B Understanding twin paradox without math

[PeterDonis]

the kink is instantaneous acceleration

Which is not physically possible; that might be why it's confusing people. The twin's proper acceleration at the turnaround can't be infinite, but that's what a kink in the worldline describes. A finite proper acceleration requires a smooth, rounded-off "corner" in the worldline. So strictly speaking, if you're going to draw a kink in your diagram, it should be because you are assuming that the proper acceleration at turnaround is so large that, at the scale of the diagram, the smooth rounded-off corner looks like a kink.

 

[obtronix]

Which is not physically possible; that might be why it's confusing people. The twin's proper acceleration at the turnaround can't be infinite, but that's what a kink in the worldline describes. A finite proper acceleration requires a smooth, rounded-off "corner" in the worldline. So strictly speaking, if you're going to draw a kink in your diagram, it should be because you are assuming that the proper acceleration at turnaround is so large that, at the scale of the diagram, the smooth rounded-off corner looks like a kink.

Literally every twin paradox example assumes infinite acceleration that should not confuse anybody, realistic acceleration just complicates the whole issue.

 

[Dale]

I think that is OK. We often assume instantaneous acceleration. I think that the discontinuity is a bigger problem.

 

[obtronix]

I think that is OK. We often assume instantaneous acceleration. I think that the discontinuity is a bigger problem.

What discontinuity?

 

[PeterDonis]

realistic acceleration just complicates the whole issue

I'm not saying you need to actually do the math or draw the diagram using a realistic acceleration. I'm just saying that, since the kink in the traveling twin worldline seems to be confusing some people, it might help to make clear that it's not meant as a claim that the twin's acceleration is literally infinite, only that it is large enough that, for purposes of this discussion, the kink in the worldline is an acceptable approximation to the actual rounded corner.

 

[Dale]

What discontinuity?

The “Space Ship” lines are discontinuous at the turn around on all of your plots except the first.

 

[obtronix]

The “Space Ship” lines are discontinuous at the turn around on all of your plots except the first.

I mean it's instantaneous so it has to be discontinuous.

 

[PeterDonis]

What discontinuity?

As @Dale says, your "space ship" worldlines have gaps in them in your second and third diagrams. That's because, in those diagrams, you did not adjust the times of the "kinks" in the space ship worldlines for relativity of simultaneity. If the two "kinks" are simultaneous in the stay at home twin's rest frames, they will not be simultaneous in either the outbound or the inbound "traveling" frames. When you correct for this, your "space ship" worldlines should be continuous (no gaps at the kinks) in all frames.

 

[PeterDonis]

I mean it's instantaneous so it has to be discontinuous.

Not in the sense @Dale meant. See my post #38.

 

[obtronix]

As @Dale says, your "space ship" worldlines have gaps in them in your second and third diagrams. That's because, in those diagrams, you did not adjust the times of the "kinks" in the space ship worldlines for relativity of simultaneity. If the two "kinks" are simultaneous in the stay at home twin's rest frames, they will not be simultaneous in either the outbound or the inbound "traveling" frames. When you correct for this, your "space ship" worldlines should be continuous (no gaps at the kinks) in all frames.

oh, ok

 

[Dale]

I mean it's instantaneous so it has to be discontinuous.

No. Instantaneous acceleration makes a sharp bend in the line, not a jump. What you have drawn is not instantaneous acceleration but teleportation.

 

[Orodruin]

Literally every twin paradox example assumes infinite acceleration

Now that’s just patently false.
It is the most common, sure, but I have seen several examples of differential aging that do not use instantaneous acceleration. The most popular being continuous acceleration.

 

[member 728827]

If we draw the lines of simultaneous events for the "left spaceship" mirror, at the turnaround event there's a velocity discontinuity, and so we can draw two lines of simultaneous events, one line connecting with T=2, and the other connecting with T=3 (so to speak, obviously).

Then, there seems that you have your "lost" year.

If you draw the turnaround not so sharply, considering an acceleration, the "left spaceship mirror" worldline will look something like the diagram I include.

 

[anorlunda]

This video from Fermilab shows an explanation of the twin paradox that has no acceleration at all. Instead, we have a fixed observer A, a traveler B moving to the right with constant velocity, and another traveler C moving to the left with constant velocity.

 

[obtronix]

This video from Fermilab shows an explanation of the twin paradox that has no acceleration at all. Instead, we have a fixed observer A, a traveler B moving to the right with constant velocity, and another traveler C moving to the left with constant velocity.

But that's not the twin paradox thought experiments because they not are together to begin with and they never meet up at the end. They would have to jump for one spaceship to another which is acceleration.

In any event to have different velocities they would have to accelerate he just kind of starts in the middle of the thought experiment after the acceleration and then claimed acceleration is not needed kind of disingenuous in my opinion.

 

[anorlunda]

But that's not the twin paradox

Dr. Lincoln thinks that it is. Or at least, it is the equivalent. Review what he says is the point at 7:21 in the video.

 

[PeterDonis]

that's not the twin paradox

The worldlines in spacetime are the same, and the "paradox" is ultimately about comparing the lengths of worldlines. There is no need for a single observer to follow the entire "traveling twin" worldline in order to have the worldlines be what they are.

 

[PeroK]

But that's not the twin paradox thought experiments

The point is that the essence of the twin paradox is the nature of Minkowski geometry and acceleration is irrelevant.

Using acceleration to explain the twin paradox is like using the physical constraints of moving a pencil to explain Pythagoras theorem.

 

[Ibix]

But that's not the twin paradox thought experiments because they not are together to begin with and they never meet up at the end. They would have to jump for one spaceship to another which is acceleration.

To put it yet another way, Lincoln's version has three straight lines forming a triangle where the usual version has one straight line and one > shaped line. Same triangle either way.

 

[obtronix]

The worldlines in spacetime are the same, and the "paradox" is ultimately about comparing the lengths of worldlines. There is no need for a single observer to follow the entire "traveling twin" worldline in order to have the worldlines be what they are.

Yes but where/when does the thought experiment start and end in his thought experiment? To compare World lines you need a start and ending event. The twin paradox gives you those two events.

He says in the beginning of his thought experiment everyone sets their clock to zero, but they're all three in different reference frames in different locations... how can everyone set their clock to zero? All of their "now" times are different, who's "now".

 

[PeterDonis]

where/when does the thought experiment start and end in his thought experiment?

The same places as in yours. As several posters have already pointed out, the curves in spacetime are exactly the same.

To compare World lines you need a start and ending event.

Same as in your version.

He says in the beginning of his thought experiment everyone sets their clock to zero

You don't actually care how anyone's clocks are set, since you are measuring differences in clock readings, not absolute clock readings. All you need are the differences in clock readings between the start and end of the three line segments in spacetime: the "stay at home" line segment, the "traveling outbound" line segment, and the "traveling inbound" line segment. Then you just compare the sum of the second and third with the first; the latter will be larger.

 

[PeroK]

In any event to have different velocities they would have to accelerate he just kind of starts in the middle of the thought experiment after the acceleration and then claimed acceleration is not needed kind of disingenuous in my opinion.

But, you do get the same numerical answer, which ought to make you stop and think.

In other words, if we measure the elapsed time on clocks on the outbound and inbound journeys we get the equivalent result as the idealised physical experiment with a single traveling clock where the acceleration is extremely large.

It also doesn't matter whether we have 0, 2, or 4 acceleration phases. That should also make you think. Whether the experiment starts and ends with the traveling clock ar rest or not is essentially irrelevant to the result.

The only role acceleration plays is that it is necessary to enable the physical experiment. But, it is not itself necessary for the difference in elapsed times on the two clocks.

 

[Herman Trivilino]

They would have to jump for one spaceship to another which is acceleration.

All they need do is synchronize clocks as they pass each other.

 

[obtronix]

You don't actually care how anyone's clocks are set, since you are measuring differences in clock readings, not absolute clock readings.

Well that's not true he's saying you start the clocks simultaneously and measure time from that start. (e.g. spaceship C to event 2).

 

[PeterDonis]

that's not true he's saying you start the clocks simultaneously and measure time from that start

He's probably expecting that you will exercise some intelligence and not take everything he is saying exactly literally.

In any case, the fact that how each leg's clocks are handled was not specified correctly (if that is the case), does not mean that what is being described "is not the twin paradox". What makes the scenario "the twin paradox" is, as has already been pointed out multiple times, the geometry of the curves in spacetime. There are multiple ways to realize those curves in an actual scenario.

 

[Ibix]

Well that's not true he's saying you start the clocks simultaneously and measure time from that start. (e.g. spaceship C to event 2).

No, he has A (stay at home) and B (outbound) zero their watches as they pass, and C (returner) zero her watch as she passes B, and make a note of B's clock reading. B and C's times get reported to A, and their sum is the same as a single out-and-back traveller's total time. (At the 5.15 mark in the "no maths" video.)

 

[obtronix]

No, he has A (stay at home) and B (outbound) zero their watches as they pass, and C (returner) zero her watch as she passes B, and make a note of B's clock reading. B and C's times get reported to A, and their sum is the same as a single out-and-back traveller's total time. (At the 5.15 mark in the "no maths" video.)

Okay he doesn't say that but maybe he misspoke.

But I could assume C at rest, perform similar SpaceTime interval calculations, and have them record the longest time.

 

[Ibix]

But I could assume C at rest, perform similar SpaceTime interval calculations, and have them record the longest time.

No you can't - all those measures are invariant. Your own diagrams show you that!

 

[obtronix]

No you can't - all those measures are invariant. Your own diagrams show you that!

My diagrams have acceleration in them! He's creating a "ghost being" that is jumping from one ship to the other without accelerating. So he's kind of adding apples with oranges intervals.

If you think that's untrue what makes A so different than B and C? Relativity says motion is relative. I certainly can make a situation where events start and stop on the C World line. He just arbitrarily picked A.

 

[Motore]

My diagrams have acceleration in them!

No, they have instantaneous acceleration, which in your words would be a "ghost acceleration" as it's physically not possible.