Twin Paradox: Who Is Right, A or B?

[Ibix]

This seems overly complex. Why not just write down the coordinates of the events in the rest frame of the planets (departure is $t=0,x=0$, arrival is $t=10,x=6$, return is $t=20,x=0$) and then use the Lorentz transforms to get the coordinates in other frames? Why are you moving the origin about? That just confuses things and doesn't add anything.

Also note that the usual convention is to draw Minkowski diagrams with the time axis vertical.

 

[robphy]

I'm working on a complete and definitive deconstruction of the twin paradox, accompanying it with space time diagrams and the Lorentz transforms for each of the frames involved. Down to business -

[snip]

So we have the travelers analysis of the trip, with a total trip time of 16 years, while the twin on the near planet ages 20 years.

Sorry, I haven't been following closely
but...
Is this still the "one-way" twin paradox posed in the OP and in the title?
Or, is this now the standard twin paradox? with the one-way variation now abandoned?

 

[Will Flannery]

This seems overly complex. Why not just write down the coordinates of the events in the rest frame of the planets (departure is $t=0,x=0$, arrival is $t=10,x=6$, return is $t=20,x=0$) and then use the Lorentz transforms to get the coordinates in other frames? Why are you moving the origin about? That just confuses things and doesn't add anything.

Also note that the usual convention is to draw Minkowski diagrams with the time axis vertical.

I think my version is as simple as it can be and still cover the necessary considerations, and it does have the advantage that all the steps are detailed. But, maybe I'm wrong. If you 'filled in the details' of your approach I could get a better idea. And, generally, I've found that I think I understand something, and then set out to do the spacetime diagrams, and I find out my understanding is not as good as I thought. I think space time diagrams are am essential heuristic aid. I know I got the convention wrong, but I started wrong, tried to change, but couldn't.

 

[Will Flannery]

Sorry, I haven't been following closely
but...
Is this still the "one-way" twin paradox posed in the OP and in the title?
Or, is this now the standard twin paradox? with the one-way variation now abandoned?

Yes, I abandoned the one-way paradox long ago, after about my 3rd post.

 

[Ibix]

I think my version is as simple as it can be and still cover the necessary considerations,

I don't understand why you moved origins. That's unnecessary and confusing. If you want your origin to be the turnaround event, why not just have the turnaround occur at (0,0) in all frames?

If you 'filled in the details' of your approach I could get a better idea.

What is there to fill in? Once you've specified the coordinates of the key events, you just need to transform them into whatever frames you wish to work in. It's just far easier to specify the coordinates of all three events in the rest frame of the Earth - you can do it pretty much by inspection. If we change to using the turnaround as the origin then by simple arithmetic the coordinates are:

• Departure: $t=-10,x=6$
• Turnaround: $t=0, x=0$
• Return: $t=10, x=6$

Then you can use the Lorentz transforms to get them in the outbound rest frame:

• Departure: $t'=-8, x'=0$
• Turnaround: $t'=0, x'=0$
• Return: $t'=17, x'=15$

Likewise you can get the coordinates in the inbound frame - given the above, the answers are obvious from symmetry. The large values for the return coordinates in the outbound frame are easy to explain. In this frame the Earth is moving at 0.6c in the +x direction and has a sizeable headstart over the return leg, which is only done at 0.88c, not a huge velocity difference.

I think space time diagrams are am essential heuristic aid.

I agree. I don't understand your diagram, though, since the three black lines, which I take to be the worldlines of the traveling and Earthbound twins don't meet. They should form a triangle, isosceles in the Earth frame.

I know I got the convention wrong, but I started wrong, tried to change, but couldn't.

I'd note that using a non-standard convention will put you at a significant disadvantage in trying to understand other people's diagrams. If you truly find switching axes challenging, I strongly suggest you put in the effort to overcome it or you will struggle with interpreting almost every spacetime diagram you encounter.

 

[Will Flannery]

I don't understand why you moved origins. That's unnecessary and confusing. If you want your origin to be the turnaround event, why not just have the turnaround occur at (0,0) in all frames?

What is there to fill in? Once you've specified the coordinates of the key events, you just need to transform them into whatever frames you wish to work in. It's just far easier to specify the coordinates of all three events in the rest frame of the Earth - you can do it pretty much by inspection. If we change to using the turnaround as the origin then by simple arithmetic the coordinates are:

• Departure: $t=-10,x=6$
• Turnaround: $t=0, x=0$
• Return: $t=10, x=6$

Then you can use the Lorentz transforms to get them in the outbound rest frame:

• Departure: $t'=-8, x'=0$
• Turnaround: $t'=0, x'=0$
• Return: $t'=17, x'=15$

Likewise you can get the coordinates in the inbound frame - given the above, the answers are obvious from symmetry. The large values for the return coordinates in the outbound frame are easy to explain. In this frame the Earth is moving at 0.6c in the +x direction and has a sizeable headstart over the return leg, which is only done at 0.88c, not a huge velocity difference.

I agree. I don't understand your diagram, though, since the three black lines, which I take to be the worldlines of the traveling and Earthbound twins don't meet. They should form a triangle, isosceles in the Earth frame.

I'd note that using a non-standard convention will put you at a significant disadvantage in trying to understand other people's diagrams. If you truly find switching axes challenging, I strongly suggest you put in the effort to overcome it or you will struggle with interpreting almost every spacetime diagram you encounter.

# The origins followed the traveling twin, which seems like the natural thing for me. He was changing frames, he was at the origin of his frame, so the new frame should have the same origin. The origins of all the frame changes is (0, 0) in the traveler's frame, and the new frame has the same origin. I assumed that was the norm.

# "What is there to fill in?" Believe it or not, I cannot follow your method at all. Maybe if I studied it a lot, but, I'm happy with my method. :) ... All this stuff seems very confusing to a newbie, I think, and I spent 10 years doing inertial nav for rockets !

# I didn't provide sufficient explanation. I'll explain the three black lines in more detail. The spacetime diagram is when the traveler changes frames when he starts the trip home. I thought there was an error when I saw that the coordinates of the original departure from the near planet were (-17, -15) in the new frame. I thought it had to be a mistake and after much gnashing of teeth I decided to tweak my spacetime diagram program to handle negative velocities so I could draw the diagram. And after a while I was able to draw the diagram and saw that the (-17,-15) was right ! So, that's the only purpose of the lines labeled -15 and -17 in the diagram. The line labeled 8 is the traveler's world line on the return trip and it intersects the near planet's worldline after 8 years in the traveler's frame.

 

[Ibix]

The origins followed the traveling twin, which seems like the natural thing for me.

Note that the frame origins are events - they can't follow anyone. In his rest frame(s) the traveller's spatial coordinates are constant and you can choose his location to be the spatial origin, yes. My point was thhat if you pick the origin event of all frames to be the turnaround event then you get the traveller at the spatial origin on both legs of the journey for free, without having to introduce a fourth frame (the one you called A') that's the same as another frame (the one you called A) except for a translation.

"What is there to fill in?" Believe it or not, I cannot follow your method at all.

I did the same thing you did, but with a better choice of origin to achieve the "traveller at rest at the spatial origin" effect you wanted. I used $x=x_0+vt$ to write down the coordinates of the three important events (departure, turnaround and return) for a ten year journey at -0.6c and another ten year journey at +0.6c given an instantaneous turnaround at the origin. Then I used the Lorentz transforms to get the coordinates in the frame where the traveller is at rest on the outbound leg. I didn't do the calculation for the other leg - you can do so explicitly or, as I suggested, just cheat because this problem is highly symmetric.

On closer inspection, I see that this is in fact a description of the second spacetime diagram in your post #90. So all I'm saying is that your first diagram is unnecessary and confusing. You also haven't drawn the journey in any other frame, only the Earth twin's frame. You may find it enlightening to draw those frames' versions of events.

The spacetime diagram is when the traveler changes frames when he starts the trip home.

As written, this doesn't make sense. A spacetime diagram covers - in principle - all of spacetime, so "when" is meaningless applied to the diagram. I think you mean that the line marked 8 is (approximately, since it doesn't connect the dots marking the events) the traveller's worldline on the return leg. The other two lines don't make sense in that context. The 15 and 17 are coordinate differences in the traveller's frame, so lines referring to them ought to be parallel to the traveller's frame's coordinate grid, which they manifestly are not.

I thought there was an error when I saw that the coordinates of the original departure from the near planet were (-17, -15) in the new frame

As I attempted to explain in my last post this is because, in the traveller's rest frame for the second part of the journey, the Earth is moving towards him at 0.6c and the first leg of the journey was at speed 0.88c, not that much faster. So the departure event has to be a long time in the past for the first leg to have built up enough of a lead over the Earth that it takes eight years sitting still for the Earth to catch up.

I note that I actually explained it last time with reference to the second leg as seen from the first leg rest frame, so the explanation was the other way around.

 

[Nugatory]

He was changing frames, he was at the origin of his frame, so the new frame should have the same origin. The origins of all the frame changes is (0, 0) in the traveler's frame, and the new frame has the same origin. I assumed that was the norm.

i do not believe it is possible to write a clear and complete description while using phrases like “He was changing frames”, and similar. The traveller isn’t switching frames, you are in your description of the problem.

There is an inertial frame in which the Earth is at rest. There is an inertial frame in which the Earth is moving to the left with a constant speed $v$ and the traveller is at rest until the turnaround. There is an inertial frame in which the Earth is moving to the right with constant speed $v$ and the traveller is at rest after the turnaround.

We can use any of these frames to describe the situation and calculate the elapsed time between separation event and reunion event for both twins, and we’ll get the same answer. We can use any of these frames along with the Doppler effect to describe what the twins see when they watch each other’s clocks through a telescope and we’ll get an intuitive description of their experience throughout.

That’s the clear way. There’s also an unclear way, which is what you’re trying to do: Use one frame for the first half of the analysis and a second frame for the second half. If so we are the ones changing frames, not the traveller - the traveller is “in” all frames throughout. This approach is complicated and error-prone and if we are not careful with the relativity of simultaneity will lead to results that appear paradoxical.

It is worth noting that when Einstein introduced differential aging in the 1905 paper, he did so without using the language of changing frames... and was able to dismiss the subject in a few sentences without any paradox to explain.

 

[PeroK]

Just to re-emphaise that the one-way journey was analysed, for an arbitrary speed, with a few lines of algebra in post #89. It was only one application of the Lorentz Transformation.

 

[Will Flannery]

Note that the frame origins are events - they can't follow anyone. In his rest frame(s) the traveller's spatial coordinates are constant and you can choose his location to be the spatial origin, yes. My point was thhat if you pick the origin event of all frames to be the turnaround event then you get the traveller at the spatial origin on both legs of the journey for free, without having to introduce a fourth frame (the one you called A') that's the same as another frame (the one you called A) except for a translation.

I did the same thing you did, but with a better choice of origin to achieve the "traveller at rest at the spatial origin" effect you wanted. I used $x=x_0+vt$ to write down the coordinates of the three important events (departure, turnaround and return) for a ten year journey at -0.6c and another ten year journey at +0.6c given an instantaneous turnaround at the origin. Then I used the Lorentz transforms to get the coordinates in the frame where the traveller is at rest on the outbound leg. I didn't do the calculation for the other leg - you can do so explicitly or, as I suggested, just cheat because this problem is highly symmetric.

On closer inspection, I see that this is in fact a description of the second spacetime diagram in your post #90. So all I'm saying is that your first diagram is unnecessary and confusing. You also haven't drawn the journey in any other frame, only the Earth twin's frame. You may find it enlightening to draw those frames' versions of events.

As written, this doesn't make sense. A spacetime diagram covers - in principle - all of spacetime, so "when" is meaningless applied to the diagram. I think you mean that the line marked 8 is (approximately, since it doesn't connect the dots marking the events) the traveller's worldline on the return leg. The other two lines don't make sense in that context. The 15 and 17 are coordinate differences in the traveller's frame, so lines referring to them ought to be parallel to the traveller's frame's coordinate grid, which they manifestly are not.

As I attempted to explain in my last post this is because, in the traveller's rest frame for the second part of the journey, the Earth is moving towards him at 0.6c and the first leg of the journey was at speed 0.88c, not that much faster. So the departure event has to be a long time in the past for the first leg to have built up enough of a lead over the Earth that it takes eight years sitting still for the Earth to catch up.

I note that I actually explained it last time with reference to the second leg as seen from the first leg rest frame, so the explanation was the other way around.

# "if you pick the origin event of all frames to be the turnaround event"
-That is exactly what I did !

#I didn't do the calculation for the other leg - you can do so explicitly or, as I suggested, just cheat because this problem is highly symmetric.
-The purpose of my explanation was to analyze the trip from the traveler's perspective, and in particular the turnaround at the distant planet where he has to reverse direction, in London's vid he jumps from one inertial frame to another. That was the focus, but if you try to do the Lorentz transform for that jump, the velocity is - v + (-v) = -2v which is greater than c and I think the Lorentz transform fails or doesn't make sense for that case, so I had the traveler jump to a frame at traveling at -v to put him back in the original A frame but with a different spatial origin, and then hop to a frame traveling at -v to get him heading back to the original planet. I see all this as necessary ... :)

#The 15 and 17 are coordinate differences in the traveller's frame, so lines referring to them ought to be parallel to the traveller's frame's coordinate grid, which they manifestly are not.
-You're right. I've fixed it ... sort of ... and it does clarify what's going on ...

# You also haven't drawn the journey in any other frame, only the Earth twin's frame. You may find it enlightening to draw those frames' versions of events.
- Each spacetime diagram shows two frames, in the first the Earth frame and the frame for the traveler's outward journey, the 2 for the translated Earth frame and the traveler's home journey. In the traveler's frame the spatial coordinate of the near and distant planets is always 0, so drawn separately ... would that be interesting ?

 

[Will Flannery]

i do not believe it is possible to write a clear and complete description while using phrases like “He was changing frames”, and similar. The traveller isn’t switching frames, you are in your description of the problem.

I agree.

 

[Ibix]

# "if you pick the origin event of all frames to be the turnaround event"
-That is exactly what I did !

In the second half of the discussion in post #90, yes. The first half of the discussion uses a different origin, though. The whole thing will be simpler if you use the same origin throughout. Then you can get rid of the distinction between A and A'.

The purpose of my explanation was to analyze the trip from the traveler's perspective, and in particular the turnaround at the distant planet where he has to reverse direction

You can't create a "traveler's perspective" by stitching together inertial frames, though. Inevitably you end up with a situation where, according to such a Frankenstein setup, two people can shake hands with each other but be doing it at different times (potentially years apart) because of the oddities of such a description. That doesn't sound like anything I'd describe as anyone's perspective.

That was the focus, but if you try to do the Lorentz transform for that jump, the velocity is - v + (-v) = -2v which is greater than c and I think the Lorentz transform fails or doesn't make sense for that case,

Velocities don't add linearly in relativity. The velocity of the return leg as measured from the inertial frame of the outbound leg is about 0.88c (or 15c/17 to be precise). Google for "relativistic velocity addition".

I see all this as necessary ... :)

It isn't. Just define the traveller's two inertial frames with reference to the stay-at-home's frame. That makes the velocities $\pm$0.6c and if they all have a common origin at the turnaround then it's job done. Introducing frames with different origins and defining the traveller's return velocity relative to his outbound rest state just obscures a fairly straightforward setup.

Each spacetime diagram shows two frames, in the first the Earth frame and the frame for the traveler's outward journey, the 2 for the translated Earth frame and the traveler's home journey.

Each of your diagrams shows the coordinate grid from two frames, but this doesn't help to visualise the traveller's frame's description of the scenario. Draw your last diagram again with the traveller's frame's axes perpendicular instead of the stay-at-home's. The worldlines of the twins will look quite different.

 

[Will Flannery]

Velocities don't add linearly in relativity. The velocity of the return leg as measured from the inertial frame of the outbound leg is about 0.88c (or 15c/17 to be precise). Google for "relativistic velocity addition".

Good point.

It isn't. Just define the traveller's two inertial frames with reference to the stay-at-home's frame. That makes the velocities $\pm$0.6c and if they all have a common origin at the turnaround then it's job done. Introducing frames with different origins and defining the traveller's return velocity relative to his outbound rest state just obscures a fairly straightforward setup.

Well ... that would be three sets of axes on one graph. Too many. Better I think would be to do a graph similar to the one I did for the return trip, i.e. with the origin at the turnaround, except for the outbound trip, and have two graphs, which could be superimposed.

Each of your diagrams shows the coordinate grid from two frames, but this doesn't help to visualise the traveller's frame's description of the scenario. Draw your last diagram again with the traveller's frame's axes perpendicular instead of the stay-at-home's. The worldlines of the twins will look quite different.

Great Scott, I was ready to go on to 4-vectors today, got all my references lined up. But no ! This is my project for the day... wait a minute, the traveler must have 2 (or in my version 3) inertial frames, so you can't draw a single graph showing the entire trip from the traveler's perspective ?

 

[Bruce Harvey]

Surely, the point of the twins paradox was to contrast the relativity theory of Lorentz and Poincare with Einstein's theory. In the former, the slowing of clocks is a real physical effect derived from Maxwell's equations. In Einstein's theory, it is an artefact of observation depending on the two reference frames overlapping.
The argument that A's frame of reference distorts as he accelerates is spurious because it has no physical existence beyond the confines of his ship.
The first experimental observation of the slowing of a moving clock settled the case. We live in a Lorentz- Poincare universe, even if we have to drop the term "Aether" and substitute "fabric of spacetime".

 

[PeroK]

Surely, the point of the twins paradox was to contrast the relativity theory of Lorentz and Poincare with Einstein's theory. In the former, the slowing of clocks is a real physical effect derived from Maxwell's equations. In Einstein's theory, it is an artefact of observation depending on the two reference frames overlapping.
The argument that A's frame of reference distorts as he accelerates is spurious because it has no physical existence beyond the confines of his ship.
The first experimental observation of the slowing of a moving clock settled the case. We live in a Lorentz- Poincare universe, even if we have to drop the term "Aether" and substitute "fabric of spacetime".

Hey, you promised not to promote your non-mainstream views on here!

I promise not to use the forum to promote my own work or give answers contrary to the standard model

 

[vanhees71]

We have to drop the term Aether because no one could ever observe it (i.e., finding a preferred frame of reference in vacuo). "Fabric of spacetime" sounds like some PR text on the cover of (usually pretty misleading) popular-science books. What's called "relativity" is nothing else than a theory about how to describe space and time and how to define reference frames, but I don't want to get into this discussion again, because that's an unnecessarily hot topic in these forums ;-)).

 

[PeterDonis]

Surely, the point of the twins paradox was to contrast the relativity theory of Lorentz and Poincare with Einstein's theory.

No, the point of the twin paradox was to contrast relativity with Newtonian mechanics.

In the former, the slowing of clocks is a real physical effect derived from Maxwell's equations. In Einstein's theory, it is an artefact of observation depending on the two reference frames overlapping.

You are mistaken. The difference in aging of the twins when they meet up again is a real affect, not "an artefact of observation" in relativity, period.

The argument that A's frame of reference distorts as he accelerates is

Not an argument that relativity makes, so you are attacking a straw man.

The first experimental observation of the slowing of a moving clock settled the case.

Between relativity and Newtonian mechanics, yes.

@PeroK has already pointed out that you previously agreed not to promote your non-mainstream views here. If you post about them again, you will receive a warning.

 

[Bruce Harvey]

I was careful not to promote my own ideas. I just quoted from history. Lorentz's derivation of the contraction from Maxwell's equations is back in print thanks to Dover. He identifies Maxwell'e wave equation in electric potential and Possion's equation as special cases of the same equation and in a few lines derives the Lorentz contraction. The increase in mass and slowing of clocks follows.

 

[PeterDonis]

I was careful not to promote my own ideas.

Not careful enough:

Surely, the point of the twins paradox was to contrast the relativity theory of Lorentz and Poincare with Einstein's theory.

We live in a Lorentz- Poincare universe, even if we have to drop the term "Aether" and substitute "fabric of spacetime".

In any case, what you said about the twin paradox was incorrect.

 

[Elroch]

Both. With a one way trip the answer is frame dependent.

There is an objective answer to the question of whether A or B is younger (age in proper time) when they meet. Thus, with all due respect, this answer simply cannot be frame dependent.
With the planets having relative velocity zero and with the watches being synchronised (by say one twin sending the time and the other using that time to set their watch with a correction for the known lag due to transit - doesn't matter which), A (the traveling twin) is younger. The reason is clear in a space-time diagram drawn in the common initial stationary frame, where A's path is shorter according to the proper time pseudometric.
Thus, as has been explained in a slightly different way by others, B is right.

 

[Motore]

There is an objective answer to the question of whether A or B is younger (age in proper time) when they meet.

Not for a one way trip (which means they don't meet).

 

[Nugatory]

With a one way trip the answer is frame dependent.

There is an objective answer to the question of whether A or B is younger (age in proper time) when they meet. Thus, with all due respect, this answer simply cannot be frame dependent.

But in a one-way trip they don't meet so your comment does not apply to that case.

"A is younger than B" implies that at the same time that A's age is T, B's age is something less than T. If the two twins are not colocated then "at the same time" and any comaparisom of their ages is frame-dependent.

 

[vanhees71]

The aging is given by the proper time and thus frame-independent. Frame-dependent quantities are auxilliary tools to get physical observables which are all frame-independent.

 

[Dale]

There is an objective answer to the question of whether A or B is younger (age in proper time) when they meet. Thus, with all due respect, this answer simply cannot be frame dependent.

Yes, the age at reunion is frame independent. What is frame dependent is the initial age and the rate of aging.

 

[Ibix]

There is an objective answer to the question of whether A or B is younger (age in proper time) when they meet.

In the original formulation of the question on this thread the twins were born on different planets and meet up later. So yes, there's an invariant answer to the question of which one is older when they meet. There is no invariant answer to the question of whether they were born at the same time.

 

[PeroK]

In the original formulation of the question on this thread the twins were born on different planets ... There is no invariant answer to the question of whether they were born at the same time.

Hey, maybe them twins ain't twins!

 

[Ibix]

Hey, maybe them twins ain't twins!

Whether it's a twin paradox or not is frame dependent.

 

[Elroch]

In the original formulation of the question on this thread the twins were born on different planets and meet up later. So yes, there's an invariant answer to the question of which one is older when they meet. There is no invariant answer to the question of whether they were born at the same time.

Yes.
But if they are essentially stationary relative to each other there is an objective answer to the question of which was born first according to either of the clocks they carry with them (because these clocks run at the same speed and can be synchronised at any time in a unique, consistent way). The special nature of a proper time clock for each person makes it seem natural to use the frame that agrees with both.
Of course we do this all the time on our own planet. If you ask someone "how old are you?", they rarely answer "with respect to a clock moving at which speed?".

 

[PeroK]

Yes.
But if they are essentially stationary relative to each other there is an objective answer to the question of which was born first according to either of the clocks they carry with them (because these clocks run at the same speed and can be synchronised at any time in a unique, consistent way). The special nature of a proper time clock for each person makes it seem natural to use the frame that agrees with both.
Of course we do this all the time on our own planet. If you ask someone "how old are you?", they rarely answer "with respect to a clock moving at which speed?".

This is missing the point about the relativity of simultaneity.

Your age is essentially the length of your worldline (though spacetime) since you were born. That is an invariant. As long as you define the event local to you to where the worldline is measured and your age calculated.

(But, your age at the time when, say, Betelgeuse goes supernova is frame dependent. Because that event is not colocated with you.)

The amount the traveling twin has aged between leaving Earth and arriving at planet X is also an invariant.

But:

The amount that the Earth has aged between these two events is frame dependent. In the Earth's rest frame it is greater than the above. But, in a reference frame in which the Earth (and planet X) are moving, the traveling twin may have initially decelerated and aged more between these events than either the Earth or planet X.

It's only when the traveller returns to Earth that all frames agree on the differential ageing. Not all frames agree at the turnaround point.

 

[Herman Trivilino]

But if they are essentially stationary relative to each other there is an objective answer to the question of which was born first according to a clock either carries with them (because these clocks run at the same speed and can be synchronised at any time in a unique way).

They are synchronized in their common rest frame, but they will not have been synchronized in the rest frame of the traveling twin. You don't really have a twin paradox here. What you have is the paradox of time dilation, that is, how can each observe that the other's clock is running slow? The solution lies in the understanding that each of the twins have different notions of simultaneity. If the staying twin observes the twins being born at the same time, the traveling twin, once he's in motion, will observe that in the rest frame he now occupies, they were born at different times.