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If a spaceship travels close to the speed of light (say, at 0.9c), how do I calculate the time as the spaceship pilot experience it? I thought the formula was

$$t = \frac{t_0}{\sqrt{1-v^2/c^2}}$$

where $t_0$ is the time it would take if traveling at c. But when applying this formula to the speeds $0.9c$ and $0.99c$, the time of traveling at $0.99c$ is much higher. Shouldn't it be the other way around? Or am I confused?

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the pilot would not experiance any distinction between traveling at higher velocity or sitting on his couch at home only the observer would see the difference. Unless the pilot goes back to the departure point he would assume that his clock did not slow down "so to speak" –  Argus Jul 1 '12 at 18:33
    
@Argus Yes, I meant the pilot would experience the traveling going much quicker when he gets back. –  Quispiam Jul 1 '12 at 18:38
    
I think you might be seeing this as a wordage problem. The pilot would "use" more time from an observers point of view. That is as you go faster your age progresses more slowly but the amount of time that passes for the observer say if you went back goes faster. this is all perception of the individuals involved. this question might help physics.stackexchange.com/questions/4404/… –  Argus Jul 1 '12 at 18:40
    
Thanks to usumdelphini for editing my question - I'm still new to SE and didn't know I could use LaTeX. :) –  Quispiam Jul 2 '12 at 14:07
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3 Answers

up vote 6 down vote accepted

where t_0 is the time it would take if traveling at c

This is your problem.

$t_0$ by the pilot, and $t$ the time the trip takes as seen by an observer moving with respect to the pilot at velocity $v$. From the stay at home observers point of view the trip takes $t = d/v$, but the pilot experiences proper time $t_0 = t/\gamma < t$.

Unfortunately that leads to a bit of a problem because there is no unique definition of "at rest". Enter the "twin paradox" which has been on the site discussed before.

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+1 very true because "at rest" is always a "at rest Compared to" representation. Sitting in my computer room I am at rest compared to a person walking down the street but compared to the sun I am not at rest –  Argus Jul 1 '12 at 19:00
    
Ah, I see. So I would calculate how long it would take to travel for example 100 000 ly in 0.9c m/s, ignoring relativity, and use it as t_0 (i.e. what it looks like to the "twin" left on Earth)? –  Quispiam Jul 1 '12 at 19:07
    
I calculated t_0 to 111 109 years (sounds plausible, traveling 0.9c 100 000 ly). I calculated t = t_0/(sqrt(1 - (0.9c)^2/c^2) but wound up with 254 901 years, which seems really wrong. Shouldn't t < t_0? –  Quispiam Jul 1 '12 at 19:30
    
After an hour of working I realised I did everything right from the beginning, but the online test was unclear as to how the write the answer (ie I wrote 48432 years when I should have written 48000)... I calculated it with t = d/v = 100 000 / 0.9 ; t_0 = t * sqrt(1 - (v^2/c^2)) = t * sqrt(1 - ((0.9c)^2/(c^2)) Thanks a bunch for the help, dmckee! –  Quispiam Jul 1 '12 at 19:59
    
Another known result to consider, if you took two clocks on the ground, synchronized them, hauled one to the top of the tower and left it there for a while, and then hauled it down again, the two clocks would read different times. There is no rule which requires time to "close" in a round-trip. –  mwengler Jul 1 '12 at 20:39
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Just to expand a bit on dmckee's answer...

The proper time is a crucial concept in SR. In just about any situation in SR, you need a coordinate system that includes the spatial coordinates and a time coordinate $t$. The key word here is coordinate. In SR, time is a coordinate to be distinguished from a parameter.

In Newtonian mechanics, $t$ is parameter; the parameter of trajectories through space. In SR, trajectories become world lines; trajectories through spacetime. But world lines need a parameter too and, in SR, that parameter is the proper time, $\tau$.

The proper time is the elapsed time along the world line; it is, loosely speaking, the elapsed time as read by a clock on the world line (or the wristwatch of the pilot of the spaceship).

There is a relationship between the proper time along a world line and the coordinate time:

$\gamma \Delta\tau = \Delta t$

$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v^2}{c^2})}}$

Where $v$ is the uniform velocity of the object in the coordinate system with coordinate time $t$.

Now, see that the elapsed coordinate time, $\Delta t$, is the elapsed proper time $\Delta \tau$ of an observer at rest in that coordinate system. Read that again to make sure that makes sense to you.

One final thing, the proper time along a world line is invariant; it is the same in all inertial coordinate systems.

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Just to expand a bit on Alfred's answer ...

When you're trying to understand a system in special relativity it's dangerous to throw around factors of gamma and hope you get the correct answer. You need to sit down and do a calculation. This sounds a bit brutal, but the calculation is often simpler than you think. Let's do it for this case.

To make things concrete let's suppose the starship passes us at time zero, and we and the starship captain synchronise our clocks so time zero is when we pass. Let's also suppose the starship is travelling to some star that is a distance $d$ away from us.

In our frame we can identify two spacetime points. The point $(0, 0)$ is when the starship passes us, and the point $(d/v, d)$ is when the starship reaches the star. A quick explanation of that second point: we write points as $(t, x)$. When the starship is at the star $x = d$, and the time taken to reach the star is distance/speed or $d/v$, so the point the starship reaches the star is $(d/v, d)$.

Now let's consider the starship's frame. In this frame the starship is stationary, and by agreement with us on Earth the point the starship passes the earth is (0, 0). We need to work out at what point $(t', d')$ the starship sees the star pass. The value of $d'$ is easy because in the starship's frame it's not moving, so $d' = 0$. All that remains is to work out $t'$.

There are various ways of calculating $t'$, but my favourite is to use the fact that the proper time is an invarient in special relativity. Alfred mentioned the proper time in his answer, and it's given by:

$$ d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$

The key thing about the proper time is that it's an invarient, which means every observer in every inertial frame will measure the same value of $d\tau$. We can use this to calculate $t'$. In our case it's simplified because we are only considering motion in the $x$ direction so $dy$ and $dz$ are both zero.

So, in our frame we calculate $\tau$ to be:

$$ \tau^2 = c^2 \frac{d^2}{v^2} - d^2 $$

In the starship's frame $d'$ is zero so the proper time is simply:

$$ \tau'^2 = c^2t'^2 $$

Both observers must agree on the proper time so $\tau = \tau'$, and setting these equal gives us:

$$ c^2t'^2 = c^2 \frac{d^2}{v^2} - d^2 $$

and a quick simplification gives:

$$ t' = \sqrt{\frac{d^2}{v^2} - \frac{d^2}{c^2}} $$

It would be nice to have $t'$ in terms of $t$ rather than $d$, and we can do this simply by noting that $d = vt$, which gives (after a quick rearrangement):

$$ t' = t\sqrt{1 - \frac{v^2}{c^2}} $$

So the time starship captain measures to reach the star is shorter than the time we measure, as you expected. The calculation now shows you that your equation was wrong as you got the factor of $\sqrt{1 - v^2/c^2}$ in the wrong place.

Incidentally, we can also work out what distance the starship captain measures to the star. We and the captain both agree on our relative speed, $v$, so the captain can calculate the distance to the star using $d' = vt'$ and gets:

$$ d' = vt\sqrt{1 - \frac{v^2}{c^2}} $$

and of course $vt$ is just $d$ so:

$$ d' = d\sqrt{1 - \frac{v^2}{c^2}} $$

So the starship captain measures the distance to the star reduced by a factor of $\sqrt{1 - v^2/c^2}$ just like the time.

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