# Question on traveling spaceships

1. Feb 10, 2008

### SteyrerBrain

Hi,

as a highschool teacher I am stuck with the following student's question on time dilatation (SR) which could be summarized as "suppose two passing space-ships return to their meeting point?".

We tried to clarify the situation by assuming the following experiment:
Two spaceships with an atomic clock start from the earth in opposite directions with the same acceleration. After a defined time t0 (measured on the spaceship) they continue traveling with constant speed (inertial frame) for a defined time t1 (again measured on the spaceship). Then both ships slow down and accelerate back towards earth (taking the time 2*t0), travel for time t1 with the same constant speed back to earth before slowing down to land on the earth again (taking time t0).

Seen from the earth both spaceships perform the same experiment, except for the direction. So they should return at the same time with their atomic clocks showing the same time.

Seen from a spaceship we have inertial frames only during the two phases of constant speed (once away from the earth, once towards the earth). In these phases both spaceships have the "impression" that the clock on the other ship is running more slowly. Hence it must be the acceleration phases that compensate for the time dilatation, because when coming back to the earth both spaceships see that the other atomic clock shows the same time. Let us assume we are doing two such experiments, where the acceleration phases are the same (t0' = t0), but the phases of constant speed have different lengths (t1' > t1). Now without going into details of what happens during the acceleration phases (to be solved by GR?) I could not answer the question how the same acceleration phases could compensate for two different amounts of time dilatations due to different lengths of constant travels in the two experiments.

There must be a rather obvious error in that argumentation? Any hints are appreciated.

Thank you,

Wolfgang

2. Feb 10, 2008

### matheinste

Hello SteyrerBrain.

As far as i am aware the time differences are due purely to the time spent in inertial relative motion. Acceleration is not the direct cause of any differences. I am sure someone will give you a more thorough answer.

Matheinste

3. Feb 10, 2008

### JesseM

Think about the relativity of simultaneity. The rocket's plane of simultaneity in its inertial rest frame immediately before accelerating to turn around (when still moving outward at constant velocity) will be quite different than the rocket's plane of simultaneity in its inertial rest frame immediately after accelerating (when it's now moving inertially back towards Earth). Have a look at "figure 3" from this section of the twin paradox page.

Also, note that if you want to take the ship's "perspective" while it's accelerating, you're using a non-inertial rest frame and that the usual time dilation equation of SR doesn't work in this frame. But the time dilation of an accelerating clock doesn't require GR, you can calculate it from the perspective of an inertial frame--if you want to know how much proper time elapses on the accelerating clock between two coordinate times $$t_0$$ and $$t_1$$ in the inertial frame you're using, and you know that the clock's velocity as a function of coordinate time is given by v(t) in this frame, then the time elapsed can be calculated using the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$

4. Feb 10, 2008

### country boy

The turn-around portion of the trip simply lets the traveler reverse direction, so that the distance traveled outbound can be subtracted on the in-bound. This makes the integrals of the invariant interval ds different for the home-base and traveler, while keeping the starting location the same as the ending location. The acceleration times can be instantaneous in the problem. The difference in elapsed times depends only on the speed and distance (or time) traveled in the interial frames..

5. Feb 10, 2008

### matheinste

Hello again.

The spacetime interval is the same (being invariant)for stay at home and traveller because the start and, later, the end points (events) are the same for each of in space and time. The way the interval is proportioned between the space and time dimensions is, however different for the pair with the traveller having a smaller share of time. With two travellers making journeys to and from earth, leaving earth at the same starting time and returning at the same later time, the one which has travelled the greatest spatial distance (irrespective of direction in space) will show the least accumulated time when they compare clocks at the end of the journeys.

Matheinste.

6. Feb 10, 2008

### JesseM

How are you defining "spatial distance" for a non-inertial path? And wouldn't distance depend on which frame you use?

7. Feb 10, 2008

### jcsd

Okay first of all clearly from the symmetry of the situation both travellers DO return with the same time on their atomic clocks. This exact time can be worked out in both cases by performing an integral and it will be the same for both travellers.

Absolute simultaneity in relativity is local - an event is only simultaneous in absolute sense with it's self. In inertial reference frames simultaneity can be defined without any difficulty in a frame-specific way.

However both the travellers in this scenario are in non-inertial frames (it matters not that some of their motion is inertial, as we are considering their journeys as a whole they are non-inertial). Unfortunately in non-inertial frames simultaneity at distance becomes somewhat arbitary.

Jesse has tried to define simulatenity using planes (hyperplanes really in 3+1 diemsnional relativity) of simulatenity, just as would be done in an inertial frame. Unfortunately because the obserevers are non-inertial you'll get all sorts of weird effects using this defitnion for example you may find that event A is simulatanepous with event B and event C, but event B and C are not simulataneous etc. Even as a functional defitnion of simulataneity it leaves a lot to be desired. The best way is just to dump general the notion of simultanoety at distance and only leave it for the certain circumstances where it is useful to define.

When talking about what the observers actually see you've got not only to consider time dialtion, but also the doppler effect. Whilst the travellers are moving away from each other they will see each others clock slow down and whilst they are moving towards each other they will see each others clock speed up. They could compensate for the doppler effect, but as I said the notion of simulataneity for a non-ienrtial obserever is troublesome.

8. Feb 10, 2008

### matheinste

Hello JesseM.

I believe, genuinely, that any point you are going to make is going to be correct so i will not dig a deeper hole for myself. I would be grateful to have any and all of my mistakes corrected so i more fully understand the less obvious points of the subject thus allowing me not to make the same mistakes again.

Thanks Matheinste.

9. Feb 10, 2008

### country boy

Even though the start and end points are the same, the spacetime paths taken by the stay at home and traveler are different. The invariant interval is path-dependent.

10. Feb 10, 2008

### JesseM

I wasn't really trying to talk about simultaneity in a non-inertial coordinate system (I think the simplest type of non-inertial system would define simultaneity to match that of the object's instantaneous inertial reference frame at each moment, but you could define other types of non-inertial coordinate systems), I was just pointing out that if we look at the plane of simultaneity immediately before the turnaround in the inertial rest frame where the ship is at rest before accelerating to turn around, and compare it with the plane of simultaneity immediately after the turnaround in the inertial rest frame where the ship comes to rest after accelerating to turn around, then these two planes are very different, with a big gap in the Earth-twin's age in the first one and the Earth-twin's age in the second one. I think this is helpful even if you don't imagine these two planes of simultaneity as being part of a single non-inertial coordinate system.

11. Feb 10, 2008

### JesseM

Well, there could be some way of thinking about the total spatial distance for each path that hasn't occurred to me and which would make your statement work, I just couldn't see it off the top of my head. Consider the frame where the traveling twin is at rest during the first phase of the journey while the Earth moves away, and then after the acceleration the traveling twin is pursuing the Earth at even greater speed until it catches up...don't the cover the same total distance in this frame, just that the traveling twin covers it faster over a shorter period of time?

12. Feb 11, 2008

### yuiop

Matheinstate is correct in what he says.

Take the Earth as an inertial reference frame (It is not, but is often aproximated as one in Twins paradox type discussions.) So in an idealised thought expression the Earth is not rotating and we are are ignoring Earths gravity. With Earth defined as an inertial reference frame there is no difficulty defining spatial distances as measured by an observer on Earth and no problem defining time as measured by the Earth observer. So say in a stationary solar system both rockets leave at the same time (in the Earth frame) and if both rockets return at the same time (in the Earth frame) then the rocket that went to pluto and back will have experienced less proper time than the rocket that went to the moon and back. All we do is add distances in the the Earth frame as scalar values and it is obvious that the rocket that went to Pluto and back has a higher average velocity than the rocket that went to the moon and back and so the pluto rocket experiences greater time dilation as measured relative to the Earth frame. I am of course assuming an ideal Solar system with no gravity.

If a third rocket left Earth at the same time as the original two and returned at the same time as the original two but went via the moon stopping there for a while and then via Mars (also stopping) there a while and then returning to Earth after visiting Pluto than the 3rd rocket will have experienced the same proper time as the rocket that went to straight to Pluto and back. It does not matter how many segments we break the round trip from Earth to Pluto and back into or what the velocity of each leg of the journey is, the ratio of the proper time experienced by the rocket relative to the time measured on Earth is purely a function of v=(distance to Pluto and back)/(time elapsed in the Earth frame) inserted into the Lorentz transformation of the time interval.

The acceleration of the rocket at any stage makes no difference. A more formal description may be found by googling for "the clock postulate".

[EDIT] The above is clearly wrong in the form I have posted it. See JesseM's response. (That will teach me not to post dimly remembered "facts" without first checking them)

Last edited: Feb 12, 2008
13. Feb 12, 2008

### JesseM

How are you calculating the distance traveled by the rocket in the Earth frame? Just double the distance from Earth to Pluto in this frame, rather than the change in the position of the rocket from the beginning to end of the trip?

But suppose we look at things from the perspective of the inertial frame where the rocket was at rest during the outbound leg of the trip while Earth was moving away, and then when Pluto caught up to the rocket, the rocket accelerated and moved in the direction of Earth at an even greater speed than Earth was moving, eventually catching up to it. Since the rocket moves zero spatial distance during the first leg of the trip, wouldn't it's total distance traveled just be the difference in position between where it was when it began to move towards the Earth, and where it was when it finally caught up with the Earth? And since the Earth started at the same position as the rocket, isn't this exactly the same distance that the Earth traveled in this frame?
That's not true, it accumulates proper time when at rest on the moon and Mars as well. Perhaps what you mean to say is that the difference between the amount the observer on the rocket has aged and the amount the observer on the Earth has observed will be the same in this situations as it is in the situation where the rocket made no stops, but in the situation where the rocket makes stops both the rocket observer and the Earth observer will be older when they reunite than in the situation where the rocket makes no stops.
Not true either. Suppose the rocket is making a trip from Earth to a space station 3 light years away and back. If it goes there at 0.6c, immediately turns around, and returns at 0.6c, then the Earth-observer will have aged 6/0.6 = 10 years, while the rocket observer will have aged 10*sqrt(1 - 0.6^2) = 8 years, so the ratio is 10:8. Now suppose the rocket goes to the space station at 0.6c, then comes to rest there and waits 20 years, then returns to Earth at 0.6c. In this case the rocket observer has aged 8 + 20 = 28 years, while the Earth observer has aged 10 + 20 = 30 years, so the ratio is 30:28, which is not the same. And your statement isn't correct even if we assume the rocket turns around immediately instead of stopping at the space station. Suppose on the outbound leg it's traveling at 0.4c so the time in the Earth frame is 3/0.5 = 6 years, and during the inbound leg it's traveling at 0.75c so the time in the Earth frame is 3/0.75 = 4 years, meaning the total time is 10 years again. In this case, during the outound leg the rocket observer will age 6*sqrt(1 - 0.5^2) = 5.196 years, and during the inbound leg the rocket observer will age 4*sqrt(1 - 0.75^2) = 2.646 years, for a total of about 7.84 years.
The clock postulate just says the instantaneous rate of ticking for an accelerating clock at a particular moment is the same as the instantaneous rate of ticking for an inertial clock with the same instantaneous velocity at that moment. This is already implicit in the integral I gave in post #3, but it doesn't imply what you suggest it does.

14. Feb 12, 2008

### morrobay

It is my understanding that calling a spacetime interval the same is referring to its value
as in S=8 for example.
And that spacetime interval invariance means that when the corridinates in one frame are transformed to values in another frame then the value for the spacetime intervals will be equal. These are two different things.
In this example a rocket with v=.6c travels 6 light years in a round trip out and back to earth in 10 years
So the spacetime interval for the rocket is sqrt 10"-6" = S= 8
For the stationary observer with delta ct at 10 and delta x at 0 the spacetime interval is
S=10 - not the same.
However if the rockets corridinates were converted to a rest frame then the S= 8 value will be obtained , but this is a different situation than the stationary S=10 frame.

15. Feb 12, 2008

### country boy

Getting back to the original question, the acceleration phases do not compensate for the time dilations. The accelerations just let you change inertial frames. The problem can be recast with instantaneous accelerations and still give the same result.

There are three inertial frames here: home base, outbound and inbound. Let +/- v be the outbound and inbound speeds and let T be the elapsed time at home base. Using invariant intervals, for home-base dS=cT, while for both the outbound and inbound ds=sqrt(c^2-v^2)T/2. Being invariant, the observers in all three frames will get the same values for these intervals. Adding the outbound and inbound, the total invariant interval for the traveler is dS'=2ds=sqrt(c^2-v^2)T. dS' is smaller than dS, as expected. The difference depends on T and v, and not on anything about the acceleration phases.

16. Feb 12, 2008

### yuiop

Proper time is independent of acceleration but my illustration in post #12 is flawed as correctly pointed out by jesseM and morrobay. (Apologies to all, for any confusion caused)

When matheinste stated “With two travellers making journeys to and from earth, leaving earth at the same starting time and returning at the same later time, the one which has travelled the greatest spatial distance (irrespective of direction in space) will show the least accumulated time when they compare clocks at the end of the journeys.” I think what he was getting at is “spatial distance” as calculated by measuring the literal path lengths on a space time diagram. The greater the spacetime path length (calculated from $$\sqrt{t^2+x^2}$$ ) the less the proper time ( calculated from the spacetime interval $$\sqrt{t^2-x^2}$$ ) experienced by the object following the path.

If we take the example given in the opening post it is best to analyse the paths of rockets A and B by considering the point of view of a third rocket C. The third rocket should maintain a constant velocity so as to provide a valid inertial reference frame to make measurements from, with a speed that is equal to the constant velocity sections of either A or B on the outward legs. From the point of view of observer C, the paths of A and B form a parallelogram where the total path lengths ($$\sqrt{t^2+x^2}$$ are equal and so are the spacetime intervals $$\sqrt{t^2-x^2}$$.

To demonstrate how each observer sees time passing for the other observers it is instructive to draw the spacetime diagram showing light signals (drawn at 45 degrees to the x axis) transmitted from each rocket at regular proper time intervals as measured by clocks onboard the rockets. This provides a nice "visual" for students. The nice diagrams shown in the doppler shift explanation of the twins paradox at this link http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#doppler can readily be adapted to the problem here.

17. Feb 13, 2008

### Xeinstein

>>> In these phases both spaceships have the "impression" that the clock on the other ship is running more slowly.

What do you mean by "impression"? I don't think "impression" belong to physics.
Can we say that I got the "impression" that angular momentum is conserved?

Do you mean time dilation is "an illusion"?
If that's the case, you're behind the times by over a century.
Physics is all about "measurement". Time dilation can be measured and is Real in any sense of the word you can think of

>>> Hence it must be the acceleration phases that compensate for the time dilatation,
>>> There must be a rather obvious error in that argumentation
Yes, but I think it has Nothing to do with acceleration.

The key question is this: How to compare time with a moving clock?
In other words, how many stationary clocks you need to compare time with a moving clock? Once you realize that you need at least two stationary clocks to compare time with a moving clock, every thing fall into place

Last edited: Feb 13, 2008
18. Feb 18, 2008

### SteyrerBrain

Many things clarified

Many thanks you for the many answers. After going through them and looking at the explanation of the twin paradox I draw the following conclusions:

* The acceleration phases are not really necessary. In the beginning and end they can be neglected trivially, and at the return event it can be seen as a limit process.

* It doesn't take the second spaceship starting from earth to get this "paradoxon". The problem (for intuition) is already inherent in the twin paradox: Since no inertial frame is singled out, most of the time (during the phases of constant speed) the space ship's point of view is, that the earth's clock is running slower than its own and so one wonders what happens in detail. I see this reflected in the mentioned explanation of the twin paradox as the "distance dependency objection". So my student's (and also my question) could also be stated as: please help us understand how to tackle the "distance dependence objection" in more detail.

* Now I am trying to give an answer myself: what easily fools (at least my) intuition has to do with simultaneity: what happens at the return event is that the "lines" of simultaneity drastically change. To my understanding this explains the "distance dependence objection", since what is perceived (computed) as simultaneous is not only speed- but also distance-dependent. (But intuition easily oversees this fact.)

* GR is not necessary for the understanding of acceleration at the return event. But I am still lacking the precise way, how to calculate what happens from the space ship's point of view during an acceleration phase. Can this be derived from the limit process of more and more, shorter and shorter phases of piecewise constant speed? Is there any webpage explaining this in more detail?

Thank you,

Wolfgang

19. Feb 20, 2008

### morrobay

Re post 18
The drastic lines of simultaneity change at the rockets turn around point are calculated from : delta ct = gamma ( delta ct' + v(delta x')/c^
See Wikipedia- special relativity page 5/20
given;
v=.6c
Lorentz factor = .8
gamma= 1.25
For ct' = 10 , x'= 6 for total and S' =8
For ct= 10 , x=0 S=10
At turnaround point , 1/2 total, delta ct' = 4 =(.8) (5)
and delta x = vt'= .6(4) = -2.4 since this is rocket frame.
So delta ct= 1.25 (4) +.6(-2.4)/c^ = 3.2 years earth time as viewed from rocket.
At turnaround +v goes to -v and above changes to 6.8 years with instantaneous
accelaration.

20. Feb 20, 2008

### yuiop

Some nice explanations and equations of accelerating clocks (without GR) can be found

here http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

and here http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

Last edited by a moderator: May 3, 2017