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I am starting over because what I was asking was unclear. I have read many articles, such as How can time dilation be symmetric? and others suggested by people here and understand the overall concepts except for one tiny detail. Perhaps by starting over with concrete numbers my confusion (and the answer) will be clearer.

Two spaceships synchronize clocks, depart at noon and meet up at a destination. For the first spaceship it appears they have traveled for 30 minutes and because of time dilation the second ship has traveled for 25 minutes. From the first spaceship's point of view they meet at exactly 12:30.

On the second spaceship, their clock shows they have traveled 30 minutes and ship one's clock says 25 minutes.

My question is, what time do the two clocks say is the time when they meet and how does the disparity in their perceptions resolve itself? Does the 5 minutes "disappear" and the two clocks snap into place a millisecond before meeting? From my viewpoint on a spaceship what does this disparity in perceptions actually look like?

Edit: As I understand the symmetry of time dilation, the direction of travel doesn't matter. So in my earlier question I had a single ship moving between two synchronized stationary clocks and the ship would perceive both clocks as slowing down while it was in motion. (Correct me if I'm wrong)

Extrapolating that analogy, route doesn't matter. Both ships could take a semicircular route moving at the same velocity and arrive at the other side of the circle at the same time.

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    $\begingroup$ You do not get to setup a scenario that is openly in contradiction with itself like that. You have not stated a scenario in a way that we can even attempt to salvage it. $\endgroup$ Commented May 2 at 20:27
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    $\begingroup$ "Meet up at a destination" What do you mean exactly? teh faster ship arrives first, the slower later, with or without relativity. Give a clear description of your scenario! $\endgroup$
    – trula
    Commented May 2 at 20:28
  • $\begingroup$ The "symmetric" case is when you and the other ship pass each other, and you observe through your magic telescope that their clock is ticking more slowly than your own, and they observe that your clock is ticking more slowly, and then you keep going on your way, and you never meet up with them again. There's no paradox to be resolved there. It just is what it is. But, if you eventually do meet up again, and compare how much time has elapsed on each of your clocks, then that's a variation on the "twin paradox." $\endgroup$ Commented May 2 at 23:27
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    $\begingroup$ The relativity of simultaneity says that if a set of (spatially separated) clocks are in sync in some frame, then they must be out of sync in any frame that's moving relative to the first frame. Wikipedia has a nice animated diagram, which I used in this answer. $\endgroup$
    – PM 2Ring
    Commented May 3 at 2:18

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Two spaceships synchronize clocks, depart at noon and meet up at a destination. For the first spaceship it appears they have traveled for 30 minutes and because of time dilation the second ship has traveled for 25 minutes. From the first spaceship's point of view they meet at exactly 12:30.

On the second spaceship, their clock shows they have traveled 30 minutes and ship one has flown for 25 minutes.

This is a very confusing description. I gather that you are intending to make the scenario completely symmetric. With that assumption then (modulo a shift of the origin) I believe the only scenario which fulfils your criteria is the following:

I will use minutes for time and light-minutes for distance so $c=1$. I will express spacetime coordinates as $(t,x)$ in flat spacetime. There are three reference frames of interest.

$F_0$ is the inertial frame where the ships are initially at rest so that they can synchronize their clocks. In this frame ship 1 is initially located at $x=-9.95$ and ship 2 is initially located at $x=9.95$. At noon both ships clocks read $t=0$ and the ships accelerate instantaneously to $v=\pm 0.302$. Then both ships will arrive at $x=0$ at $t=31.5$. In this frame both ships are time dilated by a factor of $\gamma = 1.05$ so the clocks on both ships read $30$. There are three important events: the departure of ship 1 $D_1=(0,-9.95)$, the departure of ship 2 $D_2=(0,9.95)$ and the arrival of both ships to the middle $A=(31.5,0)$

$F_1$ is the inertial frame where ship 1 is at rest after it accelerates. Note, this frame is not the rest frame of ship 1 because ship 1 is non-inertial and the frame is inertial. Nevertheless, after the acceleration it is reasonable to use this frame as the perspective of ship 1. Applying the Lorentz transform to the three events we find that in $F_1$ the coordinates are $D_1=(3,-9.95)$ for the departure of ship 1, $D_2=(-3,9.95)$ for the departure of ship 2, and $A=(33,-9.95)$ for the arrival of both ships. In this frame ship 1 is not time dilated, the journey takes 30 minutes and ship 1's clock records 30 minutes. However, in this frame ship 2 started 6 minutes early. Ship 2 is traveling at $v=-0.553$ and is time dilated by a factor of $\gamma=1.2$ The trip takes ship 2 36 minutes and ship 2's clock records 30 minutes.

$F_2$ is the inertial frame where ship 2 is at rest after it accelerates. Applying the Lorentz transform to the three events we find that in $F_1$ the coordinates are $D_1=(-3,-9.95)$ for the departure of ship 1, $D_2=(3,9.95)$ for the departure of ship 2, and $A=(33,9.95)$ for the arrival of both ships. In this frame ship 2 is not time dilated, the journey takes 30 minutes and ship 2's clock records 30 minutes. However, in this frame ship 1 started 6 minutes early. Ship 1 is traveling at $v=0.553$ and is time dilated by a factor of $\gamma=1.2$ The trip takes ship 1 36 minutes and ship 1's clock records 30 minutes.

So the key is the relativity of simultaneity. Their clocks are synchronized in $F_0$ and they start at the same time in $F_0$. Because of the relativity of simultaneity in $F_1$ ship 2's clock is 6 minutes ahead so it starts 6 minutes before ship 1. Similarly, in $F_2$ ship 1's clock is 6 minutes ahead so it starts 6 minutes before ship 2.

Note, if I did not assume correctly that you wanted a completely symmetric setup then I apologize in advance. If you did want a completely symmetric setup then this is the only one that fits your description and still addresses time dilation.

Both ships could take a semicircular route moving at the same velocity and arrive at the other side of the circle at the same time.

This will not help you understand time dilation because in this case neither ship is inertial at any time so neither ships perspective can be represented with an inertial frame over any period.

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Let me try to interpret your question in the way that I think you intend.

Two distant spaceships (at rest with one another) synchronize their clocks. They both take off at the same speed, toward each other, and pass each other when both their clocks say 12:30.

Alice, the captain of one ship, says this: Bob's clock is ticking at (say) half the normal speed. It currently (correctly) says 12:30, so it must have said 12:00 a full hour ago, at 11:30. I know Bob took off when his clock said 12:00, so 11:30 is when he took off.

In short (according to Alice), Bob left at 11:30 with his clock saying 12:00, then traveled for an hour until his (slow) clock said 12:30. Alice herself left at 12:00 with her clock saying 12:00, then traveled for half an hour until her (accurate) clock said 12:30. Therefore both clocks said 12:30 when they met.

Bob tells another version of the same story, with his own role and Alice's reversed.

Now you might object that Alice agreed at the outset that she and Bob both took off at noon, and therefore shouldn't be allowed to change her mind and say he took of at 11:30. That is like saying that if you stand in Chicago facing north and say that Los Angeles is to your left, then you must continue to say that Los Angeles is to your left even after you turn around. We are allowed to change our perspectives, and consequently to change the way we describe the world. Accelerating in spacetime is the exact analogue of turning around in space.

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For the first spaceship it appears they have traveled for 30 minutes and because of time dilation the second ship has traveled for 25 minutes. From the first spaceship's point of view they meet at exactly 12:30.

On the second spaceship, their clock shows they have traveled 30 minutes and ship one has flown for 25 minutes.

That can't happen. The elapsed times shown on the ships' clocks are measurable physical quantities, and everyone will agree on what they are. Depending on the motion of the ships, the elapsed times may be equal, or one of them may be larger than the other, but they can't both be larger than the other.

The elapsed time on a clock is just the integral of $\sqrt{1-v(t)^2/c^2}\,dt$ computed in any inertial reference frame. The result will be the same no matter which frame you pick, because inertial frames are all equivalent.

When people say that time dilation is symmetric, they're comparing apples and oranges. "The primed frame is dilated relative to the unprimed frame" means that the difference between two readings of one clock of the primed system is smaller than the difference between readings of two different clocks of the unprimed system. Switch "primed" and "unprimed" in that sentence and you'll see that the "reciprocal" time dilation is actually a comparison of a different pair of physical quantities. Relativity never says that $a<b$ and $b<a$.

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Two spaceships synchronize clocks, depart at noon and meet up at a destination. For the first spaceship it appears they have travelled for 30 minutes and because of time dilation the second ship has travelled for 25 minutes.

is contradicted by:

Both ships could take a semicircular route moving at the same velocity and arrive at the other side of the circle at the same time.

If they both travel the same distance at the same velocity in a given reference frame they will experience the same time dilation and same elapse of time on their own clocks (proper time).

Let's alter the scenario slightly so that it is not self contradictory and highlight what happens so that you can get a clear idea of what is going on.

Lets imagine they both start at the same place and time and each travel in a full circle so that they return both return to the starting place simultaneously. The ship that records 25 minutes on its onboard clock (A) will have to travel at a higher velocity and in a larger diameter circle than the ship that records 30 minutes on its onboard clock (B).

When they return, they will find that a clock left at the starting point (C) will indicate an elapsed time that is greater than either of their onboards clocks. e.g. if the two ships departed at 12 noon, clock C could show 1 PM when they both return, while clock A would show 12:25 PM and clock B would show 12:30 PM. Clock A shows the least elapsed time because it was travelling the fastest (so it had greater time dilation), while clock C shows the greatest elapsed time because it was at rest the whole time.

If both ships A and B left simultaneously and travelled the same distance at the same velocity and return to the starting point simultaneously, then their onboard clocks would both show the same elapsed time when they return, but both the onboard clocks would still show less elapsed time than the clock that remained at the starting point.

Just for clarity, the fact that the paths are circular so that the observers inside the ships are not inertial observers and experience proper centripetal acceleration makes no difference to the elapsed proper times, because time dilation is independent of acceleration and direction. See my old answer to a related question that gives 5 counter examples to the idea that acceleration causes time dilation (as claimed by Sabine Hossenfelder on YouTube).

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  • $\begingroup$ said “the fact that the paths are circular so that the observers inside the ships are not inertial observers and experience proper centripetal acceleration makes no difference”. It makes no difference when calculating time dilation with respect to an inertial frame. It makes a huge difference when speaking of the ship’s perspective. The ship’s perspective is part of this question $\endgroup$
    – Dale
    Commented May 3 at 12:47
  • $\begingroup$ @Dale Part of the OP's confusion seems to stem from thinking of time in Newtonian terms as is illustrated by the statement "Does the 5 minutes "disappear" and the two clocks snap into place a millisecond before meeting? Clearly they expect the clocks to show the same time elapsed when they return to simultaneously to the same location, which would be true in Newtonian terms whatever their relative velocities. In Newtonian terms they would see each others clocks tick at different rates when they have relative motion if they looked at each other's clocks with telescopes. due to the ... $\endgroup$
    – KDP
    Commented May 3 at 13:44
  • $\begingroup$ ... Doppler effect, but when they both return to base, they show the same elapsed time and the apparent perception of different clock rates is revealed to be an optical illusion in the Newtonian case. However, the time dilation in SR is a real difference in clock rates of clocks with relative motion and nothing to do with perceptions or light travel times. I was trying to nail down the basic indisputable facts and one of those facts is that the clocks can show different elapsed times on meeting again in Relativity. $\endgroup$
    – KDP
    Commented May 3 at 13:47
  • $\begingroup$ I just disagree with your “makes no difference” statement. It is too broad as written. $\endgroup$
    – Dale
    Commented May 3 at 13:55
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    $\begingroup$ @Dale "just disagree with your “makes no difference” statement. It is too broad as written. – Dale". Fair comment. I will change it to "makes no difference to the elapsed proper times." $\endgroup$
    – KDP
    Commented May 3 at 13:58

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