Comments

1. #1 Neil B ☺

   January 17, 2009

   Perimeter Institute sure has some good people, like “Bee” (Sabine Hossenfelder) who writes fabulous science, science-in-society and even good snarky humor at her very cool blog http://backreaction.blogspot.com/. She also carries that on through the interdisciplinary Lightcone Institute (http://www.lightconeinstitute.org/.)

   Maybe this is a good place to ask a foundational physics question. Noether’s Theorem is a key to our understanding of physical law, so I gather.Well, with no particular topic posted may I ask here: Does Noether’s Theorem generalize to universes analogous to ours but with other number of space dimensions? It seems like it should, but I’ve seen various statements that action etc. has special properties in three-D space. tx.

2. #2 John Sidles

   January 17, 2009

   Just to say … this is a fabulous, wonderful, outstanding program … I have already encouraged a student to apply … in these words …

   “For sure, you would *never* forget it. And probably, you wouldn’t regret it.”

3. #3 joe nahhas

   January 28, 2009

   Einstein’s Nemesis: DI Her Eclipsing Binary Stars Solution
   The problem that the 100,000 PHD Physicists could not solve

   This is the solution to the “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’ = [-720x36526/ 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<

   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

4. #4 Dave Bacon

   January 28, 2009

   Um, on topic, Joe? I know cut and paste is fun, but really you’re not helping your case posting comments on a blog, are you?