# Trying to understand time dilation better

1. Feb 14, 2014

### rede96

I'm only an amateur enthusiast when it comes to physics and have no background in science at all. It's just something I enjoy thinking about.

One thing I am really finding difficult to get my head around is how time dilation works. So I've put a thought experiment below and would appreciate it if anyone would check my understanding.

There are two friends of the same age, on different space ships, both in uniform motion but travelling in different directions relative to each other. While communicating they discover that their paths will cross so decide to do a little experiment.

They agree that when they pass each other, ship A will send a single to ship B and then they will both synchronise their clocks to a pre agreed date and time. They also agree that they will send each other an update once a month at an exact agreed time using the newly synchronised clocks. The message will include the current time and date from the respective ships so they can see if their clocks remain in sync and at each message, report back any difference. They will do this for 2 years.

They also measure that relative to each other they are moving apart at 0.8c and neither ship goes through any accretion during the experiment.

As I understand it, Ship A would start to receive the messages each month from ship B with longer and longer delays. However Ship B would also be experiencing the same thing.

But what would be a constant is that the delay for each message in sequence would be the same for each ship.

So after 2 years, they stop sending the messages to each other (24 in total) and eventually they each receive the final message from the other ship. They then compare together the total delays for each month interval from the other ship's messages. What they both find is that although each ship thought the other ship's time was moving slower, because the delays were getting longer, the delays for each set of messages sent, match for each ship.

Baffled by this they decide to use their new quantum teleportation devices (which is possible at least in theory) and see who has aged more.

So my questions are:

1) If they both decide that at a given date, say after each experience 10 years after they initially passed and set their clocks, they teleport to a mid point between them, will they find that they have both aged at the same rate once together?

2) If the guy from ship A decided to teleport to Ship B, would he find that his friend on ship B was actually younger? (and the same if Ship B guy teleported to Ship A)

3) Assuming the above is correct (which I have no idea if it is) then what would happen if the guy from Ship A teleported to Ship B and sees his friend is younger. But then teleports back to his ship. Then the guy from Ship B teleports over to ship A. Would he find that miraculously his friend on Ship A is now younger them him again?

All really confusing :)

Last edited: Feb 14, 2014
2. Feb 14, 2014

### Staff: Mentor

Quantum teleportation, as you're understanding it here, is not possible even in theory.

This isn't just a quibble that you can bypass by saying "Yes, but suppose it were possible... Then what would hapen?". The basic assumption you're making is that something can be moved from one point to another instantaneously, and that assumption contains a hidden internal contradiction.

Last edited: Feb 14, 2014
3. Feb 14, 2014

### Ibix

I may be wrong on this, but I get the impression that you believe that quantum teleportation proceeds faster than light. This is not the case, so there is no difference between your "exchanging light signals" and your "teleportation" cases.

The next thing to note is that there are two separate effects at work here. The first is simply the Doppler Shift. Since the ships are moving apart, each month's signal has further to go than the last one. For this reason, each traveler expects to receive the pulses from the other ship more than a month apart. The second thing to happen is time dilation. After correcting for the Doppler Shift, each traveler finds that the other one's pulses are being emitted at intervals of longer than one month, by a factor of $\gamma=1/\sqrt{1-v^2/c^2}$, which is 1.667 for you quoted v=0.8c.

1. Both travelers have done the same thing (or at least, a reflection of the other). They must end up the same age.

2. Presumably your teleporter re-constitutes the person at the other end exactly as they were when they left. In that case, the guy who teleported would be younger because he's been "frozen" in transit. However, this is not a relativistic effect, and the details of their relative ages depend on implementation details of your teleporter.

3. Again, the details of this one depend on the details of your teleporter and exactly what sequence people teleport, since the transit time is not negligible. Basically, both travelers are going to be spending years "frozen" in transit, and their relative ages depend upon how long they spend "frozen" which depends on how far apart their ships were when they beamed over.

4. Feb 14, 2014

### rede96

No, I understood that there needs to be clasical communication between the two. But am not too sure how long it would take for one person's information to be scanned and then replicated at the other side. So didn't think about him being 'frozen'.

Ok, thanks. I did ignore doppler shift as it was just the relative delay I was looking at but should have included it to make it complete.

I agree with this but thought I couldn't make that statement in one frame or the other, as the outcome is relative. So the only way to know who has aged more would be for them to meet so they are in the same frame, but by doing that it may effect who is older than who.

Yeah missed that one, thanks

OK, but lets assume that after 2 years of travel at 0.8c the ships both stop moving apart so they are at rest wto each other and therefore keeping the symetery. Also lets assume that as they are now 1.6 light years apart, that the whole process for teleprotation, including sending the information takes 3 years.

If A goes to B, then B would expect A to be three years younger, as they only difference in aging came from being frozen.

Which means that if we adjust for teleportaion then of course they must have ages at the same rate during the 2 year trip.

And this is where I get confused with terminaolgy. As I understand it, A in his own frame of reference can't say that B aged the same during the 2 year period, as the results would prove otherwise. Neither can B in his own frame say the A has ages the same rate.

So as I understand it, the only way they can say with certanity that they have aged at the same rate is if they manipulate the process of getting together in just one frame so they have indeed aged the same. And as you said as long as the whole process from start to getting together again was symetrical, then they must age the same.

And if they process of getting back together isn't symetrical, then who ever broke the symetry by the biggest margin would have ages the less. (I think!)

Does that make sense?

5. Feb 14, 2014

### rede96

Yeah, I just did a bad job of explaining it sorry. But at least in theory it is possible to use quantum entanglement to transport matter (or more replicate matter) from one place to another. They would have need to set all this up before the experiment obviously.

6. Feb 14, 2014

### Ibix

Because they are meeting, because they are in the same place, there is no ambiguity about the definition of "simultaneous", which there is for separate events. Since their world-lines were mirror images of one another in some frame, the elapsed proper time must be the same for both. So they will be the same age.

OK - you're complicating things now. But this is enough to explain where you are going wrong: two years according to who? If both travelers fire their engines when their on board clocks read two years, both will say the other burned later because the other's clocks were slow. I suspect you are thinking in the frame in which the rockets are doing equal and opposite velocities - which is a good choice of frame - but note that two years in this frame is not two years according to the rocket's clocks.

No. I'm not sure how you would measure distance from symmetry. What you do is calculate the proper time ("wrist-watch time") along the world line of each of the travelers and compare those at the instant they meet. If you see someone move a distance $\Delta x$ in a time $\Delta t$ at constant velocity, then the proper time (the elapsed time on their wrist-watch) is $\Delta\tau$:
$$(c\Delta\tau)^2=(c\Delta t)^2-(\Delta x)^2$$
Other observers will disagree on the $\Delta t$ due to time dilation and the$\Delta x$ due to length contraction, but they will all agree on $\Delta\tau$. As long as acceleration can be treated as instantaneous, all you need to do is calculate the $\Delta\tau$s between each acceleration and add them up.

This turns out to be closely analogous to saying that a zig-zag path is longer than a straight one. It is all best explained by a space-time diagram, which I suspect is being drawn by ghwellsjr as I type...

7. Feb 15, 2014

### rede96

Yes, this is where I am getting confused. I wasn't thinking about what would happen if I looked at it from a third frame. Or any frame really. What I was thinking was more a prediction.

For example, I know Ship A and Ship B are both in inertial frames and are on a big X trajectory relative to me, as we can assume that I am at the mid point of the 'X'. I also know they are both equidistant from that mid point and travelling at the same speed relative to me. Hence how I know they will meet at that mid point. So I decided to make an experiment.

When they pass that mid point, I have asked them to synchronise their clocks, then agree that for each person, after 2 years have elapsed on their respective clocks, they will turn to ships to face each other and continue along the new trajectory until they meet the middle.

I tell as their paths are symmetrical wrt the mid-point where they synchronised their clocks, then when they meet, their clocks will have both elapsed the same amount of time.

Now I am assuming that if the above is correct, I can make that prediction in any frame of reference?

Is that correct?

8. Feb 15, 2014

### ghwellsjr

This is an excellent idea and I think I can show you how it is all they need to do to establish Time Dilation with the simple application of Einstein's second postulate.

I will also show you how they can each measure the speed of the other.

This is a very important feature of Einstein's first postulate; however slow ship A sees ship B's date/time going, ship B will see ship A's date/time going slow by the same factor.

I don't think they or anybody else should be baffled that a symmetrical situation should be, well, symmetrical. There's no need to propose any device as you suggest, they already have everything they need to establish Time Dilation.

First, let's suppose that they set their calendar/clocks to zero as they passed each other (just to make our calculations simpler). Then let's just suppose that as they are sending and receiving their messages, they each see the other ones signals coming in at one-half the rate that they are each sending them. At the end of the first month ship A will send a message to ship B saying that it is the end of A's first month. Ship B will get this message at the end of his second month so he will send a message back to A saying that at B's date/time of 2 months he got A's message sent at A's date/time of one month. Ship A will get this message at his date/time of 4 months.

So what does ship A do with this information? He applies Einstein's second postulate that his first signal took the same amount of time to get to B as B's response took to get back to him. Since the total time from sending to receiving was 3 months (4-1), he divides that time by 2 and gets 1.5 months as the time after he sent his signal, or 2.5 months as the time on his calendar/clock that ship B received the signal. Furthermore, ship A also establishes the distance away that ship B was as how far light travels in 1.5 months which is simply 1.5 light-months.

Since Time Dilation is the ratio of Coordinate Time (the same as the time on a stationary clock) to the time on a moving clock, ship A has just established that ship B's calendar/clock is dilated by a factor of 2.5/2 or 1.25.

Also, ship A establishes that since ship B was 1.5 light-months away at ship A's time of 2.5 months, then ship B is traveling away from ship A at 1.5/2.5 or 0.6c while ship A was stationary.

It should be clear that ship B could have also done the same thing and established that it was at rest while ship A was traveling away in the opposite direction at 0.6 c and that it was ship A's calendar/clock that was time dilated by a factor of 1.25.

You're right that teleporting would be miraculous so there is no way to answer your questions.

I hope you're still not confused.

You may have noted that the example that I used was not the example you described because you specified a speed of 0.8c and mine turned out to be 0.6c. But if you understood my explanation, I'd like you to work through another one where they each see the other ship's calendar/clock running at one third of there own and see if that turns out to be your example at 0.8c. Can you do that or do you need more help?

9. Feb 18, 2014

### rede96

Thanks, that does help to understand a bit of how to calculate the time dilation.

But I'm still not clear on one point.

If the two guys were in a parallel path then they would in effect be at rest wrt each other. So if one of the guys decided to change course and go see the other person and travelled at 0.6c for a distance of x, when he arrived he would be the younger of the pair due to the effects mentioned. There is no ambiguity as the two people are together.

However if the guy instead turns to travel in the opposite direction away from his friend, but still travels the same distance x at 0.6c, it is valid for either pair to state that the other's clock has moved slower. We can't say the guy who travelled in the opposite direction is younger when he stops travelling as the two guys aren't together.

So it seems direction has a bearing on outcome?

10. Feb 19, 2014

### ghwellsjr

You're welcome.

Don't confuse Time Dilation with differential aging. You can only compare the amount of aging between two guys if they first start out together, then separate, then rejoin. So in this case, there is ambiguity.

Direction has no bearing on Time Dilation (which is what you are asking about) and you don't need a new scenario to get a change in direction. Your original scenario will work just fine. The two guys start out approaching each other and then after they pass, they are receding from each other.

You said that they were communicating before they passed each other so they could have easily agreed to send the same kind of signals and done the same kind of calculations that they did afterwards. The only difference is that they will each see the other's signals coming in at twice the rate of their own signals going out, instead of half the rate which is true after they pass.

Unfortunately, it gets a little hard to conceptualize how this works but let's pretend that while they are approaching, they simply keep track of the times on their own calendar/clocks and then they make an adjustment to accommodate setting their calendar/clocks to zero when they pass. This makes all their times have negative values prior to them passing.

So 4 months prior to passing (we'll call it -4 months), ship A sends a message to ship B who gets it at -2 months and sends a message back which ship A gets at -1 month. Ship A does similar calculations as described in my previous post and comes up with the same Time Dilation of 1.25.

Does this make sense to you?

Last edited: Feb 19, 2014
11. Feb 21, 2014

### rede96

Yes, I think so. What I understand by that is that it doesn't matter if they are travelling towards or away from each other, the time dilation factor will be the same. If course the time for the messages will be different due to the different distances.

This is the bit that is confusing me.

1) Imagine A and B start off from the same point together with clocks set to zero. They head off in opposite directions, travelling at the same speed away (away from the point of origin). When their respect clocks reach 4 months, the stop so they are at rest rst the origin and hence each other.

When they stop I ask them to send their clock time back to me at the point of origin. I notice I receive their information on clock times both at the same time and that both their clocks say the same amount of time has lapsed (4 Months) However the time past for me will be longer.

But I can concluded that as 4 months have elapsed on both A and B's clocks during that time, and as they were the same age when the started off, they will be the same age now. I don't need to bring them back together to know that. (As I understand it) Although I could bring them back at the same speed and would see they have aged the same.

2) Now imagine a similar situation where A and B set off from the same point with their clocks set to zero. But this time A takes 'W' trajectory away from the point of origin and B takes a 'Z' trajectory. I arrange the trip so the total distance travelled in the W and Z trajectories is 4 light months AND that they travel at the same speed. I have also worked out that relative to me the end point of each trajectory is 2 light months (or some other equal distance) away from me at the point of origin.

I tell them to set off on their journeys and at the end of their journeys (4 months on their clocks) stop so they are at rest wrt to me at the point of origin and send back their clock times.

Now this is the part I get confused over. As A and B had both travelled the same distance in their respective frame and finish the same distance away from me at the point of origin, will I receive their signals at the same time as with the first experiment above?

Or would I receive them at different times because their journeys were not symmetrical and thus I would get the 'W' journey first as it had the most changes in direction?

Also, in the first experiment there seems to be a contradiction as during the travel time both A and B could say that the other's clock was running slower but when they finished their journey, the are both the same age.

12. Feb 21, 2014

### Bill_K

Simultaneity is the answer. A key part of SR is the fact that which events an observer regards as simultaneous will change if his rest frame changes. While A and B are in relative motion, A regards B's clocks as running slow, and says for example that "now", B is younger than A, celebrating his 25th birthday. When they come to relative rest, A's definition of "now" has changed, and A says that B is now the same age as him, age 30.

13. Feb 21, 2014

### Staff: Mentor

In your rest frame, this is true. This is not true in other frames. For instance, in a frame where you are moving to the left (in the direction of A) the distance that the light from A travels is less than the distance that the light from B travels. Therefore, the fact that they arrived at you at the same time implies that A is younger than B. This makes sense because in this frame A traveled faster than B.

The problem with comparing the ages of distant objects is not that it cannot be done. Clearly it can. The problem is just that the result depends on the reference frame used for the comparison.

Last edited: Feb 21, 2014
14. Feb 21, 2014

### rede96

Thanks for that. I think I get it but not sure I fully understand. Does this mean that in A's reference frame when he wants to compare ages 'at the same time' that what A thinks is a simultaneous point for A and B to compare ages, is in fact off by 5 years? (in the example you gave) So while A and B are in relative motion, 'now' for both is relative. But when they are at rest rst each other, 'now' can be simultaneous?

15. Feb 21, 2014

### rede96

Ok, thanks. I think what I was trying to understand was if the space-time interval (which I read was invariant) is the same for both A and B for their journey then would their clocks have to read the same at the end of the journey? Then anyone in any frame if they knew the space-time interval would know that A and B's clocks would read the same relative to A and B, even though they might not agree with A and B's clocks reading the same from their frame.

I may be over complicating things, sorry if I am. But my thinking was that if I know the space-time interval (or their journey through 4d space time just in case I am not using the right terminology) is the same, then I know the net effects of any difference in time from their journey would be the same.

16. Feb 21, 2014

### Staff: Mentor

The spacetime interval is the same for both journeys, and therefore in all frames their clocks do read the same at the end of their journeys.

But the journeys only end at the same time in one frame. In other frames they end at different times and therefore they are not the same age at any one point in time.

17. Feb 21, 2014

### rede96

Ah ok. got it thanks.

18. Feb 24, 2014

### ghwellsjr

As has been pointed out by others, what you are calling "now" is only true in your rest Inertial Reference Frame (IRF) which we can illustrate in an example where the two travelers went away at 0.333c. Here is a spacetime diagram showing your scenario. You are shown as the thick black line, A is shown as the thick blue line and B is the thick red line:

Note that your clock has advanced to 5.67 months when you receive the two messages from A and B that they have reached the 4-month mark for them to stop. And by the time you get their messages, their clocks have advanced to 5.4 months.

We can use the Lorentz Transformation process to see what this scenario looks like in A's rest frame during his "travel time":

Note that you still receive the messages at your time of 5.67 months, even though it took different amounts of time for the messages to get to you and A and B are not the same age when this happens.

Yes, if you bring them back at the same speed, they will continue to age at the same rate. Here is a diagram showing this:

You have aged more than the travelers (8.48 months) but they have aged the same (8 months).

But now this is true in all other IRF's. Here's the same one we did before:

Yes, here is a diagram to depict what I think you described:

Symmetry doesn't matter. Remember in post #10 I said that direction doesn't matter? The only thing that matters is speed according to an IRF. As long as both travelers are always going at the same speed, then they will be subject to the same Time Dilation and end up with the same aging when they rejoin you.

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19. Feb 24, 2014

### ghwellsjr

If you apply the technique I described for you in post #8, sending messages back and forth, then A can make a non-inertial rest frame for himself and it will show you how even though A determines that B's clock was running slower than his at the beginning, it can still catch up at the end. To make an accurate diagram, A will have to send out continual messages and get continual responses back from B but I will only show those where B's clock increments one-month intervals. We will also have A communicating with you. Here's the signals going between A and B:

And here's the signals going between A and you:

Remember what A does with the information:

This allows A to construct this diagram:

As you can see, even though both you and B are Time Dilated at the beginning, you eventually surpass A while B just catches up.

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20. Feb 24, 2014

### rede96

Thanks for the space time diagrams, they really help make it clear.

I understand now that direction doesn't matter. But I thought change in direction did? Due to acceleration? So although the distance and speed covered by the 'W' path and the 'Z' path is the same, as the 'W' path has changed direction more than the 'Z' path doesn't that mean that the two will age differently when the come to rest?

Although as both come to rest the same distance from the origin where the set their clocks, then I guess the total acceleration (change in direction) would be the same as the angles would have to add up the same?

Does that make sense?