The Twins Paradox: A Controversial Truth or a Perplexing Paradox?

[misogynisticfeminist]

I've got a question about the twin paradox myself.

What if I establish the frame S', anchored to the spaceship as the rest frame, then frame S, anchored to the Earth is moving away from me at a certain speed. According to my reference frame, the clocks in Frame S slow down, then why is it that I'm the younger one after the journey?

 

[Hurkyl]

I've got a question about the twin paradox myself.

What if I establish the frame S', anchored to the spaceship as the rest frame, then frame S, anchored to the Earth is moving away from me at a certain speed. According to my reference frame, the clocks in Frame S slow down, then why is it that I'm the younger one after the journey?

I assume the journey you mean takes you back to Earth? What you've stated is precisely the classic twin paradox. The mistake is that you assume S' is an inertial reference frame. In particular, clocks on Earth will be running very fast according to S' while you're turning around. (Notice I didn't say clocks in S: position is an important factor for this effect)

 

[misogynisticfeminist]

I assume the journey you mean takes you back to Earth? What you've stated is precisely the classic twin paradox. The mistake is that you assume S' is an inertial reference frame. In particular, clocks on Earth will be running very fast according to S' while you're turning around. (Notice I didn't say clocks in S: position is an important factor for this effect)

I'm actually new to SR, so I have to clarify lots of stuff. So what happens if S' is not an I.R.F, i thought that non I.R.Fs is only talked about in GR about the equivalence principle, what does it mean in SR? Also, why would the clocks in S be running very fast when I am turning around??

 

[Hurkyl]

i thought that non I.R.Fs is only talked about in GR about the equivalence principle, what does it mean in SR?

That's a common misconception. One of the basic principles of SR is that the laws of physics are the same in any inertial reference frame. Now, that doesn't mean that SR cannot handle noninertial reference frames, just that the laws of physics are different.

(One thing that makes GR special is that the laws of physics are the same in all reference frames)

A good example of the difference comes directly from classical mechanics: Coriolis and centrifugal forces.


Let's think about a spatial example for a moment. You get on a merry-go round and someone starts it spinning clockwise. Now, let's consider your noninertial rest frame. You observe things far in front of you moving rapidly to the left, and things far behind you moving rapidly to the right.

Accelerations are analogous to rotations. Clocks far in front of you (assuming you're facing the way you're accelerating) start ticking really fast, while clocks far behind you are running backwards, really fast.



If you aren't drawing space-time diagrams to get a geometrical picture, then another way of seeing this fact is through the Lorentz transforms. Accelerating a reference frame is equivalent to smoothly Lorentz transforming it. When you transform, clocks in the direction of the boost jump forward, and clocks in the other direction jump backwards.

 

[JesseM]

The Earth may or may not be a preferred frame - it is certainly not unless there is something about the G field that renders it special (LR theory). But in any event, when the non rotating Earth is taken as a reference frame for measurements, we know from expereince that clocks in flight around the Earth will run slower than a stationary clock at the same height at the North Pole. There is no special sync required to check the results - we bring the clocks together at the start and set them both to zero - then fly one around the Earth - there are accelerations in the take off and there are accelerations due to the curved path the flying clock experiences - we can predict almost exactly what the difference in the clock rate will be based entirely upon the relative velocity using the LT. The clock which is moving in the space defined by its circumferential path will run slower and the two clocks can be compared on each passby.

Yes, and both an observer orbiting with the clock and the clock on the Earth would agree that the orbiting clock is running slower, if they use the method of checking the time they received a radio signal and subtracting (distance from origin of signal)/(speed of light) from that time.

If you speed up the aircraft until it has orbit velocity, the flying clock will no longer experience acceleration (at least not a G field).

"acceleration" means either changing speed or changing direction, so an orbiting clock is certainly accelerating even if its speed is constant. And it will experience some tiny g-force due to this acceleration (the 'centrifugal force').

The experiment I outlined previously is but a linear version of the same thing - instead of having the clock return by flying a circumference, the initially accelerated clock (the one that corresponds to the flying clock) simply sends a radio signal back to the stationary clock.

No, the difference is that in the experiment you outlined, if each observer uses the method of checking the time they received a radio signal and subtracting (distance from origin of signal)/(speed of light) from that time, they will both conclude the other is running slow. So there is really no way to break the symmetry here and decide whose clock is "really" running slower.

Also, what do you mean by "stationary clock"? I thought you were not arguing for a preferred reference frame--don't make me bring Zeus into this again!

As to your Q re whether one should consider a past history of how two objects that meet in space should decide which one had previously accelerated - I would say this. SR ignors all the rest of the universe - so two spaceships meeting far from any other reference can properly use Einsteins original derivation so that each can say, when I observe the other guys clock it appears to run slow. The operative word here is "observe" Obviously both clocks cannot be running slower than the other.

No, that isn't obvious at all. If you lived in the 19th century, would you also have disagreed with "Galilean relativity" because different reference frames might disagree about which of two objects has a greater velocity, and "obviously both objects cannot be moving faster than the other"? Would you also say that if we have two cartesian coordinate systems, and in one system point A has a greater x-coordinate than point B while in the other B has a greater x-coordinate than A, there must be an objective truth about the "correct" place to put the origin because "obviously both A and B cannot have a greater x-coordinate than the other"? I don't see any problem with saying that the question of which clock runs faster is analogous to the question of which of two objects has a greater velocity or the question of which of two points in space has a greater x-coordinate, in that none of these questions need have any "objective" answer and can instead depend on an arbitrary choice of which coordinate system you want to use.

SR would make no distinction between whether the Earth is moving in every direction at once so as to sweep up high speed muons

I don't understand what you mean by this--there is no inertial reference frame where the Earth is "moving in every direction at once", each frame will say the Earth is moving in a single direction at any given time.

Which is the more likely proposition. Einstein derived the LT for a situation which was observational - a subjective interpretation of lengths and times in another reference frame - then, undaunted by the fact that there was never even the slightest attempt to justify their applicability to real time differences (different rates between two clocks), he proceeded to due just that. I have read his 1905 manuscript over many times seaching for something I must have missed - but ...

I don't understand what you're talking about when you say "justify their applicability to real time differences". By "real" do you mean that you think there should be some objective answer to the question of which of two inertial clocks is running slower? If so, see my comment above.

Now Einstien must be given great credit for his bold rejection of a universal time. He was also very intuitive - he realized that actual time difference occurs when a clock is carried on a path that returns to the starting point - long before we knew of muon and pion decays - or had hi speed aircraft to test the hypothesis. Certainly, by the time he published his 1912 manuscript, he had fully rejected the notion that acceleration had anything to do with the clock paradox. But he still didn't explain it.

Didn't explain what?

So in conclusion, while both observers are on an equal footing as far as making measurments in the other frame as to appearances, actual changes in clock rates can only be brought about by some physical cause.

What do you mean by "actual changes in clock rates"? What's the difference between an actual and a non-actual change?

All the observations of the other guys clock and all of his observations about your clock can't change a thing. To my way of the thinking, H&K experiments, muon decay, and GPS provide compelling evidence that clocks in motion relative to one another will accumulate different times whether or not they are ever returned to the same point for comparison. You get answers that conform with the experiments if you consider the Earth as fixed and the high speed clock moving between two points that define a proper distance in the Earth frame. If you consider the muon frame as fixed, it will last 2 usec - so in the muon frame, the Earth could only move 600 meters between the beginning and end of the experiment. And if that is the case - how much time has passed on the Earth clock as calculated in the muon frame during the 2 usec?

It depends on the relative velocity of the muon and the earth, but it would be less than 2 usec. However, If you have different clocks which are at rest relative to the Earth and which are "synchronized" in the Earth's frame, the muon will see these clocks as wildly out-of-sync, so if it departs the Earth when the earth-clock reads t=0 usec, the clock at its point of arrival will read a time much greater than t=2 usec when it arrives there, because in the muon's frame it was ahead from the beginning.

If you are content with these appearances and believe they should be given the designation of reality, so be it. I think the flaw in SR is the failure to take into account the inital conditions - who accelerated to bring about the relative velocity - not who turned around - because whatever time is lost going out will simply be doubled when added to the time lost on the inbound journey.

You never really addressed my question about how far back we in an object's history we should go to see if it has ever accelerated, you just went off on a tangent about problems you have with relativity. Anyway, if you believe there is an objective truth about which of two clocks ticks faster, this is incompatible with the idea that we should define things in terms of who accelerated most recently. Suppose we have a space station moving inertially, and a ship accelerates to take off from it--you can't say that this means the ship's clock is "objectively" running slower than the station's clock, because what if the space station accelerated to get away from the Earth at some point further in the past, and the ship now has a lower velocity in the Earth's frame (and thus is less slowed down in this frame) than the station?

 

[yogi]

Back again - as to the issue of which clock accelerates - let's take the simple example of two clocks A and B separted by a distance d. The clocks are in the same reference frame and brought into sync (e.g. by Einstein's method). Then clock A takes off in the direction of B (A accelerates quickly up to a velocity v in a time interval that is short compared to the time it takes to travel the distance d at velocity v - then travels the rest of the distance at a constant velocity). When A arrives at B, the readings are compared. Question for Jesse - do the clocks read the same, and if not, which clock has accumulated the greater time.

 

[Hurkyl]

Clock B will read more when A arrives, no matter how A gets there.

 

[JesseM]

Back again - as to the issue of which clock accelerates - let's take the simple example of two clocks A and B separted by a distance d. The clocks are in the same reference frame and brought into sync (e.g. by Einstein's method). Then clock A takes off in the direction of B (A accelerates quickly up to a velocity v in a time interval that is short compared to the time it takes to travel the distance d at velocity v - then travels the rest of the distance at a constant velocity). When A arrives at B, the readings are compared. Question for Jesse - do the clocks read the same, and if not, which clock has accumulated the greater time.

What Hurkyl said. But acceleration isn't really relevant, all that matters is that they were initially synchronized in a frame where A was moving and B was at rest. If they had been initially synchronized in a frame where A was at rest after it accelerated while B was traveling at velocity v, then A would have accumulated a greater time when they met, even though it was A who accelerated.

 

[yogi]

So if A and B are in the stationary frame initially (yes Jesse - I said stationary - same term as used by Einstein) - and after the sync operation is completed A accelerates to v and travels at velocity v until he reaches B. We all agree that A's clock will have accumulated less time. And I assume we all agree that acceleration does not have anything significant to due with the answer - it just tells us which clock is in motion with respect to the frame in which the two clocks were brought into sync (the frame I refer to as the stationary frame).

Now if we introduce at the outset a third clock D which is initally adjacent to A, and bring it into sync with A and B, then if D remains in the stationary frame (does not change its position wrt to B), D will read the same as B thereafter (B and D will remain in sync). So when A arrives at B, the A clock will read less than the D clock (The event of arrival occurs in both frames, but not at the same time in both frames).

Now if D is the clock owned by the stay at home twin, and A is the clock carried by the traveling twin - then the one way trip results in a time differential which can be evaluated w/o having to reunite the twins (A simply flashes a light signal back to D upon arrival at B, and since D knows the distance d between himself and B) he calculates the actual time loss experienced by A.

 

[Hurkyl]

D will read the same as B thereafter

No. Such a statement is nonsensical unless you specify a coordinate chart against which this is measured.

D will read the same as B according to the reference frame in which they're stationary.

According to other reference frames, D and B will not read the same.

So when A arrives at B, the A clock will read less than the D clock

The same objection applies to this statement.

In particular, according to the reference frame in which A is stationary during its trip, the A clock will read more than the D clock.

 

[JesseM]

Now if we introduce at the outset a third clock D which is initally adjacent to A, and bring it into sync with A and B, then if D remains in the stationary frame (does not change its position wrt to B), D will read the same as B thereafter (B and D will remain in sync).

As Hurkyl said, D only is synchronized with B in the rest frame of B and D, not in other frames.

Now if D is the clock owned by the stay at home twin, and A is the clock carried by the traveling twin - then the one way trip results in a time differential which can be evaluated w/o having to reunite the twins (A simply flashes a light signal back to D upon arrival at B, and since D knows the distance d between himself and B) he calculates the actual time loss experienced by A.

That's not the "actual" time loss, just the time loss in his frame. After all, B used the assumption that light travels at c relative to himself to calculate the time loss, but in other frames light does not actually travel at c relative to B.

 

[yogi]

As always - you both want to obscure the simplicity. So I will say it again: D and B remain in sync in the stationary frame. There is nothing to be added by diversionary comments to the effect that B and D will be out of sync if viewed by any number of other frames in motion with respect to the stationary frame. "A" measures time according to the moving frame. B and D measure the passage of time in the stationary frame. The event (A's arrival at B) occurs at the same spatial point in both frames). Jesse - The proper distance between B and D is d and a light signal sent from either A or B (upon the event of A's arrival at B) will take d/c seconds to arrive at D. How can it possibly be anything else if the stationary frame is an isotroptic inertial frame?

 

[JesseM]

As always - you both want to obscure the simplicity. So I will say it again: D and B remain in sync in the stationary frame.

Einstein only used the term "stationary" for the purposes of developing his argument--in section 1 of his 1905 paper he says:

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system.''

The point here is that which frame we choose to call the "stationary system" is arbitrary, he never used the words "stationary system" to mean the frame where the physical objects you're analyzing are at rest. So, we have no obligation to define it as the frame where A and B are initially at rest in this problem.

There is nothing to be added by diversionary comments to the effect that B and D will be out of sync if viewed by any number of other frames in motion with respect to the stationary frame.

OK, I am defining the "stationary frame" as the one where A and B are initially moving at 0.99999c in the +x direction. Please, let's not have any diversionary comments about what things might look like in any other frames.

Jesse - The proper distance between B and D is d and a light signal sent from either A or B (upon the event of A's arrival at B) will take d/c seconds to arrive at D. How can it possibly be anything else if the stationary frame is an isotroptic inertial frame?

Well, if D emits a light signal, then since B is moving towards it at 0.99999c in the stationary frame as I have chosen to define it, and the distance between them is only $$d/\gamma$$ in this frame, the time for the light signal to reach B will be far less than d/c.

 

[Chronos]

By whom's clock? That is just flat wrong.

 

[JesseM]

By whom's clock? That is just flat wrong.

By clocks at rest in the stationary frame, of course (which, remember, is the frame where B and D are traveling at 0.9999c in the +x direction). How else would one define the time between two events in a given frame?

 

[Chronos]

There is no stationary frame. B, D and my reference frame are SR time dilated and length contracted in the direction of motion.

 

[JesseM]

There is no stationary frame.

Read my previous post. I am just using "stationary" in the same way Einstein did in his paper (since yogi used this term earlier and justified it by saying Einstein had also used it):

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system.''

So, the "stationary" frame is simply an arbitrary inertial reference frame which we have chosen to label as stationary, in order to distinguish this frame verbally from others.

B, D and my reference frame are SR time dilated and length contracted in the direction of motion.

Motion in whose frame? Obviously your reference frame is not moving with respect to itself.

 

[gonzo]

It seems to me like a lot of people are arguing in circles or just for the sake of arguing. Trying to find one sentence out of context in someone else's post and finding some way of looking at it that you can say is wrong.

To put it in Dr. Phil terms, are you all interested in being right or actually having some issue explained?

It's gotten to the point where I can even tell what the discussion is about anymore.

It seems to have started with something to do with the "Twins Paradox", but I can't really tell anymore.

Perhaps it would be more productive if someone tried to summarize the finer points of what the actual disagreement is here. Unless of course you all really like this type of arguing in circles, in which case I'll leave you to it with my appologies for interrupting.

 

[Hurkyl]

As always - you both want to obscure the simplicity.

Simplicity has no merit if its wrong.


As we've said over and over again, statements like:

D will read the same as B thereafter

have no meaning, except relative to a coordinate chart. Yet, for some mysterious reason, you keep saying them over and over and over...


Since you ardently reject such a correction, we are all justified in our conclusion that you are implicitly assuming an absolute notion of time. (And, thus, implicitly rejecting the relativistic notion of space-time)


We keep raising our objection in the futile hope that you'll eventually see the point of our objections.

 

[yogi]

Hurkyl - If B and D are initally brought in sync they will remain so in what I have called the stationary frame so long as they are not moved wrt to each other. If you believe something different - that is your privilege - if not, stop harping on it.

If you read what I have said - it has nothing to do with universal time - it does have to do with clocks in the same frame keeping the same time. How can you keep misconstruing this?

For those who feel this thread has wandered to far to be meaningful - I will attempt to restate the point of concern which is the root of my proposed thought experiment. Einstein in the first part of his 1905 paper derives the LT based upon how one observer views space and time in a frame that moves at velocity v wrt to his frame. And because neither can claim a preferred frame the situation is reciprocal. Then, in section 4 of that paper he draws conclusions about the physical meaning of the equations that were derived from apparent observations - specifically the exact situation posed by the clocks A and B which I have associated with twins... he states: "If at points A and B there are stationary clocks which viewed in the stationary system, are synchronous, and if clock A is moved with the velocity v along the line AB to B then on its arrival the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)t(v/c)^2 ..."
If this is correct, which we assume it is, then it should also be true that a third clock D that remained at the point where A was initially at rest in the stationary frame, can be brought into sync with B and should read the same as the B clock thereafter. Ergo, if B and D read the same in the stationary frame, D can access his brothers age upon receipt of a light signal sent by ether A or B (upon A's arrrival at B) w/o having A undergo turn around.

 

[JesseM]

Ergo, if B and D read the same in the stationary frame, D can access his brothers age upon receipt of a light signal sent by ether A or B (upon A's arrrival at B) w/o having A undergo turn around.

yogi, no one would disagree with you if you didn't use absolute terminology like "his brother's age" with no qualifiers. For example, if you said "D can access his brothers age in D's own frame upon receipt of a light signal sent by ether A or B (upon A's arrrival at B) w/o having A undergo turn around" then of course this is correct. But there is no absolute truth about what B's age is at the moment that D receives the light signal, because there is no preferred definition of simultaneity. Do you agree that any statement about their relative ages that doesn't explain which frame is being used does not provide the reader with enough information to evaluate whether it's true or false?

 

[gonzo]

Okay, Yogi, I understand what you are saying, but I don't understand what is controversial about that, aside from everyone feeling the need to nit-pick how they feel you worded things.

As far as I can tell you are just talking about the general idea of a latice work of synchronized clocks in an inertial frame, which is how SR is generally talked about.

So, what's the conflict?

 

[Hurkyl]

Hurkyl - If B and D are initally brought in sync they will remain so in what I have called the stationary frame so long as they are not moved wrt to each other. If you believe something different - that is your privilege - if not, stop harping on it.

This statement is fine, I think. You did intend "in what I have called the stationary frame" to apply to both "B and D are brought in sync" and "they will remain in sync", right?

If you read what I have said - it has nothing to do with universal time

You usually make statements like "B and D are in sync", and not statements like "B and D are in sync relative to the frame in which they're stationary".

The former only makes sense if you have an absolute notion of time. We infer that you are implicitly assuming universal time because you ardently reject our criticisms that you aren't specifying to which frame your statements are relativie.

 

[JesseM]

Okay, Yogi, I understand what you are saying, but I don't understand what is controversial about that, aside from everyone feeling the need to nit-pick how they feel you worded things.

It's not just about wording--although he said "the situation is reciprocal" in his last post, in earlier post he was never willing to admit that it is just as valid to say that in the traveling twin's frame, it is the other twin who has aged less at the moment he reaches his destination. If the situation is really symmetrical, there is no "paradox" here, because each can say the other ages less without conflict, it's only when the two twins can meet and compare ages that you might conclude there was a paradox because each should conclude the other will be younger when they meet, but only one can be right. But of course, for them to meet one has to turn around, so the situation is not actually symmetrical, unlike in yogi's example where it really is symmetrical but the two twins never meet. So, you can't really have a "twin paradox" in a situation like that where both are moving at constant velocity the whole time.

Yogi seems to understand the mathematical idea that the situation is symmetrical in each frame, but conceptually he seems to want to hold onto the idea that there is some sort of absolute truth about who has aged less. For example, in this post he says:

SR ignors all the rest of the universe - so two spaceships meeting far from any other reference can properly use Einsteins original derivation so that each can say, when I observe the other guys clock it appears to run slow. The operative word here is "observe" Obviously both clocks cannot be running slower than the other.

...Einstein derived the LT for a situation which was observational - a subjective interpretation of lengths and times in another reference frame - then, undaunted by the fact that there was never even the slightest attempt to justify their applicability to real time differences (different rates between two clocks), he proceeded to due just that. I have read his 1905 manuscript over many times seaching for something I must have missed...

...So in conclusion, while both observers are on an equal footing as far as making measurments in the other frame as to appearances, actual changes in clock rates can only be brought about by some physical cause. All the observations of the other guys clock and all of his observations about your clock can't change a thing.

gonzo, would you agree it's obvious that both clocks cannot be running slower than the other? Would you agree that the slowdown measured in different frames is just a matter of "appearances", to be contrasted with "actual changes in clock rates" which requires a "physical cause"? Do you agree that Einstein's paper is missing something because it didn't justify the applicability of "a subjective interpretation of lengths and times in another reference frame" to "real time differences" like the objective difference in ages seen between two twins who reunite after one has made an interstellar trip?

 

[gonzo]

As far as I understand it, the twins do not have to meet up again, and there is still no paradox. Is this is the debate? Here is how I understand it, and why I don't think it matters whether or not the twin turns around. I've lost all track of who is what letter, so I'll start from the beginning and see if I cover the issues correctly.

Okay, you have two Twins, Travel Twin (TT) and Home Twin (HT) for ease. They start together at START PLACE (SP) and TT then travels to END PLACE (EP).

Now, we assume the SP and EP are in the same inertial frame. We further imagine a latice of synchronized clocks throughout the whole of this inertial frame. This is a common practice when talking about SR.

As TT zooms out from SP to EP in his Rocket Ship Frame (RSF) he looks out his windows at all these clocks in the SP-EP frame and thinks they are all running slow (show less elapsed time). At each of these clocks we have a little green man in the SP-EP frame who looks in the window of the passing ship and thinks that TT's RSF clocks are running slow.

This is the first apparent problem, but this is easily resolved by issues of simultaneity.

Now, then the apparent problem arises when TT reaches EP. Right when he gets there, there seems to be a conflict. TT looks at the clock on EP and sees not much time has elapsed on that clock. While the little green man on EP looks in the window and sees that it is actually TT RSF clock that shows not much time passing.

So, then you ask what happens when TT stops, who's clock is right?

But this isn't a real symetry because TT is the one changing frames, whether he goes back, or just stops somewhere else in the SP-EP frame. He changes from his RSF frame to the SP-EP frame.

This has a weird effect. If you draw spacetime maps of the frames and draw lines of simultaneity you can see it more clearly.

When TT is still in RSF he sees the EP clock no showing much time. But we know at the end the EP clock is supposed to show a lot longer time. There are all these missing clock ticks you could say.

What happens, as I understand it, is that while TT is still in the RSF those missing clock ticks are in his future. When he changes frames back to the EP-SP frame, those clock ticks shift into his past. Easier to see on a spacetime diagram I think.

Almost like the clocks jumped ahead when he stopped, making up for lost time you could almost say.

Of course, I could be wrong in my understanding of the situation since I'm just a dabbling amateur.

 

[gonzo]

I wanted to add few comments. First I was taking SP as the origin in all inertial frames of interest.

Also, I think it is important to point out that if we imagine this latice works of synchronized clocks in one frame (all frames can have their own latice work), these clocks will not appear synchronized in another inertial frame.

In other words, we are imagining a laticework of synchronized clocks in the EP-SP frame to check time. But these clocks do NOT appear synchronized in the RSF frame. That seems an important point to mention.

 

[JesseM]

As far as I understand it, the twins do not have to meet up again, and there is still no paradox. Is this is the debate? Here is how I understand it, and why I don't think it matters whether or not the twin turns around. I've lost all track of who is what letter, so I'll start from the beginning and see if I cover the issues correctly.

Okay, you have two Twins, Travel Twin (TT) and Home Twin (HT) for ease. They start together at START PLACE (SP) and TT then travels to END PLACE (EP).

Now, we assume the SP and EP are in the same inertial frame. We further imagine a latice of synchronized clocks throughout the whole of this inertial frame. This is a common practice when talking about SR.

As TT zooms out from SP to EP in his Rocket Ship Frame (RSF) he looks out his windows at all these clocks in the SP-EP frame and thinks they are all running slow (show less elapsed time). At each of these clocks we have a little green man in the SP-EP frame who looks in the window of the passing ship and thinks that TT's RSF clocks are running slow.

This is the first apparent problem, but this is easily resolved by issues of simultaneity.

Now, then the apparent problem arises when TT reaches EP. Right when he gets there, there seems to be a conflict. TT looks at the clock on EP and sees not much time has elapsed on that clock. While the little green man on EP looks in the window and sees that it is actually TT RSF clock that shows not much time passing.

So, then you ask what happens when TT stops, who's clock is right?

There's no need to assume the TT stops, he can just whiz by EP, and as he passes arbitrarily close to the clock there, they can compare readings--both frames must agree on what each clock reads at the moment they pass, since different frames can't give different predictions about local events (if they could, different frames might disagree on whether two moving objects would collide or miss each other, for example). The key to resolving this is to notice that from TT's point of view, all these clocks at rest in the SP-EP frame are out of sync, so that although he does see each clock individually running slow, he also sees that each successive clock he passes is ahead of the previous one. So, if both his clock and the clock at SP read the same time at the moment he departs, then at the "same moment" in his frame, the clock at EP started out far ahead of the clock at SP, so despite the fact that the clock at EP was running slow throughout the trip in his frame, it still makes sense that the clock at EP is ahead of his own at the moment they pass.

If he stops at EP, then you're right that this will change his definition of simultaneity, but it will only be distant clocks whose time suddenly jumps forwards or backwards, the local clock at EP won't change if his acceleration is instantaneous.

 

[gonzo]

Sure, you are correct, I was a bit unclear. I was thinking of elapsed time. The rocket frame reads less elapsed time on the EP clock, however, as you pointed out, when he starts at SP in the rocket frame the clocks are not synchronized the clock at EP is already in the future, so when he gets there it all works out nicely so that the clock at EP reads the total time as seen from SP to EP in the EP frame.

The thing to keep in mind is that the problem is meaningless unless there is some frame changing going on somewhere. So when TT shifts frame from SP to RSF then all the SP-EP clocks jump out of synch, or anothe way of looking at it is that a lot of clock ticks that were in his future are suddenly in his past.

 

[JesseM]

The thing to keep in mind is that the problem is meaningless unless there is some frame changing going on somewhere. So when TT shifts frame from SP to RSF then all the SP-EP clocks jump out of synch, or anothe way of looking at it is that a lot of clock ticks that were in his future are suddenly in his past.

Why is it meaningless? Even if the TT flies past the EP without changing velocities, so that both twins are moving inertially throughout the entire problem, we can still ask questions like how the different clocks look in each twin's frame, or why each frame gets the same prediction for the time on the TT's clock and the time on the EP-clock at the moment they pass next to each other.

 

[yogi]

Perhaps the disagreement between our viewpoints is at least partially semantic rather than substantive - I brought up this caper with the idea of taking it a step further - As Jesse correctly surmises, and as I have more than hinted at, I feel that Einstein's great contribution was not the rederivation of the LT, but his interpretation of the meaning to be given to time as belonging to a particular frame. For those who have read different texts on the subject of the twin (or clock paradox) it is apparent there are different views - basically these views are divided into two camps - the dynamic and the kinematic. So to try to make sense out of things in the light of the fact that experimental evidence supports Einstein's physical argument, I used his example - but what is not obvious from his physical example is how he arrived at it from the concept that two passing clock frames will each measure time in the other frame as running slow. In other words, there is a jump from observation to the reality of a physical age difference between the two frames in the situation set forth...For example, we could install two clocks separated by a distance L on A's frame and measure the rate of a single clock at rest in the D-B frame (what I called the stationary frame) - and the single passing clock will always be measured to be running slow. More later -

 

[yogi]

So how does Einstein transition from observational appearances (which is what is required for reciprocity) to a statement about real age difference between two frames? - it seems that when the problem is set with Einsteins's initial conditions imposed in a frame in which the two events are separated by proper length of space, then the reality of age difference must follow irrespective of the fact that in the moving frame clocks and lengths can be constructed to show (measure) that a single passing clock in the stationary frame will appear to be running slower than the clocks in the moving A frame. It would seem then that the real age difference between the stationary frame and the moving frame is not dependent upon turn around but upon the fact that the experiment is non-symmetrical at the outset - that is, the interval as measured in the frame of A during motion comprises only time, whereas the interval as measured in the stationary frame comprises both length and time components. And while the interval is invarient vis a vis the two frames, the components of the interval in each frame are not equal.

 

[JesseM]

For those who have read different texts on the subject of the twin (or clock paradox) it is apparent there are different views - basically these views are divided into two camps - the dynamic and the kinematic.

I have never heard of a dynamic vs. kinematic view of the twin paradox--can you explain what you mean by these terms?

So to try to make sense out of things in the light of the fact that experimental evidence supports Einstein's physical argument, I used his example - but what is not obvious from his physical example is how he arrived at it from the concept that two passing clock frames will each measure time in the other frame as running slow. In other words, there is a jump from observation to the reality of a physical age difference between the two frames in the situation set forth

What is the difference between an "observation" that one twin is aging slowly vs. "the reality of a physical age difference"? Does "the reality of a physical age difference" refer to the two twins meeting up again and comparing ages in a single location, in which case every frame will agree on what their respective ages are?

...For example, we could install two clocks separated by a distance L on A's frame and measure the rate of a single clock at rest in the D-B frame (what I called the stationary frame) - and the single passing clock will always be measured to be running slow.

Sure. Would you call this a mere "observation", or would you say that the slowness of the passing clock as measured by these two clocks in A's frame is a physical reality?

So how does Einstein transition from observational appearances (which is what is required for reciprocity) to a statement about real age difference between two frames?

Again, I don't understand what distinction you're making here. If I have two clocks A and B which I have synchronized using Einstein's method, and a clock C is passing by them both, then the statements "at the moment A and C passed next to each other, A read 12:00 and C read 12:00" and "at the moment B and C passed next to each other, B read 1:00 and C read 12:30" are both statements about objective physical events which are true in every frame...but to jump from that to "C was ticking at half the rate of A and B", you have to assume that A and B were synchronized, which is only true in the AB rest frame. So would you say the first two statements are physical truths while the third statement is just about observational appearances? It's nevertheless true that if you impose the condition that every frame must see moving clocks slow down according to measurements of the times they passed clocks at rest in that frame which were synchronized according to Einstein's rule, it logically follows that if a clock travels away and then returns to another clock which doesn't accelerate, then the clock that changed velocities must have accumulated less time when they reunite. There is no way for the first idea to be true in every frame while the second idea is false.

it seems that when the problem is set with Einsteins's initial conditions imposed in a frame in which the two events are separated by proper length of space, then the reality of age difference must follow irrespective of the fact that in the moving frame clocks and lengths can be constructed to show (measure) that a single passing clock in the stationary frame will appear to be running slower than the clocks in the moving A frame.

I don't understand what "reality of age difference" is supposed to mean, if you don't actually reunite the clocks so that every frame must agree on what each one reads when they reunite. As long as they are at different locations, there can be no single objective reality about which clock has elapsed less time, since there is no objective reality about simultaneity.

It would seem then that the real age difference between the stationary frame and the moving frame is not dependent upon turn around but upon the fact that the experiment is non-symmetrical at the outset - that is, the interval as measured in the frame of A during motion comprises only time, whereas the interval as measured in the stationary frame comprises both length and time components.

The interval of what? You mean the path from taken by A? If so, that's not what I meant by "symmetrical"--it's symmetrical in the sense that the path taken by B as seen in frame A looks exactly like the path taken by A in frame B. In frame A, A's path comprises only time while B's path comproses both length and time components, while in frame B, B's path comprises only time while A's path comprises both length and time components. The symmetry here refers to the fact that if you exchange the labels of A and B, and you exchange +x for -x, then the situation is precisely identical in both frames.

 

[yogi]

No Jesse - that is point - the situation cannot be identical in both frames because of the initial conditions - if it were the comparison of clocks - as in the Einstein experiment, would never yield a different age for one twin than the other (clocks A and B read differently when A arrives at B). Einstein imposed an initial condition in which A and B were separated by a distance d in the stationary frame - and then A moves along the line AB - so the spatial interval is measured in the stationary frame of B. Since the spacetime interval is invariant, clocks in the stationary frame must intrinsically accumulate more time in order to offset the spatial distance. What I am saying, the age difference between the clocks when they are compared is not due to acceleration or turn around, or changing frames per se, but rather the reality of the time difference as between A and B is consequent to A's motion along a spatial path defined in B's inertial frame. Of course, we could have B move toward A instead of vice versa, in which case the spatial interval would be measured in A's frame, and the A clock would register more time when B arrived (if that is what you mean by symmetry - then yes, there is no difference since neither frame has any property which would render it preferred). But once you decide upon which one moves, you immediately create the asymmetry that leads to differential aging

 

[JesseM]

No Jesse - that is point - the situation cannot be identical in both frames because of the initial conditions

What initial conditions? What situation are we talking about? Isn't it the situation where the two twins are moving apart at constant velocity, without either one turning around? If so, this situation is indeed completely symmetrical.

- if it were the comparison of clocks - as in the Einstein experiment, would never yield a different age for one twin than the other (clocks A and B read differently when A arrives at B).

So now you are talking about a situation where A turns around and returns to B so they can compare clocks at a single point in space? You have to be specific about what scenario you're talking about when you switch from one to another. I agree that if one turns around, he will have aged less; if the traveling twin turns around in the Earth's frame, the traveling twin is younger when they meet, and if the earth-twin turns around in the traveling twin's frame, the earth-twin is younger when they meet.

Einstein imposed an initial condition in which A and B were separated by a distance d in the stationary frame

What does "the" stationary frame mean? Einstein said the choice of which frame you label the stationary one is arbitrary. What are the velocities of A and B in the frame you are labelling as stationary?

- and then A moves along the line AB - so the spatial interval is measured in the stationary frame of B.

So you're using "stationary frame" to mean the frame where B is at rest? That's not how Einstein used the term, as I said above. Please just say something like "B's rest frame" if that's what you mean, it'll be less confusing for both of us.

And what do you mean "so the spatial interval is measured in the stationary frame of B"--why "so"? Is there something about this problem that obligates us to use the spatial interval in B's rest frame, or do you agree that the choice of which frame's spatial interval we choose to use is a completely arbitrary one?

Since the spacetime interval is invariant, clocks in the stationary frame must intrinsically accumulate more time in order to offset the spatial distance.

What do you mean by "intrinsically?" Do you agree that we could just as easily analyze this problem in a frame where clocks in B's rest frame tick more slowly than A's clock, and we'd get exactly the same answer for the spacetime interval?

What I am saying, the age difference between the clocks when they are compared is not due to acceleration or turn around, or changing frames per se, but rather the reality of the time difference as between A and B is consequent to A's motion along a spatial path defined in B's inertial frame.

In the problem you have described, there was no initial synchronization of A's clock with B's at a single spatial location like in the twin paradox--A started out at a distance from B and moved towards it. Therefore, which clock reads a greater time when they meet depends entirely on how they were synchronized at the beginning when they were a distance d apart in B's frame. If they were synchronized in A's frame at that moment, then when A and B meet, B's clock will have accumulated less time.

Of course, we could have B move toward A instead of vice versa

How do you tell the difference? Are you assuming A and B were initially at rest at a distance d apart in their rest frame, and then A accelerated towards B? If so, you should have specified. However, this still does not mean that "A was moving towards B"--after all, there will be a frame where A and B were initially moving at velocity v, then when A accelerated its velocity dropped to zero in this frame, so it was now at rest while B moved towards it.

Again, you seem to have this crazy idea that the details of who accelerated obligates us to consider things from the point of view of one inertial reference frame rather than another. But you never answered my questions about this idea before, like what if we see two asteroids in deep space moving towards each other at constant velocity, if we want to analyze which is "really" aging more slowly do we have to know which was the last one to accelerate, even if neither has accelerated for millions of years?

in which case the spatial interval would be measured in A's frame, and the A clock would register more time when B arrived (if that is what you mean by symmetry - then yes, there is no difference since neither frame has any property which would render it preferred).

No, that's not what I mean by symmetry, I meant that if you look at some region of spacetime where two objects are both moving at constant velocity, the situation looks totally symmetrical in each frame. It's irrelevant that at an earlier point in time outside the region you're considering, the symmetry may have broken by one accelerating, that doesn't change the fact that it's symmetrical in that region of spacetime, unless you think that in order to analyze a problem taking place in some limited region of spacetime we are obligated to know the entire history of the universe leading up to the events in that region to decide which object was the last to accelerate, and we are then obligated to use the frame of the object that has been moving inertially for the longest time.

But once you decide upon which one moves, you immediately create the asymmetry that leads to differential aging

If you're using "moves" as a synonym for "accelerates", that's bogus, both for the reasons I mentioned above about needing to know the whole history of the universe to know who's "really" moving, and also for the reason I mentioned earlier, that there will always be a frame where the object was moving before the acceleration but came to rest afterwards.

 

[gonzo]

Jesse, I meant it is a meaningless comparison unless someone changes frames because otherwise you are comparing two different inertial frames, and while possibly interesting, the measurements of time and distance won't agree.

Your counter example shows that you didn't understand what I meant. If you are talking about twins, you are talking about two people starting in the same inertial frame, and then one of them shifting to a nother inertial frame, thus someone shifted frames, one of the twins. This has no bearing on what that twin later does, whether he stops or keeps moving with the new frame, he has shifted frames.

Otherwise you are just talking about measuring time and distance in two different inertial frames, and there is no even seeming paradox. They measure time differently, they measure space differently, they measure simultaneity differently. Measurements made from any two inertial frames are equally valid, it's just a different coordinate system for looking at the invariant spacetime interval between two events.

This thread I assumed by about the apparent paradox of two twins aging at different rates depending on their motion, and that whole question only has meaning if one of them moves with regard to the other.

Is that more clear?