Understanding twin paradox without math

[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.

 

[Lluis Olle]

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.

 

[Mister T]

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.

 

[PeterDonis]

He's creating a "ghost being" that is jumping from one ship to the other without accelerating.

You are quibbling. There is no need for anyone to jump from the outbound to the inbound ship at the turnaround event. All that needs to happen is that the inbound ship records the outbound ship's clock reading as they pass each other. And, as has already been pointed out, that's what the video specifies.

what makes A so different than B and C?

Spacetime geometry. As has already been pointed out, there is a triangle in spacetime with three sides, A, B, and C. The fact that A > B + C is just the spacetime version of the triangle inequality; we have > instead of < because of the minus sign in the spacetime metric.

Relativity says motion is relative.

Yes, but that doesn't imply the other (wrong) claims you are making.

I certainly can make a situation where events start and stop on the C World line.

You could define a different triangle in spacetime, sure. That would be a different scenario from the one we are discussing in this thread. And it would have a different relationship between the lengths of its sides. But that relationship would still be invariant, independent of any choice of frame.

What you can't do is have the same triangle in spacetime but then claim that "events start and stop on the C worldline". Because with that triangle in spacetime, they don't.

He just arbitrarily picked A.

If he did, then so did you when you posed the scenario in your OP.

I think you are not thinking clearly about what you are saying.

 

[Ibix]

My diagrams have acceleration in them! He's creating a "ghost being" that is jumping from one ship to the other without accelerating.

No, he's just copying information from one to the other.

If you think that's untrue what makes A so different than B and C?

The experimental setup. For example, A is the only one who sees the others as moving with equal and opposite velocities.

 

[Ibix]

I certainly can make a situation where events start and stop on the C World line. He just arbitrarily picked A.

Of course. But that would be a different scenario.

 

[obtronix]

Of course. But that would be a different scenario.

Not a different scenario a different point of view, everybody will be traveling exactly in the same way, same speed same distance, just the handoffs of times will be different.With the original twin paradox thought experiment you get the same answer from everyone's perspective.

 

[Ibix]

Not a different scenario a different point of view, everybody will be traveling exactly in the same way, same speed same distance, just the handoffs of times will be different.

But not the proper times, because they are invariant. And that's what we're measuring.

With the original twin paradox thought experiment you get the same answer from everyone's perspective.

And this one too. The triangle in the Minkowski diagram is identical to yours!

 

[PeterDonis]

everybody will be traveling exactly in the same way, same speed same distance, just the handoffs of times will be different

In which case whatever it is that you are calling "the handoffs of times" have no physical meaning and are irrelevant to the scenario.

 

[Sagittarius A-Star]

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).

According to the OP, ##\gamma = \frac{5}{4}##. Than means, the turnaround event of the "left spaceship" mirror (##\tau = 2##) has in the "left spaceship's" outbound rest-frame a simultaneity-line to the ##T= 2 \cdot \frac{4}{5} = 2- \frac{2}{5}## event of the (time-dilated) "left earth" mirror and (for symmetry reasons) in the "left spaceship's" inbound rest-frame a simultaneity-line to the ##T= 3+\frac{2}{5}## event of the "left earth" mirror.

 

[FactChecker]

IMO, a lot of the "solutions" to the paradox are dodging the real paradox by changing the scenario so that the paradox is removed. They do not really explain the original paradox.
IMO, the real paradox is due to confusing the second derivative of position with acceleration.
The second derivative of position is just as "relative" as the first derivative, velocity, is. Looking only at the first and second derivatives, the problem is completely symmetric wrt the twins. One might think that a traveling-twin-centered calculation should be just as valid as the Earth-centered calculation. Yet the two calculations give conflicting answers. The real "solution" to the original paradox is to point out that the Earth-centered calculation uses an intertial reference frame whereas the traveling-twin-centered calculation is in an accelerating frame. Acceleration is the key in distinguishing between an IRF and a non-IRF. To explain the original paradox, acceleration can not be ignored. The original paradox can be bypassed (but not explained) by changing the calculations to use only IRFs.
PS. I have seen a calculation that used the traveling-twin-centered reference frame from beginning to end, took an acceleration profile into account, and came up with the same answer as the Earth-centered IRF calculation. That was intellectually satisfying.

 

[PeterDonis]

Looking only at the first and second derivatives, the problem is completely symmetric wrt the twins. One might think that a traveling-twin-centered calculation should be just as valid as the Earth-centered calculation. Yet the two calculations give conflicting answers.

I'm not sure what "traveling-twin centered calculation" you are referring to here. Do you have a reference?

I have seen a calculation that used the traveling-twin-centered reference frame from beginning to end, took an acceleration profile into account, and came up with the same answer as the Earth-centered IRF calculation. That was intellectually satisfying.

Again, can you give a reference?

 

[PeterDonis]

IMO, a lot of the "solutions" to the paradox are dodging the real paradox by changing the scenario so that the paradox is removed.

The real "solution" to the original paradox is to point out that the Earth-centered calculation uses an intertial reference frame whereas the traveling-twin-centered calculation is in an accelerating frame. Acceleration is the key in distinguishing between an IRF and a non-IRF. To explain the original paradox, acceleration can not be ignored.

I think you are misidentifying the "paradox". The "paradox" is that the two twins have aged differently when they meet up again. But that "paradox" is resolved by spacetime geometry: the two twins follow different worldlines between the same starting and ending events, and those different worldlines have different lengths.

Proper acceleration in the original "paradox" scenario is best viewed as the means by which the traveling twin's worldline is enabled to meet up a second time with the stay-at-home twin's (since in flat spacetime it is impossible for two inertial worldlines to meet more than once).

Note, also, that proper acceleration has nothing to do with "an accelerating frame". You can perfectly well describe proper acceleration using an inertial frame, or inertial motion using a non-inertial frame. The best way to avoid confusion in cases like this is to not even mention frames at all, but to focus on the invariants: the spacetime geometry, the worldlines of the twins, and whatever invariant properties (such as the proper acceleration of the traveling twin in the standard scenario) are required to enable the two worldlines to meet a second time.