The IGNORANT falsely 'believe' there is no inward force in big bang
Once again, an explosion - in air or in space, creates an outward pressure and hence force (via F=PA where A is area), which by 3rd law of motion implies an inward pressure of the surroundings like the Higgs field or air. If you look at some of the films of tests, you see the inward pressure phase of the blast is set up before the fireball begins to rise.
Brode deals with numerically integrating the equations of motion given the conservation of momentum, mass, and energy in a shock wave.
However, what physically results as shown on his charts is an outward force at the greatest distances, with an inward force within that zone, once there is a vacuum near ground zero.
The inward force of the blast (the inner concentric sphere) is the 3rd law reaction to the outward force (outer zone of the blast).
An analogy of this to the big bang in the Higgs field gives gravity right. Personally I think it was unethical of Dr Campbell of Nature to write me that he was "not able" to publish this nearly a decade ago. However, I'm sure he will get promotion and prizes for holding back progress. His type always do.How do the horses*** artists, the string theorists, get away with their rubbish on gravity? Answer: society pays them to invent 10 dimensional horses*** speculation which cannot be tested and is not based on fact.Question: why isn't the big bang treated as an explosion, as indicated above? Why am I suppressed? Who is responsible? Answer: Jeremy Webb BSc, editor of New Scientist, and his friends like Dr John Gribbin. Also Ivor Catt, who thinks facts proved in science can be ignored using Ockham's Razor which allows him to 'accept only falsely simplistic models which bend the facts to fit them'.Now tell me that science is a matter of Popper's personal pet theories, not experimentally proved facts and testable predictions.http://nigelcook0.tripod.com/
The universe is receding; galaxies apart from a few nearby galaxies like Andromeda, all have a red shift. While there are speculations that the red shift may be tired light, there is no mechanism and no evidence of this from the spectrum of the red shifted light. In fact, the best ever experimental black body radiation spectrum was obtained by the cosmic background explorer satellite in 1992 from the 2.7 K (microwave) red-shifted 3,000 K (infrared) big bang radiation flash. This frequency spectrum was uniformly reduced by over 1,000 times by red shift, not by the effects of scattering of radiation (scattering is frequency-dependent). It was emitted about 300,000 years after the big bang. The three pieces of evidence for the big bang, namely (1) red shifts, (2) microwave background spectrum, and (3) the abundance of hydrogen, deuterium and helium in the universe are conclusive proof of the big bang in general. The purpose of this paper is to establish a fourth piece of evidence and to clarify what more we can learn from the big bang by proved experiments rather than by speculation. ...
Georges Louis LeSage, between 1747-82, explained gravity classically as a shadowing effect of space pressure by masses. The speculative, non-quantitative mechanism was published in French and is available online
(G.L. LeSage, Lucrece Newtonien
, Nouveaux Memoires De L’Academie Royal de Sciences et Belle Letters, 1782, pp. 404-31). Because gravity depends on the mass within the whole earth’s volume, LeSage predicted that the atomic structure was mostly void, a kind of nuclear atom which was confirmed by Rutherford’s work in 1911. LeSage argued that there is some kind of pressure in space, and that masses shield one another from the space pressure, thus being pushed together by the unshielded space pressure on the opposite side. Feynman explained that the major advance of general relativity, the contraction term, shortens the radius of every mass. He does not derive the equation, but we will do so below. ...
The contraction of space is by (1/3) GM/c2. This is the 1.5-mm contraction of earth’s radius Feynman obtains, as if there is pressure in space. An equivalent pressure effect causes the Lorentz-FitzGerald contraction of objects in the direction of their motion in space, similar to the wind pressure when moving in air, but without viscosity. Feynman was unable to proceed with the LeSage gravity and gave up on it in 1965. However, we have a solution.
The big bang causes an outward force (Newton’s 2nd law) that results in an equal inward force (Newton’s 3rd law) which causes gravity as an inward force, Higgs field pressure. Where partially shielded by mass, the inward pressure causes gravity. Apples are pushed downwards towards the earth, a shield. ...
METHOD 1: SIMPLE SHIELDING CALCULATION
There is strong evidence from electromagnetic theory (April 2003 Electronics World paper, reprinted near the end of this page) that every fundamental particle has black-hole properties. The effective shielding radius of a black hole of mass M is equal to 2GM/c2. A shield, like the planet earth, is composed of very small, sub-atomic particles. The very small shielding area per particle means that there will be an insignificant chance of the fundamental particles within the earth ‘overlapping’ one another by being directly behind each other. The total shield area is therefore directly proportional to the total mass: the total shield area is equal to the area of shielding by 1 fundamental particle, multiplied by the total number of particles. The earth’s mass in the standard model is due to particles associated with up and down quarks: the Higgs field. From the illustration above, the total outward force of the big bang, (total outward force) = ma = (mass of universe).(Hubble acceleration, see below), while the gravity force is the shielded inward reaction (by Newton’s 3rd law the outward force has an equal and opposite reaction): F = (total outward force).(cross-sectional area of shield projected to radius R) / (total spherical area with radius R). The cross-sectional area of shield projected to radius R is equal to the area of the fundamental particle (p multiplied by the square of the radius of the black hole of similar mass), multiplied by the (R/r)2 which is the inverse-square law for the geometry of the implosion. The total spherical area with radius R is simply four times p, multiplied by the square of R. Inserting simple Hubble law results c = RH and R/c = 1/H give us F = (4/3)p r G2M2/(Hr)2. We then set this equal to F=Ma and solve, getting G = (3/4)H2/(p r ). When the effect of the higher density in the local universe at the great distance R is included
, this becomes G = (3/4)H2/(p r (local) e3), which is accurate and identical to that obtained in the other type of proof below (which does not require the shielding area to be inserted). ...
(Symbols for density rho
and for pi
have not come out - just the letters p and r, nor has superscript, in the paragraph above, see http://nigelcook0.tripod.com/