Koizumi and Richard Gere finally meet — space time continuum remains intact… for now

OK, they look a lot less alike than they did when Koizumi first took office, but god damn:

UPDATE: For those who don’t see the resemblance, you may now shut the hell up and bask in the awe that was once the Mane of Richard Gere:
Look at all its...majesty!

It’s like an older white version of Koizumi came back to warn him about some future disaster… or maybe just dance the night away…

But seriously: what is going on? Assuming for a minute that Gere is the future Koizumi (and, just so we’re clear, I believe in this as surely as I believe in the tastiness of unagi-don), theoretically there should be a big problem. Movies have taught me that if the you of the future encounters the you of the past, there will be a rift in the space-time continuum and all existence shall be erased forever. Or could BTTF be wrong, and it’s totally OK for you to meet your future self (or even have kids with women from the past who are destined to liberate your people.. in the future…)

Or maybe it’s some combination of various sci-fi movies.. What if, for example, the BTTF warning about meeting your future self is true, but the space-time continuum doesn’t rupture instantly, instead taking 29 days to unravel a la Donnie Darko? The end may be coming folks… Feel free to share your own theories of what the Gere-Koizumi meeting portends for the survival of the universe!

6 thoughts on “Koizumi and Richard Gere finally meet — space time continuum remains intact… for now”

  1. Actually, Gere was born in 1949, and Koizumi was born in 1942.

    But it doesn’t matter. Koizumi is the man. The space-time continuum can’t touch him.

  2. according to bill and ted, you can play air guitar with your future self and save the princesses and eveything will be okay. but in star trek tng they had nothing but problems with the space time continuum and only q could navigate it without all hell braking loose. if richard gere is involved evil must be afoot. you should monitor this situation very closely.

  3. I confess my mind strayed right from the space-time continuum as soon as I clicked the link for unagi-don. I now have a new desktop wallpaper. But why do I tempt myself so.

  4. Kepler (demolish) Vs Einstein’s space jail of time
    r ————– Exp (i wt) ———–S= r Exp (ì wt) Nahhas’ Equation
    Orbit location———–Orbit light sensing ————– Visual orbit location
    Particle/Newton —————- –Visual ————————– Wave/Quantum
    Quantum – Newton=visual effects=relativistic effects=space-time confusions
    S= visual distance; r = actual distance; v = speed and c = light speed
    S = r Exp (i wt) = r [cosine (wt) + î sine (wt)]
    P =d S/d t = v Exp (ì w t) + ì r w Exp (ì w t); v=d r/d t; v=w r
    = v (1+ ì) [Exp (ì wt)] = visual velocity
    E (definition) = m/2(m v + m’ r) ²; E = mc²/2 If v = 0; m’ r=mc
    E (visual) = mp²/2 = mv²/2(1+ì) ² Exp 2(ì w t)
    E (visual) = mv²/2(1 + 2ì -1) [cos2wt + ì sin2wt]
    E (visual) = ì (mv²) [1-2sin²wt + 2i [sin (wt)] [cosine (wt)]
    If wt = (2n+1) π/4
    E (visual) = ì (mv²) [1-1 ± ỉ] = ± (mc²); v = c
    2-Central force law Areal velocity is constant: r² (d θ/d t) =h Kepler’s Law
    h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
    r² (d θ/d t) = h = S² (d w/d t)
    Replace r with S = r exp (ỉ wt); h = [r² Exp (2iwt)] (d w/d t)
    (d w/d t) = (h/r²) exp [-2(i wt)]
    d w/d t= (h/r²) [cosine 2(wt) – ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) – ỉ sin 2(wt)]
    d w/d t = d w(x)/d t + d w(y)/d t; d w(x)/d t = (h/r²) [ 1- 2sine² (wt)]
    d w(x)/d t – (h/r²) = – 2(h/r²)sine²(wt) = – 2(h/r²)(v/c)² v/c=sine wt
    (h/ r²)(Perihelion/Periastron)= [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²
    Δ w/d t = (d w/d t – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second
    Δ w/d t = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² radians
    Δ w°/d t = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² degrees; Multiplication by 180/π
    Δ w°/d t = (-720×36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years
    Δ w”/d t = (-720×3600/T) {[√ (1-ε²)]/(1-ε) ²} (v/c) ² seconds of arc multiplication by 3600
    Δ w/d t = (-720x36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century
    The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4) v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<

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