Wednesday, March 29, 2006

Outward pressure times area is outward force...

Image above is taken from Dr Samuel Glasstone's Effects of Nuclear Weapons 1957. The outward force of the blast always has an equal and opposite reaction (3rd law of motion), in this case underpressure (suction), pulling instead of pushing. See the tree stand in the middle of this video clip of the 15 kiloton Grable nuclear test. Close to ground zero, before the suction phase develops, the reaction is simply the symmetry of the blast (the reaction of the Northwards part of the blast is the Southwards moving blast, while there is still high pressure connecting them - this breaks down when a vacuum forms near ground zero, and from then on the reactive force is the inward or suction blast phase). Likewise, in a sound wave, you have to have an outward pressure followed by an inward (underpressure) force. The relationship between force and pressure is force equals pressure times area acted upon.

This whole approach to understanding sound waves, explosion blast waves, and consequently the big bang gravity mechanism, is suppressed. The logic that you get an inward force in an explosion (which by Newton's 3rd law balances the outward force) is also inherent in the implosion principle of nuclear weapons, as Glasstone explained:




If you don't have an equal and opposite reaction in a pressure wave, it isn't a sound wave.

The force you get against your eardrum isn't just a push, but a push followed by equal pull.

This mechanism explains the gauge boson inward push in the big bang, predicting gravity.

The outward force in any explosion always has an equal and opposite reaction (Newton's 3rd empirical law). If you just push air, the energy disperses without propagating as a 340 m/s oscillatory sound wave. Air must be oscillated to create sound. It delivers an oscillatory force, outward and then inward. Merely using wave equations does not explain the physical process, even where the maths happens to give a good fit to data. Sound waves are particulate molecules deep down, carrying an oscillatory force.

This makes various predictions and contains no speculation whatsoever, it is a fact based mechanism, employing Feynman's mechanism as exhibited in the Feynman diagrams - virtual photon exchange causing forces in QFT. He noted that path integrals has a deeper underlying simplicity:

"It always bothers me that, according to the laws as we understand them today, it takes a computing machine an infinite number of logical operations to figure out what goes on in no matter how tiny a region of space, and no matter how tiny a region of time. How can all that be going on in that tiny space? Why should it take an infinite amount of logic to figure out what one tiny piece of space/time is going to do? So I have often made the hypothesis that ultimately physics will not require a mathematical statement, that in the end the machinery will be revealed, and the laws will turn out to be simple, like the chequer board with all its apparent complexities." - Richard P. Feynman, Character of Physical Law, Penguin, 1992, pp 57-8.

(In the same book he discusses the problems with the LeSage gravity mechanism as per 1964.)

1 comment:

  1. Copy of a comment

    http://motls.blogspot.com/2006/11/relativistic-pinball-machine.html

    http://www.haloscan.com/comments/lumidek/116378667065851495/?a=15096#646725


    Of course matter is accelerated in all kinds of directions in shock waves. The key thing about a shock wave is that it needs a medium of gas or dust to propagate it.

    The outgoing shell or "shock front" is continuously engulfing new matter, and thus having to accelerate that new matter up to speed.

    Hence, it exerts outward force, F=ma, which is related to outward pressure by P = F/A, where A is the spherical surface area 4*Pi*R^2 for shock front radius R. See http://glasstone.blogspot.com/2006/03/analytical-mathematics-for-physical.html where we get an analytical solution of shock waves from a bit more mathematical modelling: you need to scroll down there for the proof that in a strong shock wave the strong shock front velocity is U = (2/5)R/t, where R is shock front radius and t is time, assuming that U is much bigger than the ambient speed of sound. The 2/5 factor is the deceleration factor (the shock wave velocity is ever decelerating, ambient pressure and ambient sound speed being neglected).

    The shock wave radius is there proved to be R

    = {[75E(gamma - 1)t^2]/(8*Pi*Rho_o)}^{1/5},

    where E is the energy of the explosion, gamma is the usual ratio of specific heats of the gas (generally gamma is in the range of 1.2-1.4), t is time, and Rho_o is the ambient gas or dust density.

    I'd like to get this proof on arxiv, because the only mainstream thing which comes near to it is G.I. Taylor in Proceedings of the Royal Society (v. 201A, pp. 159-86), but he uses a long-winded (27 pages) numerical integration of the equations of motion instead of the analytical solution based on causal mechanism I use. I can't get it on arxiv since they censored me for the gravity formula.

    Because of the outward force in any explosion, Newton's 3rd law says there must be an equal inward force. The region in the middle of an explosion becomes very low density very soon (which is why the fireball in an explosion is buoyant and thus rises once the shock wave departs), so the reaction force is not a symmetry effect with the outward force on one side being balanced by a reaction from the shock wave on the opposite side which is going in the other direction.

    Instead, the reaction force occurs in a second shell where gas or dust travels INWARDS, with inward force. See http://glasstone.blogspot.com/2006/03/outward-pressure-times-area-is-outward.html

    nigel cook | Homepage | 11.17.06 - 5:08 pm | #

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WHAT IS NUKEGATE? The Introduction to "Nuclear Weapons Effects Theory" (1990 unpublished book), as updated 2025

R. G. Shreffler and W. S. Bennett, Tactical nuclear warfare , Los Alamos report LA-4467-MS, originally classified SECRET, p8 (linked HERE): ...