Explaining the Twin Paradox: A Big Problem?

[Marilyn67]

TL;DR Summary

Theoretically the situation is symmetrical and the question does not make sense....
This question sounds very simple, (college) but I would like the advice of an expert.
I will explain the reason later.

Hello,

For many years, solving the twin paradox has not been a problem for me.
I understood that it is the change of Galilean frame of reference that is responsible for the age difference, and this clearly appears on a Minkowski diagram. Okay.
However, since yesterday, in wanting to explain this famous and well-known paradox (which is not one) to a student, I encountered a problem.
A big problem !

If we consider two twins moving away from each other in a uniform rectilinear motion and never seeing each other again, relativity teaches us that there is no possible answer to the question of "which clock" is really the slowest. (that's what I learned).

The situation is symmetrical and the question does not make sense.

You confirm ?

 

[PeterDonis]

I understood that it is the change of Galilean frame of reference that is responsible for the age difference

No, not a Galilean frame, a Lorentz frame. And no, just changing frames is not what is responsible for the age difference.

If we consider two twins moving away from each other in a uniform rectilinear motion and never seeing each other again, relativity teaches us that there is no possible answer to the question of "which clock" is really the slowest.

That's correct. And in this scenario, there is also no "twin paradox", because the two twins never meet up again to compare their elapsed times.

The "twin paradox" only occurs if the two twins do meet up again and can compare their elapsed times since they were last together, and find that they are different. But for that to happen (at least in flat spacetime; gravity and curved spacetime introduce further complications), at least one of them cannot remain for all time in uniform rectilinear motion.

 

[Marilyn67]

Thank you PeterDonis,

Okay.
I understood your two answers correctly.

Here is my problem:

Let us suppose our solar system and Vega as representing the same frame of reference.
Suppose a spaceship leaves our solar system (thanks to very rapid acceleration).

The spacecraft is now moving away from Earth in uniform rectilinear motion at relativistic speed with a lorentz factor equal to 10.
Vega is 25 light years away, and the ship passes Vega (without stopping) after about 25 years from Earth's perspective.

It's correct ?

 

[Ibix]

For many years, solving the twin paradox has not been a problem for me.
I understood that it is the change of Galilean frame of reference that is responsible for the age difference, and this clearly appears on a Minkowski diagram. Okay.

Changing (Lorentz!) frame is responsible for the discrepancy between a naive application of time dilation from the traveller's perspective and reality, yes. But the underlying reason for the difference in ages is effectively the triangle inequality as it applies to Minkowski geometry.

there is no possible answer to the question of "which clock" is really the slowest. (that's what I learned).

I would avoid words like "really". Rather, I would say that there is no unambiguous answer to which is slower. You can only unambiguously compare clocks if they are in the same place, and inertial clocks in flat spacetime are only at the same place once, so have no way to calculate an elapsed time.

 

[Marilyn67]

Thank you Ibix,

I agree

 

[Marilyn67]

My problem is this :

For astronauts, the spacecraft crosses Vega (without stopping) after two and a half years.
(the "rapidity" represent the speed of light multiplied by 10)

It's correct ?

 

[A.T.]

I encountered a problem.
A big problem !

Here is my problem:

My problem is this :

What is the problem exactly?

 

[Marilyn67]

The big question is this :

The fact that in one frame of reference, the trip lasts 25 years and that in the other frame of reference, the trip lasts two and a half years does not constitute proof that it is the clock of the spaceship that is lagging?
(In contradiction with the symmetry evoked before ?)

 

[Marilyn67]

In summary, the observation of the rapidity doesn't make it possible to make sense to the movement of the clocks which should be symmetrical ?

(I specify that the spaceship continues its trip and doesn't stop at Vega !)

 

[Nugatory]

The spacecraft is now moving away from Earth in uniform rectilinear motion at relativistic speed with a lorentz factor equal to 10.
Vega is 25 light years away, and the ship passes Vega (without stopping) after about 25 years from Earth's perspective.

It's correct ?

Yes, assuming as long as by "after about 25 years" you mean some time longer than the 25 years it would tke for a flash of light to make the same journey.

These problems are easier to work with if you choose the speed to be either $\frac{3}{5}c$ or $\frac{4}{5}c$ so that the Lorentz factor comes out a nice round fraction: either $\frac{5}{4}$ or $\frac{5}{3}$. So let's take the speed to $\frac{3}{5}c$ and the distance traveled to be 15 light-years (it's an example so might as well choose numbers that make the arthmetic easy).

Now the journey will take 25 years according to someone on Earth, and if we use the frame in which the Earth is at rest to assign coordinates to the three relevant events in this example:
Event 1, x=0, t=0: Traveller sets out on journey, both clocks are set to zero
Event 2, x=15,t=25: Traveller arrives at destination
Event 3, x=0, t=25: Clock on Earth reads 25

Now use the Lorentz transformations to calculate the coordinates (call them x' and t') because tht's how the transformations are usually written) of these three events using the frame in which the ship is at rest. Looking at these results, you and your student will be able to see that:
- The question of which one is "really" slow really has no answer; the answer depends on your choice of frame.
- Relativity of simultaneity is why there is no paradox in having both clocks slower than the other
- The situation really is completely symmetrical

But do note that, as @PeterDonis pointed out above, this is not the Twin Paradox. For that you'd need to bring the traveller back home again so that they could compare their clocks while at the same point in space; the paradox is that despite the symmetry of the time dilation formula the traveller ends up aged less.

 

[Marilyn67]

Thanks Nugatory,

Yes, I understood PeterDonis' answer correctly.

However, the comparison of different rapidity isn't an "exception" that allows to answer the same question, without comparing the clocks in the same place ?

I know the idea is crazy !
It is for this reason that I ask the question!

 

[A.T.]

The fact that in one frame of reference, the trip lasts 25 years and that in the other frame of reference, the trip lasts two and a half years does not constitute proof that it is the clock of the spaceship that is lagging?
(In contradiction with the symmetry evoked before ?)

What contradiction? Are you ignoring the length contraction?

 

[Marilyn67]

No, I don't ignoring the length contraction.
But that doesn't answer the question... ?

 

[Marilyn67]

By comparing the rapidity, one could break the symmetry, which is contradictory to the fact that the question of which clock is slower does not make sense.

I don't assert
I ask the question because it seems crazy to me

 

[A.T.]

By comparing the rapidity, one could break the symmetry, which is contradictory to the fact that the question of which clock is slower does not make sense.

You break the symmetry by using a third object, that is at rest in one frame, but not in the other.

 

[Marilyn67]

Why speak of a third object ?

I'm only talking about referentiel and there are only two. (?)

 

[A.T.]

Why speak of a third object ?

I'm only talking about referentiel and there are only two. (?)

Earth, Vega, spaceship

Looks like 3 to me.

 

[robphy]

My problem is this :

For astronauts, the spacecraft crosses Vega (without stopping) after two and a half years.
(the "rapidity" represent the speed of light multiplied by 10)

It's correct ?

In relativity, "rapidity" is a technical term that describes the Minkowski-analogue of an angle $\theta$ between inertial worldlines (between their 4-velocities) such that their relative-velocity is $v_{BA}=c\tanh\theta$.

Maybe you mean "time-dilation factor" $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\cosh\theta$.

 

[Nugatory]

However, the comparison of different rapidity isn't an "exception" that allows to answer the same question, without comparing the clocks in the same place ?

It is not. Rapidity is an alternative (and in some ways more mathematically convenient) way of representing relative velocities, but it doesn't eliminate the basic symmetry: take A to be at rest while B is moving, or B to be at rest while A is moving, and all physical results will come out the same.

The issue with comparing clocks at different locations is that any such comparison involves a statement of the form "at the same time that clock A reads $T_A$, clock B reads $T_B$"; the phrase "at the same time" pretty much guarantees that the result of the comparison will be frame-dependent.

 

[Dale]

And no, just changing frames is not what is responsible for the age difference.

It is not what is responsible for the age difference, but it does unambiguously show that there is asymmetry.

 

[Dale]

The big question is this :

The fact that in one frame of reference, the trip lasts 25 years and that in the other frame of reference, the trip lasts two and a half years does not constitute proof that it is the clock of the spaceship that is lagging?
(In contradiction with the symmetry evoked before ?)

This is not a contradiction. In the spaceship’s frame the Earth and Vega clocks are not synchronized.

 

[PeterDonis]

Let us suppose our solar system and Vega as representing the same frame of reference.

Meaning, they are at rest relative to each other? Ok.

Vega is 25 light years away, and the ship passes Vega (without stopping) after about 25 years from Earth's perspective.

It's correct ?

If you assume that Earth and Vega clocks are synchronized using the Einstein clock synchronization procedure, and the ship leaves Earth at Earth time zero, then when the ship passes Vega, it will observe the clocks on Vega to read time 25 years, yes.

What's the problem?

 

[PeterDonis]

For astronauts, the spacecraft crosses Vega (without stopping) after two and a half years.
(the "rapidity" represent the speed of light multiplied by 10)

It's correct ?

Meaning, if the ship's clocks are set to time zero when the ship leaves Earth, then when the ship passes Vega, the ship's clocks will read time 2.5 years, yes.

What's the problem?

 

[PeterDonis]

The fact that in one frame of reference, the trip lasts 25 years and that in the other frame of reference, the trip lasts two and a half years does not constitute proof that it is the clock of the spaceship that is lagging?

No. You have left out relativity of simultaneity. (A good rule of thumb in SR problems is that if you are confused, it's because you have left out relativity of simultaneity.)

In the ship's frame, as has been noted, Earth and Vega clocks are not synchronized. In the ship's frame, when the ship leaves Earth, Earth clocks read time zero, but Vega clocks read time 24.75 years. In the ship's frame, Earth and Vega clocks run slow by a factor of 10 (same Lorentz factor), so the trip that takes 2.5 years by the ship's clock, in the ship's frame, takes 0.25 years by Earth and Vega clocks. So when the ship passes Vega, it sees Vega clocks read 24.75 + 0.25 = 25 years. And in the ship's frame, Earth clocks at that same time (when the ship passes Vega) read time 0.25 years. So Earth clocks, according to the ship's frame, run behind Vega clocks by 24.75 years.

 

[Nugatory]

No. You have left out relativity of simultaneity. (A good rule of thumb in SR problems is that if you are confused, it's because you have left out relativity of simultaneity.)

@Marilyn67 if you go through the exercise I outlined in post #10 up above, you will see how the relativity of simultaneity comes in.

 

[A.T.]

In relativity, "rapidity" is a technical term that describes the Minkowski-analogue of an angle $\theta$ between inertial worldlines (between their 4-velocities) such that their relative-velocity is $v_{BA}=c\tanh\theta$.

Maybe you mean "time-dilation factor" $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\cosh\theta$.

@Marilyn67 Or maybe you mean proper velocity?
https://en.wikipedia.org/wiki/Proper_velocity

 

[Marilyn67]

Hello,

Excuse me for the late reply and also for my not very good English (I am French and I live in France).
I thank all the participants for their very clear answers which all show that I was wrong.
(7 messages this morning! Thank you!)

No. You have left out relativity of simultaneity. (A good rule of thumb in SR problems is that if you are confused, it's because you have left out relativity of simultaneity.)

You are right, that has been the source of my confusion.
We may well know that simultaneity is relative, our "absolute" daily life makes us continually make mistakes.

In the ship's frame, as has been noted, Earth and Vega clocks are not synchronized. In the ship's frame, when the ship leaves Earth, Earth clocks read time zero, but Vega clocks read time 24.75 years. In the ship's frame, Earth and Vega clocks run slow by a factor of 10 (same Lorentz factor), so the trip that takes 2.5 years by the ship's clock, in the ship's frame, takes 0.25 years by Earth and Vega clocks. So when the ship passes Vega, it sees Vega clocks read 24.75 + 0.25 = 25 years. And in the ship's frame, Earth clocks at that same time (when the ship passes Vega) read time 0.25 years. So Earth clocks, according to the ship's frame, run behind Vega clocks by 24.75 years.

Yes
I am ashamed !
The worst part is that I learned it !

Without considering the relative simultaneity, I deduced naively and too quickly that 2.5 years in the spacecraft corresponded to 25 years on Earth, from the point of view of the spacecraft , while it is obviously the reverse, so 0.25 years !
(I had made a calculation a long time ago with the famous hypothetical tachyon exchanges using this phenomenon to demonstrate that if tachyons existed, they violated the principle of causality by allowing information to be sent in the past ).

A question torments me, again :

I thought that by considering the Earth and Vega fixed between them, the Earth and Vega represented 2 different landmarks, but in the same frame of reference. You agree (# 22) that Earth and Vega can be considered a single frame of reference, but @A.T. sees 2, so 3 in total. However, it is theoretically possible to synchronize all the remote landmark clocks in the same frame of reference, since the landmarks are fixed between them, right ?

 

[PeroK]

A question torments me, again :

I thought that by considering the Earth and Vega fixed between them, the Earth and Vega represented 2 different landmarks, but in the same frame of reference. You agree (# 22) that Earth and Vega can be considered a single frame of reference, but @A.T. sees 2, so 3 in total. However, it is theoretically possible to synchronize all the remote landmark clocks in the same frame of reference, since the landmarks are fixed between them, right ?

The key question is what relative motion do we have?

1) The Earth and Vega are at rest relative to each other.

2) The spaceship is moving relative to the Earth and Vega.

You can study any problem in any inertial reference frame you want. In this case, however, there are two obvious frames that are the most useful:

1) The reference frame in which Earth and Vega are at rest.

2) The reference frame in which the spaceship is at rest.

All clocks at rest in an inertial reference frame can be synchronised (see Einstein synchronisation convention) - and this is the basis on which the Lorentz Transformation applies. In this case, therefore, we assume that clocks on Earth and Vega are synchronised in this way.

In the reference frame in which the ship is at rest, however, these clocks are not synchrosied - and the relativity of simultaneity applies.

 

[A.T.]

You agree (# 22) that Earth and Vega can be considered a single frame of reference, but @A.T. sees 2, so 3 in total.

No, I didn't say that Earth and Vega are two different reference frames. I said that your scenario has no symmetry between the two frames: You have two objects at rest in one frame, but only one object at rest in the other frame. To have symmetry you would for example need two spaceships at relative rest to each other, separated by the same proper distance as the planets. Only then you could claim symmetry, and would find corresponding durations which would be the same from either frame.

 

[Halc]

Let me try the original scenario, but keeping it as symmetrical as I can make it.

Earth and Vega are 25 LY apart and stationary relative to each other. They are said to have a proper separation of 25 LY. The ship (which is always inertial, never accelerating) is 25 LY in proper length and moving at v=+.995c relative to the planets, and the planets are thus moving at v=-.995c relative to the ship.

We now have 2 frames and 4 objects (Earth, Vega, Nose, Tail). The clocks on the planets are synced in the planet frame, and the clocks at opposite ends of the ship are synced in ship frame. Note that the two ship clocks are not synced in the planet frame, nor are the two planet clocks synced in the ship frame.

Lorentz factor is 10.

Planets frame:
At event A (x=0@Earth, t=0), the nose is at Earth and the Earth clock and nose clock both read t=t'=0.
The ship is contracted to 2.5LY in Earth frame, so the tail passes Earth (event B) at x=0, t=2y187d, and the tail clock reads t'=25y46d as it passes.
It takes the nose about 25 years 1.5 months to get to Vega (x=25, so at event C (nose at Vega), the Vega clock reads x=25, t=25y46d and the time dilated nose clock reads x'=0, t'=2y187d. The tail of the ship passes Vega at event D, x=0, t=27y233d.

In ship frame, the nose is at location x'=0 and the tail at x'=-25. The ship is stationary so these don't change.
At event A (x'=0, t'= 0), the Earth is at the nose and the Earth clock and nose clock both read t=t'=0.
The separation between the planets is contracted to 2.5LY in ship frame, so Vega passes the nose (event C) at x'=0, t'=2y187d and the Vega clock reads t'=25y46d as it passes.
It takes the Earth about 25 years 1.5 months to get to the tail (x'=-25, so at event B (Earth at tail), the tail clock reads x'=-25, t'=25y46d and the time dilated Earth clock reads t=2y187d. The Vega passes the tail of the ship at event D, x'=-25, t'=27y233d.

Now it seems to be completely symmetrical.

 

[robphy]

In my opinion, the physics is clearer with more convenient numerical values (like the ones given by @Nugatory)

These problems are easier to work with if you choose the speed to be either $\frac{3}{5}c$ or $\frac{4}{5}c$ so that the Lorentz factor comes out a nice round fraction: either $\frac{5}{4}$ or $\frac{5}{3}$. So let's take the speed to $\frac{3}{5}c$ and the distance traveled to be 15 light-years (it's an example so might as well choose numbers that make the arthmetic easy).

Now the journey will take 25 years according to someone on Earth, and if we use the frame in which the Earth is at rest to assign coordinates to the three relevant events in this example:
Event 1, x=0, t=0: Traveller sets out on journey, both clocks are set to zero
Event 2, x=15,t=25: Traveller arrives at destination
Event 3, x=0, t=25: Clock on Earth reads 25

The arithmetic easier when one works with pythagorean triples (like 3,4,5). This occurs when the doppler factor $k=\sqrt{\frac{1+v}{1-v}}$ is rational.
For $v=3/5$, we have $k=2$ and $\gamma=(k+k^{-1})/2=5/4$.
For $v=4/5$, we have $k=3$ and $\gamma=5/3$.
(Alternatively, using rapidities
$k=\exp(\mbox{arctanh}(3/5))=2$ and $\gamma=\cosh(\mbox{arctanh}(3/5))=5/4$.)

For $\gamma=10$, we have
$k=\exp(\mbox{arccosh}(10))=(10+3\sqrt{11})$=irrational and
$v=\tanh(\mbox{arccosh}(10))=(3\sqrt{11}/10)$ =0.9949 with help from WolframAlpha.

 

[robphy]

Apart from the equations and calculations, a good way to see what is going on is to draw a spacetime diagram.

Here's the latest version of my spacetime diagrammer (time-upwards)
(version 4) https://www.desmos.com/calculator/kq2qnojphq
(version 6) https://www.desmos.com/calculator/sk2zhnmjmm [new]

I use a relative-velocity of (3/5)c above (@Nugatory 's suggested velocity), but I invite to you to adjust the velocities.

The key idea is that "space is perpendicular to time",
a shortened slogan for
"an observer's spaceline is perpendicular to her timeline"
where perpendicularity is defined by the tangency to a "circle" centered at the tail of the displacement (at the origin).

The tangent [space]line of the red observer meets the worldline ( [time]line ) of the green observer
at the event marked by the green cross at (x=0.6,t=1). [You may wish to turn on the grid using the Desmos-wrench in the upper-right corner.]
The displacement to the green cross from the origin is 0.8 of the green radial [time]line segment to the hyperbola. (You can measure this ratio of parallel segments using any ruler.)

As expected by symmetry,
the tangent [space]line of the green observer meets the worldline ( [time]line ) of the red observer at the event marked by the red cross at (x=0,t=0.8).
The displacement to the red cross from the origin is 0.8 of the red radial [time]line segment to the hyperbola. (You can measure this ratio of parallel segments using any ruler.)

This ratio 0.8 is (4/5) = 1/(time dilation factor).
Red says the distant green-cross event occurred 1 tick after the origin.
Green says the local green-cross event occurred 0.8 ticks after the origin.
The time-dilation factor according to Red is
$\gamma_{red}=\cosh\theta$ = (adjacent to local)/(hypotenuse to distant) = (1 tick)/(0.8 ticks) = (5/4).

Green says the distant red-cross event occurred 1 tick after the origin.
Red says the local red-cross event occurred 0.8 ticks after the origin.
The time-dilation factor according to Green is
$\gamma_{green}=\cosh\theta$ = (adjacent to local)/(hypotenuse to distant) = (1 tick)/(0.8 ticks) = (5/4).

Adjust the v2-slider to see that this ratio is common to both observers.
(These are similar-triangles where similarity is defined by the "circle".)
Although I limit the range of the velocities to 0.98c,
you can type in 0.995 to override it. [You could also put "$\tanh\left(\operatorname{arccosh}\left(10\right)\right)$ "]
(You might have to zoom out to see the tip of the green-tick.)
(To get more decimal places, open the "time dilation" folder and change the number of digits in round().)

Back to v1=0 and v2=0.6,
you can drag the event on the green worldline to the green-cross event
to determine the event on the red-worldline that green says is simultaneous with the green-cross event.

(To make the arithmetic simpler, you can drag the events controlling the spacelines to 5 ticks.)

You might wish to adjust the E-slider from "1" for Minkowski-spacetime
to "0" for Galilean-spacetime and "-1" for Euclidean space.

You can also adjust both v1 and v2.

Have fun.

 

[robphy]

And here is the "rotated graph paper" version (using @Nugatory 's suggested setup).
Let each diamond represent "5 units".

See my Insights for how these are made
https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/

 

[Marilyn67]

Hello,

Once again thank you to all the participants, you helped me to see things more clearly.

@PeroK: your explanations are very clear. I just read Einstein synchronization convention and I understand where my error is coming from.

@A.T. : Ok, I misunderstood you. I understand better why Einstein used the famous wagons with their ends A and B, and respectively A'and B'.

@Halc: Thank you for this quantified demonstration. I haven't checked your calculation yet but I will, and I'm sure it's right. This will allow me to strengthen my understanding.

@robphy: Thank you for your detailed explanations and your patience !
Your software is great !
I'll use it !

Honestly, I must admit that I learn a lot more things here than on the French speaking forums.
I will recommend your site!Marilyn