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The twin Paradox revisited


Awatso

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It seems there is still an argument on whether acceleration is needed or not.  Two opposing viewpoints.

Pro

 

against

 

 

Who's right for the general layman ? 

I was thinking about this on the dog walk and it did indeed make it go quicker. Though neither of us were any younger sadly.

 

Edited by Awatso
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The explanations are more or less the same: both say that special relativity is valid for comparison of time (and distance) between two inertial frames only. But in both examples there are three inertial frames:

  • in both, we have the 'stationary frame' of earth and star, that is frame 1
  • in both we have a traveler heading for the distant star, that is frame 2. However, in Don Lincoln's example this traveler just goes on traveling in the same direction, in the first example the twin turns around back to earth
  • The 3rd frame in the first example is that from the returning twin, in the second it is the other traveler that flies in the direction of earth, and passes the star exactly at the moment that first traveler also passes the star.

So in both cases, there is no inertial frame for a) the travelling twin, there are two, and in b) we have two inertial frames for two travelers from the beginning. But, if we neglect the acceleration at the turning around of the twin, the argumentation is the same, because we use the same inertial frames.

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Why is the video with the shorter answer so much longer? 🙄

All the first one says is what they all say (e.g. at 2:30 in Dr. Lincoln's video): that for the standard version of the paradox (with only one spaceship), acceleration is required to change the spaceship's velocity.

The second video clarifies that acceleration is only involved indirectly. With two spaceships, you can get the same result with no acceleration.

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18 hours ago, Awatso said:

It seems there is still an argument on whether acceleration is needed or not.

Yes, and different people are arguing different things. Some are saying only that if one twin has a certain velocity and then a different velocity, there "needs" to be coordinate acceleration for that to be possible. Others will say that acceleration is the "cause" of the difference in proper time.

Some say that swapping the twin with another moving in the opposite direction is not "the Twin paradox" because it describes a different experiment even if the results are the same. Others argue any number of things (they can't be meaningfully swapped, the accelerated twin physically changes, the results would be different---I can't think of any examples I think are correct).

"Feeling" the acceleration or having proper acceleration shouldn't matter because it's not generally in the calculations, and the time dilation effect still happens in freefall, eg. with GPS satellites.

15 hours ago, Eise said:

both say that special relativity is valid for comparison of time (and distance) between two inertial frames only.

Well that's not true. It's valid in a single reference frame (the aging of both twins can be figured out using only the inertial twin's reference frame even if the other one is never inertial) and it's valid for comparing any number of inertial reference frames (eg. composition of velocities, using triplets all leaving together and returning again, etc.).

The first video says something like "SR only applies to uniform motion", and that's wrong. It generally applies only in uniformly moving reference frames, and I think that's implied by its postulates, which include that the speed of light is a constant c, and it's not constant in an accelerating reference frame. However, SR does apply to non-uniform motion of objects. For example, if the accelerating twin spends the entire trip accelerating, SR can be used to calculate its time dilation from an inertial reference frame just fine, just from its velocity.

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For me personally, it’s fair to say that I’ve been concerning myself with the theory of relativity for quite some time, and while I’m not an expert by any reasonable metric, I still dare say that my knowledge of the subject is above the average one would find in a group of randomly-chosen members of a (reasonable well-educated) public. Yet, even after all these years, I am still failing to understand why so many people consider these scenarios “paradoxical”? I get that the outcome of such experiments can appear surprising at first glance, but that’s not the same as calling it “paradoxical”.

I’m not being condescending, sarcastic or whatever else, it’s a genuine question. I just don’t get it.

Analogy(!!!): It’s a little bit like flying from, say, New York to London - you can fly eastbound, and follow a suitable geodesic (i.e. a great circle segment, ignoring vertical motion for simplicity now, all other factors being equal), and get to London in a certain amount of time. Or you can choose to fly westbound, and likewise follow a geodesic, just in the opposite direction. Or you can choose some other route that isn’t a geodesic at all, so long as they all start at the same place and terminate at the same place. No one would be surprised by the fact that for these three cases, the onboard clocks read different elapsed times - I mean, it’s rather obvious that there is exactly one, and only one, route that minimises the total in-flight time. Any route that diverges from that flight path will necessarily be of different duration, assuming all other factors remain equal.

It’s no different for a path in spacetime, that is traced out by different travellers between the same two events. In topologically trivial Minkowski spacetime (which is the stage against which this is set), there is precisely one - and only one - path between any pair of given events that extremises (minimises or maximises, depending on sign convention) its own length, which is by definition equal to the proper time recorded on a clock that physically travels this path through spacetime. Any path that varies from this one choice must necessarily be of different length, ie a physical clock travelling along it will record something other than the extremal value. If you set this up as a variational calculus problem, you unsurprisingly end up with the geodesic equation - the extremal path between any pair of events here is a geodesic of Minkowski spacetime, which physically corresponds to a traveller moving inertially. The reverse of this statement is just as true - in singly-connected Minkowski spacetime, any path between given events, the length of which differs from the extremal value obtained from the variational calculus problem, is necessarily non-inertial at least within some small region along it, since the extremal path is a unique solution to the equation.

Why is this considered “paradoxical”? I struggle to even consider it “surprising”, since it seems entirely obvious to me that this is what will happen, just like in the analogy of the planes above. You can’t - in general - take two different routes between the same points and reasonably expect them to have the same lengths (unless you cheat by introducing non-trivial topologies etc). Note that this isn’t about why SR is what it is (ie hypothetical underlying mechanisms etc), but simply about why this result should be considered surprising or paradoxical. To me it isn’t, unless I am missing something so basic that I can’t even see it - in which case I’d be grateful if someone could point it out to me.

My other issue is that I’ve seen the original scenario amended such that the travellers involved do not actually connect the same pair of events. What meaningful physical conclusion - in terms of elapsed times - would one hope to arrive at by comparing paths that don’t connect the same events? I don’t get this either. It seems even less surprising that - again in general - you get different results if you compare paths between different events. 

Edited by Markus Hanke
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1 hour ago, Markus Hanke said:

point it out to me

"each twin sees the other twin as moving, and so, as a consequence of an incorrect and naive application of time dilation and the principle of relativity, [i.e. the Lorentz transformation for times] each should paradoxically find the other to have aged less."

1 hour ago, Markus Hanke said:

I’ve seen the original scenario amended such that the travellers involved do not actually connect the same pair of events.

Surely not by anyone but trolls or clueless amateurs.

Edited by Lorentz Jr
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On 2/10/2023 at 6:37 AM, Awatso said:

Two opposing viewpoints

Not opposing viewpoints if they aren’t describing the same problem.

3 hours ago, Markus Hanke said:

For me personally, it’s fair to say that I’ve been concerning myself with the theory of relativity for quite some time, and while I’m not an expert by any reasonable metric, I still dare say that my knowledge of the subject is above the average one would find in a group of randomly-chosen members of a (reasonable well-educated) public. Yet, even after all these years, I am still failing to understand why so many people consider these scenarios “paradoxical”? I get that the outcome of such experiments can appear surprising at first glance, but that’s not the same as calling it “paradoxical”.

I’m not being condescending, sarcastic or whatever else, it’s a genuine question. I just don’t get it.

I think your typical person (even well-educated person) doesn’t know details about Einstein’s work beyond perhaps E=mc^2. They live in a world of absolute length and time. Disagreement in time is a paradox: How can it be 1 PM and 2 PM at the same time, in the same location?

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2 hours ago, Lorentz Jr said:

Ah ok…I can see how someone might be tempted to look at it in this way. That didn’t even occur to me. Thanks for pointing it out.
Obviously though, since there is exactly one geodesic (=extremal) path connecting any pair of events in spacetime, there must always be at least one local section of the journey where the two travellers are not related via a Lorentz transformation.

35 minutes ago, swansont said:

They live in a world of absolute length and time. Disagreement in time is a paradox: How can it be 1 PM and 2 PM at the same time, in the same location?

Yes, perhaps you’re right and it’s that simple.

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7 hours ago, Markus Hanke said:

Why is this considered “paradoxical”?

Beside the main paradox Lorentz Jr mentioned, people will add additional confusing elements that leave out something. You personally approach it from already knowing the solution, and how to find it, and so you can quickly figure out what's missing. Before one understands that it works out mathematically and that one is as likely to find a paradox in it as in multiplication and division, it's easier to assume that some "paradoxical" complication is valid. It's like a "paradox" that shows that 1=0 by a missed mathematical detail, which can still be puzzling if it is hidden in a lot of other equations.

For example, I could create paradoxes in a basic Twin paradox setup by adding something like, "now imagine that a uniform gravitational field is suddenly switched on everywhere in the universe...", and I'm sure you'll immediately see a problem in that, but if you don't already know the solution, it might be puzzling. I've seen intelligent people try to show that the asymmetry of the Twin paradox can be removed by adding some such detail, or a change in coordinates.

 

Another thing that happens is that some rephrase it as an unresolvable debate of "why" the twin paradox happens.

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Considering the case of the twins A, B and C but forgetting twin A for a while. Just twins B and C. It is the case of two twins making a completely symmetrical travel and so both are affected the same way by any consideration in acceleration or time dilation. Each one will see the other one aging less and so there would be an inconsistency. The two different frames of reference observe contradictory results. How the inconsistency would be solved if possible?

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34 minutes ago, martillo said:

Considering the case of the twins A, B and C but forgetting twin A for a while. Just twins B and C. It is the case of two twins making a completely symmetrical travel and so both are affected the same way by any consideration in acceleration or time dilation. Each one will see the other one aging less and so there would be an inconsistency. The two different frames of reference observe contradictory results. How the inconsistency would be solved if possible?

I don't see an inconsistency. Could you elaborate?

What do you mean, "making a completely symmetrical travel"?

Which "contradictory results" are observed?

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18 minutes ago, Genady said:

I don't see an inconsistency. Could you elaborate?

What do you mean, "making a completely symmetrical travel"?

Which "contradictory results" are observed?

A complete symmetrical travel would be for instance starting being together, travelling the same way in opposite directions for the same time and returning back crossing at the same point at the same instant without stopping, just crossing and comparing the results of the observations at this point.

Each twin (frame of reference) observes the other twin aging less. This is contradictory. Consider for instance a long enough travel for the twins having their beard growing. Each one would see the other one with less beard. To visualize that they could take photographs of themselves along the travel and interchange them at the crossing point so is something that could be verified in practice if needed. Each one would observe the other one with a shorter beard. The two different observations are inconsistent.

Edited by martillo
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2 minutes ago, martillo said:

A complete symmetrical travel would be for instance starting being together, travelling the same way in opposite directions for the same time and returning back crossing at the same point at the same instant without stopping, just crossing and comparing the results of the observations at this point.

Each twin (frame of reference) observes the other twin aging less. This is contradictory. Consider for instance a long enough travel for the twins having their beard growing. Each one would see the other one with less beard. To visualize that they could take photographs of themselves along the travel and interchange them at the crossing point so is something that could be verified in practice if needed. The two different observations are inconsistent.

OK. Got it. There will not be a contradiction, because at the crossing point, i.e., when they meet again, they will be at exactly the same age.

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20 minutes ago, Genady said:

OK. Got it. There will not be a contradiction, because at the crossing point, i.e., when they meet again, they will be at exactly the same age.

You didn't get the point. They would be the same age if observed from the frame in twin A who observes both the same way. From the frames on each of the twins B and C the situation is different. The both observe the other one aging less. It is not a matter of determine the three observations on the three frames and pick up a "right" one. The problem is precisely that there are different observations each one contradicting the other ones and this is an inconsistency.

Edited by martillo
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1 minute ago, martillo said:

You didn't get the point. They would be the same age if observed from the frame in twin A who observes both the same way. From the frames on each of the twins B and C the situation is different. The both observe the other one aging less. It is not a matter of determine the three observations on the three frames and pick up a "right" one. The problem is precisely that there are different observations contradicting the other ones and this is an inconsistency.

Let's see it from the frame of B observing C.

By the time they reach the turning points, B aged 10 while C aged 5. After B turns, let's say almost momentarily, B will still be 10, but he will observe C being 15. By the time they reach the crossing point, B is 20, and C is 20.

The C will find all the same symmetrically.

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6 minutes ago, Genady said:

Let's see it from the frame of B observing C.

By the time they reach the turning points, B aged 10 while C aged 5. After B turns, let's say almost momentarily, B will still be 10, but he will observe C being 15. By the time they reach the crossing point, B is 20, and C is 20.

The C will find all the same symmetrically.

How is that twin C aged 10 just in the "momentary" or "instantaneous" turn at some point? I don't get this.

Edited by martillo
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10 minutes ago, martillo said:

How is that twin C aged 10 just in the "momentary" or "instantaneous" turn at some point? I don't get this.

One way to see it is to consider that when B turns around, he experiences an acceleration. This acceleration affects his clock similarly to a gravitational field. A clock in gravitational field runs slower than the clock which is not. Thus, during the turning around the clock of B will advance little compared to the clock of C. In my example, this makes C to age 10 years more than B while B is turning.

There are other ways to analyze this situation, without involving gravity. In any case, after turning around B is in a different inertial frame than he was before, and this difference makes for the extra difference in their ages.

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49 minutes ago, martillo said:

How is that twin C aged 10 just in the "momentary" or "instantaneous" turn at some point?

It's just the relativity of time, which depends on relative velocities. When the spaceships are moving away from each other, the Lorentz transformations indicate that each astronaut perceives or calculates the other astronaut's clock to be behind by [math]\gamma v D / c^2[/math], where v is their speed relative to each other and D is the distance between them. Then, as soon as they both turn around, each perceives/calculates the other's clock to be ahead by that amount (i.e. to have jumped forward by twice that amount).

Edited by Lorentz Jr
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41 minutes ago, martillo said:

The both observe the other one aging less. It is not a matter of determine the three observations on the three frames and pick up a "right" one. The problem is precisely that there are different observations each one contradicting the other ones and this is an inconsistency.

There is no inconsistency; the situations are symmetrical for the two frames so getting the same answer should be expected. For any frames the time is given by the Lorentz transformations. There is nothing inconsistent in the math.

The unspoken assumption is that there is an absolute frame that shows the “real” time.

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34 minutes ago, Genady said:

One way to see it is to consider that when B turns around, he experiences an acceleration. This acceleration affects his clock similarly to a gravitational field. A clock in gravitational field runs slower than the clock which is not. Thus, during the turning around the clock of B will advance little compared to the clock of C. In my example, this makes C to age 10 years more than B while B is turning.

There are other ways to analyze this situation, without involving gravity. In any case, after turning around B is in a different inertial frame than he was before, and this difference makes for the extra difference in their ages.

But I can consider the travels without accelerations involved. Consider B and C far away travelling in opposite directions with constant velocities to cross at some point. Consider they start observing the things at some instant synchronized with a signal from the middle crossing point. Each one will observe the other one age less and that would be inconsistent observations.

25 minutes ago, swansont said:

There is no inconsistency; the situations are symmetrical for the two frames so getting the same answer should be expected. For any frames the time is given by the Lorentz transformations. There is nothing inconsistent in the math.

If the same answer is "the other age less" there's a problem. Both cannot age less than the other one...

Edited by martillo
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15 minutes ago, martillo said:

Consider they start observing the things at some instant synchronized with a signal from the middle crossing point. Each one will observe the other one age less and that would be inconsistent observations.

If a synchronizing signal is sent from a point in the middle in the two directions, then, in the B frame, it will reach C before it reaches B, because C moves toward B in twice the speed with which the middle point moves toward B. Thus, by the time the signal reaches B, the C clock will be already on 5. Then by the time they meet, the B clock advances 10, while the C clock advances 5, which makes them equal, 10.

Edited by Genady
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23 minutes ago, Genady said:

If a synchronizing signal is sent from a point in the middle in the two directions, then, in the B frame, it will reach C before it reaches B, because C moves toward B in twice the speed with which the middle point moves toward B. Thus, by the time the signal reaches B, the C clock will be already on 5. Then by the time they meet, the B clock advances 10, while the C clock advances 5, which makes them equal, 10.

But you mean each twin would never observe the other one aging less... I wonder if the same reasoning would apply in the classic twins' paradox with just one twin travelling...

Edited by martillo
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1 minute ago, martillo said:

So each twin would never observe the other one aging less...

No. In the last scenario, after B receives his signal, B ages by 10 while he observes C aging only by 5. The point is that when B receives his signal, C is already 5 in the B frame, because in this frame C received the signal before B.

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