At 5:41 PM 12/7/4, John Fields wrote: [snip] > >>The real beauty of your idea (is it original?) > >--- >As far as I know, it is. The lightbulb going off was due to something >of Fred Sparber's or Frank Znidarsik's (sp?) that I read a few years >ago on vortex, and since then I've been looking but haven't been able >to find anything quite like it. A month or so ago I talked to Hal >Puthoff about it and he also thought it was novel, so maybe it is.
You may want to check out the thread: "Qball: Electrostatic Sphere Field Evaluator". A sample post late in the thread follows below. I wrote (and posted) a basic program to integrate the effects of a spherical distribution of charge, and sample data. We learned a few things from the exercise if you recall... - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - At 10:27 AM 6/19/3, Horace Heffner wrote: >At 9:58 AM 6/19/3, John Fields wrote: > >>Yes, I agree. However, the equipotential surfaces look like infinitely >>thin nested spherical shells surrounding an undisturbed charged >>particle, and my analogy merely picked a balloon to represent a >>particular potential. Bringing two otherwise undisturbed charged >>particles together, then, should create a region between them (a plane >>perpendicular to a line drawn between the charges and extending to >>infinity) where the potential between them is equal and diminishes as >>the distance from the line drawn between their centers increases. Place >>a third charged particle on this plane where the line intersects it and >>it should stay there; place it anywhere else on the plane and it will go >>zooming off, but will be confined to the plane. > > >Note, however, that the force outward in the plane lying between them is >smaller than if both charges were located at their midpoint. Their >effectiveness is in effect diluted by their separation 2*r. Their >effectiveness will be further diluted if we place two more charges in the >central plain in another axis. At any point outside the central plain, >these extra charges lying on the plain *repel* the test charge away from >the plain to some degree. The centering force is reduced. There are >still channels of confinement, but the field magnitudes in every channel >is reduced with the increasing adding of charges. As the number of >channels is increased, the field strength in each channel is reduced. >However, understanding this slight reduction is only half the intuition >problem. The fact that the channel strengh is reduced does not in itself >guarantee that it goes to zero. In fact, I think as long as the number of >points if finite, as you add more charge points, each with fixed charge Q, >there will always exist channels of confinement and repelling channels. >What was surprising to me was the comparatively large magnitude of the >fields in these channels when the uniformity of the distribution is only >fraction of a percent off. This was not intuitive to me. But let's get >on with finding out why this analogy fails in the limit ... > > >> >>Why? Construct two planes parallel to the plane centered between the >>charges and cutting the centers of the charges, and insert a charge >>anywhere in the region bounded by the outer planes. What do you think >>will happen? I think the inserted charge will be repelled by the charge >>closest to it until it crosses the central plane, then be repelled by >>the particle now closest to it and will thereafter be shepherded by both >>charges until its oscillations are damped to the point where it can be >>effectively considered to be confined to the central plane. > > >Let's ignore the details about the above and assume it is completely >correct with the caveat's I already noted above. > >> >>Of course no analogy is perfect, but that's what mine was about. > > >I think this analogy is very good except for consideration of charge >dilution. To meet the criteria of the original problem we must assume >that a fixed charge is distributed around the sphere. We have a charge >density Q/A = rho. We have a fixed charge on the spherical shell, and a >fixed area A in which the charge is (or in our case is to be) distributed. >Note that we can not assume an infinite charge density, or we have a >discontinuity, and all logic it would seem falls apart - but we can deal >with that also by solving finite rho and then summing on finite rho's to >infinity. > >I think what you have intuitively shown is that when the charge Q is >distributed around a finite number of points, there will be volumes >containing confining fields separated by volumes of repelling fields. >These volumes are in the form of "channels". > >The maximum field strength in these channels is diluted by the fact the >charges are separated from each other by distance 2*r, and by the fact >that addition of more charge Q around the sphere adds more channels but >reduces the fields in each of the channels. There is one other major >dilution factor to considered. > >In your analogy, you keep adding more point charges. This is a false >analogy in that the final charge Q aroud the sphere must be Q, a fixed >value. As you add charge points in your scenario, the charge in Q_i each >point must be diminshed. This is an error I made and then discovered and >corrected between runs 2, and runs 3 of Qball. Runs 1-2 keep adding >charge as the number of points is increased. Runs 3-5 in effect keep the >total charge on the sphere constant. In runs 3-5 you can clearly see the >field intensity drop across the entire sphere interior as the number of >points goes up, the average charge separation distance goes down. > >Now to try to see if we can intuit why the field intensites go to zero as >the number of charge location points goes to infinity. Let's imagine an >initial state of N charge points spread around sphere in an approximately >grid like fashion (this is technically not feasible on a sphere, but we >are only approximating, and in fact the Qball program roughly does this, >except the "rectangles" are mostly parallelograms.) The maximum field >stength in any channel is Emax. The charge Q_i distributed at each point >is Q/N. > >At each step j that follows, we double the number of points to 2*N, and >distribute the new points in the centers of the old rectangles (or >parallelograms.) We have thus doubled the number of channels and very >slightly reduced the channel fields. However, we must now divide all the >field strengths by 2. Emax_j <= Emax_(j-1)/2. Therefore E_max -> 0 as >j-> inf. That division of charge at each point by 2 must be done to make >the final total charge distributed around the sphere constant at Q. At >each step j, we diminish the channel field strength by (1/2)^j. Since lim >j->0 (1/2)^j = 0, the channel field strengths all go to zero. > >Now let's attempt to look at your initial problem of the bubble universe >and an infinitely dense surrounding. The confinement force is now an >expansion force, due to gravity being attractive, but all else is really >the same, except for dealing with the infinite nature of the infinitely >dense volume outside the bubble. To begin this we can examine a single >shell, but only a finite portion of mass m in the shell, uniformly >distributed. This gives us a finite mass density in the shell. Such a >shell has zero force inside it, as we saw earier. We can now sum >(integrate) an infinite number of such shells, all of the same fixed >radius, each having mass m, so that the total mass in the resulting shell >is infinite. Since the force of each shell everywhere is zero, the same >is true of the final infinite sum. We can then sum (integrate) the shells >at every radius r, with r-> inf. Since the force inside every shell is >everywhere zero, this must be true inthe final sum. There is thus no >force of gravity exerted by the external volume on the universe. > >Of ourse one tiny change in density out in that infinitely dense volume ... Regards, Horace Heffner
