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Let's say there are 2 points X & Y and a spaceship is moving from X to Y at some very high speed. According to X & Y, let's say the spaceship takes 800 seconds to go from X to Y. According to the spaceship, points X & Y move past, and since there is length contraction, the journey from X to Y takes less time (let's just say 400 seconds).

Here is where I get confused. In X & Y's frame, they are recording dilated time relative to the time the spaceship records in its own frame because the spaceship is moving BUT in the spaceship's frame, X & Y are moving so if we follow the previous logic, the spaceship is now recording dilated time relative to the time X & Y record in their own frame. SO, X & Y should then be recording a time for the journey that is LESS than 400 seconds. But I know this is wrong. My question is how?

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The problem can be described in two different frames and the results in both frames are consistent. The first frame is the rest frame for the two points X and Y (let's call it $\Sigma_{XY}$). The second frame is the rest frame of the spaceship (let's call it $\Sigma^\prime_{S}$).

In $\Sigma_{XY}$:

for concreteness, let the relative speed between the two frames be $v=\frac{\sqrt{3}}{2} c\approx 0.866 c$ and the distance between $X$ and $Y$ be $s_{XY}=800\cdot \frac{2}{\sqrt{3}} $ lightseconds $\approx 693 cs$ (note that this is the proper distance, since $X$ and $Y$ are at rest in $\Sigma_{XY}$). Consequently, in this frame the journey of the spaceship takes $t=\frac{s_{XY}}{v}=800 s$.

In $\Sigma^\prime_{S}$:

The distance between $X$ and $Y$ is contracted to $s^\prime_{XY}=s_{XY}\cdot \sqrt{1-\frac{v^2}{c^2}}=\frac{1}{2}s_{XY}\approx 346.5 cs$. Therefore it takes $t^\prime = \frac{s^\prime_{XY}}{v}=400s$ for the distance $s^\prime_{XY}$ to move by the spaceship. This is the proper time of a person inside the spaceship. The person in the spaceship also knows, that a person in $\Sigma_{XY}$ will measure a dilated time $t=\frac{t^\prime}{\sqrt{1-\frac{v^2}{c^2}}}=800 s$.

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  • $\begingroup$ In Σ′𝑆 though, Σ𝑋𝑌 is moving, and in a moving frame, time runs slower, so Σ′𝑆 could argue Σ𝑋𝑌 has proper time (shorter time). What exactly makes this argument wrong b/c I know it's wrong. $\endgroup$ – user532874 Apr 19 at 10:02
  • $\begingroup$ The clean way to talk this: There are two events (A: spaceship at X, B: spaceship at Y). These two events get coordinates in $\Sigma$: $A=(0|0)$ and $B=(693|800)$ and $\Sigma^\prime$: $A=(0|0)$ and $B=(0|400)$. The coordinates in both frames are related by Lorentz transformation (the only non-obvious transformation $t^\prime_B=(800-\frac{v}{c^2}\cdot 693)\cdot \gamma = 400$). If you wish to go the other way round, just flip the sign infront of $v$. $\endgroup$ – p6majo Apr 19 at 10:55
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    $\begingroup$ @user532874 - you should not take one observer's perception of time and apply it to another observer's perception of distance if the two observers are moving with respect to each other. There is no absolute measure of how long it takes the spaceship to go from A to B. There is A's sense of how long it takes (using A's observations of space and time) and there is the spaceship's sense of how long it takes (using the spaceship's sense of space and time. But mixing one's sense of time with the other's sense of space is not something any observer would measure. $\endgroup$ – Paul Young Nov 19 at 19:18
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This is a common cause of confusion. It arises because people consider the effects of time dilation while forgetting to take into account the relativity of simultaneity. When you take both into account, you will see that the effects together are entirely reciprocal across the two reference frames, so there is no conflict.

To see this, you need to set-up the scenario is a way that is properly symmetric across the two reference frames.

Note that your original version isn't symmetric, because in one frame you have a single observer looking at the elapsed time locally to the spaceship, while in the other frame you have two people, one at X and one at Y, comparing the times on their clocks from a distance.

To make the scenario symmetric, imagine a second spaceship trailing behind the first at a distance equal to that between X and Y. So now you have two observers in each frame, and their respective separations are identical. Suppose that the clocks on the spaceships are synchronised with each other, as are the clocks at X and Y. To make life simple, let's suppose the leading spaceship's clock and the clock at X both read zero when the spaceship leaves X.

When the leading ship later arrives at Y, the pilot again checks the times, and finds that 400 seconds have elapsed on the ship's clock but the local time is 800 seconds. That seems asymmetric, however...

Meanwhile, the lagging ship arrives at X and compares times. According to the lagging ship the time is 800 seconds, but the ground station clock reads only 400 seconds.

You will see now that the effects are entirely symmetric, but not perhaps in the simple way you expected. From the perspective of the ground crew, the leading ship has moved between the two stations in a time of 800s in their frame but only a time of 400s on the ship. From the perspective of the ships, station X has moved from the leading ship to the lagging one in 800s in their frame of reference, but only a time of 400s at the station.

The reason why the differences arise is that the clock on the lagging ship and the clock at Y are not in synch with each other when the leading ship leaves X, so there is a disagreement about when the journey actually started.

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If you want not to be confused, always define events in spacetime and deal with them. First of all, let's analyze your problem by math, then we can give it physical intuition as well. Here there are 2 observers, one has stayed in x, let's call it frame X, and the other is moving via spaceship, this is frame S. There are 2 events in spacetime, the first one happens when spaceship starts its departure, and the second one is when spaceship arrives at y.

For the first event, according to frame X we have: $E_1=(ct_1,r)=(0,0)$.

And as for second event: $E_2=(ct_2,L)$ where $t_2=800$ and $L$ is the distance between x and y, according to the problem.

Now in frame S for the first event we will have: $E'_1=(0,0)$ this goes without saying, because X and S are at the same location at that moment, and they can prepare their clocks such that $t_1=t'_1=0$ . Of course, using Lorentz transformations is fine too.

For second event however, we have to use Lorentz transformations. $t'=\gamma(t-vr/c^2)$ and $r'=\gamma(r-vt)$ this leaves us with: $E'_2=(c\gamma(t_2-vL/c^2),\gamma(L-vt_2))$ but we already know that $vt_2=L$ because $t_2$ is the time of spaceship's arrival in y. So $E'_2=(c\gamma(t_2-v^2t_2/c^2),0)$. This makes sense, because spaceship and second event are in the same location at that moment. Now back to the problem, from above $t'_2=t_2\gamma(1-v^2/c^2)$ and by simplifying it we would have $t'_2=t_2/\gamma$ which as you said implies that $t'_2<t_2$ (Just choose $\gamma=2$ and we have $t'_2=400$)What if we use inverse Lorentz transformation, to see that whether still $t'_2<t_2$ is true?

$t=\gamma(t'+vr'/c^2)$, and from $E'_2$ we know that $r'_2=0$ so $t=\gamma(t_2/\gamma+0)=t_2$! It means $t_2=800$ nevertheless. What is that supposed to mean? According to S, X is the moving observer, why the measured time in X is not less than 400?! Well you see X has stayed in x, that means while event $E_2$ is happening at y, X won't observer it immediately, there is distance $L$ between him and that event after all! You see time interval is not only depend on the relative speed between observers, it's also related to position of events. On the other hand, S and $E'2$ are at the same location, Of course this observer will see this event right away. Hence the measured time interval between these two events at S has to be less than X. If you look closely to the math itself, it's because of $r_2=L$ that makes $t'=t_2/\gamma$ in the end. I don't know whether i answered your question clear enough, but i gave you an upvote because your question is good.

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  • $\begingroup$ Oh I didn't factor in the position of events. In one frame the events are at the same position while in the other they are not. Why is it called special "relativity" then? One frame always experiences less time in a given situation, so it seems to me that time is absolute. Maybe I'm confusing readings of time vs notions of time because both observers still think the other's clock is running slow even though in the end, one observer will record less time elapsed for a given situation. $\endgroup$ – user532874 Apr 19 at 16:59
  • $\begingroup$ It's called "relativity" because both of 800 "and" 400 are true at the same time! In Newtonian mechanic However, they would measure the same time interval, 400 "or" 800, not both. The thing is, it doesn't matter how much a clock become slow in one frame (in viewpoint of other observers), in the end, what matter most is that "when" that observer stops his clock. X's clock becomes slow in the spaceship's point of view, sure but that doesn't imply that X won't stop his clock much more later than spaceship you see. What is confusing here? $\endgroup$ – Paradoxy Apr 19 at 18:56
  • $\begingroup$ Even when you are dealing with twin paradox, you can see that both observers has to meet each other at some certain location in the end. Why is that? because this will eliminate the effect of position, both start at the same location and being ended up at the same location again. Which leaves only effect of velocity/acceleration in time dilation. $\endgroup$ – Paradoxy Apr 19 at 19:09
  • $\begingroup$ @user532874 Consider two events taking place on the spaceship. They have a certain distance in time, measured on the spaceship, while an observer moving with X will measure a dilated time. Now consider two events taking place at X. This time it will be the spaceship the one that measures a dilated time. These two situations are perfectly simmetric: once that two (time-like separated) events are chosen, there is always a frame in which they take place in the same place, but no frame works for every couple of events. There is no "preferential" frame that is different from the others. $\endgroup$ – HicHaecHoc Nov 19 at 22:18

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