Twin Paradox Problem: Do Twins Age Differently?

[robinpike]

I didn't see this the first time it was posted - I saw it quoted in another post. I believe that the conclusion drawn is somewhere between ill-specified and downright wrong in i'ts deduction that "the traveling twin's clock must have slowed down at some point" .

Specifically, this deduction seems to presuppose some sort of absolute time, to which the travelling's twin time can be unambiguously compared. But there isn't any such absolute time. So what is the travelling's twin's time being compared to, and how is the comparison being made?

Relativity teaches us that the process of time comparison is frame dependent, and not absolute.

The deduction does not presuppose some sort of absolute time!?

The traveling twin simply compares the time on his clock to his stay at home twin's clock.

On arriving back on earth, the twins come back to being in the same reference frame, and therefore the traveling twin's clock is now running at the same rate as the stay at home twin's clock, but the traveling twin's clock as lost time.

The deduction this leads to, is that the traveling twin's clock at some point in the journey experienced a slower rate of time. If that deduction is false - please explain why...

 

[PAllen]

The deduction does not presuppose some sort of absolute time!?

The traveling twin simply compares the time on his clock to his stay at home twin's clock.

On arriving back on earth, the twins come back to being in the same reference frame, and therefore the traveling twin's clock is now running at the same rate as the stay at home twin's clock, but the traveling twin's clock as lost time.

The deduction this leads to, is that the traveling twin's clock at some point in the journey experienced a slower rate of time. If that deduction is false - please explain why...

Draw a straight line on a piece of paper. Draw a squiggly line between the same two endpoints. Which part of the squiggly line is the 'extra length'? Obviously, there is no unique reasonable answer - it is longer, but you have no basis to claim which part is 'extra'. Every method of comparison (match starting from one end, the other end, the middle, etc.) will put the extra in different places.

The same is true for time along two different spacetime paths connecting the same two events. The non-inertial path is shorter in elapsed time, but there is no objective unique way to say where the 'missing' time is.

Note that there is no way to directly compere separated clocks. You need to send signals; then you need a model of how to account for signal delay. Since distances and simultaneity are relative as well, there is no unique to do this. You can specify some set of conventions you will use for distant clock comparison. There are many such choices. For each one, it will be true that the non-inertial twin will consider the inertial twin clock running fast for some period of time. However, when this occurs will be different for each convention you might choose for comparing distant clocks.

 

[robinpike]

The same is true for time along two different spacetime paths connecting the same two events. The non-inertial path is shorter in elapsed time, but there is no objective unique way to say where the 'missing' time is.

Thanks PAllen for this example. The problem is that the deduction does not need to know where the 'loss' in time occurred (or even how the loss in time happened).

All that is necessary is to note that, once the traveling twin arrives back on earth, the traveling twin's clock has fallen behind in time compared to the stay at home twin's clock.

The only activity that caused this to happen is that the traveling twin performed acceleration.

The deduction that is being challenged, is that the traveling twin's rate of time slowed down at some point in his journey (the deduction's validity does not rely on where or how that happened).

If that deduction is false, then please describe the correct deduction as to how the traveling twin's clock results in a retarded time at the end of his journey?

 

[ImaLooser]

Thanks PAllen for this example. The problem is that the deduction does not need to know where the 'loss' in time occurred (or even how the loss in time happened).

All that is necessary is to note that, once the traveling twin arrives back on earth, the traveling twin's clock has fallen behind in time compared to the stay at home twin's clock.

The only activity that caused this to happen is that the traveling twin performed acceleration.

The deduction that is being challenged, is that the traveling twin's rate of time slowed down at some point in his journey (the deduction's validity does not rely on where or how that happened).

If that deduction is false, then please describe the correct deduction as to how the traveling twin's clock results in a retarded time at the end of his journey?

Special relativity won't tell you. You have to use general.

 

[Rishavutkarsh]

Thanks PAllen for this example. The problem is that the deduction does not need to know where the 'loss' in time occurred (or even how the loss in time happened).

All that is necessary is to note that, once the traveling twin arrives back on earth, the traveling twin's clock has fallen behind in time compared to the stay at home twin's clock.

The only activity that caused this to happen is that the traveling twin performed acceleration.

The deduction that is being challenged, is that the traveling twin's rate of time slowed down at some point in his journey (the deduction's validity does not rely on where or how that happened).

If that deduction is false, then please describe the correct deduction as to how the traveling twin's clock results in a retarded time at the end of his journey?

how would you describe the case mentioned earlier
both start at the same time when they both reach 90%C one returns back while other keeps going ,
they both went through same acceleration and retardation but still the moving one is younger when they meet after 10 years they both experienced same acceleration (to 90%C)
and same retardation but the Earth twin experienced this earlier while the moving one later.
how will you describe this?

 

[robinpike]

how would you describe the case mentioned earlier
both start at the same time when they both reach 90%C one returns back while other keeps going ,
they both went through same acceleration and retardation but still the moving one is younger when they meet after 10 years they both experienced same acceleration (to 90%C)
and same retardation but the Earth twin experienced this earlier while the moving one later.
how will you describe this?

As I said, for the deduction to be true, it does not rely on having to know how, where or when the effect on the traveling twin's rate of time occurred...

However, in answer to your point above, it is not how long a traveller experiences acceleration that effects how much time he falls behind, but how long the traveller spends in the different inertial frame after the acceleration.

 

[PAllen]

If that deduction is false, then please describe the correct deduction as to how the traveling twin's clock results in a retarded time at the end of his journey?

I don't disagree with this. As long as you realize you can't specify the 'correct place/time' where one clock is faster, it is certainly true that any particular way assigning simultaneity between the two world lines will show one going slower on average (though portions may be faster).

 

[PAllen]

Special relativity won't tell you. You have to use general.

This is just wrong. Without gravity, there is no reason you need GR. All you get from the equivalence principle is the ability to apply what you derive for non-inertial motion in pure SR to situations involving gravity in GR. The explanatory arrow goes the opposite way from what you imply: SR result about accelerated motion implies what GR must predict about a gravitational situation.

 

[PAllen]

However, in answer to your point above, it is not how long a traveller experiences acceleration that effects how much time he falls behind, but how long the traveller spends in the different inertial frame after the acceleration.

While you seemed to understand that you can't uniquely localize the time differential, here you propose to do it. The flaw in this is shown by asking: what if there are no inertial travel portions for either traveler. One may still age much less. All you can say is one ages less, and the any particular simultaneity mapping will place the differential somewhere along the paths; and there are infinitely many ways to do this, producing different conclusions about time rate differential along the respective paths.

 

[robinpike]

While you seemed to understand that you can't uniquely localize the time differential, here you propose to do it. The flaw in this is shown by asking: what if there are no inertial travel portions for either traveler. One may still age much less. All you can say is one ages less, and the any particular simultaneity mapping will place the differential somewhere along the paths; and there are infinitely many ways to do this, producing different conclusions about time rate differential along the respective paths.

Thanks, but it is the specific twin traveling example that I am applying the deduction to. That is, one twin does the traveling and then when he comes home, his clock has a retarded time.

 

[ghwellsjr]

Thanks, but it is the specific twin traveling example that I am applying the deduction to. That is, one twin does the traveling and then when he comes home, his clock has a retarded time.

Did you understand my simple explanation in post #5?

 

[phyti]

how would you describe the case mentioned earlier
both start at the same time when they both reach 90%C one returns back while other keeps going ,
they both went through same acceleration and retardation but still the moving one is younger when they meet after 10 years they both experienced same acceleration (to 90%C)
and same retardation but the Earth twin experienced this earlier while the moving one later.
how will you describe this?

The post 175 doc. is intended to show the 2-part path (returning twin) accumulates the least time.

 

[phyti]

While you seemed to understand that you can't uniquely localize the time differential, here you propose to do it. The flaw in this is shown by asking: what if there are no inertial travel portions for either traveler. One may still age much less. All you can say is one ages less, and the any particular simultaneity mapping will place the differential somewhere along the paths; and there are infinitely many ways to do this, producing different conclusions about time rate differential along the respective paths.

The twins A and B are reunited, and only B took a different path and returned. If the B clock reads less time than the A clock, doesn't that mean the B clock lost more time than the A clock?

 

[PAllen]

The twins A and B are reunited, and only B took a different path and returned. If the B clock reads less time than the A clock, doesn't that mean the B clock lost more time than the A clock?

And where is that in dispute? (Also, what on Earth to you mean by only B took a different path? You have two paths between events - different/return are not objective. You can talk about inertial, non-inertial, but one twin being inertial is a special case; the core explanation cannot rely on it. The most general case is easily analyzed in any inertial frame in SR).

So, at the end, one ages less. That is given. What is important to clarify is that you cannot say which part of the younger one's path is where they aged slower by any unique or preferred criterion. Just among all inertial frames (let along more general simultaneity conventions), where one clock is going slower, and by how much, can differ for every one one of these frames.

 

[harrylin]

The deduction does not presuppose some sort of absolute time!?
[..]
The deduction this leads to, is that the traveling twin's clock at some point in the journey experienced a slower rate of time. If that deduction is false - please explain why...

It depends a bit on definitions and way of phrasing, as stevendaryl already mentioned. After that discussion I thought to have settled the matter in post #127 by starting with a slight reformulation with which I assumed that you would agree, but you did not respond and also Pallen seems to have missed it. I there also pointed out that you use more or less the same kind of phrasing as Einstein used in this context.

 

[Dale]

In modern physics, definitions are practical implementations, but I won't go along any further with a discussion about classical mechanics in this thread.

OK..

 

[robinpike]

You are only half correct. When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his Earth twin will age 2.2942 times as much. Simple, isn't it?

That is a statement of how to calculate the time lost by the traveling twin - that is not the same thing as an explanation?

 

[robinpike]

As I noted in post #103, likely you would agree with a change rephrasing as follows:

"it can be deduced that according to any inertial coordinate system the clock rate of the traveling twin's clock must have slowed down at some point during the journey."

I don't understand what the problem is. In 1905, right from the start, it was already deduced that:

"we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."*
- section 4 of http://www.fourmilab.ch/etexts/einstein/specrel/www/

And in 1911 it was remarked that the space traveller is :

"without possibility of coming back to inform us of the result of his voyage, since any attempt of the same kind could only transport him increasingly forward [in time]".
- p.50 of http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time

Perhaps the "problem" that you mean is that SR only explains the consequences of the "special" relativity principle; SR gives no interpretation of what "really" is happening. Thus it could happen that, despite both using Minkowski spacetime, Langevin explained it with the ether model while stevendaryl here explains it with the block universe model.

* this SR analysis did of course not account for the gravitational potential

"we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

This is the problem:

The clock at the equator moves faster than a similar clock at one of the poles, and as a consequence, what is observed is that the clock at the equator runs slow compared to the clock at the pole. (Although this example isn't as 'clean' to analyse as the traveling twin example.)

The deduction is that the rate of time of the clock at the equator is running slower than the rate of time of the clock at the pole.

All well and good...

The problem lies in explaining how the rate of time of a clock running slow can ever be made to increase from its current slower rate of time - for any acceleration or de-acceleration are equivalent actions - the only difference being the direction of the accelerating force / de-acclerating force being applied to the clock.

 

[Austin0]

"we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

This is the problem:

The clock at the equator moves faster than a similar clock at one of the poles, and as a consequence, what is observed is that the clock at the equator runs slow compared to the clock at the pole. (Although this example isn't as 'clean' to analyse as the traveling twin example.)

The deduction is that the rate of time of the clock at the equator is running slower than the rate of time of the clock at the pole.

All well and good...

The problem lies in explaining how the rate of time of a clock running slow can ever be made to increase from its current slower rate of time - for any acceleration or de-acceleration are equivalent actions - the only difference being the direction of the accelerating force / de-acclerating force being applied to the clock.

there are two points here:

You are making the assumption that dilation is directly an effect of acceleration.
While it is possible there is some such effect it has been shown by experiment that all observed dilation is totally attributable to instantaneous velocity. This suggests the possibility that the sole effect of acceleration is simply changing velocity. In this case reducing velocity relative to some frame would obviously simply have the effect of reducing the dilation factor relative to that frame.
Secondly you seem to assume that acceleration necessarily means slowing clock rate.
Consider the traveling twin accelerating to 0.8c from earth. If you assume the clock is slowing down relative to Earth as a result of that acceleration you also have to assume it is speeding up relative to a frame already traveling with that velocity 0.8c relative to earth. Yes??
SO which is really occurring?

While acceleration is a frame independent observation and so, real in that sense, whether it is speeding up or slowing down the system in which it is occurring is completely frame dependent, as is the resulting change in clock rate.

Linear acceleration to a new inertial velocity is exactly equivalent to transporting a clock from the equator to a pole. Passing along a path of gradually decreasing angular velocity, with an increasing resulting rate to arrive at the pole with the same periodicity as a resident clock. Likewise a reverse trip would result in an equivalent slowing of the clock. DO you imagine that acceleration or velocity causes some kind of permanent deformation?

 

[ghwellsjr]

You are only half correct. When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his Earth twin will age 2.2942 times as much. Simple, isn't it?

That is a statement of how to calculate the time lost by the traveling twin - that is not the same thing as an explanation?

Nobody lost any time. That's your problem. You're trying to find where the lost time went.

Einstein promoted the idea that time is what a clock measures. Therefore, if one clock measures a different time than another, then time is really different for the two clocks. It wouldn't be correct to think that one of them was correct and the other one lost [or gained] time.

 

[harrylin]

"we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

This is the problem:

The clock at the equator moves faster than a similar clock at one of the poles, and as a consequence, what is observed is that the clock at the equator runs slow compared to the clock at the pole. (Although this example isn't as 'clean' to analyse as the traveling twin example.)

The deduction is that the rate of time of the clock at the equator is running slower than the rate of time of the clock at the pole.

Yes. It suffers from a similar imprecision as your earlier statement. I see two ways to interpret it (both correct for equal gravitational potential), either relating to a single frame, or as a statement about an average:

1. As determined with a reference system in which the Earth is in pure rotation (e.g. the ECI frame), a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.

2. If one defines relative clock rate as the observed difference per rotation, then a balance-clock at the equator must go more slowly, by a very small amount, and for any valid reference system, than a precisely similar clock situated at one of the poles under otherwise identical conditions.

All well and good...

The problem lies in explaining how the rate of time of a clock running slow can ever be made to increase from its current slower rate of time - for any acceleration or de-acceleration are equivalent actions - the only difference being the direction of the accelerating force / de-acclerating force being applied to the clock.

To the contrary: the assumption here is that forces have no effect on time dilation. Einstein assumes in his clock example that a continuous slight difference in force has no effect. In the space traveler example of Langevin there is even no force felt at all at turnaround. And if you understand the calculation, it is not a problem. So, contrary to what I first thought, your problem is purely with understanding the math.

Using the perspective of interpretation 1, the clock at the equator is continuously ticking slower. With other perspectives that is different, but everyone agrees about the average effect. Same with the twins: from the perspective of a solar system reference, the traveler is practically all the time aging slower; from other perspectives this is not the case, but the average effect is what we can compare and every system agrees.

 

[Dale]

The problem lies in explaining how the rate of time of a clock running slow can ever be made to increase from its current slower rate of time - for any acceleration or de-acceleration are equivalent actions - the only being the direction of the accelerating force / de-acclerating force being applied to the clock.

It is not correct that acceleration and deceleration are equivalent. One increases the speed and the other decreases the speed. Since time dilation is a function only of speed in an inertial frame, an acceleration which increases speed is different from one that decreases speed or one that leaves it constant.

 

[robinpike]

It is not correct that acceleration and deceleration are equivalent. One increases the speed and the other decreases the speed. Since time dilation is a function only of speed in an inertial frame, an acceleration which increases speed is different from one that decreases speed or one that leaves it constant.

If that is the case, then I have a question:

A spaceship is in deep space, sitting motionless in its inertial frame, when it fires its rockets.

Has the spaceship increased its speed or decreased its speed?

 

[harrylin]

If that is the case, then I have a question:

A spaceship is in deep space, sitting motionless in its inertial frame, when it fires its rockets.

Has the spaceship increased its speed or decreased its speed?

According to original inertial frame A it has increased, and according to frame B it has decreased, let's say to zero (B is then the new cruising frame).
Thus according to A the clock in the spaceship is getting behind and according to B it is now getting ahead on a clock that is left behind.

After some time the spaceship fires its rockets to go back, and now according to A the clock is still getting behind, while according to B the clock is now very rapidly loosing time.
The total effect is that when the the clock gets back to the clock that was left behind, A and B not only agree that the spaceship's clock will be the one that is behind, but even by how much.

Sorry but I now have to ask it: did you ever calculate these things??
No calculation = no understanding, and no explanation can compensate for that.

It's similar to comparing classically the times that it takes for two cars from A to B and back, one going at 100 km/h both ways, the other car the first leg 80 km/h and the second leg 120 km/h, all according to the road rest frame. No matter how you analyze it and from what frame, the time difference will be the same.

 

[robinpike]

No calculation = no understanding, and no explanation can compensate for that.

Excuse me? Calculating and understanding are completely different things. (I hope you are not trying to suggest that I am not intelligent enough to continue with discussing this?)

It's similar to comparing classically the times that it takes for two cars from A to B and back, one going at 100 km/h both ways, the other car the first leg 80 km/h and the second leg 120 km/h, all according to the road rest frame. No matter how you analyze it and from what frame, the time difference will be the same.

You seem to think that I am questioning the calculation - I am not.

I am pointing out that if the calculation is used as the explanation for the loss in time in the traveling twin's clock, then as an explanation it fails for this reason...

The round trip journey effects the traveling twin's clock - it loses time compared to if he had not gone on the journey.

I have suggested that the reason for the loss in time is because his rate of time slowed down at some point in his journey. If that is the reason, then how does Relativity explain this (and please note I am not asking for the calculation).

If the deduction that his rate of time slowed down at some point in his joureny is incorrect, then all I ask is that the correct explantion for the loss in time is stated for me to see.

 

[ghwellsjr]

I am pointing out that if the calculation is used as the explanation for the loss in time in the traveling twin's clock, then as an explanation it fails for this reason...

The round trip journey effects the traveling twin's clock - it loses time compared to if he had not gone on the journey.

I have suggested that the reason for the loss in time is because his rate of time slowed down at some point in his journey. If that is the reason, then how does Relativity explain this (and please note I am not asking for the calculation).

If the deduction that his rate of time slowed down at some point in his joureny is incorrect, then all I ask is that the correct explantion for the loss in time is stated for me to see.

You don't like the Doppler analysis because you say it is not an explanation. You don't like Special Relativity's inertial reference frame explanations because different frames assign the "loss of time" differently. The only thing you like is your own private theory that "the loss in time is because his rate of time slowed down at some point in his journey".

You're defending the ideas that existed prior to Einstein's concept that a clock is what measures time. You're saying that just because two clocks differ in the rate at which they tick, one (or both) of them is wrong, there is only one right answer, an absolute time. But I'll bet you don't realize that.

 

[harrylin]

Excuse me? Calculating and understanding are completely different things. (I hope you are not trying to suggest that I am not intelligent enough to continue with discussing this?)
You seem to think that I am questioning the calculation - I am not.

In physics, a precise understanding of how the math works is usually necessary for a good understanding. It is not sufficient.
You came with a remark that nobody who understands the equations would make:

"The problem lies in explaining how the rate of time of a clock running slow can ever be made to increase from its current slower rate of time - for any acceleration or de-acceleration are equivalent actions - the only difference being the direction of the accelerating force / de-acclerating force being applied to the clock."

I am pointing out that if the calculation is used as the explanation for the loss in time in the traveling twin's clock, then as an explanation it fails for this reason...

The round trip journey effects the traveling twin's clock - it loses time compared to if he had not gone on the journey.

I have suggested that the reason for the loss in time is because his rate of time slowed down at some point in his journey. If that is the reason, then how does Relativity explain this (and please note I am not asking for the calculation). [..]

I already answered to that question in post #127. SR simply predicts what necessarily the observations will be, which are a logical consequence of the phenomena-based postulates; it has on purpose no physical model (=metaphysical interpretation) relating to unobservables.

And I mentioned two interpretations that can be found in the literature.
BTW, perhaps most popular one which I did not mention (stemming from Feynman?) is "shut up and calculate".

 

[phyti]

And where is that in dispute? (Also, what on Earth to you mean by only B took a different path? You have two paths between events - different/return are not objective. You can talk about inertial, non-inertial, but one twin being inertial is a special case; the core explanation cannot rely on it. The most general case is easily analyzed in any inertial frame in SR).

So, at the end, one ages less. That is given. What is important to clarify is that you cannot say which part of the younger one's path is where they aged slower by any unique or preferred criterion. Just among all inertial frames (let along more general simultaneity conventions), where one clock is going slower, and by how much, can differ for every one one of these frames.

robinpike

Thanks, but it is the specific twin traveling example that I am applying the deduction to. That is, one twin does the traveling and then when he comes home, his clock has a retarded time.

If there is no dispute, then agree with his conclusion.
As he says, it's a specific scenario, not a generalized problem. No one cares if there are many qualified frames. It's not necessary to know what part of the longer path the most time was lost. For the basic 'twin' case, the longer path looses more time, period.

Also, what on Earth to you mean by only B took a different path?

If you can't figure that out, try making a sketch.

 

[phyti]

The problem lies in explaining how the rate of time of a clock running slow can ever be made to increase from its current slower rate of time - for any acceleration or de-acceleration are equivalent actions - the only difference being the direction of the accelerating force / de-acclerating force being applied to the clock.

If that was true, a ship with a propulsion unit at each end, could alternately accelerate from each end, and slow the clock to any desired rate (reduce the ship length toward zero).
The phenomena of td & lc are not direction independent. If you return to a previous speed, the clock returns to the previous rate, because td is a function of v/c, i.e. speed.

 

[phyti]

Nobody lost any time. That's your problem. You're trying to find where the lost time went.

Einstein promoted the idea that time is what a clock measures. Therefore, if one clock measures a different time than another, then time is really different for the two clocks. It wouldn't be correct to think that one of them was correct and the other one lost [or gained] time.

I agree with you, the same amount of time (number of events in the universe) does not change for any observers.
If we say a clock measures the rate of activity for each observer, it would correspond more to reality, i.e. subjective time.

 

[ghwellsjr]

Nobody lost any time. That's your problem. You're trying to find where the lost time went.

Einstein promoted the idea that time is what a clock measures. Therefore, if one clock measures a different time than another, then time is really different for the two clocks. It wouldn't be correct to think that one of them was correct and the other one lost [or gained] time.

I agree with you, the same amount of time (number of events in the universe) does not change for any observers.
If we say a clock measures the rate of activity for each observer, it would correspond more to reality, i.e. subjective time.

I never said any of that stuff you are supposedly agreeing with me about. I don't even know what you're talking about.

What's so hard about saying that the time that a clock displays and measures is legitimate and not compromised just because it doesn't agree with what another clock with a different history displays? Einstein says they're all correct. You never have to make an excuse for one of them saying it lost some time and the other one is correct. They're all correct. Furthermore, all the objects and observers that are local to a clock, experience objective time (and subjective, if we can identify that) the same as the clock. This doesn't have anything to do with the number of events in the universe, which is infinite, by the way.

 

[PAllen]

robinpike
If there is no dispute, then agree with his conclusion.
As he says, it's a specific scenario, not a generalized problem. No one cares if there are many qualified frames. It's not necessary to know what part of the longer path the most time was lost. For the basic 'twin' case, the longer path looses more time, period.

If you can't figure that out, try making a sketch.

I am interested in avoiding so called explanations that are tied to a specific formulation of differential aging. A good understanding will apply equally to any formulation.

What is 'longer path'? The only invariant along timelike paths through spacetime is proper time. If you mean 'distance traveled' each twin may prefer to think they have traveled zero distance and the other has done the traveling. Travel distance has really nothing to do with differential aging, and is strictly frame or coordinate dependent quantity.

Maybe you mean longer as drawn on a conventional spacetime diagram? If so, I missed where you stated this. It is true that for different (timelike) paths between two given events drawn on conventional SR spacetime diagram, the longer one has less proper time.

 

[harrylin]

[...] If you mean 'distance traveled' each twin may prefer to think they have traveled zero distance and the other has done the traveling. [..]

Sorry but if you are talking SR than that is erroneous, as I just tried to clarify to robinpike: in all inertial frames it is the traveller who traveled most. In none of them does the traveller travel zero distance.

 

[PAllen]

Sorry but if you are talking SR than that is erroneous, as I just tried to clarify to robinpike: in all inertial frames it is the traveller who traveled most. In none of them does the traveller travel zero distance.

Who said inertial frames? I was not referring to inertial frames in that sentence. Earlier, I pointed out you can make invariant statements about inertial, non-inertial, etc. but not about who is traveling, or who traveled farther.

Further, it's not necessarily true anyway, even for inertial frames. Consider the classic abrupt turnaround twin scenario, from the inertial frame of the outgoing travel leg. Then both twins travel the same distance in this frame. The one who ages less, travels faster for the second leg, but both twins travel the same distance in this frame.

 

[harrylin]

[..] Consider the classic abrupt turnaround twin scenario, from the inertial frame of the outgoing travel leg. Then both twins travel the same distance in this frame. The one who ages less, travels faster for the second leg, but both twins travel the same distance in this frame.

Ah yes - hehe I overlooked that one. Thanks!

Anyway, SR relates to inertial frames, and probably so did the statements you were commenting on, as the conversation went on in the context of SR. Only in GR can the traveller choose to have traveled zero distance. And then you get on slippery grounds - the one of Einstein's 1918 paper.

 

[PAllen]

Ah yes - hehe I overlooked that one. Thanks!

Anyway, SR relates to inertial frames, and probably so did the statements you were commenting on, as the conversation went on in the context of SR. Were you talking GR instead? Then be prepared for the slippery ground!

I was referring to SR. You can have accelerated observers in SR. You can compute what they observe in an inertial frame. You can also construct various 'non-inertial' coordinates. Mainly, I was arguing in favor of terminology with invariant meaning: ( inertial vs non-inertial; path of shorter/longer proper time) for example. Not traveling vs not traveling vs traveling more. Travel distance is frame or coordinate dependent, and has nothing whatsoever to do with differential aging, as I understand it.

 

[Austin0]

Further, it's not necessarily true anyway, even for inertial frames. Consider the classic abrupt turnaround twin scenario, from the inertial frame of the outgoing travel leg. Then both twins travel the same distance in this frame. The one who ages less, travels faster for the second leg, but both twins travel the same distance in this frame.

Yes in fact this is true in ALL frames where the Earth is in motion as long as the traveler does not actually reverse direction relative to the observing frame. A case being if the Earth has a rel v of -.5c and the traveler takes off in the +x direction at +.5c for a traveler-earth rel velocity of 0.8c. Here the traveler does cover more spatial distance in the observer frame, not just more spacetime distance.

 

[Dale]

If that is the case, then I have a question:

A spaceship is in deep space, sitting motionless in its inertial frame, when it fires its rockets.

Has the spaceship increased its speed or decreased its speed?

It has increased its speed in the specified frame.

 

[kamenjar]

...You're saying that just because two clocks differ in the rate at which they tick, one (or both) of them is wrong, there is only one right answer, an absolute time. But I'll bet you don't realize that.

Why do you say that we need that one or both are wrong? There is no need for defining such thing as a right or wrong clock or absolute time. Why can't we say both clocks correct and that there's nothing wrong with any of them?

In fact, even if we had to establish absolute time rate we could do that. We could pick it to be the fastest-ticking clock in the universe - and that clock is the clock at which CMB radiates equally in all directions. all other clocks tick at equal or slower rate than that one.

 

[Dale]

We could pick it to be the fastest-ticking clock in the universe - and that clock is the clock at which CMB radiates equally in all directions. all other clocks tick at equal or slower rate than that one.

This is only true locally and even then only in the frame where the CMB is isotropic.

 

[kamenjar]

This is only true locally and even then only in the frame where the CMB is isotropic.

Which is what I said - a frame with the "absolute clock rate". No?
A frame has to have a coordinate zero, that's a different topic though.

 

[ghwellsjr]

...You're saying that just because two clocks differ in the rate at which they tick, one (or both) of them is wrong, there is only one right answer, an absolute time. But I'll bet you don't realize that.

Why do you say that we need that one or both are wrong? There is no need for defining such thing as a right or wrong clock or absolute time. Why can't we say both clocks correct and that there's nothing wrong with any of them?

Are you thinking that I'm agreeing with your quote of me? These are things that I'm claiming that robinpike is saying, not me.

In fact, even if we had to establish absolute time rate we could do that. We could pick it to be the fastest-ticking clock in the universe - and that clock is the clock at which CMB radiates equally in all directions. all other clocks tick at equal or slower rate than that one.

You have totally missed the whole point. Please don't associate me with these ideas. Please read my posts carefully before quoting me and commenting on what I'm saying.

 

[PAllen]

In fact, even if we had to establish absolute time rate we could do that. We could pick it to be the fastest-ticking clock in the universe - and that clock is the clock at which CMB radiates equally in all directions. all other clocks tick at equal or slower rate than that one.

There is no such thing as 'fastest clock in the universe'. Consider, for example, two distant clocks, each of which sees isotropic CMB. Each would see the other clock running slow.

 

[kamenjar]

Are you thinking that I'm agreeing with your quote of me?

I thought that I was opposing you... I was suggesting that in rder to come up with a simpler solution to the twin paradox, we could establish a universal clock rate.

There is no such thing as 'fastest clock in the universe'. Consider, for example, two distant clocks, each of which sees isotropic CMB. Each would see the other clock running slow.

When they are at rest? Because of the rate of expansion of the universe? If we were go into that topic, it would stray the OP's discussion.

 

[PAllen]

Ah yes - hehe I overlooked that one. Thanks!

We can go further. It is easy to construct a twin scenario where the twin who travels farther ages more, all in an inertial frame.

Alice: travels 100 km west and back very slowly (e.g. takes 1 year).
Bob: sits around stationary in this frame except briefly moving 10 km east and back at .999999c.

When Alice meets Bob again, having traveled a total of 200 km, their clock will be ahead of Bob's who has traveled only 20 km.

Reiterating: travel distance is a compete red herring in trying to understand differential aging.

 

[PAllen]

I thought that I was opposing you... I was suggesting that in rder to come up with a simpler solution to the twin paradox, we could establish a universal clock rate.

Impossible according to relativity.

When they are at rest? Because of the rate of expansion of the universe? If we were go into that topic, it would stray the OP's discussion.

Because of cosmologic redshift, yes. But then where is your fastest clock? Every inertial clock considers all clocks in motion in its rest frame as moving slow.

 

[kamenjar]

We can go further. It is easy to construct a twin scenario where the twin who travels farther ages more, all in an inertial frame.

Alice: travels 100 km west and back very slowly (e.g. takes 1 year).
Bob: sits around stationary in this frame except briefly moving 10 km east and back at .999999c.

When Alice meets Bob again, having traveled a total of 200 km, their clock will be ahead of Bob's who has traveled only 20 km.

Reiterating: travel distance is a compete red herring in trying to understand differential aging.

I don't think that's it's a valid example. I thought someone "devised" a formula that you integrate velocity times the traveled distance over the duration of time to deduce reduction in aging over a period of time...

 

[ghwellsjr]

Are you thinking that I'm agreeing with your quote of me?.

I thought that I was opposing you... I was suggesting that in rder to come up with a simpler solution to the twin paradox, we could establish a universal clock rate.

And how would that be simpler than establishing a coordinate clock rate in any inertial frame you want to choose, such as the one in which the stay-at-home twin remains at rest? We can't even get everyone to agree with that simple explanation. How do you expect to get everyone to agree with your proposal of a universal clock rate?

 

[kamenjar]

Impossible according to relativity.

I thought that OP was looking for an answer. Not what's possible or impossible. It doesn't mean that we can't use a third reference frame, which for all purposes would be the Earth twin's reference frame because CMB is quite isotropic in his.

Because of cosmologic redshift, yes. But then where is your fastest clock?

I said, anywhere where CMB is isotropic. It's definitely not in inertial observer's frame. Or is there something else I didn't specify.

Every inertial clock considers all clocks in motion in its rest frame as moving slow.

What the observer sees/measures is different what it actually is. We see all kinds of things that aren't. We see/measure some galaxies that appear to move at speeds faster than speed of light, but it's not what it is.

 

[kamenjar]

And how would that be simpler than establishing a coordinate clock rate in any inertial frame you want to choose, such as the one in which the stay-at-home twin remains at rest? We can't even get everyone to agree with that simple explanation. How do you expect to get everyone to agree with your proposal of a universal clock rate?

I actually don't. Convincing anyone anything with exception of having them only understand your point of view, is in my personal opinion wrong. And that task in itself is quite difficult. We all have different beliefs and are not equally open to other's views.

I thought that it would be helpful to break the symmetry. I don't know yet how exactly, but I'm thinking along the lines of:
1) Each observer looks at CMB, compensates for clock rate (symmetry is broken here) to establish their clock rate.
2) Each observer looks at angular size of the object they are observing. "Sizes" are known at rest.
3) Each observer looks at the light signal's doppler shift and clock reading.
4) Given the above three values, the observer projects their location and the other's path in space WRT the CMB frame.
5) Given above value, the observer can compensate for the delay of the signal received form the other twin and only then make a conclusion about the other's clock rate.