# Time dilation in accelerating reference frames

1. Jun 1, 2015

### rede96

Sorry for the lame question, but I was wondering if someone could help answer the following.

I have two synchronised clocks which I place on two different space ships which then accelerate away from me at the same rate and time until they reach a given speed.

Ship A then slows down to be at rest again wrt to me again in time t and stops his clock. Ship b slows down until he is at rest wrt to me again but in time 2t (twice as long) and then stops his clock.

If they then both return their clocks to me would the same time have passed on both clocks?

2. Jun 1, 2015

### Staff: Mentor

Rate and time as seen by you, or as seen by the crews of the ships?

Same question.

3. Jun 2, 2015

### Staff: Mentor

I've removed several posts that were just serving to further confuse a poorly framed question. Let's try this again, starting from the top....

@rede96
Try posting your question again, but be careful to specify exactly who is measuring each time interval. I know you think it doesn't matter... but it does, which is why PeterDonis has been asking you for this clarification.

4. Jun 2, 2015

### rede96

Ok, I'll give it go.

Let say that Ship A and Ship B have the same set of instructions to accelerate away from me at 9.8 m/s2 for say 4 hours according to their clocks. They have synchronised clocks that read 12pm when they start.

Then after the 4 hours Ship A is asked to slow down at the same rate which would mean he would have stopped to be at rest with respect to me in 4 hours as measured by his clock.

Ship B was also asked to slow down also after 4 hours but at a rate that would take 8 hours as measured by his clock.before he was at rest with respect to me.

When each ship has finally come to rest wrt me, they stop their clocks so no further time is registered, They then return to me in their own time.

The question is if I was measuring both journeys by my clock (until I saw them stop again wrt to me) then who's clock would show more time dilation?

Last edited: Jun 2, 2015
5. Jun 2, 2015

### Staff: Mentor

So A and B both show 4 hours elapsed for this portion of the trip.

So A shows 4 hours elapsed for this portion of the trip for a total of 8 hours.

So B shows 8 hours elapsed for this portion of the trip, for a total of 12 hours.

Which means that A will show 8 hours elapsed and B will show 12 hours elapsed, because that's how you specified the problem. I think your specification is missing something.

6. Jun 2, 2015

### rede96

Sorry, I posted the wrong reply, which I have just had to edit. Does it make more sense now?

7. Jun 2, 2015

### Staff: Mentor

What you see or measure doesn't matter; your problem specification already stipulates that A's clock stops after 8 hours and B's clock stops after 12 hours. There's no variable at all.

If you want to make your question meaningful, you need to change your specification of when at least one clock, A's or B's, stops. (I would recommend A's, but you'll need to think carefully about how you change the specification.)

8. Jun 2, 2015

### pervect

Staff Emeritus
We need to make two fairly modest additional assumptions to answer the question. One is that we compare the clocks after they have all come to rest using the common frame in which they are all at rest. I'll call this frame the symmetry frame.

The other assumption is less clear, but I'm not sure it makes a difference to the answer. One interpretation of the question is that the rockets come to rest in a proper time of t and 2t, respectively, where proper time is the time as measured by clocks carried on the rockets themselves. The second interpretation is that the clocks come to rest in a coordinate times t and 2t as measured in the symmetry frame as I've defined it.

The first interpretation was what I thought when I initially read the problem, however, knowing the lack of popularity of proper time with non-physicist posters I wouldn't be terribly surprised if the OP had in mind the second interpretation.

In both cases I believe the answer should be 'no'. The argument is clearest if we assume that the proper time is t and 2t respectively. By symmetry, the proper time to accelerate up to "v" is the same for both clocks. We add to this the proper time to deaccelerate, which was defined as unequal. We are asked to compare the proper time of each clock, which is the sum of the proper time required to reach "v" and the proper time required to deaccelerate.

The second case is harder to analyze in detail, but I believe it is true that if you accelrate at a lower rate from a given velocity, it will always take you more proper time to come to a stop.

[add] That's how I read the question, it seems that Peter thought that because the answer was trivial there was some misunderstanding of what the question was - I wish I could say he was wrong, but perhaps there is still some confusion over what the question was?

9. Jun 2, 2015

### rede96

Sorry for the poorly worded post. What I was thinking is that if I measured Ship A's 8 hour journey by my clock (until he stopped) and also measured Ship B's 12 hour journey by clock (again until he stopped), which journey would I measure to have taken longer.

10. Jun 2, 2015

### rede96

Yes, that is what I was trying to explain.

I think this is what I was trying to understand. I was sort of thinking of the equivalence principle and if the time dilation was the same for someone who spends 4 hours in a gravitational force of 9.8 m/s2 (wrt to me elsewhere) as it is for someone else who would spend 8 hours in a gravitational force of 4.9 m/s2

11. Jun 2, 2015

### pervect

Staff Emeritus
If you only use inertial frames (which I'd strongly recommend), you don't need to worry about the equivalence principle at all. The idea is to understand special relativity in inertial frames first, for which you don't need to concern yourself with the equivalence principle yet.

The twin paradox is not any more puzzling than the "triangle paradox" when the correct viewpoint is used. Adopting this viewpoint mostly consists of forgetting the notion of universal time, which seems to be difficult to get people to do.

The "triangle paradox" (which is usually not regarded as a paraodox, but rather as the triangle inequalitiy), says that if you go from A to B directly along a straight line and measure your distance with an odometer, you always get the lowest reading, while if you go from A to C and then to B, taking two legs of a traingle rather than a direct route, you will always get an equal or longer distance on your odometer, with the "equal" condition applying only when your triangle is degenerate.

The similarity with the twin paradox is quite marked, we replace "odometer" reading with a "proper time reading", we measure the proper time along a path with a clock that we carry along with us, much like we'd do with an odometer (but simpler). If you go from A to B directly, you wind up with the longest proper time, while if you go from A to C to B, you'll always get a shorter or equal proper time. There is a tricky sign reversal here, the straight line path is the longest proper time path, while it's the shortest distance path.

A key issue, as always, is distinguishing proper time, which is what a clock measures, from coordinate time. Coordinate time should ideally be viewed as a tool without any direct physical significance.

Another key issue is that straight-line motion is an absolute, and distinguishable experimentally from not-in-a-straight line motion.

Note that we don't need to introduce the concept of "at the same time" or "at the same height" at all when we solve either the twin paradox or the triangle paradox. We can introduce such notions, but it serves to make the discussion longer and more complex. There may be a small increase in understanding when we introduce these notions, but a lot of extra work. Being lazy, it's work I'd rather avoid, though I wouldn't discourage anyone from thinking about the similarities in depth.

12. Jun 2, 2015

### A.T.

Gravitational time dilation is a function of potential difference, not of the local gravitational acceleration. Same for clocks in "artificial" gravity of an accelerated frame.

13. Jun 2, 2015

### Staff: Mentor

That's obvious: Ship B's journey would take longer. Up until the point where both ships start to decelerate relative to you, their motions relative to you are identical, by construction. Once they start decelerating, their journeys are not identical, but it should be evident that A is decelerating more, relative to you, than B is, so he will come to a stop sooner relative to you.

14. Jun 2, 2015

### Maxila

Can you clarify this please. Ignore the case of a degenerate triangle and assume an inertial frame, I'm reading that as traveling a longer path (A to B to C), resulting in a lower proper time? If the clock is on board and the frame is inertial I don't see how that is possible?

15. Jun 2, 2015

### Staff: Mentor

No; the path that follows two sides of the triangle is shorter. Remember that the geometry of spacetime is Minkowskian, not Euclidean; there is a minus sign in the analogue of the Pythagorean theorem when the sides of the triangles are timelike segments.

16. Jun 2, 2015

### ash64449

Is the concept of 'straight line' defined in Minkowski space-time like this:
It is the largest interval between two points(events) of space-time where elapsed proper time is the largest?

EDIT: it sounded to me like this because any other way by which an observer is able to register at these two events in space-time like a curve in Minkowski is small and has smallest elapsed proper time.

Only an observer whose path is along the straight line in space-time registers latest possible elapsed proper time and in Minkowski would become a largest distance between the two points.

Last edited: Jun 2, 2015
17. Jun 3, 2015

### Staff: Mentor

This is one way of defining it, yes.

18. Jun 3, 2015

### Maxila

What I still can’t reconcile is we were discussing a single clock “proper time”, time measured by clock in an inertial frame. We can use the proper time of the clock on board the ship, or the proper time of an outside observer at rest relative to the ship; however we are talking about the clock measurement from a single inertial frame, aren’t we?

As far as I understand it, the minus sign in Minkowskian space shows distance is invariant for both observers; however when on board the ship if I were to travel the coordinates A to C to B of a triangle rather than from A to B ( at the same rate) my clock would show more time when traveling the three coordinates of a larger distance. The same is true of the clock for the observer at rest. What am I overlooking in your earlier statement that a single proper time (clock time) on the ship, or observer at rest, would be measured smaller going from A to B to C?

19. Jun 3, 2015

### Staff: Mentor

Proper time is the time measured by any clock; the clock does not have to be inertial. The key point is that the proper time measured by a clock depends on the path it takes through spacetime. Different clocks following different paths through spacetime will, in general, measure different proper times.

In the particular case under discussion, all three sides of the triangle are segments of inertial worldlines, so calculating the proper time elapsed along each is easy since in any inertial frame, the speed of a clock following each worldline is constant. In the more general case, where the clock can be accelerating, you have to do integrals to obtain the proper time.

It shows that you have to subtract squared spatial distance from squared time. In Euclidean space, you add squared x distance (the analogue of "distance" in spacetime) and squared y distance (the analogue of "time" in spacetime).

If by "distance" you mean "distance in spacetime" (usually called the spacetime interval to avoid confusion), i.e., squared time minus squared spatial distance, then yes, that is invariant; but what is invariant is the spacetime interval along a specific worldline. If two observers are following different worldlines, the spacetime interval along their worldlines (which is measured by their clocks as proper time) can be different.

This is not possible. Remember that we are talking about paths through spacetime, not space. Traveling from A to B means simply floating in free fall, never doing anything to change your state of motion. Traveling from A to C to B means being in free fall, but in relative motion compared to the ship traveling from A to B; then, at some point, firing your rockets to reverse your direction of motion, then being in free fall again until you meet up with the ship that traveled from A to B (i.e., that was in free fall the whole time). So it's not possible for these two paths (A to B, vs. A to C to B) to be traveled "at the same rate"; that makes no sense.

Hopefully the above clarifies things, but just to recap: the path from A to B is a different path from the path from A to C to B. Different paths can have different lengths. Because of the geometry of Minkowski spacetime, the path from A to B is longer than the path from A to C to B; the minus sign in the formula for the spacetime interval is why the word "longer" occurs here, instead of the word "shorter", which is what we would expect in ordinary Euclidean space.

20. Jun 4, 2015

### Maxila

I’m sorry but your worded explanation is still not clear; best you show me the proper time calculation for each segment. For simplification use an equilateral triangle and instantaneous acceleration. I read your point about acceleration however it is a common practice when merely illustrating SR examples. Besides if the ship accelerates the same for each vector of equal length, that won’t change anything.

21. Jun 4, 2015

### Staff: Mentor

The triangle can't be "equilateral"; remember we're dealing with spacetime, not space. It's impossible to make a triangle in spacetime with three timelike sides all the same length.

We'll use an inertial frame in which the "stay at home" twin (the one who goes from A to B directly) is at rest. We'll assume that the "traveling" twin (the one who goes from A to C to B) moves at speed $\frac{\sqrt{3}}{2}$ (in units where $c = 1$) on both legs (outbound and inbound). And we'll assume that the turnaround point for the traveling twin is one light-year away from the starting/ending point (where the stay at home twin remains).

Then we have the following coordinates for the three points in spacetime (in units of years and light-years):

A: $t = 0$, $x = 0$.

B: $t = \frac{4}{\sqrt{3}}$, $x = 0$.

C: $t = \frac{2}{\sqrt{3}}$, $x = 1$.

The lengths of the three segments are just $\sqrt{\Delta t^2 - \Delta x^2}$, so we have:

AB: $\sqrt{\frac{16}{3} - 0} = \frac{4}{\sqrt{3}}$

AC: $\sqrt{\frac{4}{3} - 1} = \frac{1}{\sqrt{3}}$
BC: $\sqrt{\frac{4}{3} - 1} = \frac{1}{\sqrt{3}}$

Obviously, we have AB > AC + BC. Hence, the stay-at-home twin experiences more elapsed time than the traveling twin ($\frac{4}{\sqrt{3}}$ years vs. $\frac{2}{\sqrt{3}}$ years). And it should be evident that the same inequality will be satisfied by the sides of the triangle regardless of the speed at which the traveling twin moves, or the distance to the turnaround point; the only difference will be how big the disparity is. (It should also be evident, from this analysis, why it's impossible to have a triangle of this kind in spacetime with all three sides the same length.)

22. Jun 4, 2015

### pat devine

Hello! I was just reading your question and whoever gave it to hasn't given you all the information required so there is no answer. The two ships accekerate away from you then coast and then decelerate at different rates. By now they are a long way away from you. You then say that the clocks are returned to you. But don't explain how they get back. I've the have to accelerate along some unknown profile to get back to you. Or perhaps I've misinterpreted. Hope this helps. P

23. Jun 4, 2015

### Staff: Mentor

Somewhere back towards the beginning of the thread, rede96 specified that the clocks would be stopped before we returned them to a common location for comparison. Thus, the return path won't change the outcome; indeed, we could have written the clock readings on pieces of paper and sent the paper back for comparison.

24. Jun 4, 2015

### pat devine

25. Jun 4, 2015

### pat devine

Thanks for that. When you speak of t and 2t whose time are we speaking of (yours or theirs) for if you mean time as you measure it then the space ship that decelerates in time t will will show the largest time dilation for the shortest time ( a slow clock observed for a shotrt time) produces a lower clock reading than the ship decelerating over time 2t ( a faster clock observed for a longer time). If the times (t and 2t) are their times then it's the same answer, the clock decelerating for time t will show a shorter elapsed time (x +t) the other ships clock will show a time (x +2t) as its implicit that they decelerate with their clocks for t and 2t.
P