# Time dilation (what else)

1. Dec 20, 2008

### daytripper

So you guys are probably tired of answering this question but I can't find a resource that properly explains this. Two twins in spaceships in flat, empty space are converging at a rate of 0.8c (for instance). They pass each other (each assuming that he is the one at rest) and as they're flying, they turn to the left to make a circle so that they cover a distance but come back to the same place. Which twin is younger? Why?
If the spaceships were made of glass and the twins had supervision, other than this being a bad spin-off of the justice league, each would see the other aging at a slower rate than himself. This is similar to a thread posted recently (which made this question come to mind) but it removes the idea of "acceleration" and other technicalities. If the fact that neither is in an inertial reference frame (because they're turning) is a problem, then change the question from "making a circle" to "they both stop and put it in reverse, backing up to meet one another". I assume there's some way to make it so that the experience of both twins is identical and yet, time dilation dictates that they are both younger than the other.
Resolution?
Thanks.

edit: here is a diagram that I didn't understand (though I looked at it and read the explanation carefully), if that helps at all with your explanation - http://web.comhem.se/~u87325397/Twins.PNG [Broken]
here is the explanation: https://www.physicsforums.com/showpost.php?p=1881599&postcount=3
thanks again.

Last edited by a moderator: May 3, 2017
2. Dec 20, 2008

### JesseM

Assuming they move in a circle at the same speed in some some inertial frame, they'll both be the same age, because in inertial frames time dilation is just a function of speed.
Not necessarily--the rate you see a clock ticking is not the same as the rate it's ticking in your inertial frame, because of the Doppler effect (which is itself just a consequence of the fact that for a clock moving relative to you, the light from different ticks will be emitted at different different distances from you and thus take different amounts of time to reach your eyes, so the visual rate of ticking will be different from the rate the clock is 'really' ticking in your frame). See the example I gave on this recent thread.
Nope, moving in a circle at constant speed is a form of acceleration--physically you can tell you're accelerating because you feel G-forces (the 'centrifugal force'). Velocity refers to both speed and direction, so to be moving at constant velocity (moving inertially) you must be traveling in a straight line at constant speed as viewed in any inertial frame. And the standard time dilation equation only works in the rest frame of an observer moving inertially.

3. Dec 20, 2008

### daytripper

I've been using the assumption that the twins have to be in the same location to see each other. My question doesn't so much involve the real-time perception of the other as much as it involves the end result. So let's say that two twins are just traveling in a straight line past each other at, say, 0.9c. After x seconds, they stop, pull out their telescopes and observe each other. Who's younger?
Sorry, my previous question was overly complicated.

Last edited: Dec 20, 2008
4. Dec 20, 2008

### JesseM

But in this case they aren't in the same location, so visually each one can see the other as younger without it being contradictory (after all, we see the stars as they were years ago, not as they are today). Both the inertial twins' rest frames would predict the same thing about what each twin would see here.

Last edited: Dec 20, 2008
5. Dec 20, 2008

### daytripper

Gr! Something's missing from my understanding and I can't figure out what!... Back to the drawing board. Thanks for your help.

6. Dec 20, 2008

### JesseM

Maybe instead of asking about what each sees, you could ask about what is true in each one's rest frame? Suppose we have two buoys in space, A' and B', and in the rest frame of the buoys the two twins A and B start out the same age midway between them, and travel in opposite directions at 0.8c in this frame. Then in this case, at the moment when twin A passes buoy A', in his own rest frame he is older than twin B is at that moment (and in this frame B has not even reached his own buoy at this moment), but the rest frame of twin B it is B that is older when B reaches buoy B' (and in this frame A has yet to reach A' at this moment). This is just an illustration of the relativity of simultaneity, which says that the different frames disagree about what events are simultaneous if they happen at different locations in space.

7. Dec 21, 2008

### daytripper

Ah... alright, something's starting to click. I'm going to write my understanding of it. Let me know if this is correct. So just because the two twins reach the bouys at the same time in a reference frame C (the bouy rest frame), does not necessarily mean that these events occur at the same time in reference frame A and B... in fact, they CAN'T occur at the same time in either A or B's frame. And this is not simply a consequence of the doppler effect (which is an observational phenomenon, rather than a relativistic one); their speeds are warping their perspective of space-time. Were they to stop the moment they reached the bouys, they would assume the reference frame C, in which they have both just reached their bouys at the same time. God, I thought I had it but there's an error there somewhere. Would B speed up in frame A upon A's deceleration, making him "catch up" to A as A gets to A'? Same as (in the classic thought experiment) the earth twin's age jumps decades during the acceleration (toward earth)? I'll figure it out with just a bit more help. =]

edit: if A stopped instantaneously, would B "jump" to B'? I realize infinite acceleration isn't possible, but hypothetically? The it would follow (or precede, i guess) that as A slows down (non-instantaneously), his reference frame is constantly changing to one in which B is closer and closer to the bouy, until he comes to a full stop at the bouy and sees that B has done the same. In this final frame, A=B=C, the twins are the same age, even though they see each other as younger simply because the image hasn't arrived yet. Correct?

Last edited: Dec 21, 2008
8. Dec 21, 2008

### JesseM

Yes, all that's right.
All reference frames are just different perspectives on the same spacetime, none more "warped" than any other (and of course speed is relative to your frame too, there is no objective truth about whether a ship is at rest and the buoys are moving or vice versa). They are just different coordinate systems, like different cartesian coordinate systems on a 2D plane with their axes rotated relative to one another. And I'm not sure if you know this, but each inertial coordinate system is physically defined in terms of local measurements on a system of rulers and clocks at rest in that system, with the clocks synchronized according to Einstein's synchronization convention--so if I see an explosion happen next to the 15 light-second mark on my x-axis ruler, and the clock at that mark read 3 seconds at the moment the explosion happened next to it, then I would assign the event coordinates x=15 l.s, t=3 s. Also important to understand is that the Einstein clock synchronization convention is based on each frame assuming light moves at the same speed in all directions in that frame, so if I set off a flash at the midpoint of two clocks in my system, they are defined as "synchronized" if they both read the same time when the light from the flash reaches them. This definition of clock synchronization naturally means that clocks that are synchronized in one frame will be defined as out-of-sync in another (which is why different frames disagree about simultaneity). After all, suppose I am on a ship moving forward relative to you, and I want to synchronize clocks at the front and back of the ship in the ship's rest frame by setting off a flash at the midpoint of the ship and setting both clocks to the same time when the light reaches them. In your frame, the front of the ship is moving away from the point where the flash was set off, while the back is moving towards it, so naturally if you define simultaneity in your own frame using the assumption light moves at the same speed in both directions relative to you, you must conclude that the light caught up with the back of the ship before it caught up with the front.
Inertial frames are defined in terms of ruler/clock systems which move inertially for all time. If the ships change velocity when they reach the buoys, it's true they have a new inertial rest frame C after this change in velocity, but in this frame they were not at rest prior to reaching the buoys (note that when studying a problem, there is no rule that says you have to use an inertial frame where any physical object being analyzed is actually at rest). If you try to combine what would be said about the other twin's age in a twin's inertial rest frame before changing velocity with what would be said about the other twin's age in the same twin's inertial rest frame after changing velocity, the result is a mess of different perspectives that is not consistent with any one inertial frame (in particular, the frame where twin A was at rest before reaching the buoy says something very different than the frame where twin A is at rest after reaching the buoy about twin B's age at the moment twin A reaches the buoy, because these frames define simultaneity differently). If you want to use the time dilation equation to calculate the elapsed time on a given clock between two events on its worldline (like a twin leaving his brother and reaching a buoy), you really need to analyze the whole thing from the perspective of a single inertial frame.

9. Dec 21, 2008

### daytripper

I made an edit to my previous post (thinking that you weren't going to respond tonight). Could you comment on what was said there? I think what I need to do is just sit down with a pen and paper and write out the spacetime diagram. I'll work on that in the meantime.

10. Dec 21, 2008

### Lord Einstein

Hey JesseM can u tell me more about the Einstein Synchronization convention? i would like to learn more

11. Dec 21, 2008

### JesseM

It seems like you're thinking of frames too physically--they're just coordinate systems, no physical event can force you to "jump" from one to another. It is true that in the inertial frame where A is at rest before accelerating, B was far from B' at the moment A accelerates, and in the inertial frame where A is at rest after accelerating, B is at B' at the moment A accelerates. But there's no physical sense in which any "jumping" happens, and nothing obligates you to switch coordinate systems at the moment A accelerates, or to use either of these coordinate systems when analyzing this problem, for that matter (you're perfectly free to use one where neither A nor B nor the buoys are at rest at any moment).
If you choose to define "his" reference frame at any given moment as the one where he's instantaneously at rest, sure.
Yes, in the inertial frame in which the buoys are at rest, the twins were the same age throughout the trip, even before they arrived at the buoys.

12. Dec 21, 2008

### daytripper

Thank you for taking all this time to explain this to me. Right now I'm drawing out a Minkowski diagram (my first) for this situation. I'm using 0.5c as the speed so that the angle $$\beta$$ between the axes is an even 30º, but... I'll take a picture.

I changed the scenario: A is simply there for reference because I'm weak-minded (I have strange labels on my axes for the same reason). B moves to B' at 0.5c. B starts off next to A, 5 light-min away from B'.

I have the feeling that perhaps the line next to the t' axis is supposed to be ON the t' axis and my geometry is just a little off (I'm using a folded piece of paper for my 30º =P)... I suppose that would mean that A sees B as experiencing 8.66 minutes, which works out. This would mean that in B's reference frame, B' came to B (not shown) and B's just been sitting there, not moving on the x axis, while in A's reference frame, it moved 5 light-minutes in 10 minutes. But then if I plot the path of A in B's reference frame (or coordinate system... I knew this was a good idea), would I be plotting into the negative x' axis (landing on (t,x) = (8.66,0) or (t',x') = (8.66, -5))?
My god... did I just do it? is this correct? I have the creeping suspicion that this is correct... but then, this is just a graph and I really was more interested in exactly how this occurs. Why do the axes rotate?

edit: oops. that should be "arcsin" on that paper

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Last edited: Dec 21, 2008
13. Dec 21, 2008

### daytripper

I just realized how tangent this thread has become to its original purpose. I'll get back to the beginning though.

14. Dec 21, 2008

### daytripper

Alright, I've gathered my thoughts. So, (close to) the original point: If A waits 10 minutes, B will have arrived at B' and would have aged 8.6 minutes. On the other hand, if B travels for 10 minutes, he will have reached his goal at the 10 minute mark and would see that A (having moved away from him) has only aged 8.6 minutes. If you combine these two scenarios, though... there lies my confusion.

In B's reference frame: would A be sitting there, waiting for B to finish when B gets to B' after 10 minutes? Would it only register with A that B has arrived at B' after B's... 11.54 minute mark (when A has aged 10 minutes in B's frame)? (we're ignoring doppler effects here. let's assume that A just "knows" when B gets there, so that we don't have several phenomenon to worry about).

...my mind is tired. let me know if I'm right about B's frame and I'll think about A. haha

edit: Would it be correct to say that you are opting to travel through space in lieu of time? The same way if I travel at 30º for 30 feet, I've only moved 10 feet in the y direction and 20 feet in the x direction. It's not that I'm walking any less, it's simply that I've chosen a different dimension in which to travel through.... edit again: wow, those numbers are way off. my trig is RUSTY. haha. I meant to say 15 feet in the y direction and ~26 feet in the x direction. The explicit analogy being: I haven't experienced less space-time, I've just chosen to travel through space, rather than time.

I thought you should know, you've got me pacing my room in anticipation of your reply. I REALLY want it to work like I described immediately above... that would be beautiful.

Last edited: Dec 21, 2008
15. Dec 21, 2008

### JesseM

If in the rest frame of A (who you said is also at rest relative to the buoy B') the distance between A the buoy B' is 5 light-minutes, then in this frame it takes 10 minutes for B to reach B'...but in this frame the clock of B is also slowed down by a factor of 0.866, so the clock will only read 8.66 minutes when B reaches B'. In the B rest frame, the buoy is only 4.33 light-minutes away from A due to length contraction, and the clock of B ticks at a normal rate as B' approaches at 0.5c, so the clock will read 8.66 minutes when B' arrives.
Again, in B's rest frame the distance between A and B' is only 4.33 light-minutes, and A is moving away from B at 0.5c while B' is moving towards B at 0.5c. It wouldn't make any sense for this frame to predict it takes 10 minutes for B to reach B', because the clock time of B will match the coordinate time in this frame, and we already know that the clock time of B when it reaches B' must be 8.66 minutes based on looking at things in A's frame (all inertial frames must agree on predictions about local events like what a given clock reads when it passes next to a certain landmark).
In the rest frame of A, the event of B reaching B' is simultaneous with A's clock reading 10 minutes. In the rest frame of B, the event of B reaching B' is simultaneous with A's clock reading 8.66*(0.866) = 7.5 minutes. You don't need to assume A or B knows anything instantly, you can just talk about the time-coordinates they assign to events in retrospect (for example, if I look through my telescope and see an event which is 50 light-seconds away according to my ruler when my clock reads 200 seconds, I'll say in retrospect that this event had a time-coordinate of 150 seconds when I factor out the light delay).
I don't really understand your ideas here.

16. Dec 21, 2008

### daytripper

Yea, I have a nasty habit of posting before refining, so I end up editing 100 times over. I think there might have been some small shred of validity to the star argument but whatever was there just complicated things. Go back and read the new "edit" section

17. Dec 21, 2008

### daytripper

That did it. I got it. I was trying to seperate the ideas of time dilation and length contraction but if you ignore one while trying to understand the other, SR become inconsistent. Thank you. By the way, could you look at that diagram and see if I constructed it correctly? Am I interpreting all of that correctly?

18. Dec 21, 2008

### JesseM

Well, put it this way--in 3D space if you move between two points with coordinates (x1, y1, z1) and (x2, y2, z2) the distance you travel is given by the pythagorean formula $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$ (and if you hold z constant this reduces to $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$). In 4D spacetime if you move between two events with coordinates (x1, y1, z1, t1) and (x2, y2, z2, t2) then the amount of time elapsed on your clock (the 'proper time' along your worldline between these events) will be $$\sqrt{(t_2 - t_1)^2 - \frac{1}{c^2}(x_2 - x_1)^2 - \frac{1}{c^2}(y_2 - y_1)^2 - \frac{1}{c^2}(z_2 - z_1)^2}$$ (and if you hold y and z constant and use units where c=1, this reduces to $$\sqrt{(t_2 - t_1)^2 - (x_2 - x_1)^2}$$). So you can see a close "family resemblance" between the formulas even though they're a bit different (even when you set c=1 in the proper time formula, you still have the fact that the time coordinate is positive while the spatial coordinates are negative, whereas in the pythagorean formula for ordinary space all the spatial coordinates have the same sign).

Hope this sort of answers your question--in any case, I'm going to bed, so I'll get back to this later.

19. Dec 21, 2008

### daytripper

You don't know how thankful I am. I've been thinking about this stuff for a long time on a conceptual level without really understanding it. Sweet vindication. haha. I still don't truly understand it, but that's ok (it's a better grasp than I need).
One last question, though. I assume you're familiar with Minkowski diagrams? That picture I have of the sketch? I attached it to one of my previous posts. I intend to use those in the future to better understand these relationships and I'd like to know that I'm implementing them correctly. If you could quickly glance over it, that'd be very helpful.
Thanks again.
-DT

20. Dec 21, 2008

### yuiop

I had a quick look at your sketch and I think you may be on the wrong track. You show the the x' axis and the t' axis both rotating anti-clockwise when they should be rotating in opposite directions to each other.

See http://casa.colorado.edu/~ajsh/sr/construction.html

http://www.rolfmuertter.com/philosophy/twin_paradox.html

and http://hubpages.com/hub/Minkowski-Diagram

21. Dec 21, 2008

### daytripper

key,
thanks for looking at the sketch. Perhaps you were thrown off by the scribble above the "x" on the x-axis, and maybe the line above the x' axis isn't clear enough, but the order of the axes, going clockwise, is t, t', x', x.
I took a note from wikipedia with respect to this stuff. Here's a link to the "prototype" for mine: http://en.wikipedia.org/wiki/File:Minkowski_diagram_-_simultaneity.png" [Broken] (where $$\beta = \arcsin \left( \frac{v}{c} \right)$$)

I noticed there's also another one, which is like the ones you posted. url: http://en.wikipedia.org/wiki/File:Minkowski_diagram_-_asymmetric.png" [Broken] (where $$\alpha = \arctan \left( \frac{v}{c} \right)$$).

However, if I'm reading this wikipedia article correctly, these two things can be used interchangeably. And, having slept on it, they seem to work out. Are you sure that this other representation isn't also correct? It makes sense, as if you were to take my graph and "squeeze" it (rotationally) so that t is verticle, t' has positive slope, x' has a lesser positive slope and x is horizontal, I'm pretty sure you would have the other representation that you're talking about. One thing that gives me confidence in this is that c is the same for both coordinate planes.
Thanks for the feedback

Last edited by a moderator: May 3, 2017
22. Dec 21, 2008

### yuiop

Your right, I misread your axis labels. Without investigating to a very deep level I now think you are on the right track. If the x' axis is rotated clockwise from the x axis and the t' axis is rotated anticlockwise from the t axis then that is OK (opposite directions) and you end up with something like your diagram or the 1st Wikipedia diagram. If you squeeze the t' axis and x' axis together, so that they are right angles to each other, you effectively transform to the x't' frame and the original x axis and t axis end up as an acute angle as shown in the second Wikipedia diagram which is the more normal representation.

Last edited by a moderator: May 3, 2017
23. Dec 21, 2008

### daytripper

That's how I understand it. Though, I think you've mixed up my t and t' and my x and x' respectively. I mean... it doesn't really matter, as it's just a naming convention, but in case that throws anyone else off. Rewriting your statements, replacing these variables, you get statements which I'm in full agreement with:

I understand the other one's more standard as I've noticed them more frequently used and if I ever made one, I'd simply do that squeeze thing (by constructing it with orthagonal axes with t' and x' at arctan(v/c) degrees from the t and x axes respectively (rotated towards the lightspeed line). I did this one becaue it seemed much more natural to me. I'm only now getting into coordinate transformations and the idea of a "skewed" coordinate system is much simpler conceptually than a "squeezed" one. Anyway, I'm glad to know that I finally understand this stuff. Those diagrams were key. Thanks.
-Tim

24. Dec 21, 2008

### Fredrik

Staff Emeritus
The only difference between the two options is the sign of the velocity difference between the two frames.

Draw the x and t axes first. Then draw the word line of the second observer. That line is the t' axis. Then you have to draw the x' axis so that the speed of light is 1 in the primed frame too.

25. Dec 21, 2008

### daytripper

So then to draw mine "normally", I don't need to find arctan(v/c)? If t and x are orthogonal axes and B travels at 0.5c, then the t' axis will have a slope of 2 on the tx-plane? Then the x' axis would have a slope of 1/2? Am I understanding this correctly?

edit: at first I didn't understand the first part of your post. I see now though. Thanks (still a little confused about the questions above though)

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