To keep the calculations simple, say that a traveler had a spaceship that could instantly accelerate to 1/2 of light speed (without killing him) and he started a journey to a point in space 10 light years away to finally instantly decelerate at the moment of arrival.

How many time would it pass from the point point of view of the traveler inside the spaceship?

Intuitive thinking may make one think it'd be 20 years but I know relativistic speeds don't work that way. What would be the equations to calculate that "t"?

For the given values it would be:


10 / ( Root( 1-( (1/2)^2 / 1) ) ) = 11,5470


11 years and a half?

From the traveler's perspective, the calculation is simple... Time = Distance / Speed... which is 10 / 0.5 = 20 years.

The time dilation effect occurs when you compare what an external observer sees versus what the traveler sees. In this example, more than 20 years will have passed for the rest of the universe. The reason for this is that the non-moving reference frame sees the traveler's clock running slow (the actual amount is given by the Lorentz factor as stated by Emergent). So 20 years for a slow running clock means more than 20 years will have passed for the rest of us.

However, no matter what the speed of travel is, the time taken for the journey from the traveler's point of view is the simple formula : Time = Distance / Speed.

Edit: The value of the Lorentz factor for the given setup is about 1.1547, so about 23.094 years will have passed. In your calculation you used the distance (10 light years) not the time (20 years).

no, wait I meant 10 years traveling at speed c (which still makes no sense I realise). The resulting value is the elapsed time for an observer outside the ship then.

[font="arial, verdana, tahoma, sans-serif"][s]no, wait I meant 10 years traveling at speed c (which still makes no sense I realise). The resulting value is the elapsed time for an observer outside the ship then[/s]

No I got that right. [/font]

Not sure about the formula Emergent provided though. As the traveler approaches the speed of light, the flow of time in his local frame decreases by which the object in which he is traveling, to him, appears to be moving faster than what it really is (covering more external distance in less time). So, his perception of the time passed since he started traveling will be inferior than the time that passes for an external observer.

I double checked this in Isaac Asimov's "Extraterrestrial Civilizations" non-fiction book (Chapter "The speed of light"). He mentions that at 293,800 km/s or 98% of the speed of light, time inside a spaceship will flow at 1/5 of the rate it would if it was at rest. He states then than for a distance of 10 light years at that speed, from the point of view of an external observer (at rest in relation to the ship, say the Earth) the travel would seem to last a little longer than 10 years but for the traveler on board the ship the travel would seem to have lasted just one week.

So, again, what's the formula with which I can calculate those numbers?

Well, if he travels at 0.5*c, then 10 lightyears means 20 years from the observers POV. But the equation should be dt'/gamma because the distance is measured in the observer's coordinate system (which is considered standing: we defined distances in Space like that).

You can verify it, because the time from the traveler's CS should be less than the time in the observers CS, because we know (from sci-fies....) if the traveler would travel with c his time would be zero. (if you substitute v = 0, dt'/gamma will be zero as expected.)

so 17.32 years is the correct answer.

Another verification:

v/c = 0.98, dt' = 1 ->

dt = 0.198997 -> 1/5, see Asimov's

EDIT: edited dt' dt confusion. Emergent's formula is perfect, it's just we are not calculating dt', but dt.

Thanks, szecs. Could you write the formula for me in pseudo-code so I can see it better? Just pretend I'm scientific-notation blind

[font="Courier New"]20 * ( Root( 1-( (1/2)^2 / 1) ) ) = 17.32[/font]

or

[font="Courier New"]time_for_traveler = time_for_observer * sqroot(1-(v/c)^2)[/font]

just pretending I'm not having sense of humor (which I'm really not having..)