# Twin Paradox - Question on Space Time Diagram Analysis Resolution

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1. May 20, 2015

### controlfreak

I started a thread previously (https://www.physicsforums.com/threads/time-dilation-happens-on-moving-frame-but-which-one.814622/) around this concept but it got closed by one of the staff after helpfully pointing me to http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html.

I read that link and it does nicely explain how Stella's elapsed "proper" time is lesser when analysed using the lens of Minkowski spacetime, from the frame of reference of Terrence (assuming Terrence is at rest). But suppose if we do the same calculation from the frame of reference of Stella (assuming Stella is at rest), wouldnt we get different worldlines? Infact the worldlines will switch between them. So from Stella's frame of reference, she will calculate that the elapsed "proper time" of Terrence is less than Stella and he would have aged less. So as I see it the paradox isnt avoided and infact reinforced by this elaborate calculation.

What am i missing?

2. May 20, 2015

### Mentz114

Worldlines are actual things and they cannot be altered. They will look the same from any prespectve ( for instance if a ball bounces off a wall - that is a fact and the worldlines express this).

The other thing is that the proper time measured along a section of a worldline is an absolute invariant. It is always the same. This is because the metric $ds^2= -c^2dt^2+dx^2 + ...$ which defines the proper interval is invariant inder Lorentz transformations.

The metric may also be written $-c^2d\tau^2= -c^2dt^2+dx^2 + ...$

3. May 20, 2015

### stevendaryl

Staff Emeritus
Let me give an analogy: You have two roads that both go from city A to city B. One road starts at A, goes north 100 miles, makes a right-angle turn, and goes east 100 to reach city B. Another road goes northeast from A to B. The roads have different lengths. It doesn't matter how you turn the map, the straight road is not going to have the same length as the bent road.

A change of reference frame in SR is analogous to rotating a map in Euclidean space. You can't change a "straight" path into a "bent" path. There is no inertial reference frame in which Stella is always at rest.

4. May 20, 2015

### Staff: Mentor

No. That is the point of the Minkowski spacetime approach. Choosing different frames of reference only changes the grid lines that we draw on spacetime. The geometric figure remains the same regardless of how we choose the grid lines.

Stella's line is geometrically bent and Terence's world line is geometrically straight independent of any grid lines. Even using weird bent grid lines does not change that fact. Just as you can draw a straight line and a bent line on a piece of paper and, independent of any grid, determine that the straight line is straight and the bent line is bent and determine the length of each line, similarly with Minkowski spacetime.

Last edited: May 20, 2015
5. May 20, 2015

### ghwellsjr

What you are saying is true for the beginning of the scenario but not for the entire scenario. Let's suppose that Stella travels away from Terrence at 60%c for 8 years then turns around and takes another 8 years to get back. Here is a spacetime diagram showing this where Stella's worldline is red and Terrence's is blue and the dots mark off 1-year increments of Proper Time for both of them:

Note that after 5 years of Coordinate Time, Terrence has aged 5 years (because this is Terrence's rest frame) while Stella has aged only 4 years.

Now we transform to the rest frame of Stella during the first part of the scenario:

Note how after 5 years of Coordinate Time, Stella has aged 5 years while Terrence has aged only 4 years just as you pointed out. However, when we look at the entire scenario from either spacetime diagram, Terrence ages 20 years while Stella ages 16 years.

Does this clear up what you are missing?

Last edited: May 21, 2015
6. May 20, 2015

### controlfreak

ghwellsjr - Thank you for this answer. Infact if you see before the turn around (the first part of the journey) from Stella's rest frame, Stella has aged 8 years while Terrence has aged only 6+ years of proper time.

The turn around is the main reason. In this case, if you see, after the turn around, the reference frame used by your diagram here is no more the "turned around" Stella's reference frame but the original reference frame of stella during the first part of the journey which is a reasonable and right approach, as you dont want to change your frame of reference midway just because Stella jumped unto a different reference frame for the second part of the journey. If you actually plot the reverse journey with the "turned around" Stella's reference frame as rest frame and then paste it with the first part of the journey (though not correct to do it as you need to stick when plotting space time diagram with one reference frame) it would be an opposite of the space diagram as seen from the Terrence's reference frame (but lengths would be different as the longest side will be 16 instead of 20 but nevertheless the bents would be switched resulting in a smaller triangle). But ofcourse I realize I followed an illegal approach as I changed the reference frame midway because I was sticking to Stella. But unfortunately she changed her reference frame by turning around. That was what I think I was missing - thank you for the insight and clarity!

[mentor's note: some discussion quoting a subsequently post has been removed]

Last edited by a moderator: May 20, 2015
7. May 20, 2015

### controlfreak

But how did you calculate when using Stella's rest reference frame in the onward journey that she is at rest for 8 years? If we assume Stella is at rest then the space station (at the turn around point) has to travel to her. In that case, I am assuming the space station is travelling at 0.6 c towards Stella who is at rest. The distance between Stella and space station in 6 light years. So shouldnt space station take 10 years to reach Stella according to Stella? Why 8 years? Shouldnt the bend happen after 10 years of rest (travelling in time)?

8. May 20, 2015

### controlfreak

I think in this diagram, I feel there should be a correction made. The red line should go from (0,0) to (0,10) and not to (0,8) and the bend for the red line happens at (0,10) and then the line goes to (-15,25). At coordinate time 10 yrs, Terrence location will be -6 in x-axis which is correct as at the turn around point Terrence and Stella are separated by 6 light years. And also Terrence would have also spent proper time of 8 years (check graph) and so 2 years slower than Stella as we expected. So it is not 8/6.5 but 10/8 between Stella and Terrence at the turn around point as suggested by the above figure, an exact reverse of the world lines as seen from Terrance side (atleast for the first part of the journey) like I initially pointed out in my first post. But after that for the second part of the journey lines slope in such a manner we get back to original answer of 16 years for Stella and 20 years for Terrence at (-15,25) which tallies well with the answer we expected at the end of the journey. No confusion there if we dont change frames midway following Stella.

Considering this to be the case, then I can go back to my original thought experiment which I believe is now valid if the above reasoning is valid.

Here is a thought experiment.

Restating the problem which we have been discussing - Assume relative velocity between the two frames (Terrence & Stella) is 0.6 c. Stella travels from earth to a space station which is 6 Light years away and then turns around and comes back. In the first part of the journey symmetry is not broken and so both Stella and Terrence's frames are equivalent. Both think the other person's time is slower. As per Stella she reaches the space station in 10 years, Terrence thinks she reaches the space station in 8 years.

Say Stella is a cancer patient and will die around 9 years if left untreated. Suppose the medicine for treatment is present in the space station at the end of the first part of the journey and suppose they get the medicine when they pass that station during the turn around and give it to Stella after that, Stella will be cured. Now the question if Stella is alive when she returns, then she reached the space station before 9 years which means the time was running slow in the space ship which is what Terrence will argue if the journey is viewed from his reference, otherwise if she is dead when she returns then the time was running faster in space ship which is what Stella herself will argue if she viewed the whole journey from her reference.

What is the truth? She cant be both dead and alive?

9. May 20, 2015

### Staff: Mentor

We have two events: Stella leaves earth; and Stella arrives at space station.

In a frame at which Terence is at rest, the (t,x) coordinates of these two events are (0,0) and (10,6): Terence set his watch to zero at the time of Stella's departure and says that Stella arrives at the station 6 light-years away at the same time that Terence's watch reads 10 years. Clearly Terence ages ten years while this is going on, but we can double-check by calculating the proper time between the events "Stella's departure" and "Terence celebrates his tenth birthday and says to himself 'wow - she should be arriving at the space station right about now'". That event has coordinates, in the frame in which Terence is at rest, of (10,0) and the proper time comes out to be $\sqrt{(10-0)^2-(0-0)^2}=10$.

Now we use the Lorentz transforms to see what the coordinates of these two events are using the frame in which Stella is at rest on the outbound leg. Her departure event is still (0,0), but Stella coordinates for the second event are (8,0) - and we get this not by using the time dilation formula, but by using the Lorentz transforms which relate Terence's x,t coordinates to Stella's and vice versa.

You can calculate how much Stella aged during her trip by using either Terence's coordinates ($\sqrt{(10-0)^2-(6-0)^2}=8$ or using Stella's coordinates $\sqrt{(8-0)^2-(0-0)^2}=8$ and we get the same eight years as the answer. It would be a good exercise to transform Terence's coordinates (10,0) for the event of his tenth birthday party to Stella's coordinates - note that although Stella arrives at the station at the same time as the party starts according to Terence, the party and Stella's arrival at the station are not(!) simultaneous according to Stella - and verify that Stella also calculates that Terence has aged ten years between the departure event and the party event.

Last edited: May 20, 2015
10. May 20, 2015

### Staff: Mentor

No, the diagram is right. The event "Stella arrives at the station" happens at time 10 and position 6 in a frame in which Terence is at rest - but the diagram is showing coordinates in which Stella is at rest, and in those coordinates the arrival event happens at time 8 and position 0. An arbitrary point on the Terence diagram (the one in which Terence's worldline moves straight up the page) with coordinate x and t is the same point as the point on the Stella diagram with coordinates x' and t' given by the Lorentz transformations:
$x'=\gamma(x-vt)$
$t'=\gamma(t-vx)$

Try it, and you'll find that Terence's (10,6) is Stella's (8,0).

11. May 20, 2015

### Staff: Mentor

Terence is the one who finds that outbound leg is ten years. This is just basic distance=speed*time and time=distance/speed: Stella has to cover six light years while moving at .6c relative to Terence so he divides the distance by the speed (6/.6) and gets ten years. As far as Stella is concerned, however, she is at rest while Terence and the station are moving backwards at .6c. Because they are moving relative to her, the distance between them is length-contracted to 24/5=4.8 light years. So she has a space station that is 4.8 light years away, racing towards her at .6c even as the earth and Terence disappear in her rear view mirror at the same speed. She calculates 4.8/.6 to get 8 years for the station to reach her.

12. May 20, 2015

### controlfreak

Yes. This is the main point of contention wrt to (0,10) or (0,8) and Terrence is not even in the picture at this stage and is all between Stella and the space station. If we solve that without bringing Terrence into picture everything will follow and the contradiction will vanish. So the question is whether it takes the space station to reach Stella 10 or 8 years viewed from the reference of Stella's rest frame.

I was about to write it but you had added another post addressing the same. I believe Lorentz contraction happens only for moving objects in the direction of motion. So in the rest frame of Stella, she might possibly see the length of the space station moving towards her to contract in its length as it is moving but would she also see the space between her and the space station contract so that the space station travelling at 0.6 c cover that contracted space (4.8 Light years) in 8 years? Is the space moving towards her for it to contract as per Lorentz contraction? What does that mean? I am missing something here.

In other words, what is the difference between a space craft moving away from you at 0.6 c (in case of Terrence) and a space station moving towards you at 0.6 c (in the case of Stella). Why should both behave differently wrt distance covered. I still feel there should be a correction to the diagram as I suggested. Am I missing something?

13. May 20, 2015

### A.T.

My advice : Try to understand the Lorentz transformation as a whole, instead of thinking about it's individual effects like contraction, dilation etc.

On the top right you have a slider "Observers velocity in A's frame" that allows you to smoothly switch between different inertial frames.

Here also a good visualization of Lorentz transformation:

14. May 20, 2015

### Mentz114

Yes. You ignore any mention of proper time. The diagrams are correct, and the time on a clock between any two events $(t1,x1)$ and $(t2,x2)$ is $\sqrt{ (t1-t2)^2 - (x1-x2)^2 }$ which you can easily calculate. As A.T. has said - forget about length contraction and time dilation. It is all in the proper times.

15. May 20, 2015

### Staff: Mentor

They don't. In a frame in which the station and Terence are at rest the six light-year initial distance between Stella and the station is closed in ten years; in a frame in which Stella is at rest and Terence and the station are moving the 4.8 light-year distance between them is closed in eight years.

And as for why I keep saying that the distance between Stella at the start of the journey and the station at the start of the journey is 4.8 light-years using Stella's coordinates? Well, the length of something is defined as the distance between its endpoints at the same time. (If both endpoints are at rest relative to us we don't have to worry about the "at the same time" part because nothing is changing, but as the station is moving relative to Stella she has to be a bit more careful). So Stella is somewhere (the point x=0 in her coordinates, to be precise) when the station starts moving towards her at .6c at time t=0. Where is the station at the same time that Stella's clock reads t=0, using Stella's coordinates? It's at the point (0,4.8) - play some with the Lorentz transforms and you'll find that that is the only value that is consistent with Stella being at the origin of the frame in which Terence and the station are at rest at time zero and the distance between that origin and the station is six light years in that frame.

16. May 20, 2015

### Staff: Mentor

The diagram is correct. You are missing the relativity of simultaneity. You need to use the full Lorentz transform, not just bits and pieces.

Or you can take Mentz114's suggestion and forget coordinates and frames altogether and focus on the invariant quantities like proper time.

17. May 20, 2015

### Ibix

This is key. Up here in #8 you (@controlfreak) said:
But, as Nugatory points out, the frames are never symmetric. The origin and destination points are stationary in only one of them - they are moving in the other. There is symmetry only if we don't specify a destination point.

18. May 21, 2015

### controlfreak

Here below I list both paths in the diagram - the path I suggested and the path originally which was suggested by ghwellsjr. This diagram is drawn with the reference of Frame 2 (the frame which travels at 0.6c away from earth). Stella travels with Frame 2 in the first half of the journey and then the second half she turns around and travels in direction opposite to Frame 2's movement with the same speed (0.6c). Frame 1 is the one assuming the rest frame as earth.

I have been contending in the above ST diagram, the path taken by Stella is S2 and not S1 as originally proposed by ghwellsjr. The key point of contention is in Frame 2, whether the turn around point happens at tp1 (0,8) or tp2 (0,10)? I still hold Lorentz contraction is not for space but for moving objects. In Michelson Morley experiment or above cart video, the experimental set up itself contracted to reduce the length traveled by light to ensure you dont need to increase the speed of light for the light to do the round-trip in the given time. There distance in space outside the experiment didnt contract which I believe if I am not wrong was what was suggested above by Nugatory in reducing the space to 4.8 light years instead of 10 which I am not still able to understand convincingly though I get the calculation from the formulaes.

I appreciate the specific answers provided by ghwellsjr and Nugatory giving the relevant arithmetic (using problem parameters) in addressing the problem in a specific manner. Thank you for the same as it clarified my doubts well and also enabled to articulate my point of view well.

Say you forget about this problem and digress a bit about another. Say Stella is at earth and then there is a space station which is at 6 light years away and is moving towards her with 0.6 c, then with rest frame of Stella what would you conclude as to when the Space station will reach earth? 10 years or 8 years? I would say 10 years and I would think you will agree with that. Now what is the difference between this formulation and the original scenario. Both scenarios Stella is at rest and distance covered is measured from Stella's frame of reference and space station is moving towards Stella at 0.6 c and so why in one case space contracts and in the other case it doesnt and why in one case it should take 8 years coordinate time and 10 years coordinate time in the other case?

But having said all that, I see there is a way to prove that my contention about the path S2 is wrong. The slope of the line S2a is 45 degrees which means Stella is travelling in Frame 2 at light speed which is not true. The relative velocity based on the formula of relative velocities is :

v = (u - w)/ (1 - uw/c^2)

In this case when Stella is taking the reverse journey at .6c, the relative speed of Stella with respect to Frame 2 (which is moving in the opposite direction to Stella at 0.6c ) can be calculated by putting u = -0.6c and w = 0.6c and we get -(1.2/1.36) c. Considering we have ct coordinates. The slope should be (1.2/1.36) which is the slope of S1 and not S2a. So that proves my contention is wrong and S1 wins. Please note when calculating slope (tan theta) we need to take the angle between the line and time (vertical) axis.

But someone may question why should T and S meet at frame 2 coordinates (-15,25). Say if I draw a line S2a' with the slope (1.2/1.36) at tp2 which will be parallel to S1, it will intersect with T at a different point in the frame 2 coordinate system. If we calculate the coordinate time and distance at that point we get (-18.75,31.25). Can we not say that they meet at (-18.75,31.25) instead? This way we would have maintained the bend happens at tp2 and then slope is also now the right slope (1.2/1.36).

One problem is with respect to total proper time elapsed for T and S.

Calculating for T: At coordinate time 10, the proper time elapsed for T is $\sqrt(10^2-6^2)$ which is 8. And at coordinate time 25, the proper elapsed time for T is $\sqrt(25^2-15^2)$ which is 20. At coordinate point (-18.75,31.25), the proper elapsed time for T is 25 (if you do the calculation)

Calculating for S through different paths S1 and S2a' :

Path S1 - Till tp1, the proper time elapsed as per S1 path is 8 years which is also the coordinate time. The additional proper elapsed time for S if we follow S1 further till (-15,25) we will have that as $\sqrt(17^2-15^2)$ which is 8 years. So all is fine at (-15,25) - T's total elapsed is 20 yrs and S's elapsed is 16 (8+8) yrs just as we would expect. So Path S1 seems to be on the right track.

Path S2a' - Till tp2, the proper time elapsed as per S2 path is 10 years which is also the coordinate time. The proper elapsed time for S if we follow S2a' till the supposed meeting point with T which is at (-18.75,31.25) we get the proper elapsed time of S as $\sqrt(21.25^2-18.75^2) + 10$ which is 20 years.

So summarizing if we use the original path suggested by ghwellsjr, it all works out fine in terms of elapsed time at 20 for T and 16 for S. But if we use the S2a' path we get the elapsed time to be 25 for T and 20 for S which I am not sure what that means. If we assert that elapsed proper times should be same for T & S in all reference frames then we might need to go with path S1. But that is the assertion we started out to validate when twin paradox put that in question. If suppose that assertion is not valid then S2a' also seems to weirdly valid and Terrence and Stella meet when they are 25 and 20 instead of 20 and 16. I am not sure what that means.

In anycase this discussion has been quite useful to think deeper about these concepts. I am sure there is something not falling in place for me in mind and paradox is not fully resolved in my mind but these calculations did clear some confusions.

19. May 21, 2015

### Ibix

A standard derivation of the Lorentz transforms is the light clock, which is simply a light pulse bouncing back and forwards between two mirrors. You do not need to assume that the mirrors are attached to one another; I have no idea how you would introduce such a requirement into the maths.

Alternatively consider four spaceships at rest in some frame. Spaceships A1 and A2 are next to each other, and spaceships B1 and B2 are next to each other, but some distance away from A1 and A2. A1 and B1 are connected by a long piece of string.

Now view this from a frame where the ships are in motion. If Lorentz contraction applies to objects only, A1 and B1 (which are connected) must be closer together viewed from this frame since the string is contracted; however A2 and B2 (which are not connected) are not closer together. So you end up with a situation where an observer at rest with respect to the spaceships accepts that someone can lean our of A1's airlock and knock on A2's airlock, but according to a moving observer the ships are kilometers apart.

20. May 21, 2015

### Mentz114

You've taken a big step forward by working out those proper times yourself. You will find if you repeat that on one of the boosted diagrams that you get the same answers. I still think that is pretty amazing and is at the heart of special relativiry.

21. May 21, 2015

### controlfreak

Possibly it should be reworded and was said in a context where one body is at rest and the other is moving and we were contracting the space in between to calculate the time needed to cover the distance between them which I wasnt so comfortable with and indicated the space doesnt contract in that context.

But having said that I understand what you are referring to. Say there are two aircrafts moving at a similar velocity one behind the other and not connect by any physical string. Say a light signal is sent from one air plane to another in the direction of motion, we would say that the distance between the aircrafts will need to contract in the direction of its motion to calculate the light's motion from one aircraft to another and to calculate the time needed to cover the distance. Possibly in this case we are considering the two aircrafts put together as a super object virtually connected by similar velocities. Here they are not physically connected but gives the feeling of being connected (a continuum). If you look at it at the atomic scale nothing is physically connected as there is always space in between atoms, electrons, neutrons etc. An atom is just a super object of electrons and protons etc. They are connected through forces and bonds (generated through forces). But in the aircraft situation there is pretty much no appreciable force of attraction between them and so it is a case being "virtually" connected.

But the interesting question is what happens if one is an helicopter and is stalling and the other is an aircraft and moves at a velocity, will the distance between the aircrafts contract in the direction of the motion of the first aircraft. What happens when one aircraft is moving at a certain velocity and there is another aircraft which is moving in a different velocity behind it - how does the contraction work then when the light travels from first aircraft to another? How does the space contract then? Does it contract based on the relative velocity as calculated by SR.

Thank you. I dont think I have fully understood the concept to my satisfaction but someday it will happen I suppose but I am glad there has been an improvement in my understanding through this dialogue with various experts here in this forum.

I am not sure if this below problem has already been discussed in this forum but when I searched I wasnt able to find a reference to this formulation.

Problem : Assuming Terrence and Stella are in earth and a space station 6 Light years away respectively. They both start at the same time and then approach each other. Each is travelling at a speed of 0.5 c in opposite directions. How does it play out? We see that the relative velocity works out to 0.8 c. Both of them see that the other person ages slowly as the other person is moving at -0.8 c as seen from the rest frame of the other person. What happens when they meet midway. Who will be aged?

22. May 21, 2015

### A.T.

If by "see" you mean visually with a telescope: That will vary before and after they see the other start..

If by "see" you mean what they compute in their rest frame to happen with the distant twin: The other twin will age faster during the acceleration of your own ship, then slower once you move initially.

In both cases the clock rate differences will cancel out, so both meet after aging the same amount since start.

23. May 21, 2015

### PAllen

As far as special relativity is concerned, they will end up aged the same. Accounting for general relativity (gravitation), the one leaving earth will have aged very slightly less. Barring gravity, this problem is completely symmetric, so the answer must be 'the same'.

24. May 21, 2015

### Mentz114

If you drew a diagram and it was symmetric then you can guess that the proper times of the paths will be the same. Any problem of this kind can be resolved by working out the proper times.

I see PAllen has made this point while I was thinking.

25. May 21, 2015

### Staff: Mentor

The interplay of of time dilation and length contraction that we're describing has actually been experimentally observed - not with a cancer patient on a spaceship speeding towards the space station that holds the miracle cure as in your thought experiment, but with subatomic particles. These have short enough lifetimes that the entire journey can play out here on earth, and large particle accelerators would not work if Lorentz contraction applied only to moving objects.

Perhaps the best way of thinking about it is that Lorentz contraction applies to all distances, not just the distance between the ends of single object.

Imagine five meter sticks, touching one another end-to-end. This gives us one rod five meters long in a frame in which the rod is not moving. An observer moving at .6c relative to the assembly will measure the distance between the ends of the rod to be four meters long so will say that is length-contracted. Now suppose I remove the meter stick in the middle - this doesn't change the position or length of the other four, so I now have two rods two meters long and separated by a one meter gap and the distance between the left-hand end of the left-most meter stick and the right-hand end of the right-most meter stick is five meters.

But what about the observer moving at .6c relative to the assembly? Nothing has changed with the four meter sticks that we didn't touch, so he sees two rods, both 1.6 meters long and separated by a gap of .8 meters where the middle meter stick used to be, and of course the distance between the left-hand end of the left-most meter stick and the right-hand end of the right-most meter stick is four meters.

There are three kinds of distance here: the length of single objects (the meter sticks, the five-meter rod, the two two-meter rods); the length of an empty space gap (the gap between the two two-meter rods); and the distance between points on two different objects (the five meters between the left-hand end of the left two-meter rod and the right-hand end of the right two meter rod). All of these have length-contracted in the same way.

Last edited: May 22, 2015