Twin Paradox Problem: Do Twins Age Differently?

[Dale]

In SR, we have to artificially ascribe the velocity to the traveling twin by considering that he accelerates and is not in an inertial frame. In that case, I feel that the solution goes outside the domain of SR, since we are bringing in reasons like 'feeling acceleration' which were specifically left out by Einstein when deriving SR, and included in GR.

Why do you believe that "feeling acceleration" is outside of the domain of SR? Clearly, the concept of an inertial frame is part of the foundations of SR, so how would you define an inertial frame without reference to "feeling acceleration" or its equivalent?

 

[arindamsinha]

Why do you believe that "feeling acceleration" is outside of the domain of SR? Clearly, the concept of an inertial frame is part of the foundations of SR, so how would you define an inertial frame without reference to "feeling acceleration" or it's equivalent?

Dalespam, I was thinking of the equivalence principle and considering (perhaps incorrectly) that feeling 'acceleration' and feeling 'gravity' are one and the same thing. In SR, I thought relative velocities are the only thing considered, and we look at it kinematically, and any acceleration is mutual and translates into only the instantaenous relative velocity (which again is mutual). I am ready to stand corrected in this respect if I have misunderstood.

However, I would value your opinion on the overall thought process in my previous thread, not just this particular statement.

 

[Dale]

Dalespam, I was thinking of the equivalence principle and considering (perhaps incorrectly) that feeling 'acceleration' and feeling 'gravity' are one and the same thing. In SR, I thought relative velocities are the only thing considered, and we look at it kinematically, and any acceleration is mutual and translates into only the instantaenous relative velocity (which again is mutual). I am ready to stand corrected in this respect if I have misunderstood.

I think you have misunderstood. The concept of "kinematic" does not enter into the postulates, which are the core of SR, but the concept of an "inertial frame" does. To me it seems that you have exaggerated the ancillary concept of kinematic to the point that you have lost track of the defining concepts of SR. You cannot have SR without a definition of inertial frames, and so you can certainly use inertial frames to resolve paradoxes and distinguish between observers.

However, I would value your opinion on the overall thought process in my previous thread, not just this particular statement.

Regarding the overall thought process, I don't really get what you mean by the Schwarzschild metric without gravitation. I think that is SR in spherical coordinates.

 

[djy]

The advantage I see in using the GR Schwarzschild solution for explaining experimental observations and paradoxes is this - velocities considered here are from the CG of the two-body system under consideration, rather than absolute relative velocities between the two components. (At least, that is how I have interpreted the GPS time dilation, though I may be wrong).

This neatly explains why the traveling twin has velocity, why he should time dilate, and why GPS clocks get slower compared to Earth ones (again ignoring the gravitational TD part). This also avoids the somewhat magical 'clock jumping' that happens at the point of reversal of the traveling twin, which I have seen in some SR solutions.

The Schwarzschild metric is the solution of spacetime around a spherical, uncharged, non-rotating mass. Yes, it does closely explain the difference in passage of time on GPS satellites compared to the Earth's surface. And, as you said, there are components of time dilation due both to velocity and gravitational potential. However, you cannot use this metric to analyze the traditional twin paradox, which takes place in flat spacetime. Since the metric does not describe flat spacetime, using it is simply wrong.

In SR, we have to artificially ascribe the velocity to the traveling twin by considering that he accelerates and is not in an inertial frame. In that case, I feel that the solution goes outside the domain of SR, since we are bringing in reasons like 'feeling acceleration' which were specifically left out by Einstein when deriving SR, and included in GR.

It's a common misconception that acceleration is outside the domain of SR. On the contrary, acceleration is handled equally well by both SR and GR. In both cases, the proper time experienced by an observer is simply the length of its world line in spacetime. In GR the spacetime may be curved (but is not necessarily); in SR it is always flat.

To analyze the traditional twin paradox, simply substitute the metric of flat spacetime for the Schwarzschild metric, and integrate the lengths of the world lines of the two twins to find the difference in proper time passage.

 

[Jeronimus]

Why do you believe that "feeling acceleration" is outside of the domain of SR? Clearly, the concept of an inertial frame is part of the foundations of SR, so how would you define an inertial frame without reference to "feeling acceleration" or its equivalent?

Isn't free fall, being in a force field, also accelerating? We cannot feel that.

If we were able to create a force field into any direction which is local, then we would not feel any acceleration as all parts would accelerate evenly. That is, for an infinitesimal small volume. Unfortunately we cannot magically create a force field to freely fall towards, but instead rely on accelerating "one part" of an object, which then pushes against other parts (electromagnetic forces). By doing so, the structure of the body changes and this is why we believe to feel acceleration imo.

Do accelerometers measure acceleration or do they merely measure a chance in the structure of an object caused by different parts of the object being accelerated differently?

I also have a hard time understanding how someone can "feel acceleration". I understand how someone or something can detect a change in the structure of it's body like accelerometers do.edit: "feeling acceleration" is not required in order to arrive at the formulas of SR. The concept of changing the inertial frame of reference an object is at rest in is required. Who is responsible for changing an object's rest frame? Acceleration we say. What causes acceleration or a force field? Energy or mass (or objects "pushing against each other?). What is energy or mass? Now that's a difficult one.

 

[stevendaryl]

The Schwarzschild metric is the solution of spacetime around a spherical, uncharged, non-rotating mass. Yes, it does closely explain the difference in passage of time on GPS satellites compared to the Earth's surface. And, as you said, there are components of time dilation due both to velocity and gravitational potential. However, you cannot use this metric to analyze the traditional twin paradox, which takes place in flat spacetime. Since the metric does not describe flat spacetime, using it is simply wrong.

Well, actually, in the limit in which the gravitational field of the Earth is negligible, the Schwarzschild metric reduces to the Minkowsky metric in spherical coordinates:

ds^2 = (c dt)^2 - dr^2 - r^2 (d\theta^2 + sin^2(\theta)d\phi^2)

You can certainly use this metric to calculate the ages of the two twins in the twin paradox, and get the same answer as using the usual Minkowsky coordinates.

 

[Dale]

Isn't free fall, being in a force field, also accelerating? We cannot feel that.

Here you need to distinguish between coordinate acceleration, which is frame variant and which we cannot feel, and proper acceleration, which is frame invariant and which we can feel. Accelerometers measure proper acceleration.

"feeling acceleration" is not required in order to arrive at the formulas of SR. The concept of changing the inertial frame of reference an object is at rest in is required.

Then please define an inertial frame without "feeling acceleration" or its equivalent. I would be very interested in hearing such a definition since I cannot think of it myself.

 

[harrylin]

[..]
Then please define an inertial frame without "feeling acceleration" or its equivalent. I would be very interested in hearing such a definition since I cannot think of it myself.

Jeronimus wrote: "Isn't free fall, being in a force field, also accelerating? We cannot feel that."

This is implied in post #156. In the original space travelers example, the traveler feels no acceleration at turn-around but he is nevertheless not in uniform rectilinear motion. Originally SR was defined wrt to Newtonian (or "Galilean") reference systems; that is what is meant with "inertial frames" in the context of SR.

 

[Dale]

. Originally SR was defined wrt to Newtonian (or "Galilean") reference systems; that is what is meant with "inertial frames" in the context of SR.

And how are 'Newtonian (or "Galilean") reference systems' defined? The only way I know is through "feeling acceleration" or something equivalent.

 

[harrylin]

And how are 'Newtonian (or "Galilean") reference systems' defined? The only way I know is through "feeling acceleration" or something equivalent.

In any case, the space traveller who is in free fall is accelerating in such a frame, just as a stone in free fall is accelerating in classical physics. There is no disagreement about the working of SR between Einstein and Langevin

For a real discussion about classical reference frames it's a good question to ask in the classical physics forum; but here are my "2cts", for the case that no such thread is started.
Newton defined it as in uniform straight line motion wrt the "fixed stars", and for the traveller who falls around a star that would work rather well in practice (replacing "fixed stars" by apparently fixed distant stars).
No doubt this can be replaced (and probably was) by the definition of inertial motion at places far away from massive bodies, and/or simply correcting for the acceleration due to gravitation if a massive body is nearby; and that also works fine for the space traveller, just as elaborated.

 

[arupel]

Riskavutkarsh's statement is good. I read a book about this. Basically both twins see the same time dilation. However for the twin in which the velocity reverses, the time-space line takes a jump. Immediately before the reversal he sees his twin graduating. Immediately after, he sees his twin on his death bed.

 

[Dale]

In any case, the space traveller who is in free fall is accelerating in such a frame, just as a stone in free fall is accelerating in classical physics.

arindamshina is explicitly neglecting gravitation, so we don't have to worry about that.

Newton defined it as in uniform straight line motion wrt the "fixed stars", and for the traveller who falls around a star that would work rather well in practice (replacing "fixed stars" by apparently fixed distant stars).
No doubt this can be replaced (and probably was) by the definition of inertial motion at places far away from massive bodies, and/or simply correcting for the acceleration due to gravitation if a massive body is nearby; and that also works fine for the space traveller, just as elaborated.

That was OK for Newton, but now we know that the stars aren't fixed wrt each other, so they cannot be used to define a single reference frame.

 

[Austin0]

The problem with Relativity's explanation for the Twin paradox, is that, once back on earth, for the traveling twin's clock to have a lesser time than the stay at home twin's clock, it can be deduced that the rate of time on the traveling twin's clock must have slowed down at some point during the journey.

Once a clock has had its rate of time slowed down by acceleration, Relativity has no mechanism to return the rate of time back to 'normal' - since any further acceleration can only cause the clock's rate of time to slow down even more...

Hi
i certainly agree with your logic, that accumulated time difference requires an assumption of a difference in instantaneous rates over the course of the exercise but there is no means to determine what the relative rates are for any time interval whatsoever during transit.
As for the acceleration: Yes the term acceleration applies equally to any change of velocity irrespective of direction, but in this circumstance there is a fundamental difference between acceleration away and the deceleration required at turnaround.
The dilation factor changes with the instantaneous velocity relative to Earth on the way out.
On turnaround the initial negative acceleration reverses those changes wrt Earth up to the point where the traveler is instantly at rest wrt earth, where there is no difference (initial condition) From there the dilation factor from acceleration (instantaneous velocity) again begins to increase in magnitude to the final inertial velocity gamma of the return leg..

 

[pervect]

The problem with Relativity's explanation for the Twin paradox, is that, once back on earth, for the traveling twin's clock to have a lesser time than the stay at home twin's clock, it can be deduced that the rate of time on the traveling twin's clock must have slowed down at some point during the journey.

Once a clock has had its rate of time slowed down by acceleration, Relativity has no mechanism to return the rate of time back to 'normal' - since any further acceleration can only cause the clock's rate of time to slow down even more

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.

 

[zonde]

Riskavutkarsh's statement is good. I read a book about this. Basically both twins see the same time dilation. However for the twin in which the velocity reverses, the time-space line takes a jump. Immediately before the reversal he sees his twin graduating. Immediately after, he sees his twin on his death bed.

Welcome to PhysicsForums, arupel!

You should be careful with your usage of word "sees". Your statement would be rather sensible if you would replace "sees" with "calculates" or "interprets".
But if you want to stick to statements about what twins see about the other twin then you can look at one example in my post #176 (hopefully clear enough).

 

[zonde]

That was OK for Newton, but now we know that the stars aren't fixed wrt each other, so they cannot be used to define a single reference frame.

You mean because of redshift?

 

[harrylin]

arindamshina is explicitly neglecting gravitation, so we don't have to worry about that.

The OP is Rishavutkarsh, and you commented on Jerominus who mentioned free fall. Free fall is used in the first full (two-observer) discussion, which is not (and never was) a problem in the context of SR (post #188 once more).

That was OK for Newton, but now we know that the stars aren't fixed wrt each other, so they cannot be used to define a single reference frame.

I next gave you three simple practical means and with modern technology there are more; please start a topic in classical physics if you don't understand how to apply any of them.

You mean because of redshift?

Please don't elaborate here; if three practical ways (not including Newton's) result in zero "clicks", then it would certainly require a long discussion to explain the reference systems of classical mechanics - and that is not the topic here.

 

[Dale]

I next gave you three simple practical means and with modern technology there are more; please start a topic in classical physics if you don't understand how to apply any of them.

It isn't a question of practical implementation, but one of definition. The distant stars are not fixed wrt each other, so they cannot be used to define a reference frame, even if from a practical standpoint the difference can be neglected.

Furthermore, suppose that you used some specific distant stars, and defined a frame where each of your set of distant stars had some well defined velocity. How would you know where or not the frame so defined is inertial? The stars cannot do it, so instead you look for the absence of fictitious forces. That is the only way I know to define an inertial frame.

http://en.wikipedia.org/wiki/Fixed_stars#The_fixed_stars_in_classical_mechanics

 

[Dale]

You mean because of redshift?

Redshift demonstrates radial motion, but there is also proper motion (transverse to us).

http://en.wikipedia.org/wiki/Fixed_stars#The_fixed_stars_are_not_fixed

 

[harrylin]

It isn't a question of practical implementation, but one of definition. [..]

In modern physics, definitions are practical implementations, but I won't go along any further with a discussion about classical mechanics in this thread. If your new practical definition of inertial frame works for Newton's falling apple as well as for Langevin's space traveler example (which I guess it does), then it is OK for classical mechanics and SR.

 

[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.