# Simultaneity is directional while time dilation is not?

1. Oct 14, 2011

### universal_101

As time-dilation is stated in STR,

1.) it is an intrinsic decrease in rate of passing of time, with relative motion(radioactivity half-life change).
2.) it is non-directional, that is, time-dilation depends only on the magnitude of relative speed.

Now, please consider the following scenario,

Three spacecrafts A, B and C are parked in space in the sequence (A-B-C) with perfectly synchronized clocks on-board, where B always stays stationary in a particular observing reference frame while A and C starts moving away to left and right direction respectively and after some time they stop and return back to their original position. Now, considering that the spacecrafts A and C are programmed identically for their motion, that is, in the observers reference frame their motion is always equal and opposite.

Now, my question is, How can one explain the time-dilation of the other two spacecrafts,
1.) from the rest frame of reference of B and
2.) from the frame of moving space crafts (be it A or C).

Because in the rest frame the time-dilation of the two spacecrafts should be equal therefore time should also be equal, while in frame of any moving spacecraft the time-dilation of other moving spacecraft should be even bigger than that of in rest frame, that is, according to STR, time on moving spacecraft A' w.r.t C' should be different at the end of the journey. But at the end of the journey both the moving frame and the rest frame are one and the same. How can time be different at a particular instant in just one single reference frame ?

2. Oct 14, 2011

### Naty1

correct...but B (stationary) experiences more time.....

incorrect overall....clock difference is bigger as they move apart,as relative velocities are greater, but smaller as they converge to again rejoin "B"....

Last edited: Oct 14, 2011
3. Oct 14, 2011

### ghwellsjr

You actually have described 3 different inertial reference frames:

1) Rest frame of all three spacecraft at the beginning and end of your scenario.

2) Rest frame of A during the first half of the scenario and of B during the last half of the scenario.

3) Rest frame of A during the last half of the scenario and of B during the first half of the scenario.

If you analyze all three spacecraft using any one of these three inertial rest frames, they will all indicate at the end of the scenario that A and C have accumulated less time than B.

In frame 1, both A and C experience time dilation.

In frames 2 and 3, B experiences a constant time dilation but A and C experience no time dilation during half of the scenario but a greater time dilation than B during the other half of the scenario. A and C experience the same time dilation as B before and after the scenario but this doesn't affect the final result.

This is really no different than the normal twin paradox with just two spacecraft where one never accelerates while the other one is at rest in two different inertial frames but at different times. The mistake is using two different inertial rest frames for the accelerating spacecraft and assuming that it is as straightforward and simple as a single inertial rest frame.

4. Oct 14, 2011

Synchronised to what?

5. Oct 14, 2011

### ghwellsjr

To each other. Why is this an issue to you?

6. Oct 14, 2011

### universal_101

Does it mean that at the end of the journey the time on the clocks of the spacecrafts A and C will be same no matter which frame we choose ?

If it is so then how can one account for the time-dilation of A' w.r.t C' or vice-versa, because they are always moving relative to each other during the whole journey.

Last edited: Oct 14, 2011
7. Oct 14, 2011

### ghwellsjr

If A and C start out at the same location with B and end up there (the normal way of presenting the Twin Paradox), then all frames will agree that A and C have the same time on them at the beginning and end of the scenario.

However, if A and C are separated from B at the start and end of the scenario, then other frames will not agree that A and C have the same time on them either at the start or at the end but they will agree that both clocks have accumulated the same amount of time and that it is less than what B has accumulated.
Are you now asking about the conventional Twin Paradox where A and B start out and end up at the same location? If so, please do a search on this forum for "Twin Paradox" and you will find lots of explanations.

But generally, the same issue applies: you need to use one inertial frame for the whole scenario and the easiest one to use is one where they both start out at rest and end up at rest. In this case, B remains stationary and experiences no time dilation while A travels and experiences all the time dilation. It's very simple.

If you want to complicate things by using one of the other two frames I described earlier, then you will have to do some computations to see that all three frames end up with the same final solution, but generally what I said there still applies--B will experience time dilation the entire time while A starts out and ends up with that amount of time dilation but during half the trip experience no time dilation and during the other half of the trip experiences a greater time dilation that B.

8. Oct 14, 2011

### universal_101

You are correct it is just extended Twin Paradox,

And I guess there is no point discussing it because above paradox is already a established theory.

Anyways , thanks for your time and your kind replies

9. Oct 14, 2011

I'd just like to know how you would synchronise spatially separated clocks is all?

10. Oct 14, 2011

### ghwellsjr

That's fundamental to Special Relativity. It takes time to do it and there are many ways to do it. One way is for B to send a message via a light signal to A and C asking them to send him back a message of what time it was when they got his message. When B gets these two messages, he looks at his clock compared to his first reading to see how long the round trip light messages took. The readings on A and C's clocks should be half way between the two readings on his clock. He calculates how far off their two clocks are and sends them another pair of messages, telling them how much adjustment they need to make to their clocks to get them synchronized.

Of course, for this to work, the turn-around time for the messages must be instantaneous or fixed at some known interval to make the correct adjustment or carried out with automatic equipment to take this all into account.

EDIT: I'm assuming that you already know that only clocks that are at rest with respect to each other can be synchronized by this method.

11. Oct 14, 2011

There's no such thing as instantaneous in relativity. Any spatial separation takes time for a signal to cross as c is the fastest anything can travel. For simplicity (and to expose the problem clearly) let's say that the 3 clocks form an equilateral triangle with sides of 1 light second each. ie. it takes 1 second for any of the clocks to transmit a signal to the other two clocks.

A sends a signal to B and C. B and C set their clocks accordingly. B looks at C's clock and says, "Hey, your clock is 1 second behind mine". C looks at B's clock and says, "No, your clock is 1 second behind mine". A looks at both clocks and says, your clocks are both 2 second behind mine!

This cannot be overcome and this is precisely why we have relativity. There is no universal time. As such you cannot synchronise clocks. The best you can do is agree on a time to a defined margin of error.

12. Oct 14, 2011

### ghwellsjr

I was referring to the turn-around time from when A or C receives a message until they send out their response, not the time it takes for the message to travel between the spacecraft and I explained what they can do to overcome this technical issue. It has nothing to do with relativity.
If you will read carefully what I said you will see that it is totally different from what you said. Of course what you said won't work because there is no agreement among the three spacecraft on who is going to be the master clock that the other two will synchronize to.

The reason why we have relativity is not because there is a delay in the transmission of signals over long distances, it's because we cannot know what that delay is even if we know the distance. You say that if the distance between the spacecraft is 1 light second then it takes 1 second for a signal to travel between them. If that were the case, then there would be no need to synchronize clocks because we would already know what the delay is.

But the fact of the matter is that when two spacecraft are separated by 1 light second, they can measure the round-trip delay and it will be 2 seconds but they cannot know how to divide that time interval into the two segments for each portion of the trip until and unless they synchronize their clocks using the convention outlined by Einstein which is to divide the time interval equally. The purpose of this process is to define what the time is on the remote clocks, not merely to account for the light transmission delay.

So I don't know where you got the idea that you cannot synchronize clocks. That's really what Special Relativity is all about. Maybe you should read Einstein's 1905 paper introducing SR.

13. Oct 15, 2011

With respect, that wasn't clear from how you expressed it, but you have clarified your meaning so this is not an issue.

You can only align them to the signal. When clock B aligns to the signal of clock A it is immediately out of synch as the person standing next to that clock A will testify to. He will say, your clock is slow. The difference being twice the spatial separation. All you are saying is that we are aligning the clock to the signal from the other clock. I don't know about you, but I was taught that when clocks are synchronised they show the same time. Well, they do to one person, but not the other. Maybe you can explain how this makes them synchronised?

There's a delay at any distance.

You neglected to take into account the very next sentence - there is no universal time. That's why there is relativity. As you say above, we cannot even agree on distance, and you can't agree that your clocks are running at the same rate either. They can only approximate to each other to within a defined error margin. We live in a Universe where any two points spatially separated do NOT have the same length measuring sticks/clocks.

!!

I could have easily left out defining the distance between them and the argument would be the same. You and me know the clocks are 1 light second apart. The people in the spaceships don't. They're not privy to information defined by the example.

Very good. Now tell me how the two clocks can say the same time to both people.

So, clock A sends out a signal. Clock B synchronises to that signal. Clock B sends their signal back. Clock A says, your clock is 2 seconds slow. So clock B sets his clock foward 1 second and you set your clock back 1 second. Clock B now says, your clock is slow by 2 seconds. Where is the synchronisation?

Which we can do without synchronising. Clock A says 12:15:02, clock B says 12:15:00 according to clock A. According to clock B, both clocks say 12:15:02.

http://en.wikipedia.org/wiki/Einstein_synchronisation

14. Oct 15, 2011

### DrGreg

Slinkey, you have misunderstood how Einstein synchronisation works. If two clocks A and B are a constant 1 light-second apart (determined by reflecting a signal A-B-A and timing it to be a constant 2 seconds), then B's clock should be adjusted so that each visually sees the other as being 1 second slow. When they each take account of the 1 second flight-time delay, they agree they are synchronised.

(But a third observer C who is moving relative to A and B will disagree and say they are not synchronised.)

15. Oct 15, 2011