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

[Will Flannery]

TL;DR Summary

The twin paradox is modified to consist of a one-way trip to a distant location.

In this version of the twin paradox one twin, A, is located on earth, and the other twin, B, is located on a distant planet, which is at a fixed location in A's frame of reference.

At the beginning time t = 0 the twins are stationary and their clocks are synchronized at 0.

A gets on a rocket and travels to B's planet. During the trip A notes that B's frame of reference, frame B, is moving at a high velocity with respect to his, A's, frame of reference, which is fixed with A at its origin. So, aware of time dilation, he computes that his, A's, clock is running faster than B's clock. Thus when A arrives at B's location A expects to see a younger B.

B is also aware of time dilation, and during A's trip B notes that A's frame of reference is moving at a high rate of velocity with respect to his, B's, frame and calculates that his, B's, clock is running faster than A's clock. Thus when A arrives at B's location B expects to see a younger A.

Who is right, A or B?

 

[Dale]

Summary:: The twin paradox is modified to consist of a one-way trip to a distant location.

Who is right, A or B?

Both. With a one way trip the answer is frame dependent.

 

[Ibix]

Summary:: The twin paradox is modified to consist of a one-way trip to a distant location.

Who is right, A or B?

If they synchronised their clocks in the rest frame of the planets, the traveller will be younger when he arrives. If he thinks otherwise, it's likely because he forgot (as you did) to account for the relativity of simultaneity when he switched from the "stationary" inertial frame to the "moving" one.

As Dale notes, though, the answer is dependent on how they initially synchronised their clocks to call themselves "twins". I'm assuming they used Einstein synchronisation in their initial mutual rest frame, but it is not obligatory to do so.

Edit: how many typos is it possible to make in one post? I think I've corrected them all now...

 

[Ibix]

Both. With a one way trip the answer is frame dependent.

I think you have the scenario back to front. The twins start on separate planets and finish on a common one, so there's a unique answer to which one is older at the end. What there isn't is a unique answer to which was older before the experiment - so the OP calling these people twins is a bit misleading.

 

[Will Flannery]

I'm assuming they were born twins (but it's not important). And, at some point in time one migrated to the outer planet.

They synchronize clocks with A setting his clock to 0, and flashing a light at the same time. B sees the light, and knowing the distance between A and B sets his clock to be in sync with A.

What does 'account for simultaneity when he switched from the "stationary" inertial frame to the "moving" one' mean?

 

[Janus]

As indicated by Ibix, the key here is to note that Twin A has to transition from being at rest with respect to both B and Earth, to having a motion relative to them. In order to do this, there will be a period (however short) where he is making his measurements from an non-inertial frame of reference. In other words, he will be accelerating.
For him, the Earth clock and B's clock will go out increasingly out of sync as he accelerates towards B, with B's clock advancing much faster than his own.
So A will say that B aged very rapidly as A accelerated up to speed, and then aged slowly once the acceleration phase ended. The amount B aged during the acceleration (according to A) more than makes up for the period where he ages slowly(according to A) and B ends up accumulating more age over the duration of the trip than A.
The only effect that the acceleration has on A according to B ( or the Earth) is that as A increases in speed, the rate at which A ages decreases.

Both A and B will agree on their respective age differences once they meet up, they will however disagree as to just how that age difference came about.

 

[Ibix]

I'm assuming they were born twins (but it's not important). And, at some point in time one migrated to the outer planet.

In that case it's a two way trip, just with a rest at turnaround. The traveller will be younger.

They synchronize clocks with A setting his clock to 0, and flashing a light at the same time. B sees the light, and knowing the distance between A and B sets his clock to be in sync with A.

Depends on how you define "distance". The traveller will be younger as noted, but depending on how you do the synchronisation (your choice of definition of distance, or your choice of lightspeed anisotropy) his clock, which only traveled one way, may be ahead or behind.

 

[Herman Trivilino]

In this version of the twin paradox one twin, A, is located on earth, and the other twin, B, is located on a distant planet, which is at a fixed location in A's frame of reference.

At the beginning time t = 0 the twins are stationary and their clocks are synchronized at 0.

How would they go about synchronizing their clocks? However you do it, once one of them starts moving he will no longer observe that they are synchronized.

Thus when A arrives at B's location B expects to see a younger A.

But you are using the clock at B's location to reach that conclusion. If you use the clock at A's original location then you will reach a different conclusion. The conclusions are contradictory because they are based on a contradictory premise, namely that the clocks stay synchronized in A's rest frame.

 

[PeroK]

Who is right, A or B?

$B$ remains in the same inertial reference frame throughout and his analysis is valid. $A$ changes from one inertial reference frame (the common initial rest frame of $A$ and $B$) to a different inertial reference frame (this was detectable through the real, proper acceleration). It is invalid, therefore, for $A$ to pretend he/she stayed in the same inertial frame throughout.

$B$ is right.

 

[Nugatory]

At the beginning time t = 0 the twins are stationary and their clocks are synchronized at 0.

You are making an additional assumption here, namely that they are the same age when they synchronize their clocks. Without that assumption there's no point in comparing their ages. You should think about exactly how you establish the truth of that assumption.

(Also note that "they are the same age" requires a careful definition. Suppose at birth they're both given wristwatches that follow them around for the rest of their lives; we can they say that their age is whatever amount of time has elapsed on their lifetime wristwatches, or if their heartrates were fixed we could count heartbeats do determine their age, or...)

 

[Will Flannery]

"For him, the Earth clock and B's clock will go out increasingly out of sync as he accelerates towards B, with B's clock advancing much faster than his own"

I had given that a lot of thought and was unable to agree or disagree with it. But now I have a way to test it. I started with wiki, which tells us ...

Starting with Paul Langevin in 1911, there have been various explanations of this paradox. These explanations "can be grouped into those that focus on the effect of different standards of simultaneity in different frames, and those that designate the acceleration [experienced by the traveling twin] as the main reason"

and which also references Einstein's answer to the very question ...

However, the reason that that line of argument as a whole is untenable is that according to the special theory of relativity the coordinate systems K and K' are by no means equivalent systems. Indeed this theory asserts only the equivalence of all Galilean (unaccelerated) coordinate systems, that is, coordinate systems relative to which sufficiently isolated, material points move in straight lines and uniformly. K is such a coordinate system, but not the system K', that is accelerated from time to time. Therefore, from the result that after the motion to and fro the clock U2 is running behind U1, no contradiction can be constructed against the principles of the theory.

On the other hand the modern sources I've been reading attribute time dilation to relative velocity and not acceleration.

So, which is it? For time dilation due to relative velocity we have a formula to calculate it. Is there a formula for computing time dilation due to acceleration? And, it seems to me that the acceleration at the beginning of the trip would be offset by the deceleration at the end of the trip. And, there are now explanations that claim that acceleration/deceleration is not the reason ... for example ... this vid by Don Lincoln of Fermilab:

So ... the test would be a formula for calculating the time difference that includes the effects of both acceleration and relative velocity. I haven't seen one.

 

[PeroK]

... to be precise, you are talking about differential ageing. Time dilation itself is symmetric aspect of measurements between reference frames in relative motion. The difference in ages is, in fact, not really due to time dilation at all. In this case, the differential ageing is due to $A$ changing their notion of simultaneity when they changed reference frames.

Acceleration does not "cause" time dilation. Nor does it cause differential ageing. But, it does signify that you are changing inertial reference frames and you can no longer consider yourself to be an inertial observer for a whole experiment.

That's why $A$'s analysis is invalid. $A$ changed their simultaneity convention and pretended they hadn't.

 

[Ibix]

Is there a formula for computing time dilation due to acceleration?

No. There isn't any time dilation from acceleration. If you want to talk about the rest frame of an accelerating twin then what you have is a non-inertial frame, and time dilation isn't uniquely definable in such a frame. That's the meaning of the Einstein quote you provided.

If you specify how you wish to define simultaneity in the accelerating frame then you can establish a relationship between clocks at rest in this frame and those at rest in the inertial frame. That would be a time-varying time dilation, in some sense, although not the commonly accepted one.

 

[Will Flannery]

" Nor does it cause differential ageing. But, it does signify that you are changing inertial reference frames and you can no longer consider yourself to be an inertial observer for a whole experiment. "

In the Don Lincoln vid linked in my previous post he 'explains' the twin paradox without using any non-inertial frames. He does this by ignoring acceleration and deceleration at the beginning and end of the trip, and doing a handover from one inertial frame to another at the distant planet.

He also makes a bold assertion at the beginning of the vid ... at 1:37 "Everyone can say they aren't moving". Given his credentials, I am not inclined to contradict him.

 

[PeroK]

" Nor does it cause differential ageing. But, it does signify that you are changing inertial reference frames and you can no longer consider yourself to be an inertial observer for a whole experiment. "

In the Don Lincoln vid linked in my previous post he 'explains' the twin paradox without using any non-inertial frames. He does this by ignoring acceleration and deceleration at the beginning and end of the trip, and doing a handover from one inertial frame to another at the distant planet.

He also makes a bold assertion at the beginning of the vid ... at 1:37 "Everyone can say they aren't moving". Given his credentials, I am not inclined to contradict him.

Yes, that's the best way to analyse the paradox in my opinion (without acceleration). But, you asked "who is right"? You provided an analysis from an observer ($A$) who changed reference frames by accelerating and I've explained why that analysis is invalid.

 

[Ibix]

In the Don Lincoln vid linked in my previous post he 'explains' the twin paradox without using any non-inertial frames.

You can explain it without any frames at all if you want. The problem with your analysis is that you are failing to account for the change in simultaneity definition being used by the traveling twin when he starts to move. You either need a non-inertial correction or a separate correction for this.

 

[Halc]

So, which is it? For time dilation due to relative velocity we have a formula to calculate it. Is there a formula for computing time dilation due to acceleration?

There is no dilation directly due to acceleration.
I can have one clock on Earth's equator, accelerating minimally (about 0.034 m/sec²) as the Earth spins, and another in a centrifuge at one of the poles accelerating at an insane g force as it spins at the same linear speed (about 460 m/sec). The only difference is one accelerates thousands of times more than the other. They will remain in sync indefinitely, so acceleration alone does not cause dilation.

The change of B's synchronization as A accelerates toward's B is due to relativity of simultaneity coupled with B's significant separation from A. A completely different time on B's clock is simultaneous with A's clock in one inertial frame vs another. Hence any synchronization procedure between these separated clocks must assume an inertial frame of reference.

 

[Will Flannery]

Yes, that's the best way to analyse the paradox in my opinion (without acceleration). But, you asked "who is right"? You provided an analysis from an observer ($A$) who changed reference frames by accelerating and I've explained why that analysis is invalid.

I've been working on this thing for several days, and each day I have a new problem - now I have today's problem - try to formulate the 1-way paradox with inertial frames only.

 

[Ibix]

I've been working on this thing for several days, and each day I have a new problem - now I have today's problem - try to formulate the 1-way paradox with inertial frames only.

That's easy. Just remember to work out the time on Earth that's simultaneous in the traveling inertial frame with the travveller setting out. It'll be later than zero by exactly enough to account for the age difference once you also account for time dilation.

 

[PeroK]

I've been working on this thing for several days, and each day I have a new problem - now I have today's problem - try to formulate the 1-way paradox with inertial frames only.

That's the best way. You need $A$ to hop (instantaneously) aboard an interstellar shuttle service that is already traveling at relativistic speed towards $B$.

As others have said, when $A$ hops aboard the prior calculation for the age of $B$ "now" is no longer valid. The ship's crew will say that $B$ is older than $A$ "now" in the ship's reference frame. That's the critical aspect.

 

[Ibix]

Draw a Minkowski diagram in the frame of the planets.

Using this frame's simultaneity convention, shade the region that is before the departure of the traveller. Shade the region that is after the arrival of the traveller.

Let the inertial frame of the traveling twin be S'. Use the inverse Lorentz transforms to determine the form of a line of constant t' in terms of the x and t coordinates of the planets' frame.

Draw a line of constant t' through the departure of the traveling twin. Draw a line of constant t' through their arrival. Shade the area between these lines.

The shaded areas represent the time before, during, and after the trip according to inertial frames. You will see that a portion of Earth's worldline is not in any shaded region. This is the bit that you forgot about when you changed from the planet frame to the traveling frame. You will find that it is $(1-1/\gamma)\Delta t$ long, where $\Delta t$ is the travel time in the Earth frame.

 

[Dale]

I think you have the scenario back to front. The twins start on separate planets and finish on a common one, so there's a unique answer to which one is older at the end. What there isn't is a unique answer to which was older before the experiment - so the OP calling these people twins is a bit misleading.

Yes, I had assumed that twins start the journey together. My misunderstanding.

I'm assuming they were born twins (but it's not important). And, at some point in time one migrated to the outer planet.

Then it is just a standard two way twins scenario.

 

[Herman Trivilino]

I've been working on this thing for several days, and each day I have a new problem - now I have today's problem - try to formulate the 1-way paradox with inertial frames only.

Just have B already in motion (relative to A) when the thought experiment begins. Of course, "when the thought experiment begins" is a relative statement. A and B won't agree on it. You want to say it happens at t=0 when they synchronize their clocks, but A won't agree that B did it correctly and B will not agree that A did it correctly.

How is it possible that while they're in relative motion each will observe the other's clocks running slow? In my opinion, if you can answer that question you will be able to understand the twin paradox.

 

[Dale]

So, which is it? For time dilation due to relative velocity we have a formula to calculate it. Is there a formula for computing time dilation due to acceleration?

The acceleration does not produce the time dilation. It breaks the symmetry so that the accelerated reference frame is non inertial so the usual formulas don’t apply.

 

[Sagittarius A-Star]

The Minkowski diagram describes it in an inertial reference frame. It can be also described in A's reference frame.

As long as A's rocket accelerates towards B, this is a reference frame with a proper acceleration, which contains pseudo-gravity. A clock at the front end of A's rocket ticks by a factor ≈ 1 + ΔΦ/c² faster than a clock at the rear end of the rocket with the pseudo-gravitational potential Φ = a * L, with a = proper acceleration of the clock at the rear end of the rocket and L = length of the rocket. While this acceleration, B is in a much higher pseudo-gravitational potential than both rocket-clocks because of the large distance between A and B, so the dominating factor is pseudo-gravitational time-dilation. B's clock is running much faster than A's clock(s). When the rocket finished it's acceleration and continues the fligth with no acceleration, B's clock is ticking slower than A's clock. Integrated over the complete travel, B is at the end older than A.

 

[Ibix]

While this acceleration,

Note that care is needed over the definition of "while" here. It likely means the patch of spacetime outside the past lightcone of the start of the acceleration and the future lightcone of the end of the acceleration, but I have not checked that.

 

[Dale]

He also makes a bold assertion at the beginning of the vid ... at 1:37 "Everyone can say they aren't moving". Given his credentials, I am not inclined to contradict him.

I agree with him that everyone can say they aren’t moving. However, not everyone can say they are inertial. The ones who cannot say they are inertial also cannot use the inertial-frame formulas in a frame where they are not moving.

 

[Will Flannery]

That's the best way. You need $A$ to hop (instantaneously) aboard an interstellar shuttle service that is already traveling at relativistic speed towards $B$.

As others have said, when $A$ hops aboard the prior calculation for the age of $B$ "now" is no longer valid. The ship's crew will say that $B$ is older than $A$ "now" in the ship's reference frame. That's the critical aspect.

I like this idea, and it is apparently what is normally done at the beginning and end of the journey, and in the Lincoln vid A even jumps from one inertial frame to another with no ill effects.

However, A estimates B's age using his (A's own) clock and nothing unusual happens to his clock when he jumps on the shuttle. I see that his 'now' line changes, but ... so what?

Then A jumps off the shuttle when he reaches B's planet. Again his now line changes again, but nothing happens to his clock. Because of time dilation B's clock has run faster and A is surprised that B is older.

The above I think matches up with the normal analysis.

And, from B's perspective how are things different? Nothing that I can see. The relative motion is exactly as if B had jumped on a shuttle and headed toward's A's location, and hence B expects A to be older.

(Note: I've also most of my time today with spacetime diagrams and in that context things get confusing... but I've done enough for today !)

 

[Ibix]

I see that his 'now' line changes, but ... so what?

So he's made the same mistake I would make if I stepped across a time zone boundary and wondered why it had taken an hour to take a single step. It didn't. I just changed from one set of clocks to another and failed to account for the offset.

In this case, "now, on Earth" goes from the time when Earth clocks show zero (simultaneous with the traveller's departure in the Earth frame) to the time when they show $(1-1/\gamma)\Delta t$ (simultaneous with the traveller's departure in the traveling frame). If the traveller only counts the time before zero and the time after $(1-1/\gamma)\Delta t$ and ignores the bit inbetween, of course he'll get the wrong answer.

 

[Halc]

Just have B already in motion (relative to A) when the thought experiment begins. Of course, "when the thought experiment begins" is a relative statement. A and B won't agree on it. You want to say it happens at t=0 when they synchronize their clocks, but A won't agree that B did it correctly and B will not agree that A did it correctly.

It matters not. Not in the frame of either can two approaching clocks remain in sync, so syncing them by any procedure is useless since each runs slower in the frame of the other.
Choose the frame between then, the one where each is moving at an equal but opposite speed. In that frame they'll stay in sync.

How is it possible that while they're in relative motion each will observe the other's clocks running slow?

They won't. An approaching clock will appear to run fast (blue shift), not slow. There's a difference between 'that approaching clock runs slow in my frame' and 'that approaching clock appears to run fast from my vantage'. The former involves only dilation, but the latter also adds Doppler effect.

 

[Dale]

However, A estimates B's age using his (A's own) clock

This is incomplete as stated. A estimates B’s age using his clock and a simultaneity convention. Forgetting simultaneity is the number one cause of confusion in analyzing thought experiments.

nothing unusual happens to his clock when he jumps on the shuttle. I see that his 'now' line changes, but ... so what?

So his estimate of B’s age also changes because it depends on both his clock and his simultaneity convention.

A is surprised that B is older.

Not if he does a valid analysis. The analyses you have proposed that lead to his surprise involve one of two flaws: ignoring the simultaneity convention or treating a non-inertial frame as though it were inertial.

 

[Will Flannery]

And a week ago I thought this was going to be easy (as there was basically no math). Wrong.

I attempted to reformulate my original post using only inertial frames, and I have failed. The experience has led me to several conclusions:

#1 - Einstein's objection to the criticism quoted was that non-inertial frames invalidate the analysis. Given that, I think Lincoln's use of 'hopping onto a moving frame' is not valid. It is not physically realizable, and it can be approximated by using accelerating frames, and the higher the acceleration the closer the approximation. So, it is really a way to introduce accelerating frames into the analysis, hence not valid.

#2 - All the descriptions of the twin paradox have the traveler hopping onto a moving frame at the start, decelerating and accelerating to turn around, and hopping off a moving frame at the end. Hence these descriptions are not realizable (because of the hopping) and also not valid (because of the non-inertial frames).

#3 - What experiments confirm time dilation? I've read that the GPS system accounts for time dilation in it's calculations, but satellites are continuously accelerating. Is there a realizable experiment confirming time dilation that doesn't involve non-inertial frames?

 

[PeroK]

Given that, I think Lincoln's use of 'hopping onto a moving frame' is not valid.

Given his credentials, I am not inclined to contradict him.

Is there a realizable experiment confirming time dilation that doesn't involve non-inertial frames?

Okay, a couple of hard facts. The basics of SR have been well understood for 115 years. All of modern physics is based on SR. Understanding SR is a hurdle that all students have to get over. It's not a reasonable or logical way out to start to doubt SR.

Built on time dilation, length contraction and the relativity of simultaneity is the SR theory of energy, momentum and relativistic particle dynamics. This is tested every day in particle physics experiments. In addition to GPS calculations there is also the Hafele-Keating experiment.

To get back to the question. It's a common mistake to focus on time dilation, which is only one of a trio of equally fundamental results about spacetime. In many ways the relativity of simultaneity is even more fundamental.

There are many threads on this site where someone has presented the obvious and unavoidable contradictions if we accept time dilation alone. The counterargument is always the same: they are forgetting the relativity of simultaneity.

So, perhaps, take a break from thinking about time dilation and focus on the relativity of simultaneity.

 

[Dale]

I attempted to reformulate my original post using only inertial frames, and I have failed.

It is easy to formulate it using only inertial frames: simply do the analysis in the home twin’s frame.

I assume that you mean you are trying to reformulate it using only inertial objects. That can be done in terms of what is sometimes called a “triplets scenario”. In that case you have three inertially moving clocks, say P, Q, R. In P’s frame Q is moving to the right at v and R is moving to the left at v. As Q passes P they reset their clocks to 0. Then as R passes Q we have R set their clock to match Q. Then we compare the clocks of R and P when they meet.

The difference is the same as the twin paradox, but all clocks are inertial.

#3 - What experiments confirm time dilation? I've read that the GPS system accounts for time dilation in it's calculations, but satellites are continuously accelerating. Is there a realizable experiment confirming time dilation that doesn't involve non-inertial frames?

Many experiments confirm time dilation. However, it seems like you specifically want experiments that confirm time dilation for inertial objects. Correct?

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

 

[Herman Trivilino]

I attempted to reformulate my original post using only inertial frames, and I have failed.

I can do it. Planets A and B are at rest relative to each other and separated from each other. T is a traveler that passes A on his way towards B. When T passes A he sets his clock to zero and notes that A's clock also reads zero. Prior to this happening the occupants of A and B have synchronized their clocks, so a clock at B also reads zero, according to the occupants of both A and B, when T passes A.

But T will not agree that the clock on B reads zero when he passes A. Instead, he will observe that the clock on B is well ahead of the clock on A. This effect is called the relativity of simultaneity.

You cannot understand the symmetry of time dilation (each observes the other's clocks running slow) without also understanding the relativity of simultaneity. This is the stumbling block to understanding the twin paradox.

 

[Sagittarius A-Star]

#1 - Einstein's objection to the criticism quoted was that non-inertial frames invalidate the analysis.

Non-inertial reference frames can be used in SR, but it is more complicated to calculate in non-inertial frames than in inertial frames.

#2 - All the descriptions of the twin paradox have the traveler hopping onto a moving frame at the start, decelerating and accelerating to turn around, and hopping off a moving frame at the end. Hence these descriptions are not realizable (because of the hopping) and also not valid (because of the non-inertial frames).

In 1918, Einstein described the twin paradox in both frames, without hopping:
https://en.wikisource.org/wiki/Translation:Dialog_about_Objections_against_the_Theory_of_Relativity
(Remark: When he speaks in flat spacetime about "gravitational field", this is named by today's physicist mostly "pseudo-gravitational field".)

#3 - What experiments confirm time dilation?

For example experiments with myons in storage rings:

Bailey et al. (1977) measured the lifetime of positive and negative muons sent around a loop in the CERN Muon storage ring. This experiment confirmed both time dilation and the twin paradox, i.e. the hypothesis that clocks sent away and coming back to their initial position are slowed with respect to a resting clock.

Source:
https://en.wikipedia.org/wiki/Experimental_testing_of_time_dilation#Twin_paradox_and_moving_clocks

Is there a realizable experiment confirming time dilation that doesn't involve non-inertial frames?

Almost without acceleration are myons in the upper atmosphere (moving inertially with almost c):
http://www.atmosp.physics.utoronto.ca/people/strong/phy140/lecture32_01.pdf

 

[Will Flannery]

It is easy to formulate it using only inertial frames: simply do the analysis in the home twin’s frame.

Not quite. The problem is that in the original post A stopped at B. He didn't note B's age as he passed by. I regarded having the traveler stop as part of the scenario. It's a question of physical realizability.

"Thus when A arrives at B's location A expects to see a younger B."

However, it seems like you specifically want experiments that confirm time dilation for inertial objects.

Inertial object? What is an inertial object?

I assume that you mean you are trying to reformulate it using only inertial objects. That can be done in terms of what is sometimes called a “triplets scenario”. In that case you have three inertially moving clocks, say P, Q, R. In P’s frame Q is moving to the right at v and R is moving to the left at v. As Q passes P they reset their clocks to 0. Then as R passes Q we have R set their clock to match Q. Then we compare the clocks of R and P when they meet.

The difference is the same as the twin paradox, but all clocks are inertial.

This is an improvement to the Lincoln explanation in my view as it doesn't have the hopping. But it still lacks the realizable angle. In the twin paradox the twin (as it is usually formulated) the twin has to arrive home, not flash by at fraction of the speed of light.

So, the twin paradox scenario as stated, with the twin arriving back and stopping, has not been analyzed at all as far as I know. To do so would require analyzing the effect of acceleration/deceleration on the traveling twin's clock. What if he arrived and his clock matched that of his twin ?

 

[Dale]

Inertial object? What is an inertial object?

It is an object which is moving inertially. I.e. an object where an attached accelerometer would read 0.

The point is that inertial objects are physical. You can physically take an object and attach an accelerometer and determine if it reads 0. In contrast an inertial frame is a mathematical construct that requires an arbitrary synchronization convention and which has no impact on any physical measurement.

But it still lacks the realizable angle.

Why not? It is perfectly realizable. Of course it is not identical to the twin scenario, but that is obvious: since you wanted to eliminate acceleration then you must make some changes.

 

[Will Flannery]

Why not? It is perfectly realizable. Of course it is not identical to the twin scenario, but that is obvious: since you wanted to eliminate acceleration then you must make some changes.

I didn't want to eliminate acceleration, instead all the analyses we've seen have eliminated it without mentioning the fact. I harbor the suspicion the the twins' clocks would read the same :).

 

[Dale]

I didn't want to eliminate acceleration, instead all the analyses we've seen have eliminated it without mentioning the fact

Huh? Most analyses of the twin paradox do not eliminate acceleration. That is, in fact, the standard approach. You are making no sense.

I harbor the suspicion the the twins' clocks would read the same :).

Your suspicion is unfounded. Accelerations have been tested up to about 10^18 g. No additional acceleration-based time dilation effect was found.

 

[Ibix]

I didn't want to eliminate acceleration, instead all the analyses we've seen have eliminated it without mentioning the fact. I harbor the suspicion the the twins' clocks would read the same :).

We've been assuming acceleration was instantaneous because that way you don't need calculus. Finite acceleration is fairly straightforward to include if you want to do it, and it will yield the same result that the traveller is younger (assuming that the twins' ages are compared by Einstein synchronisation before the experiment starts).

 

[Will Flannery]

Huh? Most analyses of the twin paradox do not eliminate acceleration. That is, in fact, the standard approach. You are making no sense.

? Analysis of acceleration/deceleration is entirely absent from the standard approach.

Your suspicion is unfounded. Accelerations have been tested up to about 10^18 g. No additional acceleration-based time dilation effect was found.

I base it on this (but it's too late in the day for me to check) - Need to rethink - I'll go over this tomorrow (or not).

 

[Nugatory]

I didn't want to eliminate acceleration, instead all the analyses we've seen have eliminated it without mentioning the fact. I harbor the suspicion the the twins' clocks would read the same :).

There are several acceleration-free ways of stating the problem. Here’s one, described using a frame in which the Earth is at rest:
We start with spaceship A somewhere to the left of the earth, moving right at speed v relative to earth, and spaceship B somewhere farther to the right of earth, moving left at speed v relative to earth. A will pass by Earth and then eventually meet B moving in the opposite direction; and then B will continue on past the Earth while A sails off into the distance.

As A passed Earth we zero a clock on Earth and on ship A; this works because A and Earth are at the same place at the same time. Later A passes B and and as they pass the clock on B is set to $T_1$, the time on A’s clock as they pass one another; again this works because A and B are colocated. B’s clock continues ticking as B approaches earth. The time on B’s clock as B passes Earth will be the total time along the outbound leg and the inbound leg; it’s the time that would be registered by a hypothetical acceleration-tolerant clock that was transferred from A to B as they passed one another.

When we compare B’s clock and the Earth clock as they pass, we will find that more time passed on Earth than on the two legs of the round trip.

 

[Ibix]

An object undergoing constant proper acceleration $\alpha$ has coordinate acceleration $dv/dt=\alpha/\gamma^3$. You can integrate this to obtain its coordinate velocity:$$v=\frac{\alpha c t}{\sqrt{c^2+\alpha^2t^2}}$$That allows you to write the Lorentz $\gamma$ factor:$$\gamma=\sqrt{1+\frac{\alpha^2}{c^2}t^2}$$Noting that $d\tau/dt=1/\gamma$, you can substitute the expression for $\gamma$ and integrate to determine the proper time elapsed on the ship while coordinate time $t$ elapses:$$\tau=\frac c\alpha\sinh^{-1}\left(\frac\alpha ct\right)$$If a ship, initially at rest, accelerates at some constant proper acceleration $\alpha$ for some coordinate time $T$ then decelerates at the same proper acceleration for the same amount of time, we can calculate the elapsed time for the ship to be$$\begin{eqnarray*}
\tau&=&\frac c\alpha\sinh^{-1}\left(\frac\alpha cT\right)+\frac c{-\alpha}\sinh^{-1}\left(\frac{-\alpha}cT\right)\\
&=&2\frac c\alpha\sinh^{-1}\left(\frac\alpha cT\right)
\end{eqnarray*}$$You can confirm for yourself that $\sinh^{-1}(x)\neq x$ and hence $\tau\neq 2T$. In fact $|\sinh^{-1}(x)|<|x|$, and hence $\tau<2T$ and the traveling twin will be younger on arrival.

Note that this flight profile has no inertial phase at all - it is under acceleration all the way, and the result is qualitatively the same as the purely inertial case. This is, in fact, obvious from examining a spacetime diagram in the rest frame of the planets the ship flew between. The ship's worldline is slanted (albeit not a constant slant); its elapsed time is, therefore, lower.

 

[Dale]

? Analysis of acceleration/deceleration is entirely absent from the standard approach.

What are you talking about. The standard scenario is that the traveling twin accelerates at the beginning, middle, and the end! And the most common variant is that the traveling twin accelerates in the middle. How can you say it is absent from the standard approach?

I base it on this (but it's too late in the day for me to check) - Need to rethink - I'll go over this tomorrow (or not).

Bailey et al., “Measurements of relativistic time dilation for positive and negative muons in a circular orbit,” Nature 268 (July 28, 1977) pg 301. They confirmed the clock hypothesis (no additional acceleration effect for time dilation) to 10^18 g. The acceleration effect is 0. The same experiment is a direct test of the twin paradox also.

Your suspicion has no basis, neither in theory nor in experiment.

 

[Sagittarius A-Star]

Thus when A arrives at B's location A expects to see a younger B.

That's wrong. This can be easily understood at the example of myons in a storage ring. Myons can be used as clocks, because they have a defined half-life period.

Assume, myon A ist rotating in the storage ring and myon B is at rest in the center of the ring. A theoretical model of this is the rotating disk, which Einstein described in his popular book from 1917 (Appendix III c):

“From this it follows that
ν = ν₀ (1+ Δϕ/c²)
In the first place, we see from this expression that two clocks of identical construction will go at different rates when situated at different distances from the centre of the disc. This result is also valid from the standpoint of an observer who is rotating with the disc.
Now, as judged from the disc, the latter is in a gravitational field of potential ϕ “

Source:
https://en.wikisource.org/wiki/Rela...isplacement_of_Spectral_Lines_Towards_the_Red

Einstein showed there, why A and B do not disagree.

 

[Ibix]

What are you talking about. The standard scenario is that the traveling twin accelerates at the beginning, middle, and the end!

I suspect that @Will Flannery has observed that if you can write a standard twin paradox with no accelerations (as you can by replacing the traveller with two inertial travellers moving in opposite directions) then something may have been left out by using infinite acceleration. That's why I wrote down the maths for a constant acceleration case in #44, to show that there are no qualitative differences with finite acceleration.

More generally, @Will Flannery, the elapsed time for any worldline between (inertial) coordinate times $t=T_1$ and $t=T_2$ is given by$$\tau=\int_{T_1}^{T_2}\sqrt{1-\frac{v^2(t)}{c^2}}dt$$which follows from the definition of proper time. This is clearly maximised by $v=0$, and always smaller for any $v(t)$ that is non-zero at any time. Thus the traveling twin is younger, whatever their velocity profile, and we haven't had to consider any non-inertial frames.

Edit: furthermore, as Dale noted, this theoretical argument has been tested to some pretty high accelerations.

 

[SiennaTheGr8]

OP might find it useful to read the "Trip to Canopus" chapter of Taylor & Wheeler's Spacetime Physics. In particular, this page may help develop some intuition for how the relativity of simultaneity factors in.

 

[Will Flannery]

I suspect that @Will Flannery has observed that if you can write a standard twin paradox with no accelerations (as you can by replacing the traveller with two inertial travellers moving in opposite directions) then something may have been left out by using infinite acceleration.

No, I said that the standard scenario didn't consider acceleration/deceleration not because the traveling twin didn't accelerate/decelerate, but because the analysis of the traveling twin never analyzed the acceleration/deceleration. E.g. in Lincoln's vid the traveler hops on and off inertial frames traveling at different velocities, which seemed to me to be avoiding acceleration/deceleration.

But, now I see that it didn't need to be mentioned because it doesn't make much difference, there are no extraordinary effects from acceleration/deceleration.

But then, I abandoned my original version of the paradox because of the unanalyzed acceleration/deceleration, which I (now see) correctly thought didn't make much difference in terms of time dilation.

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.

So, that's what I'm puzzling over now. And I'm wondering ... was Lincoln wrong when he said, quoting "All observers can completely accurately claim that they're the single unmoving person in the universe, and everyone is moving around them'? Because if he is right then any observer can consider himself at rest in an inertial frame (?), and if that's true then I think my original version of the paradox is OK !

 

[Ibix]

I figured you should be able to analyze the motion in both frames ignoring the acceleration/deceleration.

You can analyse the trip in any frame you like. However, naively patching together the inertial frames associated with inertial segments of a worldline does not produce a valid frame - the result has gaps and overlaps. The overlaps aren't relevant here, but the gaps are, because the worldline of the Earth crosses one such gap. The time elapsed in that gap is not accounted for by such a naive approach. Failing to realize this is how your analysis of the traveller concluding that the Earth twin is younger goes wrong. It's not that you can't do the analysis in a non-inertial frame - it's that you aren't doing the analysis in a frame at all. You are doing it in a kludge that doesn't account for all of the relevant portion of the Earth twin's history.

Lincoln is correct that you can always regard yourself as at rest. You cannot always regard yourself as inertial, and if you are not inertial you cannot construct a rest frame by patching together parts of inertial frames.