Time Dilation when they meet

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $b<a$.

Answer 4

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).

Answer 5

Yours is a common misunderstanding and it arises because you assume, incorrectly, that time dilation involves the comparison of one moving clock with one other clock. Time dilation isn't symmetrical in that sense. The time dilation equation of SR applies where you have one clock moving inertially between two other clocks that are stationary relative to each other. In that equation, you have a time interval t' recorded by the one clock moving between the other two. You never compare that directly to a reading on a single clock in the other frame. Instead you compare it with the difference between the reading on one clock and the reading on another. That point is a crucial one, because it brings in the relativity of simultaneity, which means that in the frame of the moving clock the other two clocks are out of synch with each other, and that's what causes time dilation.

So, to go back to your set-up. Let's suppose the two spaceships are stationary some distance apart with synchronised clocks showing 12:00, and then they accelerate towards each other, meeting after thirty minutes has passed on their respective clocks. If each of their clocks shows that 30 minutes have passed, how can that be consistent with relativity and time dilation? The answer is that as soon as the ships started moving, time became out of synch between their respective travelling frames. If I was in one of the ships, I would say that I accelerated at 12:00 in my frame, and the other ship accelerated at 11:55 in my frame. That is, the other ship started its journey five minutes before I started mine-its clock shows thirty minutes have passed, but it actually has been travelling for thirty-five so it is time dilated. Likewise, if you were in the other ship, you would say that I started travelling at 11:55 in your frame, that I have been travelling for thirty five minutes and therefore I am time dilated. The effect arises because we each think the other took off earlier, and that is because what is simultaneous in one frame is not simultaneous in the other.