Today we got an interesting all-roe email which I though I would share with the blogosphere. You have probably noticed that this is the International Year of Astronomy. Tomorrow there is an official Opening Ceremony in Paris. The press release is here. But much more fun, you can follow it on a webcast, available Jan 15th from http://www.astronomy.org/webcast.

There is also a fun new “Cosmic Diary” web site. This includes a blog by Staff Writer Lee Pullen, and you can track the Opening Ceremony action on his blog. (Is there a Twitter, does anybody know ??)

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This entry was posted on Wednesday, January 14th, 2009 at 7:13 pm and is filed under Astronomy. You can follow any responses to this entry through the RSS 2.0 feed.
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The main Twitter stream for the IYA this one, but you get a more complete picture by doing a Twitter search for the #iya2009 hashtag once a day or so. And being on the German IYA Committee, I also twitter about the IYA rather frequently …

Hello, thanks for the mention of my liveblog! There isn’t a Twitter feed, unfortunately. We didn’t think of that, but the truth is that it would probably be a bit much to handle as well as everything else. Maybe for the Closing Cerenony! Now I should stop trying here and get back to liveblogging!

Actually I am twittering the opening ceremony right now with high frequency: The live feed from Paris includes both audio, moving pictures of the speakers – and (sometimes) their slides in moderately high resolution. In my Twitter stream – just click on “Daniel Fischer” – I’m including occasional screenshots of the latter.

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)

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

Golly gosh. Shocking that Newton and Kepler failed to explain those quantum relativistic effects. Maybe Isaac was having an off day. Trooble at ‘t Mint I guess.

Einstein’s Physics Dollar Store on Campus
MIT Harvard Cal-Tech
Sponsored by NASA
Why Relativity theory is not Physics and why Einstein’s “thought” = 0
Walking the walk and talking the talk taking on all space-time confusion of physics by
MIT Harvard and Cal-Tech and all other Physics dollar stores departments
And why LHC burned itself

Visual Effects and the confusions of “Modern” physics

r ——— Light sensing of moving objects ——- S
Actual object—– Light ——— Visual object
r – ——-cosine (wt) + i sine (wt) – S = r [cosine (wt) + i sine (wt)]
Newton– Kepler’s time visual effects — Time dependent Newton
Particle ————– Visual effects ——————– Wave

Line of Sight: r cosine wt

r ——————- r cosine (wt) line of sight light aberrations

A moving object with velocity v will be visualized by
light sensing through an angle (wt);w = constant and t= time
Also, sine wt = v/c; cosine wt = √ [1-sine² (wt) = √ [1-(v/c) ²]

A visual object moving with velocity v will be seen as S

S = r [cosine (wt) + i sine (wt)] = r Exp [i wt]; Exp = Exponential

S = r [√ [1-(v/c) ²] + ỉ (v/c)] = S x + i S y

S x = Visual along the line of sight = r [√ [1-(v/c) ²]

This Equation is special relativity length contraction formula
And it is just the visual effects caused by light aberrations of a
moving object along the line of sight.

In a right angled velocity triangle A B C: Angle A = wt; angle B = 90°; Angle C = 90° -wt
AB = hypotenuse = c; BC = opposite = v; CA= adjacent = c √ [1-(v/c) ²]

Err… this post seems to have become the site of some kind of strange parallel universe discussion. Now I do have an open comment policy … but maybe this counts as spam, as it seems to have no connection to this post or any other post ????

p.s. if regular readers comment on these comments, do stay polite…

This blog has an open comment policy. Please respect this by keeping comments the pleasant side of insult and libel ... and do be aware that the comments don't necessarily reflect my own opinions.

RT @BeffernieBlack: Bring facebook into public ownership? Are you... are you suggesting we [dramatic pause] seize the memes of production?… 16 hours ago

The main Twitter stream for the IYA this one, but you get a more complete picture by doing a Twitter search for the #iya2009 hashtag once a day or so. And being on the German IYA Committee, I also twitter about the IYA rather frequently …

Hello, thanks for the mention of my liveblog! There isn’t a Twitter feed, unfortunately. We didn’t think of that, but the truth is that it would probably be a bit much to handle as well as everything else. Maybe for the Closing Cerenony! Now I should stop trying here and get back to liveblogging!

– Lee

Actually I am twittering the opening ceremony right now with high frequency: The live feed from Paris includes both audio, moving pictures of the speakers – and (sometimes) their slides in moderately high resolution. In my Twitter stream – just click on “Daniel Fischer” – I’m including occasional screenshots of the latter.

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’ = [-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

Golly gosh. Shocking that Newton and Kepler failed to explain those quantum relativistic effects. Maybe Isaac was having an off day. Trooble at ‘t Mint I guess.

Maybe His wife was a bad cook!

Einstein’s Physics Dollar Store on Campus

MIT Harvard Cal-Tech

Sponsored by NASA

Why Relativity theory is not Physics and why Einstein’s “thought” = 0

Walking the walk and talking the talk taking on all space-time confusion of physics by

MIT Harvard and Cal-Tech and all other Physics dollar stores departments

And why LHC burned itself

Visual Effects and the confusions of “Modern” physics

r ——— Light sensing of moving objects ——- S

Actual object—– Light ——— Visual object

r – ——-cosine (wt) + i sine (wt) – S = r [cosine (wt) + i sine (wt)]

Newton– Kepler’s time visual effects — Time dependent Newton

Particle ————– Visual effects ——————– Wave

Line of Sight: r cosine wt

r ——————- r cosine (wt) line of sight light aberrations

A moving object with velocity v will be visualized by

light sensing through an angle (wt);w = constant and t= time

Also, sine wt = v/c; cosine wt = √ [1-sine² (wt) = √ [1-(v/c) ²]

A visual object moving with velocity v will be seen as S

S = r [cosine (wt) + i sine (wt)] = r Exp [i wt]; Exp = Exponential

S = r [√ [1-(v/c) ²] + ỉ (v/c)] = S x + i S y

S x = Visual along the line of sight = r [√ [1-(v/c) ²]

This Equation is special relativity length contraction formula

And it is just the visual effects caused by light aberrations of a

moving object along the line of sight.

In a right angled velocity triangle A B C: Angle A = wt; angle B = 90°; Angle C = 90° -wt

AB = hypotenuse = c; BC = opposite = v; CA= adjacent = c √ [1-(v/c) ²]

Err… this post seems to have become the site of some kind of strange parallel universe discussion. Now I do have an open comment policy … but maybe this counts as spam, as it seems to have no connection to this post or any other post ????

p.s. if regular readers comment on these comments, do stay polite…

my first post

thank you for having me! 😀