B Fewer seconds or shorter seconds?

[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

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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.

 

[Kairos]

they do not agree, at least not in the naive sense that you are assuming.

I used "during the flight" to shorten "somewhere between the spacetime point of separation and the spacetime point of reunion of the brothers". I am sure the brothers would have agreed on this term and that you also perfectly understood. The time elapsed on the brother's clocks between these points is the subject of the twin experiment.

 

[jbriggs444]

I used "during the flight" to shorten "somewhere between the spacetime point of separation and the spacetime point of reunion of the brothers". I am sure the brothers would have agreed on this term and that you also perfectly understood. The time elapsed on the brother's clocks between these points is the subject of the twin experiment.

If you tighten that up to be "in the intersection of the future light-cone of the separation event and the past light-cone of the reunion event" then I have no objection.

"Between" is rather more ambiguous than that.

 

[Kairos]

of course! thank you for this precision; unfortunately not always repected in the literature such as for example in reports on atomic clocks in airplanes in which I read "during the trip". It takes very long sentences to speak about relativity in this forum but fortunately equations are simpler (... provided their variables are clearly defined )

 

[vanhees71]

Yes, write equations! Math is the only language adequate to express what one really means!

 

[robphy]

Yes, write equations! Math is the only language adequate to express what one really means!

and write words and draw diagrams (also aspects of Math) !

"A spacetime diagram is worth a thousand words."