Shayne T said:
From what I've gathered, this would be true for any outside observer who were not traveling at near light speeds. I guess a better way to put it, would be how close to the speed of light would you need to go in order to achieve a time dilation factor of 94billion:80 (traveller experiences time at a decelerated rate of 1.175 billion times less than a stationary observer left home at earth)
Formally - you have proper distance d in frame S, and you want observer in frame S' traveling with relative speed v (wrt S) to traverse that distance in S in coordinate time t' or less. The time in S to traverse d is going to be ##t=d/v = \gamma t'## so solve for v.
Regarding your comment on the frame of a photon, I've always had trouble wrapping my brain around the following. If special relativity states that time approaches 0 as velocity approaches c, and photons are the only thing able to actually reach c, do photons not experience time?
This question is meaningless in special relativity.
And if something that is moving, which does not experience time, wouldn't it, from its own frame of reference, instantaneously exist everywhere in the universe?
Everything is stationary in their own reference frame. So to which observer do you think something could be considered "everywhere in the Universe"? Surely all observers agree that the photon exists in only one place at a time?
... consider that the length contraction in this limit is 100%, so the "photon" does not exist everywhere in space, but everywhere in the "direction of travel" is the same place - so there is only one place for the photon to be.
But like I said before - the question is actually meaningless. The confusion above is due to this.
To see why, try constructing the usually time-dilation derivation using light-clocks but putting v=c in at the start ... then add the postulate that all observers measure the same value for c. You should see a logical contradiction pretty much right away. Therefore, the equation you were taking the limit in before cannot even be constructed.
You can get at the contradiction much quicker by considering the reducto ad absurdum: pre-suppose an observer that is stationary in the frame of a photon, the photon has v=0 in this frame. Now apply Einsteins postulates to bring the physics into SR: all frames/observers measure the same speed for the photon.
Then we note that not all observers measure v=0 for the photon ... which is a logical contradiction. Either the initial presupposition is untrue or special relativity is untrue.
It follows that a photon cannot be an observer.
There's probably a more rigorous way to put this but you get the idea.
