>>21
>The traveler always sees his own natural time.
This is true for any speed less than exactly the speed of light.
Because it does not hurt me exercise my english a little bit, I will explain it to you.
Fact n1: the speed of light is the same for everyone.
Fact n2: there is no experiment that you can do on a non-accelerating spaceship that will prove that you're moving (the thing that moves by outside the window could be moving and you're the one that is standing still).
Imagine a photon based clock, two mirrors amidst which a photon jumps up and down. If the mirrors are distant 1 meter then 150 milion jumps (up+down) equals 1 second. So when you're standing still the outside observer sees that the photon needs 1/150000000 of a second for a jump. But if you move the photon will travel diagonally for the outer observer, this means the photon has traveled a longer distance. That means it has spent more time (the observer's one) to do one second (the spaceship's).
I think you knew all that, now imagine that same clock on a spaceship traveling at the speed of light. If the photon would try to travel diagonally (jumping up and down for the ones on the spaceship) it would stay behind the mirrors, which would be against the fact n2 (dunno how it's actualy called :/). Therefore the personal time of the spaceship is not moving. The ones on it would not move in time until the spaceship would slow down.
t(personal) = t(observers's)*sqrt(1 - v^2/c^2)
if v=c
t(p) = t(o)*sqrt(1 - c^2/c^2)
t(p) = t(o)*sqrt(1 - 1)
t(p) = t(o)*sqrt(0)
t(p) = t(o)*0
t(p) = 0
>jump outside the universe
>Funny how I don't see photons doing this very often.
You got me there. Sorry for the mistake. :P