Understanding the Twin Paradox: Exploring Special Relativity and Time Dilation

[NoahsArk]

Hello. Thank you to everyone who helped me with my previous post where I had general questions on special relativity. Several users suggested that I start by trying to understand the relativity of simultaneity which I did. Although I haven't mastered it, I understand it better now. Also, someone on another post recommended the book "Relativity Visualized" by Lewis Epstein which was also very helpful and uses R.O.S. as the foundation.

Regarding this post, I have been stuck on trying to understand the twin paradox. I've been trying to work up to an understanding of it. However, I've watched and read probably over ten different online lessons explaining the twin paradox, but I still don't understand it. Here are a few things that I'm trying to clarify:

1. In Epstein's book and in other lessons that I've seen, he uses a space time diagram to try and illustrate why one twin ages less than the other when they meet again. I'm not sure how to draw a space time diagram in here (I'd appreciate help on that as well), but the diagram is very simple so I'll just describe it: Earth twin has his X and Y coordinate system. He see's space twin get off into a rocket. On Earth twin's diagram, he will see space twin moving at some angle upwards and also to the right. That will be represented by a diagonal line moving in the positive X and Y directions. Then, on the return trip, the Earth twin will see the space twin moving in a diagonal line upwards and to the left. When the space twin returns to earth, the Earth twin will have aged more because he went straight through time up the Y axis, whereas space twin went in a bent line to the right and to the left back to the Y axis.

That all sounds logical, but the question is how would space twin's diagram of Earth twin's "journey" look like (since Earth twin is the one moving from space twin's perspective). It seems to me that it would be a mirror image of Earth twin's diagram. Space twin would see Earth twin first moving up to the left, and then, on the return trip, up and to the right. If that's the case they should be the same age when the meet again. If we drew side by side a space time diagram for each twin, how would those diagrams look compared to one another?

2. Can this all be explained with special relativity? If so, are there explanations that don't involve acceleration and the Doppler effect?

3. As a related question to question 1, I watched the minute physics video on the twin paradox:
Firstly, he says, as an answer to the asymmetry problem, that time rotates. I am very confused by what it means for time to rotate?? When he draws the diagram from the space twin's perspective, the lines are at a different angle then the lines from the Earth twin's perspective. Why is that? Also, he draws the diagram from the perspective of how the Earth twin perceives the time of the space twin, and then compares that to how the space twin views his own journey's time (proper time). Shouldn't we be comparing the Earth twin's perspective of space twin's time with space twin's perspective of Earth twin's time, and not with space twin's perspective of his own time?

Thanks

 

[PeroK]

First, I'd forget about that video and all like it. It's impossible to learn anything subtle from a video in the style of a corn-flake advert, with everything gabbled at break-neck speed.

Second, you can only really understand the twin paradox once you have nailed the basic concepts of SR. It's not the other way round: you can't (IMO) use the twin paradox to help learn the basic concepts of SR.

Third, if you can't understand the basic concepts of SR (spacetime diagrams, ROS, Doppler effect) from a formal text, then it's hard for anyone on here. Your question seems to me to reveal (I'm sorry to say) that you haven't really grasped the basics.

The key point you're missing is that the twin who accelerates is consantly changing their (inertial) reference frame (during the acceleration and deceleration phases), so they cannot use the same inertial reference frame/spacetime diagram throughout. The twin who stays at home can, of course, use the same inertial frame throughout.

One simple approach to the paradox is to trust the stay-at-home twin's calculations (you know you can trust them, because you've nailed the concepts of inertial reference frames and time dilation). And, you can simply say that you don't know how to do the calculations for the moving twin because of the periods of accelerating and decelerating reference frames.

It's nice, of course, to be able to analyse the problem from the traveling twin's perspective. But, logically, that isn't necessary. If you can prove it from the stay-at-home twin's perspective, you don't have to repeat the proof another way. One proof is enough.

However, ...

In my view, one simple way to understand things from the traveling twin's perspective is to analyse what they actually see (this is essentially just the Doppler effect calculated by hand, as it were). For example, if B moves at $\frac{4}{5}c$ for a distance of $4$ light years, hence $5$ years (in A's frame), then B sees only $1$ year pass on A's clock on the outward journey and $9$ years pass on A's clock on the return journey. And, because of length contraction, the journey time out and back for B is only 6 years in their own frame.

If your algebra and arithmetic are up to it, you can work all that out on a bit of paper for yourself. Or, you can trust the Doppler equations, which are effectively what you are verifying.

In any case, this verifies what you know from the analysis in A's frame: that 10 years have passed on A's clock and only 6 years on B's during the out and back journey that B made.

 

[Vitro]

That all sounds logical, but the question is how would space twin's diagram of Earth twin's "journey" look like (since Earth twin is the one moving from space twin's perspective).

The space twin cannot draw one Minkowski diagram of Earth twin's trip because the space twin is not inertial for the whole trip. He has to draw two separate diagrams, one covers the outbound trip until right before the turnaround, the second will cover the inbound trip right after the turnaround. But to figure out how to put the two together you'd need some math. Do you know how to use the Lorentz transformations? Try a numeric example.

 

[PeroK]

PS perhaps the most fundamental asymmetry between A and B lies in the distance between the starting point and the turnaround point. A will measure the distance between these as $4$ light years throughout. Whereas, B will initially measure this distance as $4$ light years, but after his/her acceleration to $\frac{4}{5}c$ will measure the distance as only $2.4$ light years. When he/she stops at the turnaround point, the distance is once more $4$ light years and so on. Hence, there is no doubt that B is changing his/her IRF and A is not.

And, perhaps, that alone shows why the situation between A and B is not symmetrical and there is actually no paradox.

 

[robphy]

Warning: Epstein doesn't draw the standard Minkowski spacetime diagrams. (While it may have some advantages for specific situations, it is unlikely to be as versatile as the Minkowski diagram.)

Have you looked at some of the PhysicsForums Insights?

• https://www.physicsforums.com/insights/geometrical-view-time-dilation-twin-paradox/
• https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/ [this one is mine]

 

[FactChecker]

The key point you're missing is that the twin who accelerates is consantly changing their (inertial) reference frame (during the acceleration and deceleration phases),

I think this is the point that the OP is alluding to that is unsatisfying within SR. Yes, the traveling twin accelerates because it's motion is referenced to the rest of the universe. But how does measuring velocities wrt the rest of the universe make the "stationary" twin's inertial reference frame better? Yes, the traveling twin can feel his acceleration and deceleration because his velocity is changing wrt the rest of the universe. But why does that happen? Einstein wrote about these type of questions as motivation for GR. I have seen a thread somewhere in this forum that discussed this in depth.

 

[robphy]

I think this is the point that the OP is alluding to that is unsatisfying within SR. Yes, the traveling twin accelerates because it's motion is referenced to the rest of the universe. But how does measuring velocities wrt the rest of the universe make the "stationary" twin's inertial reference frame better? Yes, the traveling twin can feel his acceleration and deceleration because his velocity is changing wrt the rest of the universe. But why does that happen? Einstein wrote about these type of questions as motivation for GR. I have seen a thread somewhere in this forum that discussed this in depth.

No need to look outside at the rest of the universe.
Suppose each ship has within it a ball sitting on a level frictionless table.
While the stay-at-home twin's ball will remain in place, the traveling twin's ball will not when that twin turns to return home.

In the most general version of this twin paradox, the point is that
elapsed wristwatch time between separation and reunion events depends on the worldline path between those events.
The inertial observer will log the longest elapsed time.
(The Euclidean analog of this is that the arc-length along a path joining two points is shortest for the straight-line path.)

 

[NoahsArk]

For example, if B moves at $\frac{4}{5}c$ for a distance of $4$ light years, hence $5$ years (in A's frame), then B sees only $1$ year pass on A's clock on the outward journey and $9$ years pass on A's clock on the return journey. And, because of length contraction, the journey time out and back for B is only 6 years in their own frame.

When I worked out this example, at 4/5c, γ would be 1.667. So, if 5 years went by for A, 3 would go by for B. On the way back, I think, another 5 years would go by for A, and another 3 for B. That's all from A's perspective.

From B's perspective, if 3 years went by for him, then 1.8 years will have gone by for A during the outward journey if I did it right. On the way back, another 3 will have gone by for B, and he'll see another 1.8 pass for A. So, when they come back together (and I'm sure I'm wrong on this based on how it sounds!), A will think he's aged 10 years and that B has aged 6, and B will think that he's aged 6 years and that A has aged 2.6 years! When A looks in the mirror, 10 years have gone by, but when B looks at him, he's only aged 2.6 years. That doesn't make sense to me but when I do the calculations it's what I get.

How could B see 9 years pass on A's clock on the return journey when he only saw three years of his own time passing? This is the opposite effect of time dilation. Should he be seeing less time pass for A since A is moving relative to him? If the answer involves the fact that B is accelerating, I thought the twin paradox could be understood without taking the acceleration factor into account (e.g. if we assume the period of acceleration was very small). If I'm wrong, and we do need to understand the effects of acceleration on the problem (which would involve changing gamma factors during the trip), please let me know. Wouldn't that be getting into general relativity?

The space twin cannot draw one Minkowski diagram of Earth twin's trip because the space twin is not inertial for the whole trip. He has to draw two separate diagrams, one covers the outbound trip until right before the turnaround, the second will cover the inbound trip right after the turnaround. But to figure out how to put the two together you'd need some math. Do you know how to use the Lorentz transformations? Try a numeric example.

But then how can the Earth twin draw one diagram? The space twin is accelerating away from him just like the Earth twin is accelerating away from the space twin?

Have you looked at some of the PhysicsForums Insights?

I just took a look. I think at my level it will take me time to understand. It uses hyperbolas which is complicated for me. Even the two dimensional graphs present me some problems! For example, in a stationary frame's coordinate system, the X and Y axis are at a 90 degree angles from one another like we usually see. If another frame is moving relative to the first coordinate system, its X and Y axis will be bent inward. Wont the person moving, however, see their X and Y axis the same way the person in the first frame saw his axis (i.e. at 90 degree angles from one another)? I'm assuming its the person in the stationary frame that sees the person in the moving frame as having tilted axis?

Epstein, for the first few chapters, helped a lot. I stopped understanding when he got into the twin paradox.

 

[robphy]

I just took a look. I think at my level it will take me time to understand. It uses hyperbolas which is complicated for me. Even the two dimensional graphs present me some problems! For example, in a stationary frame's coordinate system, the X and Y axis are at a 90 degree angles from one another like we usually see. If another frame is moving relative to the first coordinate system, its X and Y axis will be bent inward. Wont the person moving, however, see their X and Y axis the same way the person in the first frame saw his axis (i.e. at 90 degree angles from one another)? I'm assuming its the person in the stationary frame that sees the person in the moving frame as having tilted axis?

Epstein, for the first few chapters, helped a lot. I stopped understanding when he got into the twin paradox.

The hyperbolas are there, just like circles are there in Euclidean geometry.
However, the insights I referred to don't make explicit use of the hyperbolas.
In my case, I have developed an alternate structure to encode the hyperbolas.
The result is an accurate, physically-motivated visualization of the time elapsed along a piecewise-inertial worldline
without explicitly using the Lorentz Transformation formulas.

In my opinion, 2D graphs are essential. (Doing geometry with just words is difficult.)

In fact, you can play with a GeoGebra file I created for a talk:
https://www.geogebra.org/m/HYD7hB9v#

 

[Nugatory]

How could B see 9 years pass on A's clock on the return journey when he only saw three years of his own time passing? This is the opposite effect of time dilation. Should he be seeing less time pass for A since A is moving relative to him? If the answer involves the fact that B is accelerating, I thought the twin paradox could be understood without taking the acceleration factor into account (e.g. if we assume the period of acceleration was very small).

It can be, and indeed that is by far the easiest way to understand it. Once you understand it under the assumption that the period of acceleration is very small (essentially, B's path in the space-time diagram is two straight diagonal lines forming a sharp angle at the turnaround point), you understand it. Including the time spent accelerating (so that B's path is a hairpin curve around the turnaround point) complicates the calculation without adding any new physical insight.

If you haven't seen http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gap.html, take a look at it now. Pay particular attention to the "time gap" and "spacetime diagram" sections.

And here is a question/hint for you... B sees three years pass on his clock before he reaches the turnaround, so if we assume near-instantaneous acceleration, one nanosecond before the turnaround B's clock reads three years minus one nanosecond. One nanosecond after the turnaround, B's clock reads three years plus one nanosecond. Using the frame in which B is at rest, what does A's clock read at the same time that B's clock reads three years minus one nanosecond? Using the frame in which B is at rest, what does A's clock read at the same time that B's clock reads three years plus one nanosecond? Note that these are different frames because B's velocity relative to A is different in the two cases.

(And note also the appearance of the phrase "at the same time", which suggests that relativity of simultaneity is involved. It usually is).

If I'm wrong, and we do need to understand the effects of acceleration on the problem (which would involve changing gamma factors during the trip), please let me know. Wouldn't that be getting into general relativity?

You are not wrong, and no general relativity is involved.

 

[PeroK]

But then how can the Earth twin draw one diagram? The space twin is accelerating away from him just like the Earth twin is accelerating away from the space twin?

I really don't know what it will take to make you understand that the situation is not symmetrical and only one twin is accelerating. Acceleration is absolute, not relative. The Earth twin remains in a single IRF throughout; the space twin does not.

Until you understand that, you will just keep repeating the same fallacy that each twin can measure the other's motion in the same way. And then patently there is no way out of the paradox.

How you measure things from the space twin's perspective is problematic. But you really must stop insisting that the space twin remains in a single IRF throughout.

 

[PeroK]

Whether yet another explanation will help, I'm not sure, but here's my take on how ROS (leading-clocks lag) allows you to analyse the motion from the space twin's perspective.

First, you have A & B at rest on Earth and C at rest a distance of 4 light years away at the turnaround point. Initially, A, B & C all share the same IRF.

But, after B has accelerated to $\frac{4}{5}c$, B has changed his IRF and in B's reference frame things are no longer the same. A and B are still (approximately) colocated at the start and still at time $t = 0$ (approximately) in each of their reference frames. This is "when" B starts the journey, and in B's new IRF.

1) C is now only 2.4 light years away.

2) The time at C in the A-C IRF is not $t=0$. From B's perspective, when he starts his journey, the time in the A-C frame at C is $t_C = \frac{16}{5} = 3.2$ years.

In other words, from B's perspective, the clocks at A and C are no longer synchronised. And the time at C (in the A-C IRF) has already advanced by 3.2 years. If you don't understand this, you need to go back and master ROS.

Now, B's journey takes 3 years in his IRF and the clocks at A and C advance only by 1.8 years according to B (normal time dilation). So, when he arrives at C (and before he slows down), the clocks at A and C read 1.8 and 5 years respectively.

If he slows down quickly, then his clock reads 3 years when he stops, C's clock reads 5 years. And now, because once again B has changed IRF's, he finds that A's clock also reads 5 years, as A, B and C once again share the same IRF.

Exactly the same thing happens on the way back, except the roles of A and C are reversed. So, when B arrives back at A and stops, he finds 6 years have elapsed by his clock and 10 years on the clocks at A and C.

You can also see from this that you don't need the return journey to resolve the paradox. By changing IRF's twice, B has experienced less time than A & C during the outward journey.

Loosely speaking, the effects of time dilation from B's perspective were outweighed by the effects of ROS. So, although A & C's clocks ticked slower throughout from B's perspective, there was the ROS over the distance between A and C to take into account, and when you account for this, you find that more time passed on A and C's clocks than B's during the journey. (This is all from B's perspective.)

 

[NoahsArk]

And note also the appearance of the phrase "at the same time", which suggests that relativity of simultaneity is involved. It usually is

Loosely speaking, the effects of time dilation from B's perspective were outweighed by the effects of ROS.

This is actually starting to make sense! I'm going to need to think about it more over the next few days because it's still so strange. At the beginning, A, B, and C all saw their clocks at t = 0. Then, once B starts moving at .8c, C's clock all of a sudden jumps from t=0 to t=3.2 years! Can you please also post the general equation for calculating the lag between two clocks which are positioned along the direction of motion?

 

[Nugatory]

Then, once B starts moving at .8c, C's clock all of a sudden jumps from t=0 to t=3.2 years!

C's clock doesn't really jump. All that is happening is that when B starts moving B's definition of "at the same time" changes, and along with it his notion of what C's clock reads "at the same time" that B's clock reads some given value..

 

[PeroK]

This is actually starting to make sense! I'm going to need to think about it more over the next few days because it's still so strange. At the beginning, A, B, and C all saw their clocks at t = 0. Then, once B starts moving at .8c, C's clock all of a sudden jumps from t=0 to t=3.2 years! Can you please also post the general equation for calculating the lag between two clocks which are positioned along the direction of motion?

The "leading clocks lag" rule means that (using this case as an example) as B travels from A to C at speed $v$, clocks that are synchronised in the A-C frame are not synchronised in B's frame and:

$\Delta t = \frac{vD}{c^2}$

Where $\Delta t$ is the time difference and $D$ is the proper distance (i.e. the distance in the A-C frame) between the "synchronised" clocks.

In particular, if $t=0$ in the A-C frame when B leaves A, and $t_b = 0$ (in B's IRF) when B leaves A, then $t = +\Delta t$ at C when $t_b = 0$ (at C).

What this means is:

If B could carry out an experiment to determine what C's clock read "when" (according to B's IRF) he left A, he would find $t_C = +3.2$ years.

Whereas, A would conclude that when (according to A's IRF) B left him $t_C = 0$

This is a key aspect of ROS and, as has been mentioned before, this perhaps the most difficult aspect of SR to grasp.

I'm sure a good textbook would explain this much better than I have in a few lines here!

 

[Mister T]

Firstly, he says, as an answer to the asymmetry problem, that time rotates. I am very confused by what it means for time to rotate?

Time doesn't rotate. The time axis rotates as shown in the diagrams. The reason has to do with the way you draw lines of constant time, they take into account relativity of simultaneity. Two events that are simultaneous are connected by a line of constant time. But if you are moving relative to me, my line of constant time and your line of constant time are rotated relative to each other because events that are simultaneous to me are not simultaneous to you.

 

[NoahsArk]

The "leading clocks lag" rule means that (using this case as an example) as B travels from A to C at speed vv, clocks that are synchronised in the A-C frame are not synchronised in B's frame and:

Δt=vDc2

Thank you.

I'm sure a good textbook would explain this much better than I have in a few lines here!

Introducing C into the picture was very helpful.

C's clock doesn't really jump. All that is happening is that when B starts moving B's definition of "at the same time" changes, and along with it his notion of what C's clock reads "at the same time" that B's clock reads some given value.

Well the example got me thinking: Say A and C are two 30 year olds in their IFR, and they are standing the same distance apart like A and C in PeroK's example. When B is at rest relative to A and C, he will also see them as both being 30. Now, if B starts going at at a fast enough speed, and the time it takes him to reach that speed is near instantaneous, then it's possible that he'll see C as a 90 year old and A as still being 30. Also, if I understand it right, C himself would see himself as a 30 year old when B would see him as 90! Does that mean that if someone flew by me fast enough they'd see me as a 90 year old??

Time doesn't rotate. The time axis rotates as shown in the diagrams. The reason has to do with the way you draw lines of constant time, they take into account relativity of simultaneity. Two events that are simultaneous are connected by a line of constant time. But if you are moving relative to me, my line of constant time and your line of constant time are rotated relative to each other because events that are simultaneous to me are not simultaneous to you.

Thank you for clarifying this. I am going to need to learn the rules for drawing the diagrams (e.g. what the relationship is between the speed of the moving frame and the angle that the axis shifts.

 

[PeroK]

Well the example got me thinking: Say A and C are two 30 year olds in their IFR, and they are standing the same distance apart like A and C in PeroK's example. When B is at rest relative to A and C, he will also see them as both being 30. Now, if B starts going at at a fast enough speed, and the time it takes him to reach that speed is near instantaneous, then it's possible that he'll see C as a 90 year old and A as still being 30. Also, if I understand it right, C himself would see himself as a 30 year old when B would see him as 90! Does that mean that if someone flew by me fast enough they'd see me as a 90 year old??

You're still making two fundamental mistakes. The first is that these things are not what you "see", but the results of measurements. The second is that the lack of synchronicity is due to the distance between A and C: B cannot be in both places at the same time. So, B can never directly observe A (locally) and C (locally) at the same time.

The way I would think about this is to have a fourth person, D, at rest with respect to B. So that B and D share an IRF and D passes C "when" (in the B-D IRF) B passes A. Let's move away from the twin paradox for a moment and assume that B and D have established that they are at rest with respect to each other and (from their perspective) A and C are 2.4 light years apart and moving towards them at $\frac{4}{5}c$.

Now, B records the time on A's clock when he passes A; and D records the time on C's clock when he passes C. Then, they communicate the results to each other.

B communicates: "I passed A at $t_B = T$ (say) and A's clock read $t_A = 0$.

D communicates "I passed C at $t_D = T$ (this establishes that B was at A when D was at B - in the B-D frame) and C's clock read $t_C = 3.2 years$

This constitutes a "measurement" in the B-D frame of the lack of sychronicity of the A-C clocks. Neither B nor D can do it easily by themselves, but between them they can carry out this measurement.

Meanwhile, A and C measure their clocks as synchronised in their frame and, in their IRF, B passing A and D passing B were events that were 3.2 years apart. In other words: simultaneity is relative.

I suggest you go back to your textbook and master ROS. I don't think one can "half" understand ROS: you either understand it completely or you don't!

 

[Nugatory]

Well the example got me thinking: Say A and C are two 30 year olds in their IFR, and they are standing the same distance apart like A and C in PeroK's example. When B is at rest relative to A and C, he will also see them as both being 30. Now, if B starts going at at a fast enough speed, and the time it takes him to reach that speed is near instantaneous, then it's possible that he'll see C as a 90 year old and A as still being 30.

Go back and read Einstein's train example showing the relativity of simultaneity again. Instead of the lightning strikes, we have passengers A and C celebrating their 30th birthdays at each end of the train, and B is the observer on the train, in the middle. B accelerates by jumping off the train and ending up at rest on the platform.

And then back to your spaceships...
We may say that the events "A celebrated his 30th birthday" and "C celebrated his 90th birthday" happened at the same time using the frame in which A and C are moving.

We also may say that the events "A celebrated his 30th birthday" and "C celebrated his 30th birthday" happened at the same time using the frame in which they are at rest.

Let's consider exactly what those statements mean. No matter where you are and how fast you are moving, you will eventually see all three birthday celebrations in your telescope. If your clock reads T when you see one of these celebrations and it happened at a distance Q light-years from you, you will say that the celebration happened at time T-Q; you're just correcting what you see for the light travel time.

Even observers at rest relative to A and C may see the two 30th birthday celebrations in their telescopes at different times, because they may be different distances away. If A and C are one light-year apart, someone 1000 light years away will see one celebration happening when his clock reads 1001 years and the other happening when his clock reads 1000 years. That doesn't mean that the two events happened at different times, it means they happened at the same time and the light took longer to cover the greater distance between the more distant celebration and the telescope.

However, no one who is moving relative to A and C will ever find that these two celebrations happened at the same time. As in the at-rest case, they will eventually see all three celebrations in their telescope, but when they allow for the light travel time as they calculate what their clock read at the same time that the celebration happened, they'll get different results for the two 30th birthdays.

Furthermore, if we choose B's post-acceleration speed properly, observers who are at rest relative to B after the acceleration will find that A's 30th birthday celebration and and C's 90th happened at the same time after they allow for the light travel time. It's a good exercise to construct such a situation. (Hint: A and C must be at least 60 light-years apart in their rest frame).

 

[NoahsArk]

B communicates: "I passed A at tB=Tt_B = T (say) and A's clock read tA=0t_A = 0.

D communicates "I passed C at tD=Tt_D = T (this establishes that B was at A when D was at B - in the B-D frame) and C's clock read tC=3.2years

However, no one who is moving relative to A and C will ever find that these two celebrations happened at the same time. As in the at-rest case, they will eventually see all three celebrations in their telescope, but when they allow for the light travel time as they calculate what their clock read at the same time that the celebration happened, they'll get different results for the two 30th birthdays.

Since the two examples are similar, I'll combine them to see if I'm understanding R.O.S correctly.
1. There are two frames: A-C and B-D. In the A-C frame, D's clock is the lagging clock, and in the B-D frame, A's clock is the lagging one.
2. The proper length between B and D is less than the proper length between A and C. That must be true because in PeroK's example, according to the B-D frame, when the two frames past each other they were of equal length (A lined up with B and C lined up with D). Since B and D are measuring a contracted length of A-C and getting an = measure, A-C's proper length must be bigger.
3. If, in the A-C frame, A and C celebrate birthdays on the same day, then in the B-D frame C must be older. When I say "on the same day" I am taking into account the time it takes for light to travel.
4. In the B-D frame, there comes a point where the nose and tail end of the two frames line up (A with B and C with D). On a two by two checker board, A and B would be the bottom row and C and D would be in the top row. In the A-C frame, however, A and B will cross each other first, and then only later will C and D cross. That's because according to the A-C frame, B and D are compressed together, and it takes longer for D to get to C.
5. All four of them have cameras. A and C, who are the exact same age in their frame, decide that exactly when they turn 30, they will take pictures of themselves blowing out the candles on their cake. We don't need to worry about light travel time because they took the pictures up close. It just so happens that as A is blowing out the candles, B is passing by and appears in the photo. When C takes his photo, D hasn't passed him yet and doesn't appear in the photo. A and B get together and compare photos. A appears in his photo with B (who is also taking a picture of A as described below), and C appears in his photo alone.
6. B and D, whose clocks are synchronized in the B-D frame, agreed to take simultaneous photographs when B sees A blowing out the candles (they happen to know A's birthday is coming up and they know when that will be in their frame). They then get together and compare photos. B's photo will show pretty much the same thing (with differences only due to the angle the photo was shot from) that A's photo showed: a thirty year old A blowing out candles. D's photo will show C, but C will be older than 30. How much older depends on the velocity of the B-D frame in relation to the A-C frame.
Is any of this right so far?
7. Here is the most confusing part for me: Now say all for of them are in the same reference frame at rest with respect to each other. A and C are 29 and have their birthdays tomorrow. B and D, since they are both at rest with respect to A and C, measure A and C to be 29 also (after adjusting for light travel time). That same evening B and D suddenly start moving with respect to A and C at a speed close to the speed of light. The very next day, A and C will measure each other's age at thirty, but B and D will have measured A at 30, and C at 90. Is this what could happen if B and D went fast enough? If that's actually what could happen, then how can we reconcile that for B and D, C's clock, due to time dilation, is moving slower than their own, but yet, due to R.O.S, it's moving faster than their own?

 

[PeroK]

Most of what you say is correct, although some of it confuses me. Then, point 7 undoes much of the good work.

In 7, you've introduced a complexity without noticing. If B and D simultaneously accelerate in their reference frame, then they do not simultaneously acclelerate in the A-C frame. They begin to accelerate simultaneously in the A-C frame, but once B & D have a relative velocity wrt A & C, then the simultaneity is lost. A & C will measure B accelerate increasingly ahead of D (becauses D's clock gets increasingly behind B's in the A-C frame).

Part of the reason you missed this is that you are thinking in terms like: has a birthday the next day. The ROS means that phrases like that are universally meaningless. The next day in whose reference frame?

My advice is to put to one side all the everyday language and start thinking about SR in terms of events with time and space coordinates in each reference frame. That's dry and mathematical, of course. But, with everyday language there are too many pitfalls.

You have events and in each IRF an event has a time $t$ and a location $x$. In another IRF, the same event has a time $t'$ and a location $x'$. These are the coordinates of the event. You must specify these for each event, otherwise you are simply not being precise enough.

 

[NoahsArk]

PeroK, in your earlier example, when there was just A,B, and C in the picture, you said that if they are all stationary relative to each other, then all their clocks will be synchronized (at T=0). Then you said the instant that B reaches the speed of 4/5c relative to A and C, then B will measure C's clock at 3.2 years and A will still be at T=0 (if B's acceleration to 4/5c was instantaneous). That's what's so strange to me- that in one instant B's measure C's clock went from 0 to 3.2 years.

Part of the reason you missed this is that you are thinking in terms like: has a birthday the next day. The ROS means that phrases like that are universally meaningless. The next day in whose reference frame?

Going back to the last example, if B is traveling at 4/5C, and if one day went by in the A/C frame, then just over a day and a half went by B's frame. Yet for B, a day and a half earlier he measured C to be 29 years old, and now he's measure him to be 90. This is also surprising for the same reason that it was surprising in your original example where C went from T= 0 to T=3.2 years in an instant for B. .

In your first example, where B measure C's clock at 3.2 years, I'm assuming it doesn't matter where B's position was relative to A and C when he reached the speed 4/5C. In your example he was positioned next to A, but even if he was positioned close to B, the instant he started moving at 4/5C he would measure C's clock at 3.2 years? Thanks

 

[PeroK]

In your first example, where B measure C's clock at 3.2 years, I'm assuming it doesn't matter where B's position was relative to A and C when he reached the speed 4/5C. In your example he was positioned next to A, but even if he was positioned close to B, the instant he started moving at 4/5C he would measure C's clock at 3.2 years? Thanks

Your assumption is wrong (because you haven't taken the time to learn the "leading clocks lag" rule). This is my last attempt.

If you have a sequence of synchronised clocks at rest with respect to each other. Let's say there are 9 clocks, all 1 light-year apart. And you have an observer (B) moving at $\frac{4}{5}c$ with respect to this row of clocks, then in B's reference frame, the clocks will read:

-3.2 years, -2.4 years, -1.6 years, -0.8 years, 0 years, +0.8 years, +1.6 years, +2.4 years, +3.2 years

Assuming B is passing the middle clock. In other words, every clock in the "stationary" frame is at a different time, depending on how far it is from B. Clocks in front of B (in B's direction of motion) are ahead; clocks behind B are behind.

If B started at rest next to the middle clock and accelerated almost instantantaeously, then those are the times he would measure on the stationary clocks.

If B started next to the furthest clock on the right, then after the acceleration he would measure:

-6.4 years, -5.6 years, -4.8 years, -4.0 years, -3.2 years, -2.4 years, -1.6 years, -0.8 years, 0 years

 

[Mister T]

Then you said the instant that B reaches the speed of 4/5c relative to A and C, then B will measure C's clock at 3.2 years and A will still be at T=0 (if B's acceleration to 4/5c was instantaneous). That's what's so strange to me- that in one instant B's measure C's clock went from 0 to 3.2 years.

That's a 3.2 year shift in the reckoning of simultaneous events, not a 3.2 year shift in a clock reading. Focus on the fact that C is distant from B. Awareness at B of events at C occurs much later. It's only after B accounts for this delay that he can reckon when the event occurred. He cannot be present for it.

 

[NoahsArk]

Your assumption is wrong (because you haven't taken the time to learn the "leading clocks lag" rule).

I just ordered Introduction to Special Relativity by Helliwell- you recommended a book by him. I also ordered one by Smith. To be honest, when I look at the books online, I don't see much written about R.O.S. I got the Wheeler book which is good, but he gives R.O.S. one page of discussion. If you or anyone else knows of something specifically good for working on R.O.S., with problems, please let me know. Nugatory you initially emphasized that R.O.S. is key for understanding S.R., which got me thinking a lot more about it.

When we were talking about R.O.S. by itself, without changing IFRs, I got the basics- e.g. leading clocks lag, and there is relativity of simultaneity in the direction of motion, and that even though clocks may be out of sync for an observer moving relative to them, they will still tick at the same rate. I also understand that R.O.S. has nothing to do with when someone "sees" an event happening. Whenever I say that A "observes" or "measures" two events happening at the same time, I mean that A already adjusted for the time it took light to travel to his eyes before concluding that they happened at the same time.

That's a 3.2 year shift in the reckoning of simultaneous events, not a 3.2 year shift in a clock reading.

That's the point that is throwing me off. If there was a 3.2 shift in the reckoning of simultaneous events, then there must have been a shift in clock reading. All of the 9 clocks in the above frame were in sync for the person stationary with respect to them. As soon as they went out of sync it means they had different readings for the observer that they went of sync for. If they still had the same readings then they wouldn't be out of sync (at least that's one of the things causing me trouble).

Here's one last example for now which hopefully pin points my misunderstanding better. I hope this also clarifies for me a little the relationship between time dilation and relativity of simultaneity. Take a very long rocket ship (one light year in length) at rest on Earth getting ready for take off. I'm standing outside the rocket waiting for it to leave. At both ends of the rocket, there are clocks- clock A is at the tail, and clock B is at the nose. I also have my own wrist watch. I observe the minute hand of all three clocks to strike 11:55am at the same time (the light from clock B will take longer to get to me then the light from the other two clocks, but I'll still conclude, after adjusting the time it took for clock B's light to travel to me, that all three clocks struck 11:55am at the same time). The clocks are all ticking harmoniously, and a minute later they all strike 11:56, then a minute later they all strike 11:57 etc. At exactly 12 noon, the rocket instantaneously starts flying at a speed of .8c. Now we have two different IFRs.

What happens now? As far as time dilation goes, in my IFR a minute later I'll observe the minute hand on my wrist watch strike 12:01pm. While a minute passed for me, I'll observe that about only 35 seconds have passed in the rocket frame.

But what about the effects of R.O.S? As a side note, one subtle point of confusion is the distinction between clocks on the rocket going out of sync vis a vis other clocks on the rocket (here A and B), and clocks on the rocket going out of sync with my wrist watch or with any other clock in another IFR. Are there separate rules I need to learn for that? Regarding the clocks in this example, at 12:01pm in my IFR, if I left out the effect of R.O.S., I'd assume that I'd observe both clock A and clock B would read 12:00 and 35 seconds. What would they actually read, though, when my watch reads 12:01? If I use the equation for R.O.S., I'd get a time difference of .8 between clocks A and B. Does that mean I'd observe clock A to read 12:00:35 when clock B to reads 12:00:28? Phrased more in R.O.S. terms, would I observe the events "A's minute hand strikes 12:00:35" and "B's minute hand strikes 12:00:28" to be simultaneous? If so, which of those two events is simultaneous with the event "my minute hand on my wrist watch strikes 12:01"?? Or are both events simultaneous with my minute hand striking 12:01? That gets more into the confusion about when two events in different frames are simultaneous. As another poster pointed out, the concepts of S.R. are very subtle. It took me a long time to even think of how to phrase my question and articulate what's confusing to me.

Again, thanks for the help.

 

[Mister T]

Take a very long rocket ship (one light year in length) at rest on Earth getting ready for take off. I'm standing outside the rocket waiting for it to leave. At both ends of the rocket, there are clocks- clock A is at the tail, and clock B is at the nose. I also have my own wrist watch. I observe the minute hand of all three clocks to strike 11:55am at the same time (the light from clock B will take longer to get to me then the light from the other two clocks, but I'll still conclude, after adjusting the time it took for clock B's light to travel to me, that all three clocks struck 11:55am at the same time). The clocks are all ticking harmoniously, and a minute later they all strike 11:56, then a minute later they all strike 11:57 etc. At exactly 12 noon, the rocket instantaneously starts flying at a speed of .8c.

You'll be close enough to see the tail of the rocket lift off, but you'll still have almost a year of time to wait to verify that the clock at the nose read noon at take-off. By that time the rocket will have traveled 0.8 light years.

You seem to be ignoring the fact that you have to wait for the awareness of distant events.

 

[NoahsArk]

"You seem to be ignoring the fact that you have to wait for the awareness of distant events."

I did not ignore that. I specifically stated that we are adjusting for the time it take light to travel. My question relates to what the measurements will be when I get them.

 

[Nugatory]

At exactly 12 noon, the rocket instantaneously starts flying at a speed of .8c. Now we have two different IFRs.

1) There's no "now we have two" here - you ALWAYS had two different inertial frames (actually, you always have an unlimited number of inertial frames, but in this problem only two are being used). There is a frame (#1) in which the surface of the Earth is at rest, and there is a frame (#2) in which the surface of the Earth is receding at .8c. If you choose to describe the problem using the first frame, then the rocket starts at rest and accelerates to a speed of .8c. If you choose to describe the problem using the second frame, then the rocket was moving at speed -8c and accelerated until its speed was zero and it was at rest. Either way, both the rocket and the Earth are in both frames all along (as well in the frame in which an observer on Mars is at rest, and a frame in which a spaceship zooming towards the Earth at .99c is at rest, and a frame in which ...).

2) You have specified that at the same time that the liftoff was initiated all the clocks at various positions along the ship read 12:00 noon using frame #1's definition of "at the same time". However, because of relativity of simultaneity, the "this clock reads 12:00 noon" events for the various clocks along the length of the ship do not happen at the same time, nor at the same time that the liftoff was initiated, if you use frame #2 (or any other frame).

3) When you say "the rocket instantaneously starts flying at .8c" you are basically saying that all parts of the rocket start moving at the same time (and there is that phrase again!). If all parts of the rocket start moving at the same time using one frame (#1, the way you've set the problem up), then they do not start moving at the same time if you use any other other frame to assign time values to the "starts moving" events.

(You may think you can evade #3 because the rocket is a rigid object, so if one part of it moves all parts of it must move at the same time because that's what "rigid" means. In fact there are no perfectly rigid objects; when we say that something is "rigid" that's an approximation that only works when relativistic effects are not significant. You can no more make one end of a rod move at the same time that you push the other end than you can make a wave appear on the African coast at the same time that you disturb the water on a South American beach - you have to wait for the disturbance to travel the breadth of the ocean).

 

[Vitro]

@NoahsArk, take a look at your watch right now. Is the reading advancing steadily one second at a time? Why isn't it shifting back and forth randomly since there are countless "observers" in the Universe changing speed relative to it? In fact shouldn't it show many different readings all at once, a different one for each observer? But it doesn't do that, it only has one reading advancing one second at a time, regardless of any observer. So a clock's reading doesn't change as a result of an observer changing speed.

What if the clock itself chages speed? Well, its rate will change but the instantaneous reading won't; if it undergoes a boost (instantaneous change from one constant speed to another) it will show the same reading right before the boost and right after.

Let's take three clocks, all at rest and synchronized, one at x = 0, a second one at x = -1 (light-year) and the third at x = 1.

You are at rest next to the clock at x = 0, and when it shows t = 0 you boost yourself to v = 0.6c towards the clock at x = 1. We have established that from the three clocks' point of view the fact that you changed your speed has no effect on their readings, they will all still show t = 0 right after your boost. From your perspective, after the boost, you'll ask what do the three clocks read "now"? The clock right next to you will still read 0 but the clock at x = 1 (per the previous IFR) now reads 0.6 (years) and the one at x = -1 now reads -0.6. But that's not because the two distant clocks jumped forward/backward in time or their reading shifted, they always ticked one second every second their entire existence, no jumps at all, it's your notion of "now" that changed and those are the normal readings on the clocks when observed at the same time in this other IFR.

Let's reset, you are again at rest next to the middle clock but this time we make the three clocks undergo a boost to v = 0.6c at the same time, when all read t = 0, while you remain stationary. In your IFR, right after the boost, they all still read t = 0; they are still synchronized in your IFR which means they are not synchronized in their new rest frame. Taking the clock at x = 0 as the reference it means the other two clocks are showing the wrong time in their new IFR, so they should be pre-programmed to adjust their readings immediately after the boost, the clock at x = 1 has to set itself to -0.75 and the one at x = -1 must set its reading to 0.75.

If you found that ROS doesn't get a lot of indepth treatment I would think it's because there's really not that much to say about it. All you need to remember is that two space-like separated events must happen simultaneously in one IFR and only that IFR. Beyond that just identify any relevant events and frames and apply the Lorentz transformations.

 

[Stephanus]

@NoahsArk. My belated welcome to PF Forum
If I may share my experience with understanding SR. You have to solved it yourself.
For solving Twins Paradox, you only need very basic Lorentz Factor and high school math.
The very basic rules to solve Twins Paradox are these:
1. The parties should travel in constant speed, no acceleration involved. After you have mastered Twins Paradox, then you can study SR with acceleration involved.
2. Rest and travel are abstract things in SR. A "rest" observer will see that the "traveling observer" is (of course traveling), and the "traveling observer" will see that it is the rest observer who is traveling.
3. A "rest" observer will see the "traveling" observer's clock dilated.
4. Rule 3 will also appliy to the "traveling observer". The traveling observer will see that the "rest" observer's clock dilated.
5. Ignore Time dilation for this problem, first. We only use Doppler effect. But later to understand Twins Paradox, time dilations play a very important role.

Okay..., so let's begin our thought experiment. No Time dilation for now.
Use doppler's effect or your logic.
Two observers A and B
A stays and B moves 0.6c to the west. And in B frame, it's B who stays and A travels 0.6c to the east.
The distance between B and A2 is 225 millions KM (in A frame).

A1 keeps sending signal every 1 milli second (in A Frame) to B and B keeps sending signal every 1 milli second (in B Frame) to A1.
When B reach A2 (both frame agree with this event) B turns around and heads back to A1 at 0.6c.

So the question is this:
1. Before B reaches A2 how many ticks per second (ignore time dilation, ignore time dilation) that B receives from A1 and how many ticks per second that A1 receives?
2. The moment B reaches A2 and turns back to A1, how many ticks per second that B receives and so does A1? (ignore time dilation, ignore time dilation)
You might want to solve this yourself, to under Twins Paradox I think.
Then we analyze further the Doppler effect, then the combination with time dilation.

 

[NoahsArk]

When you say "the rocket instantaneously starts flying at .8c" you are basically saying that all parts of the rocket start moving at the same time

That's something I didn't consider, and that fact, that the bottom of the rocket starts moving first, makes my example more complicated:)

But that's not because the two distant clocks jumped forward/backward in time or their reading shifted, they always ticked one second every second their entire existence, no jumps at all, it's your notion of "now" that changed

That's the subtle point that I'm trying to wrap my head around, that my notion of "now" has changed, but so far have not been able to do. I have the notion (which is false based on the responses here) that when I read X = .6, after it earlier read X=1, that I've traveled into the past of that clock! That probably sounds silly, but its how I'm seeing it. Likewise, when the clock went from -1 to -.6, I feel like I've jumped into that clock's future.

With time dilation, I can at least kind of see that the clocks in the other IFR are just running slower relative to mine, and it doesn't seem as far fetched as what's happening in the scenarios above.

Stephanus in your example it seems like both A and B will have received the same number of signals. They both moved the same distance relative to each other.

 

[Nugatory]

With time dilation, I can at least kind of see that the clocks in the other IFR are just running slower relative to mine, and it doesn't seem as far fetched as what's happening in the scenarios above.

Even though your clock is also running slower than the clock in the other frame? That's a question, not an argument, but as I've pointed out above... Time dilation is derived from relativity of simultaneity, so it's hard to see how you can accept time dilation yet find relativity of simultaneity more far-fetched.

 

[Mister T]

Even though your clock is also running slower than the clock in the other frame? That's a question, not an argument, but as I've pointed out above... Time dilation is derived from relativity of simultaneity, so it's hard to see how you can accept time dilation yet find relativity of simultaneity more far-fetched.

Well, strictly speaking, it's the symmetry of time dilation that requires relativity of simultaneity for it to make sense.

It's very easy to start with the derivation of time dilation (using the light clock) and not address the issue of relativity of simultaneity until later. This is of course a mistake, because it leaves the symmetry of time dilation either ignored or a mystery to be solved later.

 

[NoahsArk]

Even though your clock is also running slower than the clock in the other frame? That's a question, not an argument, but as I've pointed out above... Time dilation is derived from relativity of simultaneity, so it's hard to see how you can accept time dilation yet find relativity of simultaneity more far-fetched.

Time dilation at least makes visual sense for me- yes, even though my clock is also running slower in the other frame. A helpful comparison for me from the "Relativity Visualized" book was the comparison to perspective in art, or how I see someone standing far away to be shorter, but they also see me to be shorter.

Relativity of simultaneity makes visual sense too, and the example you used with the two painters and the brushes in the earlier post was helpful. Its when someone is in one frame and has one view about the simultaneity (or lack thereof), between two events, and then switches frames and has another view about the simultaneity of the events that is hard to accept. If events A and B happened at the same time in my frame, its hard to accept then when I switch frames I'll conclude they happened at different times.

 

[m4r35n357]

That's the subtle point that I'm trying to wrap my head around, that my notion of "now" has changed, but so far have not been able to do.

Changing of "now" is just an inevitable consequence of the concept of simultaneity changing with velocity (boost). Are you familiar with the Andromeda Paradox? That is about as clear an explanation as I can find ATM.

It is what it is, and doesn't necessarily describe any sort of "reality". Think of it as accounting if that helps, and just accept it ;)