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When calculating the amount of time elapsed for an inertial reference frame over the course of its travel at constant velocity between two points, are the effects of both length contraction and time dilation taken into account (i.e. the distance seems shorter to the traveler and the time for the traveler seems to pass more slowly to a stationary observer), or is only one or the other effect used to calculate the amount of time elapsed in the inertial reference frame in motion?

I believe the answer is only one or the other, since the effects are really two sides of the same coin, one applying to an observer within the frame and the other applying to an observer outside the frame.

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    $\begingroup$ Which article is that? How about a proper citation and a reason why you think that it's false? $\endgroup$
    – CuriousOne
    Mar 21, 2016 at 4:07
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    $\begingroup$ Also I'm not sure but wouldn't it matter how fast you accelerate and what the overall speed (fast or slow) that you average. Just asking $\endgroup$ Mar 21, 2016 at 4:12
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    $\begingroup$ You haven't specified the problem until you say who measured the 6 ly. yrs. Presumably that is the person with respect to whom our observer is making 0.6c, but it is important to know ... $\endgroup$ Mar 21, 2016 at 4:13
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    $\begingroup$ Why do you want to compare with the article?! IMO, @lostinthecloud is perfectly right in asking without making you biased by stating the article $\endgroup$
    – Ilja
    Mar 21, 2016 at 7:08
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    $\begingroup$ Well, this for sure as heck was homework, but congrats, you got them to do it for you. :-) $\endgroup$
    – CuriousOne
    Mar 21, 2016 at 7:54

3 Answers 3

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Suppose the starting point is $A$ and the ending point is $B$. I will assume the question is stating that the distance between $A$ and $B$ in a frame where both $A$ and $B$ are stationary (we assume such a frame exists) is 6 light years.

Now, a more practical scientist would consider acceleration and deceleration, but as a theoretical physicist, I will do nothing of the sort (spherical cow!). Let us start at $A$ and instantly accelerate to $v = 0.6c$. In this new frame, the distance $AB$ is length contracted and the new length is

$$ L = \frac{ 6 }{ \gamma } ~ \text{light-years} = \frac{ 24 }{ 5 } ~ \text{light-years} $$ Further in this frame, the "rod" $AB$ is moving towards me at speed $v = 0.6c$. Then, then time taken for the end point $B$ to reach me is $$ t = \frac{24}{5\times0.6} ~\text{years} = 8~\text{years} $$ Once I reach $B$ I instantly decelerate and viola!

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  • $\begingroup$ Thank you very much for your response. Your answer agrees with the answer proposed in the article. For some reason (I must not have been thinking straight), when I was analyzing the solution in the article last night, I mistakenly thought their solution used both length contraction and time dilation together to shorten the time doubly (i.e. 80% the distance at 80% the time), as opposed to using one or the other. $\endgroup$ Mar 21, 2016 at 17:19
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For problems like this use the Lorentz transformations:

$$ t' = \gamma \left( t - \frac{vx}{c^2} \right ) $$

$$ x' = \gamma \left( x - vt \right) $$

In this case we'll choose our origin so the spaceship leaves at time zero so the starting point is $(t=0,x=0)$. We observe the spaceship to travel a distance $d$ at a speed $v$, so the time we observe it to take is $t=d/v$ and the ending point is therefore $(t=d/v, d)$.

The starting point $(0,0)$ is the same in both frames so we just need to find out where the point $(d/v, d)$ is in the spaceship frame. If we plug these values of $t$ and $x$ into the Lorentz transformations we get:

$$\begin{align} t' &= \gamma \left( \frac{d}{v} - \frac{vd}{c^2} \right ) \\ &= \gamma \frac{d}{v}\left( 1 - \frac{v^2}{c^2} \right ) \\ &= \frac{1}{\gamma}\frac{d}{v} \end{align}$$

$$ x' = \gamma \left( d - v\frac{d}{v} \right) = 0 $$

In this case we have d = 6 lyrs and v = 0.6$c$, and plugging these values into the equation for $t'$ gives:

$$ t' = 8 \text{years} $$

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The answer of John Rennie uses the Lorentz transformations, which is correct and most general, but you maybe expected contraction/dilatation-arguments?

Those are two special cases of the constancy of the spacetime-distance: you have contraction of spatial distances in a moving frame, and dilatation of time spans (you get this by using the special cases of $\Delta t = 0$ or $\Delta x = 0$ for the distance between the events).

Prahar uses the first argument of contracting the distance, you could equivalently argument the other way round: in the outside frame the voyage takes 10 years, but in the moving frame time runs not so fast, and you have to divide by $\gamma = 1.25$

Giving the same result of 8 years.

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  • $\begingroup$ I very, very strongly recommend that beginners to special relativity use the Lorentz transformations because they always work. It's very tempting to wave factors of $\gamma$ around when looking at problems like this, but that's a dangerous strategy as unless you know what you're doing it's very easily to make mistakes. Of course Ilja knows this, but that's because he's already spent several years studying relativity :-) $\endgroup$ Mar 21, 2016 at 8:16
  • $\begingroup$ yes. I didn't want to recommend my method, the answer was meant to give an overwiev, to explain the connection between the three seemingly different viewpoints. $\endgroup$
    – Ilja
    Mar 21, 2016 at 8:21

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