I Twin Paradox: Who Is Right, A or B?

[Ibix]

#1 - and I suppose that the frame for an object as it progresses through time has to cover all of spacetime, that is its now line has to sweep out all of spacetime? (that makes sense, but noth

It's possible to have coordinate systems that don't cover all of spacetime. But they must cover the region of interest once and only once for them to be useful for a given problem.

#2 - I do believe that the Lincoln vid was while not technically wrong it was misleading, on his insistence that acceleration did not account for the distinction in the twins perspect

Not sure without re-watching. It certainly seems to have mislead you.

#3 - I just re-watched a Ted vid on the subject which I do think is technically wrong in that it says that the traveling twin sees the time passing slowly for the stationary twin on the outbound flight and quickly on the inbound flight.

I haven't watched the video, but it is possible that they are referring to what you literally see. If you travel towards a clock, the Doppler effect makes it appear to tick faster and if you travel away it makes it appear to tick slower. If you subtract out the changing lightspeed delay, you will be left with time dilation, which is the same in both directions.

Note that this analysis shows another asymmetry. The traveling twin will see the apparent rate of the stay-at-home's clock switch from slow to fast half way through the journey when they turn around. The stay-at-home will see the traveller's clock tick slowly until they see the turnaround, which will be a long time aftef half way through the trip.

 

[Janus]

#3 - I just re-watched a Ted vid on the subject which I do think is technically wrong in that it says that the traveling twin sees the time passing slowly for the stationary twin on the outbound flight and quickly

#3 - I just re-watched a Ted vid on the subject which I do think is technically wrong in that it says that the traveling twin sees the time passing slowly for the stationary twin on the outbound flight and quickly on the inbound flight. Note that both these vids are relatively recent and have 1+ million views, no mean achievement for a physics vid I think. Yes? No?

I'm pretty sure what is being considered here is what they would see due to relativistic Doppler shift. The "speed up" seen during the return leg is just due to the decreasing travel distance for the light. The traveling twin would still conclude that time was passing more slowly during the return leg.
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[Herman Trivilino]

However, the acceleration/deceleration does make a big difference in which frame is considered inertial, so that was the problem in my original version, as I figured you should be able to analyze the motion in both frames ignoring the acceleration/deceleration.

Did you not read Post #35 where I showed that both of your "twins" are inertial? At the end, T whizzes past B and compares his clock to B's clock.

 

[Halc]

It is, of course, possible to construct non-inertial frames in a non-naive manner so that they do cover all the relevant regions. My favourite is Dolby and Gull, but plenty of others are possible.

As for choice of systems (hard to call them frames) that do cover all regions, and only once, my favorite is the one the popular media often uses: What you see is what's happening now.

So for instance, a headline of 'Betelgeuse is expected to go supernova next week' is more digestible by the public than 'Betelgeuse is expected to have gone supernova 640 years ago'.

So A and B (not twins) look at each other from 20 light year distance, and each looks 20 years younger than himself, so each considers the other to be 20 years younger.
When A is 40, he embarks on a trip to B at .8c, when B appears to be 20. During the 15 years of the trip (ship clock), B's age accelerates to 3x normal, so B is 45 years older, or 65 (20 + 45) at trip end, and A is 55 (40 + 15).
From B's perspective, A looks 20 years younger until B's apparent departure at A's age 60 when A appears 40. B then appears to age 3x normal for the 5 year duration of the trip. At arrival 5 years later, A has aged 15 years during the 5 apparent traveling years, and B has aged 5 more, so they're 55 and 65 respectively.

All nice and consistent, symmetrical even, and no white or double grey areas. No discontinuities of anybody's age just because they're not present at some acceleration event. The rule is: for any observer, all of local spacetime is foliated by his past light cone. It lacks commutativity: If A is 40, B is 20 to A, but A is not 40 to 20 year old B, but that jives with reality. B cannot communicate with A anyway without the 40 year round trip lag, just like talking to the Mars rover.

#1 - and I suppose that the frame for an object as it progresses through time has to cover all of spacetime

Actually, there's no such rule, but sometimes it's nice to keep it that way locally. For instance, the inertial frame of the solar system hardly covers all of spacetime. Only the local stuff, not things say beyond the radius of the observable universe.

 

[Will Flannery]

Did you not read Post #35 where I showed that both of your "twins" are inertial? At the end, T whizzes past B and compares his clock to B's clock.

There was no 'T' in my scenario. I do understand that simultaneity is frame dependent.

 

[kuruman]

In the twin paradox there is a time in which A and B are at the same age at the same location (Earth). That's the starting point. This scenario has B already on the distant planet before he is joined by A. Obviously, if the two traveled together to the planet, there would be no difference in their ages throughout the trip and thereafter.

Here is my simple thought: If the two travel separately, assuming that the two trips are identical in terms of accelerations and velocities, when the twins meet on the distant planet, they will have zero age difference relative to each other. The relative aging process should be time-invariant and cannot depend on departure delays.

@Will Flannery's mistake is in A's reasoning. Suppose we start with a triplet instead of twins. Sibling C stays on the Earth, B leaves first and is followed by A. When A and C receive word that B has reached his destination safely, they know that B is younger than both of them. When A reaches his destination safely, he knows that he has aged relative to C exactly the same way that B has. Therefore, he expects to see B as aged as he is but younger than C. Similarly, when he lands on the planet, B knows that he is younger than A and C and that he will age at the same rate as them thereafter. Thus, when B is joined by A on the planet, he expects him to be as aged as he is, with both of them younger than C.

 

[Will Flannery]

On the one hand what Lincoln is saying is trivial: every observer, regardless of their state of motion, has a valid rest frame. The question is what laws of physics apply in an arbitrary frame of reference?

In SR, we have the Lorentz transformation between inertial reference frames (this unifies TD, LC and RoS into one neat mathematical package). In the case of the twin paradox, A changes from the initial rest frame into a new rest frame. It's valid to do this. But, he/she must apply the Lorentz transformation. Any data that A had from the initial rest frame (e.g. the location and time at planet B), must be transformed to the new "space shuttle" frame. This transformation changes the time "now" at planet B into some future time. Thereafter, time dilation applies to clocks on planet B, but they were already well ahead when A joined the space shuttle. And, when you do the maths, A gets the same answer that $B$ got using the initial rest frame: A's clock shows less time.

I missed this when you posted it, but now I see that that's it exactly, and I think it applies to Lincoln's version of the twin paradox where the traveling twin begins his return trip by hopping from his inertial frame heading outward to an inertial frame heading back to the home planet. When he hops he changes the basis for the spacetime vector space, and therefore he needs to apply the appropriate coordinate transformation to his current measurements, and if he does that he will 'get the right answer' when he returns home and will not be surprised by the age difference. So it's not the acceleration/deceleration per se that creates the problem, the problem is that the acceleration/deceleration produces a change in velocity and hence a basis change for the traveler's measurements, and he must apply the appropriate coordinate transformation before proceeding. Sweet!

 

[Herman Trivilino]

There was no 'T' in my scenario.

That has no effect on the issue. I re-stated your one-way twin paradox using only inertial observers and participants.

I do understand that simultaneity is frame dependent.

So you understand why the time on T's clock doesn't match the time on B's clock when T passes B. That should resolve the issue you raised in your original post.

 

[Will Flannery]

In the London vid he says
"All observers can completely accurately claim that they're the single unmoving person in the universe, and everyone is moving around them?"

And this seems to me to be clearly wrong. If I can say that I am not moving then it naturally follows that I am not accelerating, as if I were accelerating I would have to be moving. However, whether or not an object is accelerating is not 'mathematical' property only, it is a physical property that is frame independent. If I can say that I am not accelerating then I can say that I'm the center of an inertial frame. Thus, if London is right every observer can say he is at the origin of an inertial frame. It doesn't get any more wrong than that. ... :).

 

[PeroK]

In the London vid he says
"All observers can completely accurately claim that they're the single unmoving person in the universe, and everyone is moving around them?"

And this seems to me to be clearly wrong. If I can say that I am not moving then it naturally follows that I am not accelerating, as if I were accelerating I would have to be moving.

Absolute (proper) acceleration is not the same as absolute motion. For example, those of us on the surface of the Earth are accelerating upwards from the force of the surface of the Earth. But, we can still consider ourselves legitimately at rest in our frame of reference - which is, therefore, not an inertial frame. Meanwhile, objects in freefall towards the Earth have no forces on them, hence no proper acceleration.

Proper acceleration implies you do not have an inertial rest frame, but it does not imply some sort of absolute motion. You can measure the absolute value of the acceleration, but you cannot measure any absolute value of velocity at any time.

 

[Nugatory]

If I can say that I am not accelerating then I can say that I'm the center of an inertial frame.

That is not correct. For a counterexample, consider me sitting at rest in my chair right now: It is very convenient to describe myself as at rest, choosing a frame in which my coordinate velocity and acceleration is zero. Yet that frame is not inertial; the proper acceleration is non-zero.

 

[Sagittarius A-Star]

If I can say that I am not moving then it naturally follows that I am not accelerating, as if I were accelerating I would have to be moving.

No. The following "hopping" is only choosen as "twin paradox" scenario to avoid calculus:

where the traveling twin begins his return trip by hopping from his inertial frame heading outward to an inertial frame heading back

A smooth turn-around of the traveling twin A with a finite acceleration towards the "stay at home"-twin B is also allowed. While the turn-around, twin B is aging in A's frame faster than twin A. Twin A can regard himself as beeing at rest.

The following animation of a Minkowski diagram demonstrates this:

https://www.geogebra.org/m/M3YVE6eT

The green dashed line, that is a tanget line to the world line of A, is A's ct'-axis.
The other green dashed line is A's x'-axis. It is also a "simultaneity-line" in A's frame.

Please move the red "Time" control button on the upper right corner from left to right, with constant speed on your screen.

Then you can observe in parallel, how fast the intersection between A's green "simultaneity-line" and B's black t-axsis is moving up. You will see, that B ist aging fastest in A's frame while A's smooth turn-around.

A's acceleration can be regarded as a sequence of infinitesimal small "hoppings" from one "co-moving" inertial frame to the next "co-moving" inertial frame.

 

[Sagittarius A-Star]

In the London vid he says
"All observers can completely accurately claim that they're the single unmoving person in the universe, and everyone is moving around them?"

Here is another video. It shows the traveling twin at rest in his non-inertial frame:



(Remark: The title "Twin Paradox in General Relativity" is misleading, because the video shows a twin paradox in flat spacetime, but with "pseudo-gravitational time-dilation" in an accelerated frame. That is a topic of SR.)

 

[Ibix]

If I can say that I am not moving then it naturally follows that I am not accelerating, as if I were accelerating I would have to be moving.

This does not follow. The point is that there is no absolute meaning to "not moving". I have to say "not moving with respect to something", and if the something I pick is not inertial then I also need to accelerate.

There is an absolute meaning to "inertial" and "not inertial". But that isn't the same thing.

 

[Will Flannery]

The point is that there is no absolute meaning to "not moving".

There is no absolute meaning to not moving, but surely if I'm accelerating I cannot claim to be not moving.

 

[Ibix]

There is no absolute meaning to not moving, but surely if I'm accelerating I cannot claim to be not moving.

See the sentence after the one you quoted. Or @Nugatory's response.

 

[PeroK]

There is no absolute meaning to not moving, but surely if I'm accelerating I cannot claim to be not moving.

Have you studied uniform circular motion? The acceleration is centripetal (towards) the centre. But, there is no "motion" towards the centre. The acceleration is at right angles to the direction of motion .

The analogy isn't perfect, as you are moving in a circle, but never in the direction of the acceleration.

You are accelerating; you are changing your inertial reference frame. But, at any instant you will be at rest relative to some inertial observer. The key point is that you cannot give an absolute value to your state of motion. Perhaps forget about imprecise statements like "claim to be moving". The precise statement is that you may attribute no absolute velocity at any time. You may, however, attribute an absolute proper acceleration.

 

[kuruman]

Folks, tell me I am wrong if I am, but isn't it true that the twins at some point were together on Earth and were aged identically until B left? Isn't it also true that whatever affected the aging of B during his trip relative to A is time-invariant? In other words, B's aging process relative to A while traveling does not depend on whether B leaves for his trip today or a month later. The same applies to A. When A decides to leave, if his trip is identical in all respects to B's, he will age identically with B while traveling.

Obviously, the twins would have the same relative age if they traveled together. Unless delaying the departure of A somehow affects A's aging process differently from B's, when the twins are reunited on the planet they will have the same age.

I posted as much in #66, but there were no replies. If I am naively wrong, I would like to know. Thanks.

 

[PeroK]

Folks, tell me I am wrong if I am, but isn't it true that the twins at some point were together on Earth and were aged identically until B left? Isn't it also true that whatever affected the aging of B during his trip relative to A is time-invariant? In other words, B's aging process relative to A while traveling does not depend on whether B leaves for his trip today or a month later. The same applies to A. When A decides to leave, if his trip is identical in all respects to B's, he will age identically with B while traveling.

Obviously, the twins would have the same relative age if they traveled together. Unless delaying the departure of A somehow affects A's aging process differently from B's, when the twins are reunited on the planet they will have the same age.

I posted as much in #66, but there were no replies. If I am naively wrong, I would like to know. Thanks.

That's a good answer, but I think we are assuming that the twins have ended up on different planets after perhaps several space journeys and, by a process of Einstein clock synchronisation, established that, lo and behold, they are still approximately the same biological age in their common rest frame.

PS I have in the past tentatively suggested that using twins is a singularly unscientific way to measure time!

 

[Nugatory]

PS I have in the past tentatively suggested that using twins is a singularly unscientific way to measure time!

It is indeed - identically constructed clocks would be much better. But when constructing example for people unfamiliar with the concepts, biological aging feels more "real", maps better to what somene's notion of proper time ought to be.

 

[PeroK]

It is indeed - identically constructed clocks would be much better. But when constructing example for people unfamiliar with the concepts, biological aging feels more "real", maps better to what somene's notion of proper time ought to be.

I know my views are not shared, but when I was learning SR I decided to clear all the extraneous material away: twins, pole vaulters and barns, lightning strikes. Everything I considered "garbage". Not only did I consider these things extraneous, but that they were designed to deceive, cloud the mind and obscure the physics.

We have events, particles, measurements of position and time, reference frames, clocks and metre sticks. What austerity!

 

[kuruman]

That's a good answer, but I think we are assuming that the twins have ended up on different planets after perhaps several space journeys and, by a process of Einstein clock synchronisation, established that, lo and behold, they are still approximately the same biological age in their common rest frame.

PS I have in the past tentatively suggested that using twins is a singularly unscientific way to measure time!

Thank you for your response. The discussion may have digressed to different planets and multiple journeys, but the original post states that "A gets on a rocket and travels to B's planet."

I agree with your sentiments about the twin paradox. IMO it is as scientific as Schrodinger's cat.

 

[PeterDonis]

surely if I'm accelerating I cannot claim to be not moving

As you sit at your computer typing your posts, you are accelerating, because you are on the surface of the Earth and you feel weight. Are you moving?

 

[Shirish]

I'll take a stab at explaining this through spacetime diagrams. Other members can feel free to correct if there are some things I got wrong.

The thick axes correspond to $A$'s planet's frame, and the orthogonal axes on the right are those of $B$'s planet. The above spacetime diagram is from $B$'s perspective. The origin clocks on both planets are synchronized. But suddenly $A$ jumps into the rocket, and you can see that $t=0$ for the rocket coincides with $t=t_0$ for the planet $B$ origin clock in the rocket frame. You see that the time elapsed in the journey in $B$'s frame is $T$.

The demarcations on the time axes represent "ticks" on the clocks in the respective frames. $B$ sees that the rocket's time axis demarcations are larger, i.e. the clock ticks in the rocket take longer, i.e. $B$ will say that fewer ticks were counted by the rocket frame clock.

When $A$ stops on planet $B$, even though $A$'s clock switches its ticking rate to what $B$'s clock's ticking rate is, that still doesn't change the fact that fewer ticks were counted by $A$'s clock so far from $B$'s planet's perspective, which is where they're finally comparing times.

Why can't we reverse this argument and switch roles? Let's look at another ST diagram from the rocket's perspective:

Remember we'd concluded that $t=0$ for the rocket coincides with $t=t_0$ for the planet $B$ origin clock? This is what we're trying to account for above and that's why planet $B$'s spacetime diagram is shifted below. So now $t_0$ of $B$ planet's origin clock coincides with $0$ on the rocket clock, which is what we want as this ST diagram is from the rocket frame perspective.

In terms of aging, $B$ has a headstart of $t_{\text{extra}}$ on $A$. So even though $B$'s clock ticks slower from $A$'s perspective, some time has already elapsed on $B$'s clock which makes up for it.

Hope that helps!

 

[Dale]

There is no absolute meaning to not moving, but surely if I'm accelerating I cannot claim to be not moving.

Sure you can. You just have to use a non inertial frame.

 

[Will Flannery]

I'm working on a complete and definitive deconstruction of the twin paradox, accompanying it with space time diagrams and the Lorentz transforms for each of the frames involved. And in this process I've hit a snag ...
# I start with frame A stationary, frame B moving with velocity V = .6, giving a gamma of 1.25. The arrival at the distant planet is at (10, 6) in the A frame and (0, 8) in the B frame.
#When the moving twin reaches the distant planet we go from frame B stationary to frame A' moving with velocity V=-.6. The departure from the near planet is at (-10,-6) in the A' frame. All is well so far.
#Now we go from frame A' stationary to frame B' moving at velocity V =-.6 for the return trip. However, applying the Lorentz transform, I get the location of the departure from the near planet at (-

 

[Dale]

Looks like you are missing some text. Are your coordinates (t,x) or (x,t)? Do your frames all share an origin? I am a little confused about the primed frames.

 

[Will Flannery]

Looks like you are missing some text. Are your coordinates (t,x) or (x,t)? Do your frames all share an origin? I am a little confused about the primed frames.

I didn't mean to post the above, hit the wrong button. I'm still working on it, there is a conceptual error lurking somewhere in there and I haven't given up on it yet ... :)

 

[PeroK]

I didn't mean to post the above, hit the wrong button. I'm still working on it, there is a conceptual error lurking somewhere in there and I haven't given up on it yet ... :)

Let me give you a hand. Consider a journey from $A$ to $B$ in two frames. The first frame is the rest frame of $A, B$ and the second is a frame moving at speed $v$ from $A$ to $B$.

We'll take the common spacetime origin to be the departure event at $A$, event $E_0$; and the second event to be the arrival at $B$: $E_1$:

In the first frame we have, with the time coordinate first:
$$E_0 = (0, 0), \ \ E_1 = (t, x) = (\frac d v, d)$$
And, in the second (primed) frame we have:
$$E'_0 = (0, 0), \ \ E'_1 = (t', x') = (\gamma(t - \frac v {c^2} x), \gamma(x - vt)) = (\gamma(\frac d v - \frac{vd}{c^2}), 0)$$
That looks good: the second event also takes place at the $x' = 0$ and we have:
$$t' = \frac{\gamma d}{v}(1 - \frac{v^2}{c^2}) = \frac{d}{\gamma v} = \frac t {\gamma}$$

 

[Will Flannery]

I'm working on a complete and definitive deconstruction of the twin paradox, accompanying it with space time diagrams and the Lorentz transforms for each of the frames involved. Down to business -

# I start with frame A stationary, frame B moving with velocity V = .6, giving a gamma of 1.25. The arrival at the distant planet is at (10, 6) in the A frame and (0, 8) in the B frame.

The relevant Lorentz transforms are from the Stationary A to Moving B frames :
SMT(10,6,V) = 8 % A to B time of arrival event
SMX(10,6,V) = 0 % A to B location of arrival event

#When the moving twin reaches the distant planet we go from the moving frame B to a stationary to frame A' moving with velocity -V. The A' frame is the same as the A frame with the origin at (10,6). The departure from the near planet is at (-10,-6) in the A' frame. All is well so far.

#Now we go from frame A' stationary to frame B' moving at velocity -V for the return trip. However, applying the Lorentz transform, I get the location of the departure from the near planet at (-17,-15) in the B' frame ! Of course we don't want to go back to that event, but still this isn't right.

Well, it is right after all. The space time diagram for the return flight, showing the A' frame and the B' frame, is

The Lorentz transforms are:
SMT(-10, -6, -V) = -17
SMX(-10, -6, -V) = -15 % this was a big surprise to me, and inexplicable without the diagram
The original departure event is at (-17, -15) in the B' frame, and B' traveler timeline intersects the world line for the near planet at (8,0) in the B' frame.

So we have the travelers analysis of the trip, with a total trip time of 16 years, while the twin on the near planet ages 20 years.

All that is missing is a discussion of the asymmetry of the twins journeys which has been discussed. I've just added the spacetime diagrams, but I think that's an important addition towards making everything explicit.
Notes:
SMT = @(t,x,V) gamma*(t - x*V); % Stationary to moving Lorentz transform
SMX = @(t,x,V) gamma*(x - t*V);