Post taken from EnvironmentalGraffiti.com

At Lancaster University, they’re unraveling the secrets of how to build a universe. In fact, they have already formed one, or something very much like it. This scientific breakthrough lies in the bottom of a chamber no larger than your pinky finger, filled with helium and cooled to 0.0003 degrees Fahrenheit above absolute zero.

By placing helium in a state which most closely resembles the form it held at the beginning of the universe, scientists have created an opportunity for the gas to go through several low-energy evolutions. These defects in space-time, are represented by tiny whirlpools in the helium, which are created by the rapid expansion, and equally rapid slowing of the expansion; something that it’s believed our own universe did at the big bang and in the moments thereafter.

How, then, did our universe go from whirlpools that could fit in a thimble to galaxies larger than our imaginations can properly comprehend? Physicists, ever ready with their dry wit, have deemed these phenomena “inflation.” Nobody knows how this works or why, this happened; vast amounts of energy aren’t something you’d like to replicate in a lab. Black holes and supernovas aren’t pleasant lab partners. It’s quite evident to the researchers however, that inflation, or something very much like it took place and, lacking the ability to do field research of lab trials, they have built scale models. This is where the tiny galaxies come in.

The theory being presented by the physicists in Lancaster University is that inflation is the product of violent competition: a series of collisions between universes known as “3-branes;” a term related to string theory which I’m frankly not smart enough to explain to you. Suffice to say that our universe is one, because it exists in 3-5 dimensions.

What the string theorists claim is that in a collision of two 3-branes, or two different modes of pure helium like that containing the mini-galaxy, the universe will rapidly expand and stop instantly, mimicking the halting advance of the universe’s growth. Remarkably, when super cooled helium in different phases is mixed, it does exactly that: symmetries in the solution disappear, and aberrations form; the first step in several that lead to the forming of galaxies out of nothing. The secrets of the universe it seems, aren’t safe for long.

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Einstein’s Nemesis: DI Herculis Eclipsing Binary Stars Solution

The problem that the 100,000 PHD Physicists could not solve

This is the solution to DI Her “Quarter of a century” Smithsonian-NASA Posted motion puzzle that Einstein and the 100,000 space-time physicists including 109 years of Nobel prize winner physics and physicists and 400 years of astronomy and Astrophysicists could not solve and solved here and dedicated to Drs Edward Guinan and Frank Maloney

Of Villanova University Pennsylvania who posted this motion puzzle and started the search collections of stars with motion that can not be explained by any published physics

For 350 years Physicists Astrophysicists and Mathematicians and all others including Newton and Kepler themselves missed the time-dependent Newton’s equation and time dependent Kepler’s equation that accounts for Quantum – relativistic effects and it explains these effects as visual effects. Here it is

Universal- Mechanics

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location

r = r (x, y, z). The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location

P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment

= change of location + change of mass

= m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

F = d P/d t = d²S/dt² = Force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r

= m γ + 2m’v +m”r; γ = acceleration; m” = mass acceleration rate

In polar coordinates system

r = r r(1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)

F = m[(r”-rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)] + 2m'[r’r(1) + rθ’θ(1)] + (m”r) r(1)

F = [d²(m r)/dt² – (m r)θ’²]r(1) + (1/mr)[d(m²r²θ’)/d t]θ(1) = [-GmM/r²]r(1)

d² (m r)/dt² – (m r) θ’² = -GmM/r²; d (m²r²θ’)/d t = 0

Let m =constant: M=constant

d²r/dt² – r θ’²=-GM/r² —— I

d(r²θ’)/d t = 0 —————–II

r²θ’=h = constant ————– II

r = 1/u; r’ = -u’/u² = – r²u’ = – r²θ'(d u/d θ) = -h (d u/d θ)

d (r²θ’)/d t = 2rr’θ’ + r²θ” = 0 r” = – h d/d t (du/d θ) = – h θ'(d²u/d θ²) = – (h²/r²)(d²u/dθ²)

[- (h²/r²) (d²u/dθ²)] – r [(h/r²)²] = -GM/r²

2(r’/r) = – (θ”/θ’) = 2[λ + ỉ ω (t)] – h²u² (d²u/dθ²) – h²u³ = -GMu²

d²u/dθ² + u = GM/h²

r(θ, t) = r (θ, 0) Exp [λ + ỉ ω (t)] u(θ,0) = GM/h² + Acosθ; r (θ, 0) = 1/(GM/h² + Acosθ)

r ( θ, 0) = h²/GM/[1 + (Ah²/Gm)cosθ]

r(θ,0) = a(1-ε²)/(1+εcosθ) ; h²/GM = a(1-ε²); ε = Ah²/GM

r(0,t)= Exp[λ(r) + ỉ ω (r)]t; Exp = Exponential

r = r(θ , t)=r(θ,0)r(0,t)=[a(1-ε²)/(1+εcosθ)]{Exp[λ(r) + ì ω(r)]t} Nahhas’ Solution

If λ(r) ≈ 0; then:

r (θ, t) = [(1-ε²)/(1+εcosθ)]{Exp[ỉ ω(r)t]

θ'(r, t) = θ'[r(θ,0), 0] Exp{-2ỉ[ω(r)t]}

h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity

h = 2πa²√ (1-ε²); r (0, 0) = a (1-ε)

θ’ (0,0) = h/r²(0,0) = 2π[√(1-ε²)]/T(1-ε)²

θ’ (0,t) = θ'(0,0)Exp(-2ỉwt)={2π[√(1-ε²)]/T(1-ε)²} Exp (-2iwt)

θ'(0,t) = θ'(0,0) [cosine 2(wt) – ỉ sine 2(wt)] = θ'(0,0) [1- 2sine² (wt) – ỉ sin 2(wt)]

θ'(0,t) = θ'(0,t)(x) + θ'(0,t)(y); θ'(0,t)(x) = θ'(0,0)[ 1- 2sine² (wt)]

θ'(0,t)(x) – θ'(0,0) = – 2θ'(0,0)sine²(wt) = – 2θ'(0,0)(v/c)² v/c=sine wt; c=light speed

Δ θ’ = [θ'(0, t) – θ'(0, 0)] = -4π {[√ (1-ε) ²]/T (1-ε) ²} (v/c) ²} radians/second

{(180/π=degrees) x (36526=century)

Δ θ’ = [-720×36526/ T (days)] {[√ (1-ε) ²]/ (1-ε) ²}(v/c) = 1.04°/century

This is the T-Rex equation that is going to demolished Einstein’s space-jail of time

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²—) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)

v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

v = v (center of mass); v is the sum of orbital/rotational velocities = v(cm) for DI Her

Let m = mass of primary; M = mass of secondary

v (m) = primary speed; v(M) = secondary speed = √[Gm²/(m+M)a(1-ε²/4)]

v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. joenahhas1958@yahoo.com