We have a triangle.
On one side we've got a walking Einstein, and perpendicular to him is time. He walks away from his home that is still.

so the time moves in terms ct = s, ie. t = s/c

and the vt/c is a lost time in the same proportion and dimension as time itself that Einstein have, since the photons in his pocket still moves in the speed of light, as compared to the home.

So since they are in the same dimension, I get: (s/c)²(home time) - (vt/c)²(Einsteins additional time) = (s/c)²(Einsteins time)

So: (t)²(home time) - (vt/c)²(Einsteins lost time) = (t)²(Einsteins time)

But when he goes home, his pocketphotons moves in the speed of light compared to the closing home. The cathet turn positive with einsteins distance closing in. We have:

(t)²(home time) - (vt/c)²(Einsteins lost time) = (t)²(Einsteins time)

Both these equations can be simplified into:

t²(home time) - (distance/c)²(Einsteins lost time) = (t)²(Einsteins time)

As you can see, the time is squared, so we draw a root:

√(t²(home time) - (distance/c)²(Einsteins lost time)) = (t)(Einsteins time)

So obviously, when 2 people move away from eachother, they both get a shrink in time according to that equation.

So what we see in space, the time distance is about 0.

Last edited by LQ (2010-06-14 05:09:34)