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

[Will Flannery]

TL;DR

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.