B Fewer seconds or shorter seconds?

[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."