category: MyBlog

Written by Jack Sarfatti

Published on Friday, 16 December 2011 11:40

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