Fewer seconds or shorter seconds?

[Ibix]

As the 2 is correct, then your calculation is not an average period as I imagined

It is. It's just not the clearest form. The point is that $2T$ is the total time experienced by the traveller and $2\gamma T$ is the total time experienced by the stay at home. Thus the stay at home argues that one tick of the traveller's clock is $\gamma$ times longer than his own.

Similarly, the traveller can argue that the stay at home's clock must, on average, have ticked once every $1/\gamma$ ticks of his own. Learning how to reconcile that with the fact that the teaveller also says that due to time dilation the stay at home's clock ticks once every $\gamma$ ticks of his own clock is the point of the twin paradox.

 

[Ibix]

Is this duration measured by one of the brothers or do they both find a different one?

Who is moving and who is measuring?

 

[Kairos]

Each brother (the stay at home and the traveler) can measure the duration of the trip (between the separation and the reunion of the brothers) using his own clock. Does one of them verify that this measured duration corresponds to 2D/v? where v is the speed of the rocket (on which they agree), and D is the distance between two distant planets (on which they also agree according to astronomical measurements).

 

[Ibix]

D is the distance between two distant planets (on which they also agree according to astronomical measurements).

I think you've forgotten that the planets are moving in one frame and the distance between them is length contracted by a factor of $\gamma$ compared to the other frame.

 

[Nugatory]

where v is the speed of the rocket (on which they agree)

Do they? According to the traveling twin the speed of the rocket is zero at all times, while the speed of the Earth is $v$, first receding and then approaching.

 

[Kairos]

I think you've forgotten that the planets are moving in one frame and the distance between them is length contracted by a factor of $\gamma$ compared to the other frame.

You mean that the contraction of lengths is not only valid for the rocket but also for space. So for the traveler, the distance traveled is in fact shorter than the one predicted by the terrestrial astronomers ? this would indeed answer my question.

 

[Ibix]

So for the traveler, the distance traveled is in fact shorter than the one predicted by the terrestrial astronomers ?

Well, as Nugatory notes, for the traveller the planets are moving. But yes, the inbound and outbound inertial frames regard the distance between the planets as length contracted. Imagine actually laying a physical ruler that just touches each planet. Since the ruler is moving in the traveller's frames it must be length contracted, but both frames must agree whether or not the ruler touches the planets. So the distance between the planets must be contracted too.

 

[Kairos]

Do they? According to the traveling twin the speed of the rocket is zero at all times, while the speed of the Earth is $v$, first receding and then approaching.

One can certainly consider that the whole universe shifts with respect to the rocket seen as immobile, but in the case of a terrestrial takeoff, it is the cosmonaut which experienced an acceleration.

 

[Nugatory]

but in the case of a terrestrial takeoff, it is the cosmonaut which experienced an acceleration.

This is true, but irrelevant to distance measurements made using either the inertial frame in which the cosmonaut is at rest on the outbound leg or the different inertial frame in which the cosmonaut is at rest during the inbound leg.

 

[Kairos]

Indeed, I agree that when a constant velocity is reached, one goes into a perfectly symmetrical system, although it is generally easier for an astronaut (who remembers his acceleration) to think that it is he who is flying rather than the earth.
But your remark makes me ask another question: when the system is symmetrical, why decide that it is the astronaut's path that contracts and not the Earth's one? If the two contract symmetrically, Ibix's argument is no longer valid and the 2D/v duration is indeed reciprocal

 

[Dale]

when the system is symmetrical, why decide that it is the astronaut's path that contracts and not the Earth's one?

When the system is symmetrical then each contracts with respect to the other equally, preserving the symmetry. However, that is not relevant here since the situation is not symmetrical.

If the two contract symmetrically, Ibix's argument is no longer valid and the 2D/v duration is indeed reciprocal

His argument is valid for the asymmetrical situation under discussion.

 

[jbriggs444]

both frames must agree whether or not the ruler touches the planets

To be clear, both frames agree that the front end of the ruler touches the front planet at some time and that the back end of the ruler touches the back planet at some time. They disagree on whether these two events are simultaneous.

 

[Ibix]

when the system is symmetrical

It isn't. The destination planet is at rest in the stay at home frame, and there is no analogous object at rest in the traveller's frame.

 

[Ibix]

To be clear, both frames agree that the front end of the ruler touches the front planet at some time and that the back end of the ruler touches the back planet at some time. They disagree on whether these two events are simultaneous.

I was imagining a ruler at rest with respect to the planets, so that it was length contracted in the traveller's frame.

 

[Kairos]

When the system is symmetrical then each contracts with respect to the other equally, preserving the symmetry. However, that is not relevant here since the situation is not symmetrical.

His argument is valid for the asymmetrical situation under discussion.

That's what I thought but the discussion drifted from above remarks. It is indeed because the astronaut's experience is non-inertial that there is asymmetry. This is precisely why I would have been interested in a demonstration in the style of the post #61 but from the point of view of the other brother, to show the asymmetry.

 

[Kairos]

When the system is symmetrical then each contracts with respect to the other equally, preserving the symmetry. However, that is not relevant here since the situation is not symmetrical.
His argument is valid for the asymmetrical situation under discussion.

is 2D/v the duration of the trip for the stay at home? and introducing length contraction 2D*sqrt(1-(v/c)^2)/ v for the traveler?

 

[Dale]

is 2D/v the duration of the trip for the stay at home? and introducing length contraction 2D*sqrt(1-(v/c)^2)/ v for the traveler?

Yes. Although for the traveler there is no standard reference frame since they are non-inertial. So the number for the non-inertial frame is dependent on how things are defined.

 

[Kairos]

Thank you very much. I suppose that this result holds for a simple U-turn like a rebound without velocity change. I think I will contact you later for a new thread about the twin experiment. This simple demonstration poorly convinces me because the only parameter that distinguishes the twins: the U-turn, is not introduced in the reasoning and the difference of travel duration between the brothers is proportional to D, whereas things are perfectly symmetrical between them during the phases of inertial motion.

 

[lightarrow]

In the twin experiment, the travel time is shorter for the traveling brother than for the static brother. Since the unit of time in physics is the second, is the travel duration shorter because: (1) it contains fewer seconds or (2) the number of seconds is the same for both brothers but the traveler's seconds are shorter?

(1) it "contains fewer seconds": the traveller's clock doesn't "walk slower", it travels a shorter path in spacetime.
Sorry, I haven't read all the answers, probably this has already been written more times.

--
lightarrow

 

[A.T.]

the U-turn, is not introduced in the reasoning and the difference of travel duration between the brothers is proportional to D

As an analogy: Turning with your car doesn't affect the rate of the odometer, but the total distance traveled between A and B is affected by the turns you make.

 

[lightarrow]

As an analogy: Turning with your car doesn't affect the rate of the odometer, but the total distance traveled between A and B is affected by the turns you make.

Exactly

--
lightarrow

 

[Kairos]

As an analogy: Turning with your car doesn't affect the rate of the odometer, but the total distance traveled between A and B is affected by the turns you make.

Your remark is absurd and disconnected from the subject. Quoting only a few words from a sentence is misleading. I repeat, hopefully more clearly, my 2 main concerns with this demonstration:
1- The only parameter that distinguishes the twins: the U-turn, is not introduced in the reasoning (purely SR)

and

2- The difference of travel duration between the brothers is proportional to the traveled distance, whereas things are perfectly symmetrical between them during the phases of inertial motion (which represents the majority of the trip)

 

[Ibix]

Your remark is absurd and disconnected from the subject.

It's an extremely close analogy for what's going on, actually, just in Euclidean space instead of Minkowski.

1- The only parameter that distinguishes the twins: the U-turn, is not introduced in the reasoning (purely SR)

I don't understand this. That one twin turns around is stated in the problem description, and explicitly handled in both the Doppler approach I demonstrated or the usual inertial frame approach. So I don't see how it's "not introduced in the reasoning". Or have you got the idea that special relativity can only deal with inertial motion? If so, that's incorrect, although it's one of the more common misconceptions about SR.

2- The difference of travel duration between the brothers is proportional to the traveled distance, whereas things are perfectly symmetrical between them during the phases of inertial motion (which represents the majority of the trip)

The problem here is your casual use of the word "during". The relativity of simultaneity is very important, and you appear not to be taking it into account.

 

[Sagittarius A-Star]

2- The difference of travel duration between the brothers is proportional to the traveled distance, whereas things are perfectly symmetrical between them during the phases of inertial motion (which represents the majority of the trip)

Let's take a specific example: D=4 LY and v=0.8c. Then the travel duration is 10 years for the stay-at-home twin and 6 years for the traveling twin. Let the turn-around-time be neglectable on the traveller's clock.

In the frame of the "travelling" twin, the duration of the phases of inertial motion is together 6 years, and on the (moving) clock of the "stay-at-home" twin it is (in this frame) 6 years * 0.6 = 3.6 years, which represents not the majority of the trip (of 10 years, according to that clock).

 

[Dale]

I suppose that this result holds for a simple U-turn like a rebound without velocity change.

In a U-turn, or any turn, the velocity changes. Velocity is a vector quantity, so it has magnitude and direction.

This simple demonstration poorly convinces me

Agreed. Personally, what finally convinced me was not such thought experiments, but reading and understanding the mountain of actual physical experiments that confirm relativity: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

 

[PeterDonis]

simple U-turn like a rebound without velocity change

You are confusing speed with velocity. In a "simple U-turn like a rebound", the speed does not change, but the velocity does, because the direction changes.

 

[jbriggs444]

Let's take a specific example: D=4 LY and v=0.8c. Then the travel duration is 10 years for the stay-at-home twin and 6 years for the traveling twin. Let the turn-around-time be neglectable on the traveller's clock.

In the frame of the "travelling" twin, the duration of the phases of inertial motion is together 6 years, and on the (moving) clock of the "stay-at-home" twin it is (in this frame) 6 years * 0.6 = 3.6 years, which represents not the majority of the trip (of 10 years, according to that clock).

To make this a bit more clear.

The portion of the stay-at-home twin's interval which the traveler considers to be simultaneous with the traveler's outbound leg amounts to 1.8 years. The portion of the stay-at-home twin's interval which the traveler considers to be simultaneous with the traveler's return leg also amounts to 1.8 years. The total is 3.6 years of the stay-at-home twin's 10 year snooze accounted for so far.

The twin paradox resolution that first resonated with me was the idea of the traveling twin's [hyper-]plane of simultaneity sweeping forward across the stay-at-home twin's world line as the turnaround is made. That accounts for the extra 6.4 years missed by the above accounting.

In the odometer analogy, the non-straight twin sees the same thing happening during his turn. A line drawn at right angles to his car's path sweeps backward along his twin's straight path.

 

[Kairos]

I don't understand this. That one twin turns around is stated in the problem description, and explicitly handled in both the Doppler approach I demonstrated or the usual inertial frame approach. So I don't see how it's "not introduced in the reasoning". Or have you got the idea that special relativity can only deal with inertial motion? If so, that's incorrect, although it's one of the more common misconceptions about SR.

Please read again: this comment concerned the solution that I described in post # 83 which is not a Doppler approach but just the contraction of distances in the SR. The use of the Doppler effect shows indeed that the U-turn leads to a dissymmetry in the number of signals received between the brothers over the whole travel. I disagree with your demonstration as seen by the stay at home in post# 61 but that would merit a new thread (not to confuse this one further) in which I would be happy to participate.

The problem here is your casual use of the word "during". The relativity of simultaneity is very important, and you appear not to be taking it into account.

In the present case, I am sure that the two brothers agree perfectly on "during the flight"

 

[Kairos]

Let's take a specific example: D=4 LY and v=0.8c. Then the travel duration is 10 years for the stay-at-home twin and 6 years for the traveling twin. Let the turn-around-time be neglectable on the traveller's clock.

In the frame of the "travelling" twin, the duration of the phases of inertial motion is together 6 years, and on the (moving) clock of the "stay-at-home" twin it is (in this frame) 6 years * 0.6 = 3.6 years, which represents not the majority of the trip (of 10 years, according to that clock).

As you say, the turn around time and also the acceleration and deceleration times can be set as neglectable on the traveller's clock. I didn't want to say anything else.

 

[jbriggs444]

In the present case, I am sure that the two brothers agree perfectly on "during the flight"

They do not. The details of why not matter. See #97 for some details on why not.

 

[PeterDonis]

I am sure that the two brothers agree perfectly on "during the flight"

They agree on the events where they separate and where they come back together again. They do not agree on "what time it is" in between.

 

[Ibix]

In the present case, I am sure that the two brothers agree perfectly on "during the flight"

As @jbriggs444 has already noted, they do not agree, at least not in the naive sense that you are assuming. Not realising this is the root of the vast majority of all problems people have with SR.

 

[Sagittarius A-Star]

As you say, the turn around time and also the acceleration and deceleration times can be set as neglectable on the traveller's clock. I didn't want to say anything else.

O.K.

In my example in posting #94, in the frame of the "travelling" twin, the (moving) clock of the "stay-at-home" twin ticks in the following way:

• It advances by 1.8 years while the outbound leg (inertial),
• It advances by 6.4 years while the turnaround (non-inertial),
• It advances by 1.8 years while the inbound leg (inertial).

In the frame of the "stay-at-home" twin, the same (non-moving) clock of the "stay-at-home" twin ticks in the following way:

• It advances by 5 years while the outbound leg (inertial),
• It advances by approximately 0 years while the turnaround,
• It advances by 5 years while the inbound leg (inertial).

So both twins agree, that the clock of the "stay-at-home" twin advances overall by 10 years.

 

[Nugatory]

is $\frac{2D}{v}$ the duration of the trip for the stay at home? and introducing length contraction $\frac{2D}{v}\sqrt{1-(v/c)^2}$ for the traveler?

Yes, provided that by "duration" you mean the time elapsed on their respective clocks between when they separate and when they rejoin.

These results are best calculated using the formula for the spacetime interval. Done this way the $1/\gamma = \sqrt{1-(v/c)^2}$ factor emerges naturally without messing with the length contraction and time dilation formulas, and the pitfalls involving relativity of simultaneity are avoided.

We have three events:
E0: Traveller leaves. Choosing a convenient frame in which the Earth is at rest, this event has coordinates $t=0$ and $x=0$.
E1: The turnaround. Using the same frame, this will have coordinates $t=D/v$ and $x=D$.
E2: Traveller returns, with coordinates $t=2D/v$ and $x=0$.

The duration for the traveller is the sum of spacetime intervals between E0 and E1, and between E1 and E2.

 

[jbriggs444]

In the frame of the "stay-at-home" twin, the same (non-moving) clock of the "stay-at-home" twin ticks in the following way:

• It advances by 5 years while the outbound leg (inertial),
• It advances by approximately 0 years while the turnaround (non-inertial),
• It advances by 5 years while the inbound leg (inertial).

So both twins agree, that the clock of the "stay-at-home" twin advances overall by 10 years.

The highlighted segment is inertial, not non-inertial. The fact that a remote object is accelerating has little to do with whether a frame is accelerating. All three segments are accounted for in the same, unchanging inertial reference frame.

Yes, both twins agree on the 10 year advance of the stay-at-home clock. That is a measurement of proper time, which is a relativistic invariant. That is, it is a numeric quantity that is agreed upon regardless of reference frame.