Space traveler and time dilation

[yogi]

You're missing my point. I did say

"In general relativity, ... the satellite is inertial."​

which agrees with everything you said above (subject to JesseM's correct note about local frames). But I also said

"In general relativity, the south pole pole is accelerating..."​

It is undergoing proper acceleration upwards and therefore is not inertial.

Yes - things on the surface of the Earth are subject to g - but for purposes of synchronizing GPS satellites, the non-rotating Earth centered reference frame is taken as a basis - in the proposed thought experiment, the affect of satellite height is nullified by the height of the imaginary towers - so perhaps I should have said both clocks are at the same gravitational potential at all times.

The point of the analogy is, it is not necessary to explain the paradox as being the result of the traveling twin feeling "turn-around acceleration" (as is frequently asserted - perhaps even by myself in past posts). This is not what distinguishes the two clocks - they log different amounts of time because of the invariance of the spacetime interval. The traveler travels in both space and time - the tower twin travels in time only.

 

[JesseM]

The point of the analogy is, it is not necessary to explain the paradox as being the result of the traveling twin feeling "turn-around acceleration"

But you can explain it in terms of one moving inertially while the other doesn't, and the question of which one turns can be decided by the turn-around acceleration.

This is not what distinguishes the two clocks - they log different amounts of time because of the invariance of the spacetime interval. The traveler travels in both space and time - the tower twin travels in time only.

But in a GR case, you can come up with a coordinate system where the first remains at a single point in space (traveling 'in time only', not moving in space) and a different coordinate system where the second is the one that remains at a single point in space, both coordinate systems would be equally valid in GR, there'd be no physical basis for saying one "really" moved in space while the other didn't.

 

[Jocko Homo]

Forget about "clocks running slower". That's Lorentz Ether language. Of course, A cannot run slower than B while B runs slower than A. This sort of language is incompatible with SR.
I know that's how they teach it. Forget it. Look at your diagrams. There's a triangle, and there's a triangle inequality (in our case the non-straight path being shorter). That's true no matter in which frame you view it.

Your math is correct. But there's no such things as slowing clocks, contracting metersticks, or planets jumping forward in time. SR is about relations of objects, not changes happening to them.
Take the geometric viewpoint, there are projections, slices, different paths (time dilation, length contraction, twin paradox), not broken clocks.

While Minkowski diagrams provide a powerful abstraction to the physical reality of Special Relativity, I think "clocks running slower" is a fair characterization considering the accelerating twin really does come back younger...

 

[yogi]

Make it more symmetrical by using two satellites instead, orbiting in opposite directions. Each is traveling through the same curvature and each is accelerating the same. That creates a paradox if you apply only SR. How does it manage to work out using GR? Once I understand that, I can ponder the Universe form of the question again.

If you have two circular orbiting satellites at the same height traveling in opposite directions and set both clocks to zero as they pass, when they meet again, both clocks will read the same

 

[yogi]

But you can explain it in terms of one moving inertially while the other doesn't, and the question of which one turns can be decided by the turn-around acceleration.

It will give you the right answer - but you can also get the correct age difference by doing a one way trip - setting the clocks to zero on the initial flyby and stopping the traveling clock at the instant it flies by the destination - you now have a one way trip with no accelerations and no turn around acceleration - to get the result you simple double the time difference for the one way trip - my point is not that other methods give wrong results - but rather, SR age differences are fundamentally a relative velocity problem that has been misappropriated to GR

But in a GR case, you can come up with a coordinate system where the first remains at a single point in space (traveling 'in time only', not moving in space) and a different coordinate system where the second is the one that remains at a single point in space, both coordinate systems would be equally valid in GR, there'd be no physical basis for saying one "really" moved in space while the other didn't.

Agreed, travel with respect to space is meaningless - I guess it should be clarified as travel with respect to space as measured in the fame which is chosen to be at rest

 

[Al68]

IMO - any analysis that results from dependence upon accelerating frames and the like is going to cloud the reality of what is properly explained by SR - this can be done by using the one way trip and doubling the result...

I agree that if the problem was presented and explained as two one way trips, most of the confusion would be completely eliminated. But the ship is still non-inertial for an (entire) one way trip.

Einstein confused a lot of his followers when he published his 1918 article that explained the clock paradox using a pseudo G field that gives the same answer to the aging difference - but for the wrong reason.

Why would the reason be wrong? The reason for the aging difference is exactly the same in Einstein's 1918 paper as in standard resolutions. The only difference is he uses realistic acceleration (with Earth clock running fast in ship frame) instead of an instantaneous turnaround (earth clock "jumps ahead").

 

[matheinste]

In SR if the twins part and reunite then at least one of them has been moving non-inertially at some time.


Matheinste.

If you have two circular orbiting satellites at the same height traveling in opposite directions and set both clocks to zero as they pass, when they meet again, both clocks will read the same

Assuming that they move symmetrically, that is, they meet again for the first time at the opposite "pole" to where they started as referred to the orbited body,then isn't this a case of both objects traveling non-inertially but both traveling equal spacetime paths and so clocks traveling with them will record the same elapsed proper time.

Matheinste.

 

[Ich]

While Minkowski diagrams provide a powerful abstraction to the physical reality of Special Relativity, I think "clocks running slower" is a fair characterization considering the accelerating twin really does come back younger...

IMHO, time dilation is a convenient memory and calculation aid, but has no (or even less) value for understanding SR. It should be used by those only who understand SR and know how, when, and why time dilation can be applied.
If you confront beginners with that phrase ("clocks run slower"), you inevitably push them down the Lorentzian road, where SR is the oncoming traffic. However, it is common praxis in education, and you see the result here every week with a new thread about the twin paradox.

 

[JesseM]

Agreed, travel with respect to space is meaningless - I guess it should be clarified as travel with respect to space as measured in the fame which is chosen to be at rest

OK, but in the GR case where you're dealing with non-inertial frames either way, depending on the frame you choose it may not be the twin that "travels with respect to space" who ages less, it could be the other twin.

 

[Mike_Fontenot]

Here are a couple of my responses to essentially the same question that was asked on an earlier thread:
______________________________________________________

During the constant-speed legs of the trip, BOTH twins conclude that the other twin is ageing slower. But when the trip is over, they both agree that the stay-at-home twin is older. How is that possible?

It's possible because, during the turnaround, the traveler will conclude that the home twin quickly ages, with very little ageing of the traveler. The home twin concludes that neither of them ages much during the turnaround. When you add up all these segments of ageing, you get the result that the home twin is older (and both twins exactly agree on that).

Years ago, I derived a simple equation (called the "CADO" equation) that explicitly gives the ageing of the home twin during accelerations by the traveler (according to the traveler). The equation is especially easy to use for idealized traveling twin problems with instantaneous speed changes. But it also works for finite accelerations. I've got a detailed example with +-1g accelerations on my webpage:

http://home.comcast.net/~mlfasf

And I've published a paper giving the derivation of the CADO equation:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.
____________________________________________

Here's a brief description of my "CADO" equation:
__________________________________________________ __

Years ago, I derived a simple equation that relates the current ages of the twins, ACCORDING TO EACH TWIN. Over the years, I have found it to be very useful. To save writing, I write "the current age of a distant object", where the "distant object" is the stay-at-home twin, as the "CADO". The CADO has a value for each age t of the traveling twin, written CADO(t). The traveler and the stay-at-home twin come to DIFFERENT conclusions about CADO(t), at any given age t of the traveler. Denote the traveler's conclusion as CADO_T(t), and the stay-at-home twin's conclusion as CADO_H(t). (And in both cases, remember that CADO(t) is the age of the home twin, and t is the age of the traveler).

My simple equation says that

CADO_T(t) = CADO_H(t) - L*v/(c*c),

where

L is their current distance apart, in lightyears,
according to the home twin,

and

v is their current relative speed, in lightyears/year,
according to the home twin. v is positive
when the twins are moving apart.

(Although the dependence is not shown explicitely in the above equation, the quantities L and v are themselves functions of t, the age of the traveler).

The factor (c*c) has value 1 for these units, and is needed only to make the dimensionality correct.

The equation explicitly shows how the home twin's age will change abruptly (according to the traveler, not the home twin), whenever the relative speed changes abruptly.

For example, suppose the home twin believes that she is 40 when the traveler is 20, immediately before he turns around. So CADO_H(20-) = 40. (Denote his age immediately before the turnaround as t = 20-, and immediately after the turnaround as t = 20+.)

Suppose they are 30 ly apart (according to the home twin), and that their relative speed is +0.9 ly/y (i.e., 0.9c), when the traveler's age is 20-. Then the traveler will conclude that the home twin is

CADO_T(20-) = 40 - 0.9*30 = 13

years old immediately before his turnaround. Immediately after his turnaround (assumed here to occur in zero time), their relative speed is -0.9 ly/y. The home twin concludes that their distance apart doesn't change during the turnaround: it's still 30 ly. And the home twin concludes that neither of them ages during the turnaround, so that CADO_H(20+) is still 40.

But according to the traveler,

CADO_T(20+) = 40 - (-0.9)*30 = 67,

so he concludes that his twin ages 54 years during his instantaneous turnaround.

The equation works for arbitrary accelerations, not just the idealized instantaneous speed change assumed above. I've got an example with +-1g accelerations on my web page:

http://home.comcast.net/~mlfasf

The derivation of the equation is given in my paper

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

Mike Fontenot

 

[Al68]

Years ago, I derived a simple equation (called the "CADO" equation) that explicitly gives the ageing of the home twin during accelerations by the traveler (according to the traveler). The equation is especially easy to use for idealized traveling twin problems with instantaneous speed changes. But it also works for finite accelerations. I've got a detailed example with +-1g accelerations on my webpage:

http://home.comcast.net/~mlfasf

And I've published a paper giving the derivation of the CADO equation:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

I didn't carefully analyze your web page, but will assume the results match standard lorentz transformations for a co-moving inertial frame for any given time on the accelerated clock. And I did note the phrase "the bizarre behavior of the CADO, for an accelerating observer, must be regarded as being fully meaningful and real."

I would also note that if the traveler is accelerating in a direction away from earth, the results could be even more bizarre. For example, from the accelerated traveler's point of view, the funeral of a person on Earth could be followed by that person's wedding.

I would have to say that people rising from the grave would qualify as not "fully meaningful and real" by any reasonable definition.

Note that I'm not disputing that result. We could obtain the same result by performing lorentz transformations directly during inertial motion before and after acceleration, if the ship were distant from Earth and the acceleration was away from earth. But most would disagree that the result represented reality in any sense other than assigning coordinates.

 

[stevmg]

George Jones and Ich are correct. Attached is a picture I downloaded from among the plethora of explanations on the web. You can see that on the way out, the twin on the rocket sees two years pass on Earth, while four years pass for him. On the way back, he sees 8 years pass on Earth while four years pass for him. So when he returns, 10 years have passed on Earth while 8 years have passed for him, so the twin on the rocket is younger.

JesseM gave an example using SR why the "moving" twin ages less quickly than the earthbound twin. He does it both ways - using the Earth as the inertial frame and then the spaceship as the inertial frame. In both ways the Earth twin aged more than the traveling twin. Here is the post:

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

 

[stevmg]

George Jones and Ich are correct. Attached is a picture I downloaded from among the plethora of explanations on the web. You can see that on the way out, the twin on the rocket sees two years pass on Earth, while four years pass for him. On the way back, he sees 8 years pass on Earth while four years pass for him. So when he returns, 10 years have passed on Earth while 8 years have passed for him, so the twin on the rocket is younger.

How do I post a picture like you did on this post
https://www.physicsforums.com/showpost.php?p=2753642&postcount=13

H-E-L-P

Steve G
Melbourne FL

 

[DrGreg]

How do I post a picture like you did on this post
https://www.physicsforums.com/showpost.php?p=2753642&postcount=13

While you are editing your post (in the "advanced" editor -- which you get automatically if you use the QUOTE button) press the "Manage Attachments" button not far underneath.

 

[Mike_Fontenot]

[...]
I would also note that if the traveler is accelerating in a direction away from earth, the results could be even more bizarre. For example, from the accelerated traveler's point of view, the funeral of a person on Earth could be followed by that person's wedding.

Yes, that behavior is easy to see, directly from the form of the CADO equation itself, when there are instantanious changes in the velocity v, such that v gets more positive, or less negative (v is positive when their separation is increasing). And this effect isn't limited to instantaneous velocity changes...the voyage detailed in the web page includes segments where the traveler must conclude that the home twin gets younger, with only 1g accelerations involved.

But the web page, and my paper, also make it clear that this phenomenon in no way influences the perception, by the home twin, of the normal progression of time. It is somewhat analogous to your ability to reverse the direction of a movie projector...your action doesn't bother the actors, or make any changes to the frames of the film.

I would have to say that people rising from the grave would qualify as not "fully meaningful and real" by any reasonable definition.

The critical point to understand (which is elaborated in detail in the paper), is that the traveler CAN ADOPT NO OTHER CONCLUSION, IF HE WISHES TO AVOID CONTRADICTING HIS OWN ELEMENTARY (AND CORRECT) MEASUREMENTS. It is in that sense that I use the phrase "real and meaningful". If elementary, correct measurements are not "real and meaningful", how can anyone ever hope to do any physics?

Mike Fontenot

 

[ExecNight]

Hello,

Theoretically speaking, if we get a clock that ticks a red light every 1 second on the spaceship or whatever. And it has a quantum entangled pair on earth.

So how would the Twin on the spaceship actually see the Earth clock ticking throughout the journey?

That would be interesting to know maybe explains how this works better in a way.

 

[Al68]

The critical point to understand (which is elaborated in detail in the paper), is that the traveler CAN ADOPT NO OTHER CONCLUSION, IF HE WISHES TO AVOID CONTRADICTING HIS OWN ELEMENTARY (AND CORRECT) MEASUREMENTS. It is in that sense that I use the phrase "real and meaningful". If elementary, correct measurements are not "real and meaningful", how can anyone ever hope to do any physics?

The traveler can certainly recognize that the coordinates he correctly assigns (using Einstein's simultaneity convention) in his frame to events on Earth do not represent a "real and meaningful" sequence of events requiring causality to be explained.

I might even use the word "fictional" in the same way that fictional forces are used to explain the motions of objects when using a non-inertial reference frame. (The traveler must also assign fictional forces to account for Earth's motion). These forces are not "real and meaningful", either, yet physics thrived for hundreds of years using fictional forces.

And those fictional forces arise for the same reason that bizarre time coordinates get assigned to events on earth, so why not call them fictional?

 

[Ich]

CAN ADOPT NO OTHER CONCLUSION, IF HE WISHES TO AVOID CONTRADICTING HIS OWN ELEMENTARY (AND CORRECT) MEASUREMENTS.

Sorry, that's nonsense. Go back one step to the operational meaning of the respective coordinates.

If elementary, correct measurements are not "real and meaningful", how can anyone ever hope to do any physics?

Which measurements? (This is a rhetorical question).
Describe how you take the "measurement" of something going back in time.

There is none.

What you see at this locus will preserve causality.

 

[Mike_Fontenot]

Describe how you take the "measurement" of something going back in time.

If an unaccelerated object (the "home" twin) is periodically transmitting her current age, then an inertial observer (the traveling twin), who is moving at a constant velocity with respect to her, can receive those messages. He knows that, when he receives a message, that the age being reported in the message is not her current age (because she has aged while the message was in transit). From first principles, and elementary calculations, and without knowing anything about special relativity, the traveler can compute what his twin's current age is (by properly allowing for her ageing during the transit of the message).

That calculation, although elementary, is very easy to do incorrectly. The proper way to do it is detailed in my paper. If that process is done correctly, the result is precisely what is given (much more quickly and easily) by my CADO equation.

If you want to know more, you'll have to dig up my paper...most university libraries either have it, or can obtain it from another library.

(And the reason why the above process, which assumes the traveler is also unaccelerated, is of any value in determining the conclusions of an accelerating traveler, about the current age of the home twin, is also detailed in the paper).

Mike Fontenot

 

[yogi]

I agree that if the problem was presented and explained as two one way trips, most of the confusion would be completely eliminated. But the ship is still non-inertial for an (entire) one way trip.

Why would the reason be wrong? The reason for the aging difference is exactly the same in Einstein's 1918 paper as in standard resolutions. The only difference is he uses realistic acceleration (with Earth clock running fast in ship frame) instead of an instantaneous turnaround (earth clock "jumps ahead").

Why is the ship non-inertial?


Yes as to your second comment - it is like many of the standard solutions - those based upon resolving the problem via GR - Max Born and others who thought a reason for the age diffeence other than SR was needed to resolve the clock paradox - that is why after 1918, some authors of repute began saying flatout the problem can only be solved with GR - that is properly refuted by most others - but it all started with the the subtrifuge introduced by the 1918 paper.

 

[yogi]

OK, but in the GR case where you're dealing with non-inertial frames either way, depending on the frame you choose it may not be the twin that "travels with respect to space" who ages less, it could be the other twin.

So, if we hang the satellite clock on a sky hook, and call it a coordinate frame - then rotate the Earth beneath at the speed necessary to eliminate the affects of gravity acting upon the clock at the top of the tower, then it is the tower clock that does the traveling wrt the fixed frame of the satellite clock - so the two clocks are no longer in sync when they meet after one revolution - I guess that is what you are saying - or did I miss your point.

 

[Al68]

I agree that if the problem was presented and explained as two one way trips, most of the confusion would be completely eliminated. But the ship is still non-inertial for an (entire) one way trip.

Why would the reason be wrong? The reason for the aging difference is exactly the same in Einstein's 1918 paper as in standard resolutions. The only difference is he uses realistic acceleration (with Earth clock running fast in ship frame) instead of an instantaneous turnaround (earth clock "jumps ahead").

Why is the ship non-inertial?

Because the ship can't get half way through the twins paradox scenario without accelerating. At the half way point, the ship is at rest with earth.

And if we make that the end of a one-way trip, the ship twin is younger than the Earth twin unambiguously, since both are at rest in the same inertial frame.

As far as Einstein's 1918 paper, he never claimed GR was necessary to resolve the twins paradox. He just showed that it could also be analyzed from the accelerated frame of the ship with the same result for the same reason as the standard SR resolutions.

 

[yogi]

Because the ship can't get half way through the twins paradox scenario without accelerating. At the half way point, the ship is at rest with earth.

And if we make that the end of a one-way trip, the ship twin is younger than the Earth twin unambiguously, since both are at rest in the same inertial frame.

As far as Einstein's 1918 paper, he never claimed GR was necessary to resolve the twins paradox. He just showed that it could also be analyzed from the accelerated frame of the ship with the same result for the same reason as the standard SR resolutions.

Although Einstein started with two clocks at rest in the same frame in his 1905 description -it is not necessary. The one way trip I always thought we had in mind was that of a twin initially accelerated to crusing speed thereafter passing the stay put twin at a close distance at which instant both clocks are set to zero, and the voyage begins. Same thing upon reaching the target -the flyby twin notes the time on a local clock (which we can stipulate to be in the fame of the stay put twin) and the observer at the target notes the time on the flyby clock. This eliminates any acceleration - the voyage from stay-put twin to destination is inertial all the way.

I will agree that Einstein didn't promulgate the 1918 paper as exclusive - but others have taken the position that only GR is the correct reasoning - but this may be a subjective bias of mine - perhaps the issue is at some level, simply a distinction without a difference. - GR deals with potentiall energy, SR with KE differences between moving frames - and we know one can be derived from the other - so maybe I will stop campaigning this issue

 

[stevmg]

JesseM gave an example using SR why the "moving" twin ages less quickly than the earthbound twin. He does it both ways - using the Earth as the inertial frame and then the spaceship as the inertial frame. In both ways the Earth twin aged more than the traveling twin. Here is the post:

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

I repeat, here is JesseM's post. It is short, sweet and simple.

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

The GPS satellites that were sent up thirty years ago confirm this.

 

[JDługosz]

If you have two circular orbiting satellites at the same height traveling in opposite directions and set both clocks to zero as they pass, when they meet again, both clocks will read the same

So I would think. The question is why? Each sees the other running slow as they pass (a SR effect). How does GR resolve this?

 

[Mike_Fontenot]

The one way trip I always thought we had in mind was that of a twin initially accelerated to crusing speed thereafter passing the stay put twin at a close distance at which instant both clocks are set to zero, and the voyage begins. Same thing upon reaching the target -the flyby twin notes the time on a local clock (which we can stipulate to be in the frame of the stay put twin) and the observer at the target notes the time on the flyby clock. This eliminates any acceleration - the voyage from stay-put twin to destination is inertial all the way.

You didn't say what you thought the two twins would conclude about the home twin's age at that instant when the traveler flys by the target (without changing his speed).

The answer (which perhaps you already know) is that the traveler will conclude that the home twin is younger at that instant, whereas the home twin will conclude that she (the home twin) is older that the traveler at that instant. I.e., you just get the simple time-dilation result in that case, where each twin concludes that the other twin is younger...they disagree about their corresponding ages.

And the target inertial observer (at rest relative to the home twin) will agree with the home twin. Suppose the target observer happens to be the same age as the home twin, according to both of THEM. Then the traveler WON'T regard the target observer and the home twin to be the same age.

This example can be generalized in a very enlightening way. Suppose that, when the traveler is flying by the target, that there just happen to be lots of inertial observers passing the target at that same instant...with those observers all having different constant velocities wrt the home twin. I.e., thay are all momentarily co-located at the target at that instant, but they are all moving at different constant velocities wrt one-another. In that case, those inertial observers will all come to DIFFERENT conclusions about the home twin's age at that instant.

Equivalently, you can imagine that the traveler himself, while momentarily located at the target, repeatedly makes a sequence of instantaneous velocity changes. But suppose he doesn't maintain any of those velocities long enough for his age, or for the separation between the two twins, to change. I.e., he is doing a bunch of velocity changes, but they are all packed into essentially a negligible total amount of time. (It is also necessary to say that the essentially constant separation referred to above is the separation ACCORDING TO THE HOME TWIN, at the instant when the traveler is doing all his instantaneous speed changes (whereas the traveler will conclude that their separation is instantaneously changing whenever he instantaneously changes his velocity, and that's NOT what we want to use in the CADO equation)).

In the above scenario, the traveler will be constantly CHANGING his conclusion about the home twin's current age. This is trivial to see from the structure of the CADO equation:

CADO_T = CADO_H - L*v,

where for brevity I have not shown the dependence of all the quantities in this equation on the traveler's age t, and I have omitted the factor of c*c needed in the second term on the RHS for dimensional consistency (so we must use the equation as written only with units where c = 1). Recall that, in this equation, the separation L is taken as positive, and v is taken as positive when the twins' separation is increasing.

In that equation, at the instant when the traveler is doing all his repeated instantaneous velocity changes at the target, the ONLY quantity on the RHS that changes is the velocity v. The first term on the RHS is the current age of the home twin ACCORDING TO THE HOME TWIN, at the instant of all those velocity changes by the traveler. Since the traveler is making the entire sequence of all these velocity changes in essentially zero time, the home twin will conclude that neither of them is ageing during that entire sequence of velocity changes. Likewise, the home twin will conclude that their separation doesn't change at all during the entire sequence of those velocity changes, because the total time which elapses during the entire sequence of velocity changes is infinitesimal.

So it's easy to see from the equation that the quantity on the LHS (which is the current age of the home twin, ACCORDING TO THE TRAVELER), will be going through a sequence of instantaneous changes. And it is clear that the larger the separation L is, the larger those swings in the age of the home twin will be. [ADDENDUM2]: Since the velocity v must be in the range -1 < v < 1, the CADO equation shows that the current age of the home twin, according to the traveler, can suddenly change by up to 2L years. So, for example, if the twins are 20 lightyears apart when the traveler instantaneously changes his velocity, the home twin's age can change by up to 40 years. [END ADDENDUM2]

And, as already pointed out earlier in this thread, note that any time the velocity is suddenly increased (made more positive, or less negative), the age of the home twin will suddenly DECREASE, ACCORDING TO THE TRAVELER.

With finite accelerations, the behavior is qualitatively similar, if the separation is large enough. I give a numerical example on my previously referenced webpage that provides a fairly dramatic illustration.

Mike Fontenot

 

[Al68]

Although Einstein started with two clocks at rest in the same frame in his 1905 description -it is not necessary. The one way trip I always thought we had in mind was that of a twin initially accelerated to crusing speed thereafter passing the stay put twin at a close distance at which instant both clocks are set to zero, and the voyage begins. Same thing upon reaching the target -the flyby twin notes the time on a local clock (which we can stipulate to be in the fame of the stay put twin) and the observer at the target notes the time on the flyby clock. This eliminates any acceleration - the voyage from stay-put twin to destination is inertial all the way.

Of course that works, but with your one way trip, the ship twin ages less between "events" only because you arbitrarily chose Earth's frame as the one to measure proper distance in by defining the "target" as a location at rest with earth. That's just asking for the objection: "Why don't we look at it from the ship's frame"? If you instead chose the ship's frame to measure proper distance in, for example by defining the target as a hypothetical object (buoy) trailing (at rest with) the ship, you would get the opposite result. And both twins would then agree that less time elapsed for the Earth twin between "events", if the second event is the buoy reaching earth.

So for an inertial trip, time dilation is symmetrical, and the symmetry is only broken by arbitrarily picking a frame for the proper distance between events to be measured in. That's perfectly valid, but understandably inadequate as an explanation to many.

But if the twins are at rest with each other at the end of the trip, the choice of frame to measure the proper distance in between events isn't arbitrary, so the solution is more satisfying to many.

 

[matheinste]

There are of course many alternative scenarios for this problem, many of them are aimed at removing acceleration so that it is not considered to be the cause of the differential ageing. However, they are open to the criricism of being a little complicated and not really making the point.

They often boil down to A passes B and sets clocks. A meets C and reads clocks, C meets B and compares clocks. This is often interpreted as, two people who have never met before, meet, and compare ages. I know it is a little more complicated than that and does make a point, but that is how it looks at first sight. Uninteresting.

The original, A leaves B when they are the same age. A returns to B, their ages are compared and are found to differ from each other. The point being that they are the SAME objects, be it clocks or twins or whatever, at parting and reuniting. Simple and highly thought provoking as shown by the continuing number of threads on the subject although the resolutions to the problem are well known.

Matheinste.

 

[yogi]

You didn't say what you thought the two twins would conclude about the home twin's age at that instant when the traveler flys by the target (without changing his speed).

The answer (which perhaps you already know) is that the traveler will conclude that the home twin is younger at that instant, whereas the home twin will conclude that she (the home twin) is older that the traveler at that instant. I.e., you just get the simple time-dilation result in that case, where each twin concludes that the other twin is younger...they disagree about their corresponding ages.

Mike Fontenot

No that is not the result to expect - The distance should be measured in the proper frame defined by the separation between the Earth and target - there is no motion between these two clocks and there is no time difference between them - The traveler has moved a distance relative to this length in the Earth target frame - the only spacetime distance traveled by the stay at home twin consists of a temporal increment - so 3 of the factors for the two spacetime points are known, namely the fixed twins time read by the target clock as the traveler flies over: the distance traveled by the stay put twin (which equals zero), and the proper distance traveled by the traveling twin (the separation between the start and target in the earth-target frame). All that is left to calculate is the time lapsed on the traveling twins clock - it will always be less than the reading of the stay put twin's clock and the target clock - there is no ambiguity as to which twin aged the most - this is a simple application of the principle of interval invariance.

Einstein got the same result by sync both clocks in the same frame - then accelerating one clock to crusing speed until it reached the other clock (first example in part 4 of the 1905 paper under peculiar results) All that has been done is to start and stop the travelers clock on the fly - no new physics, and no ambiguity

The situation is different if the circumstances are different - two clocks passing each other at relalive velocity v will always measure the other clock to be running slow without more information- here you have a third clock that defines that defines a proper length - and when that bases is used as the length the time difference is real, not apparent. To carry it further - once we have calculated the traveling twins time, ithen you plug back into the equation and you get the apparent distance that the traveler beleives he has traveled to the target -

A similar analogy is the hi speed laboratory generated pion that reaches top speed in a fraction of inch - then continues unabated until disintegration - start the lab clock a fraction of an instant after emission - the pion travels a few feet and disintegrates - the trip as measured from the time the pion reaches crusing speed to disintegration is a one way inertial voyage - all the clocks in the lab read the same - there is no issue about whether the pion aged less than the lab personnel

 

[yogi]

Of course that works, but with your one way trip, the ship twin ages less between "events" only because you arbitrarily chose Earth's frame as the one to measure proper distance in by defining the "target" as a location at rest with earth. That's just asking for the objection: "Why don't we look at it from the ship's frame"? If you instead chose the ship's frame to measure proper distance in, for example by defining the target as a hypothetical object (buoy) trailing (at rest with) the ship, you would get the opposite result. And both twins would then agree that less time elapsed for the Earth twin between "events", if the second event is the buoy reaching earth.

So for an inertial trip, time dilation is symmetrical, and the symmetry is only broken by arbitrarily picking a frame for the proper distance between events to be measured in. That's perfectly valid, but understandably inadequate as an explanation to many.

But if the twins are at rest with each other at the end of the trip, the choice of frame to measure the proper distance in between events isn't arbitrary, so the solution is more satisfying to many.

You can look at it from the traveling twins frame - but the target is moving toward the traveling twins clock so the traveling twin's measure of distance will not be a proper one because the target is not fixed - so immediately, the traveling twin views the distance shorter and therefore since the relative velocity is v, he will necessarily conclude that his clock logged less time because t = d/v and since his measurement of the shorter distance, is in his own frame he will also measure a shorter time - so again we know which twin aged the most and there is no ambiguity - both twins agree on the one way trip just as they agree upon the round trip

 

[yogi]

There are of course many alternative scenarios for this problem, many of them are aimed at removing acceleration so that it is not considered to be the cause of the differential ageing. However, they are open to the criricism of being a little complicated and not really making the point.

They often boil down to A passes B and sets clocks. A meets C and reads clocks, C meets B and compares clocks. This is often interpreted as, two people who have never met before, meet, and compare ages. I know it is a little more complicated than that and does make a point, but that is how it looks at first sight. Uninteresting.

The original, A leaves B when they are the same age. A returns to B, their ages are compared and are found to differ from each other. The point being that they are the SAME objects, be it clocks or twins or whatever, at parting and reuniting. Simple and highly thought provoking as shown by the continuing number of threads on the subject although the resolutions to the problem are well known.

Matheinste.

Some good observations

My idea behind proposing experiments that are "acceleration-free" is to eliminate a factor which is not excluded in Einstein's original description which involved statically sychronized clocks. To put one clock in motion , some acceleration is involved and therefore the question as to the influence of acceleration on clocks lingered. As to whether it played a part in the process - I am convinced it does not - although this has not always been my position - but while all accepted solutions give the same result - they are sometimes not satisfying if one is really trying to get a physical picture of what is happening - Its easy to be left perplexed as to "where the time has gone" and how does the clock that runs fast wind up that way without some physics at play - Lorentz worried most of his life about the problem - conjurred up many theoretical reasons that would bring about physical results, but in the end they do not seem to meet our modern understanding For my own peace of mind, i found an answer that was useful for me ... using only Minkowsky unification, i.e, the invariance of the spacetime interval. I now no longer think of clocks running slow in other frames - rather I view the spacetime distance (interval) between two points in spacetime is always the same - it then turns out to be a simple matter to decide the coordinates of the endpoints of the interval in each frame - in other words, the clock can be thought of as not running slow per se, but rather not running for as long a distance in time because it has to do some of its running in the space direction.

 

[Aaron_Shaw]

But there's no such things as slowing clocks, contracting metersticks, or planets jumping forward in time. SR is about relations of objects, not changes happening to them.

Does anyone have any references i can read through regarding the above statement? I was under the impression that time dilation and length contractions etc were REAL physical phenomena, rather than illusions or optical effects?

Thanks.

 

[Ich]

I was under the impression that time dilation and length contractions etc were REAL physical phenomena, rather than illusions or optical effects?

I'm not saying they are illusions. The classification as "real physical phenomena" is misleading as well.
In a spacetime diagram, a meterstick is two-dimensional, not one-dimensional. Rather than "the length" changing with velocity, you're talking about different lengths derived from an unchanged two-dimensional object.

"Spacetime Physics" by Taylor & Wheeler should be good for a start.
Or try "Geometry of the Theory of Relativity" from http://www.itp.uni-hannover.de/~dragon/".

 

[yuiop]

... But there's no such things as slowing clocks, contracting metersticks, or planets jumping forward in time. SR is about relations of objects, not changes happening to them...

As Galileo might have muttered if he was around when relativity was introduced "... and yet one twin biologically ages less than other."

That is a direct contradiction to teaching that "there's no such things as slowing clocks" and bound to be confusing to students. If it was carefully explained that clocks with relative motion really do physically run at different rates, but we can not tell which clock is really running slower until both clocks are brought to rest with respect to each other, then I think there would be a lot less confusion.

 

[matheinste]

As Galileo might have muttered if he was around when relativity was introduced "... and yet one twin biologically ages less than other."

That is a direct contradiction to teaching that "there's no such things as slowing clocks" and bound to be confusing to students. If it was carefully explained that clocks with relative motion really do physically run at different rates, but we can not tell which clock is really running slower until both clocks are brought to rest with respect to each other, then I think there would be a lot less confusion.

I believe that although saying that moving clocks run slow has its place in the teaching of SR and only causes confusion initially, it does not explain the reciprocal effect of time dilation very well. But anyway, differential ageing is not relly about time dilation.

The whole twin scenario is a victim of its own popular appeal. It is introduced in popular books to whet the reader's taste for relativity and make it interesting. After all, to non physicists, it is the counterintutive aspect of relativity which is its appeal. You do not need the twin scenario to explain the concept of differrent proper time being recorded along different spacetime paths, but once the twin sceanario is seen there is no escaping it.

As for clocks appearing to run at different rates for different observers, pehaps it is better to remove the emphasis from the clock and focus on the observer. If we say that for a given clock, the time it records is the projection of its spacetime path onto the time coordinate axis of the coordinate system in which the observer of the clock is at rest, it gives a better picture, is easier to visualise and does not use the mistaken and even meaningless concept of ideal clocks, similar by defintion, running physically at differtent rates. Unfortunately the spacetime geometry concepts are not always introduced at the necessarily early stage at which time dilation is introduced.


Matheinste.