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

[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?

 

[Hurkyl]

For fun, let's consider a completely symmetrical problem.


There are two pairs of clocks, A and B, and C and D.

A and B are stationary with respect to each other, and are synchronized in their rest frame, and separated by a distance L.
C and D are stationary with respect to each other, and are synchronized in their rest frame, and are separated by a distance L.

However, the two pairs are not stationary with respect to each other: there is a relative velocity between them.

The clocks travel so that:
B and C meet before B and D meet.
B and C meet before A and C meet.

(So, one might diagram it as saying they started out arranged like this:
A...B..C...D and ended like C...D..A...B)

For convenience, let us also say that B and C both happen to read zero when they meet.

---------------------------

So, what can we say about this problem?

First, we can consider the times on clocks when they meet. They can be ordered as follows:

(1) The time on B and C when they meet.
(2) The time on C when A and C meet. The time on B when B and D meet.
(3) The time on A when A and C meet. The time on D when B and D meet.
(4) The time on A and D when they meet.

In particular, when A and C meet, the time on A is greater than the time on C. And, when B and D meet, the time on D is greater than the time on B.

Now, some frame specific statements.

In the frame where A and B are stationary (and in sync):
D meets B before C meets A.
The time on D is always greater than the time on C.

In the frame where C and D are stationary (and in sync):
A meets C before B meets D.
The time on A is always greater than the time on B.


Hrm, I can't think of anything else interesting to say for this example.

(P.S. you notice that if you ignore, say, clock A, the problem is essentially the one you are analyzing?)

 

[JesseM]

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.

But you are free to only start considering the problem at the instant where the traveling twin has finished accelerating. Like I said to yogi, you don't have to consider the entire history of the system back to the the beginning of the universe.

Also, you can consider two clocks which are moving at constant velocity through space, and at the instant they pass infinitesimally close to each other, they both read the same time. Then as they move apart, this is just like the twins moving apart.

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.

Well, tell that to yogi! He doesn't seem to agree with this, and that's one of the main reasons our debate has dragged on for so long.

 

[yogi]

No - I do not agree that measurements made in all frames stand on the same footing - those made in the rest frame by an observer in that same frame are proper distances and proper times - measurments made to assess the length of a rod in a frame in relative motion is not a proper distance - it will have different values depending upon the relative velocity. Measurements of lengthts and times made in the moving frame by an observer in the moving frame are proper measurements in the moving frame - they don't vary with velocity. Measurements of lengths in another frame are apparent - Jesse and Hyrkyl - if you don't like that, again that is your right - but I am also entitled to adopt the interpretation of Eddington and Resnick and other recogonized experts on SR.

Moreover Jesse - you continue to distort what I have said - I post a few lines to clarify something and you want to write a book about it. If I ask you what time it is, your going to tell me how to build a clock. What I have said is consistent with the few paragraphs that Einstein wrote in connection with the physical meaning to be accorded the transforms that were derived by making observations of how things looked in a relatively moving frame. Moreover you have got your clocks mixed up - it is A that moves to B, and there is no turn around - A and B are together at the completion of the one way trip.

 

[Hurkyl]

those made in the rest frame

What is "the rest frame"? A universal frame of reference?

 

[yogi]

Hurkyl - Of course not - when you do a relativistic analysis, you are always free to select one frame and consider it at rest. Then by definition the other is considered moving. Does that mean that there may not be a tie between SR and the rest of the universe - no again - but since we don't have a good theory of why SR works so well as a stand alone .. we go along using it as though it is a final resolution - maybe it is !

 

[Hurkyl]

Of course not - when you do a relativistic analysis, you are always free to select one frame and consider it at rest.

But that is precisely what we mean by "equal footing". The measurements in that frame aren't special: they're simply relative to the frame you selected. If I chose to make measurements according to a different reference frame, you have no grounds to claim that your measurements are any more "proper" than my measurements.


You're entirely right that, in SR, you are free to pick a single reference frame and use that as a coordinate chart for doing the analysis. In fact, I would usually suggest it as a good practice to work in a single coordinate chart unless there's good reason to do otherwise.

It doesn't follow that doing so is any more "proper" than anything else.

but since we don't have a good theory of why SR works so well as a stand alone

What do you mean by that?

 

[JesseM]

No - I do not agree that measurements made in all frames stand on the same footing - those made in the rest frame by an observer in that same frame are proper distances and proper times - measurments made to assess the length of a rod in a frame in relative motion is not a proper distance - it will have different values depending upon the relative velocity. Measurements of lengthts and times made in the moving frame by an observer in the moving frame are proper measurements in the moving frame - they don't vary with velocity.

The concept of "proper length" as a quantity which all frames can agree on only makes sense if you're talking about the length of physical objects like rulers, it doesn't apply to the distance between two objects in space, since a distance in space doesn't have a rest frame like a ruler does.

Moreover Jesse - you continue to distort what I have said

How? Why don't you clarify what you meant, if I'm misunderstanding?

What I have said is consistent with the few paragraphs that Einstein wrote in connection with the physical meaning to be accorded the transforms that were derived by making observations of how things looked in a relatively moving frame.

Please quote those paragraphs, I don't know what you're talking about here. You are not correct that Einstein used the phrase "stationary frame" to mean the rest frame of the object being analyzed, for example.

Moreover you have got your clocks mixed up - it is A that moves to B, and there is no turn around - A and B are together at the completion of the one way trip.

I understood that A is the one who accelerates (the phrase 'A moves to B' is meaningless if you don't specify what frame you're referring to), and that there is no turnaround--where did I say otherwise? It would really help if you would quote my post when responding, rather than just making general comments that don't address most of the points I brought up or the questions I asked you.

 

[yogi]

Jesse - read my post 55. I think it will answer most of your questions

 

[JesseM]

yogi, the quote by Einstein you provided in post #55 was:

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 ...

In what way do you think this supports your argument? Do you agree that the following quote by Einstein shows that he considered it an arbitrary choice which inertial frame you denote the "stationary system" in a given problem?

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."

In particular, do you agree that he would have been perfectly happy to analyze this same physical experiment in terms of a different "stationary system" in which the clocks are not initially synchronized, and in which the clock which was moving in the other frame is now at rest, and the clock which was at rest in the other frame is now moving?

 

[yogi]

Jesse - of course - where did you get the idea that I considered the stationary frame as preferred - I called it stationary for the same reason as Einstein did - not because it is preferred, but because we can always pick one fame and consider it as stationary for the purpose of making measurments in a relatively moving frame. I think if you read my other posts, in particular # 75, I have made it clear. What Einstein denominated stationary has evolutionized to proper in modern parliance.

What I really wanted to talk about in connection with this interesting subject is how Einstien's physical explanation is arrived at from observational assumptions - going back to a point you raised in a previous post as to how would a person know - once the A clock is accelerated into motion along the AB line - what distinguishes the two frames if you don't know the history (the initial conditions). For example, let us assume a fourth clock D aboard an inbound rocket ship is also moving toward B on a line coextensive with AB and as A gets up to speed it is adjacent to D and traveling at the same speed. Since A and D are now in the same inertial frame, would you agree that D's clock and A's clock should run at the same rate. If so, then would the D clock lag behind the B clock at the time D arrived at B (we know the A clock will read less because Einstein told us). But if that is the case, how does this square with the proposition that D (who is only aware of his relative motion wrt to B) can consider himself at rest and B in motion, expect to measure a lesser time on B's clock when they meet? Jesse - try to keep your answer under 4 chapters, thanks

 

[JesseM]

What I really wanted to talk about in connection with this interesting subject is how Einstien's physical explanation is arrived at from observational assumptions - going back to a point you raised in a previous post as to how would a person know - once the A clock is accelerated into motion along the AB line - what distinguishes the two frames if you don't know the history (the initial conditions). For example, let us assume a fourth clock D aboard an inbound rocket ship is also moving toward B on a line coextensive with AB and as A gets up to speed it is adjacent to D and traveling at the same speed. Since A and D are now in the same inertial frame, would you agree that D's clock and A's clock should run at the same rate. If so, then would the D clock lag behind the B clock at the time D arrived at B (we know the A clock will read less because Einstein told us). But if that is the case, how does this square with the proposition that D (who is only aware of his relative motion wrt to B) can consider himself at rest and B in motion, expect to measure a lesser time on B's clock when they meet? Jesse - try to keep your answer under 4 chapters, thanks

It's all about whose frame the clocks are synchronized in. In your example, at the moment before A accelerates towards B, you're probably assuming that A and B are synchronized in their mutual rest frame. On the other hand, say that at the moment D catches up to A (after A has finished its acceleration), D's clock reads the same time as B in D's own rest frame. In this case, the D clock will be ahead of the A clock, so even though they both tick at the same rate, at the moment they pass B, the A clock will be behind B while the D clock will be ahead of B.

 

[yogi]

The same question arises in connection with Einstein's statement that a clock at the Earth's' equator (assume Earth is spherical so we can ignor GR, Earth's oblateness etc) will run a little bit slower than one at the pole. But does this follow from the fact that the two were initially in sync at the north pole and one is carried to the equator so that the equator clock (call it E) has changed frames - or is it a consequence of E's motion relative to the rest frame of the N (the north pole clock). What if both clocks are brought to sync at the equator and N is carried to the pole. Why can't E consider itself at rest and deduce that it is N that is in motion around him and conclude that it is N that is running slower. Einstein gave no credence to acceleration as having anything to due with why one clock runs faster - albeit E will experience some acceleration, Einstein discounted it (reaffirmed in his 1912 manuscript).

 

[yogi]

With regard to your post 81, what I had in mind was that D starts counting time on his clock when he sees A pull along side. If D logs 4 hours from this point until his arrival at B, then A's clock should also log 4 hours?

 

[yogi]

The interesting question to the above is - even though A and D now travel together in the quote moving frame - they got into the situation because of entirely different histories. If both clocks are now running at the same rate - is there an intrinsic rate of time passage assocated with every inertial frame?

 

[JesseM]

The same question arises in connection with Einstein's statement that a clock at the Earth's' equator (assume Earth is spherical so we can ignor GR, Earth's oblateness etc) will run a little bit slower than one at the pole. But does this follow from the fact that the two were initially in sync at the north pole and one is carried to the equator so that the equator clock (call it E) has changed frames - or is it a consequence of E's motion relative to the rest frame of the N (the north pole clock). What if both clocks are brought to sync at the equator and N is carried to the pole. Why can't E consider itself at rest and deduce that it is N that is in motion around him and conclude that it is N that is running slower. Einstein gave no credence to acceleration as having anything to due with why one clock runs faster - albeit E will experience some acceleration, Einstein discounted it (reaffirmed in his 1912 manuscript).

E is not in an inertial reference frame, the situation is not symmetrical, E cannot assume that N is ticking slower--in this case every possible inertial reference frame will agree that E's clock ticks less time than N's clock after a full revolution of the Earth (although there may be times when N is ticking slower at some point in the day from the perspective of a given inertial reference frame). If you're saying the fact that E is accelerating (moving in a circle) is not relevant to this problem, then I'm sure you're misunderstanding what Einstein was saying about accelerating--could you provide the quote?

With regard to your post 81, what I had in mind was that D starts counting time on his clock when he sees A pull along side. If D logs 4 hours from this point until his arrival at B, then A's clock should also log 4 hours?

They will both log an interval of 4 hours between the time A and D start traveling alongside each other and the time they meet B, regardless of whether D's clock is in sync with A. If you mean that D synchronizes his clock with A at the moment they start traveling together, then both clocks will read the same time when they meet B.

The interesting question to the above is - even though A and D now travel together in the quote moving frame - they got into the situation because of entirely different histories. If both clocks are now running at the same rate - is there an intrinsic rate of time passage assocated with every inertial frame?

I'm not sure what you mean by "intrinsic", but any two clocks which are at rest wrt each other will tick at the same rate, regardless of which frame is observing them.

 

[yogi]

If you look at the 1905 paper - Einstein simply gives the formula for the difference between the clocks - whether the path followed by A is straight or polygonal - no time is added or subtracted because A is not following a straight trajectory - See also his 1912 paper. We know there is a real age difference between A and B with A accumulating less time at the event of arrival - this is not a reciprocal situation - A accumulates less time as per B's clock - but why is D not entitled to conclude that B's clock is running slow ... he can consider himself at rest at the time A pulls along side and measure B's motion toward him. Will he not believe that B's clock is slow compared to his own.

What I meant by intrinsic - it is an issue that has been much discussed in the literature - but here we have a non-reciprocal situation where clocks in the A-D frame accumulate less time than the clock in the B frame - so either they don't really run at different rates and we have to explain the age difference at the end some other way such as 1) there is a specific rate of time passage (intrinsic) associated with every inertial frame that depends upon its motion with respect to every other intertial frame or 2) the two frames are not equal in all respects or 3) the spatial part of the interval is different for the two frames and this inequality leads to the different ages or.. (there are other ideas also). Einstein didn't profer an answer - now 100 years later, the subject is still debated and unresolved - it came up frequently in Einstein's own life, but he didn't shed any light upon it. My own view is that of 3) which I have previously stated.

 

[JesseM]

If you look at the 1905 paper - Einstein simply gives the formula for the difference between the clocks - whether the path followed by A is straight or polygonal - no time is added or subtracted because A is not following a straight trajectory - See also his 1912 paper.

I have looked at the 1905 paper, which is online here:

http://www.fourmilab.ch/etexts/einstein/specrel/www/

If you look at his derivation of the Lorentz transform in section 3, he specifies that he is talking about a system which is in a state of "uniform motion of translation" relative to the stationary system--ie an inertial frame. If you think this derivation would apply to non-inertial frames, then you aren't understanding the derivation. In section 4 he does talk about motion in a polygonal line, but he's only talking about looking at the motion of a non-inertial object from the perspective of an inertial frame, he's not saying the non-inertial object has its own frame which works just like an inertial one.

We know there is a real age difference between A and B with A accumulating less time at the event of arrival - this is not a reciprocal situation - A accumulates less time as per B's clock

Only because at the moment A finished its instantaneous acceleration, at that moment A and B were synchronized in B's frame--if they had been synchronized in A's frame at that moment, then B would have accumulated less time at the event of arrival.

but why is D not entitled to conclude that B's clock is running slow ... he can consider himself at rest at the time A pulls along side and measure B's motion toward him. Will he not believe that B's clock is slow compared to his own.

Once again, it's simply a question of what D's clock is set to read at the moment it catches up with A. Do you agree that, since A and D have never been at the same point in space before (and thus there was no initial moment when both could agree their clocks were synchronized), the choice of whether to have B and D read the same time in B's frame or D's frame when D catches up with A is a completely arbitrary one?

What I meant by intrinsic - it is an issue that has been much discussed in the literature - but here we have a non-reciprocal situation where clocks in the A-D frame accumulate less time than the clock in the B frame

No, it's completely reciprocal, because the choice of whose frame they should read the same time in at the moment D catches up with A is completely arbitrary, there is no physical reason to prefer one frame's definition of simultaneity.

To show more clearly how reciprocal it is, imagine that when A and B are at rest relative to each other, A is x meters to the right of B in their rest frame. Meanwhile, suppose that as D heads left towards A, D also has another object C which is at rest relative to D, and which is x meters to the left of D in their rest frame. At the moment D reaches A's location, A instantaneously accelerates so it's now at rest relative to D; at the moment C reaches B's location, C instantaneously accelerates so it's now at rest relative to B. Also, until C and A accelerate, A is synchronized with B in the AB rest frame and C is synchronized with D in the CD rest frame. Now, in terms of how D and A's times relate when they meet, and how C and B's times relate when they meet, you have two choices:

1. When D meets A, their clocks read the same time, and read the same time thereafter. This means C's clock will not read the same time as B's when they meet.
2. When C meets B, their clocks read the same time, and read the same time thereafter. This means A's clock will not read the same time as D's when they meet.

You can't have both, but do you agree there's no reason to prefer one over the other, since if you pick #1 the situation in B's frame looks exactly how the situation in D's frame looks if you pick #2, and vice versa? (this also means that if you pick #1, and when all four meet A and D's clocks read time x while C's reads y and B's reads z, then if you pick #2, you know that when all four meet C and B's clocks will read time x while A's reads y and D's reads z--once again, totally reciprocal).

 

[yogi]

I would agree the derivation is based upon inertial frames - but the physical explanation disregards any effect of changing direction Says Einstein: "If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which remained at rest the traveled clock on its arrival at A will be 1/2t(v/c)^2 second slow." How do you conclude this refers to a non-inertial object?

I would agree as I have previously stated that either frame can be considered at rest, and in that sense the situation is reciprocal - but once the goal posts that define the proper distance in one frame are set, the age difference is one way - the clock that moves between two spatial points separated by a proper distance defined in one frame always reads less than a clock in the frame where the proper distance is measured.

Going back to my quote about the continuous curved path - Einstien has implicitly resolved the twin paradox w/o considering acceleration, changing frames's, shifting hyperplanes ect. Specifically, if B lies to the east of A and both are initially in the same frame, and A starts out headed north, then gradually veers east and then South so as to pass by B and continues by curving west and then North to return to his original starting point so that A's path describes a large circle, A will return having aged less than some sibling that remained at A's starting point. Moreover, because the circle is symmetrical and continuous and because A travels at constant velocity, there can be no logical reason to assume the aging effect comes about at any particular point in the journey (such as turn around). Upon passing by B, A's age as judged by B will be 1/2 that lost for the round trip (no pun intended). In other words, the differential aging between the twins is a continuous linear function of the time spent in moving at a constant velocity relative to the frame in which the twins are at rest.

 

[gonzo]

I must be more confused than I thought. I was under the impression that any time you changed direction you would be required to be changing inertial frames. So that going in circle, no matter how constant your velocity around the circle was, would involve constantly changing inertial frames.

 

[Garth]

yogi Going round in a circle is accelerating, it is not an inertial frame.
Nevertheless the conclusion is valid, the person moving relative to the observer encounters her again after clocking up less proper time.

Garth

 

[JesseM]

I would agree the derivation is based upon inertial frames - but the physical explanation disregards any effect of changing direction Says Einstein: "If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which remained at rest the traveled clock on its arrival at A will be 1/2t(v/c)^2 second slow." How do you conclude this refers to a non-inertial object?

Because moving on a closed curve is by definition moving non-inertially. If an inertial observer sees an object moving inertially, with its velocity as a function of time given by v(t), then if he wants to know how much time will elapse on the object's clock between t_0 and t_1 (as measured in his own inertial frame), he must evaluate the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. Obviously, if the velocity is constant this integral is simply equal to (t_1 - t_0)\sqrt{1 - v^2/c^2}, even if the direction of the velocity is changing. On the other hand, if the object moving non-inertially sees the velocity of the inertial observer relative to himself at a given time t' as given by v'(t'), he cannot simply integrate \sqrt{1 - v'(t')^2/c^2} to see how much time passes on the inertial observer's clock--if he starts out at the same position as the inertial observer and then later meets up with him again, he would get the wrong prediction for the amount of time elapsed on the inertial observer's clock if he evaluated this integral.

One example of this--suppose I am whirling a clock around my head at some constant velocity v, how slowly will I see the clock ticking, and how slowly will an observer sitting on the clock see my clock ticking? The answer is that I will see the whirling clock running slow relative to my own by a factor of \sqrt{1 - v^2/c^2}, but unlike with inertial motion there is no disagreement here, an observer sitting on the whirling clock will agree that my clock is running faster than his own by 1/\sqrt{1 - v^2/c^2}. You can prove this by looking at how fast his clock is ticking in my frame, then considering what time light rays from each tick of my clock reach his position, and what time his own clock will read at each moment he receives light signals from a tick of my clock--again, this analysis would all be done from within my own inertial frame, no need to consider any accelerated frames to solve this problem.

I would agree as I have previously stated that either frame can be considered at rest, and in that sense the situation is reciprocal - but once the goal posts that define the proper distance in one frame are set, the age difference is one way - the clock that moves between two spatial points separated by a proper distance defined in one frame always reads less than a clock in the frame where the proper distance is measured.

Well, the only thing we ever mean by "reciprocal" is that the situation in one frame is symmetrical with the situation in another frame--of course it's true that once you make an arbitrary choice of which reference frame you want to work in, then there will be a single truth about who is aging slower, but this seems pretty trivial. Is that really all you're arguing here, that within the context of a single frame there is only a single truth about which twin is aging more slowly?

Going back to my quote about the continuous curved path - Einstien has implicitly resolved the twin paradox w/o considering acceleration, changing frames's, shifting hyperplanes ect. Specifically, if B lies to the east of A and both are initially in the same frame, and A starts out headed north, then gradually veers east and then South so as to pass by B and continues by curving west and then North to return to his original starting point so that A's path describes a large circle, A will return having aged less than some sibling that remained at A's starting point. Moreover, because the circle is symmetrical and continuous and because A travels at constant velocity, there can be no logical reason to assume the aging effect comes about at any particular point in the journey (such as turn around).

In an inertial frame where the center of the circle is at rest, the twin moving in a circle will be moving at the same speed at all times, so of course he'll be aging at an equally slow rate throughout the trip. But in an inertial frame where the center of the circle is moving, the twin will be moving faster at some times than others, so his rate of aging will vary depending on what part of the circle he's on. Either way, all frames will agree on how much time he's lost relative to another person who he departs from and then returns to after making one complete revolution.

Upon passing by B, A's age as judged by B will be 1/2 that lost for the round trip (no pun intended). In other words, the differential aging between the twins is a continuous linear function of the time spent in moving at a constant velocity relative to the frame in which the twins are at rest.

All frames will agree that if a twin takes an accelerated path away from a twin moving inertially, and later returns to meet up with the inertially-moving twin again, the accelerated twin will have aged less. But this is fundamentally different from the case where both twins are moving inertially, because in that case different frames cannot agree on who's aging less.

 

[JesseM]

I must be more confused than I thought. I was under the impression that any time you changed direction you would be required to be changing inertial frames. So that going in circle, no matter how constant your velocity around the circle was, would involve constantly changing inertial frames.

You are correct, circular motion is always non-inertial.

 

[yogi]

Quite right Gonzo and Garth - in actuality we probably cannot find a perfect inertial frame in which to conduct such experiments - at least not on earth, because we are always in elliptical motion about the Sun - but Einstein's formula for the time difference and his statements discounting the significance of acceleration lead us to conclude that A's accceleration wrt to the frame in which both A and B were initially at rest is circumstantial - it is what allows A to acquire motion with respect to B, but it does not impact the aging process nor does it affect the numerical difference. And since A according to Einstein can follow any polygonal path, the journey could be comprised of abrupt saw tooth oscillations that involve many hi accelerations and decelerations - but as long as the velocity is constant along the path (as it is in the one way trip), we can expect the age difference to depend only upon the velocity and total time. This is the basis of the path integral approach to the Twins problem. It is simple, direct and consistent with all experiments.

 

[yogi]

In the last line of part 4 of the 1905 paper - there is no reference to whether the clocks were synced together at one of the poles and then separated or if they were brought into sync at the equator and then separated - Einstein says: "...a balance-clock at the equator must go more slowly..than a precisely similar clock situated at one of the poles.." If the Earth is perfectly spherical this will be the case - so what does acceleration have to do with it? It is true that the clock at the equator will undergo continuous acceleration in the form of directional change - but as I said above this is only circumstantial - there is no acceleration term involved. Same thing is true for GPS - we Sync the satellite clocks in the frame of the rotating Earth - transform to the pole, and preset the rate to account for the orbit height and velocity before launch - lots of accelerations in the form of directional change every second thereafter but the satellite clocks behave just as though they are moving in an inertial frame

 

[yogi]

Jesse - With regard to your last statement in post 91 - in the example of A and B, after the brief impulse in getting A up to speed to move toward B, both A and B are in inertial frames - yet one continues to age more than the other. As I previously stressed, The acceleration can be discounted because If there were another clock Q in the B frame that lies on the AB line but is twice as far, when A arrives at Q, the Q clock would read twice what the B clock read at the time A passed by B. What I am saying is that A must be viewed as being in an inertial frame once he gets up to crusing speed, and the time differential between A as he arrives at each further clock will be proportional to the distance of the successive clocks in the B frame. According to SR, A's frame is equivalent to B's once A gets to up to speed - it is no longer distinguishable - and therein lies the heart of the dilemma because there is no physical justification for mechanical clocks to run at different rates in identical inertial frames. If the two frames are indentical - the A clock and the B clock and the Q clock should all run at the same rate - but experiments tell us otherwise.

 

[JesseM]

In the last line of part 4 of the 1905 paper - there is no reference to whether the clocks were synced together at one of the poles and then separated or if they were brought into sync at the equator and then separated - Einstein says: "...a balance-clock at the equator must go more slowly..than a precisely similar clock situated at one of the poles.." If the Earth is perfectly spherical this will be the case - so what does acceleration have to do with it? It is true that the clock at the equator will undergo continuous acceleration in the form of directional change - but as I said above this is only circumstantial - there is no acceleration term involved. Same thing is true for GPS - we Sync the satellite clocks in the frame of the rotating Earth - transform to the pole, and preset the rate to account for the orbit height and velocity before launch - lots of accelerations in the form of directional change every second thereafter but the satellite clocks behave just as though they are moving in an inertial frame

The situation with one clock on the equator and one clock on the pole is exactly like the situation where I am whirling a clock in a circle around my head while carrying a clock of my own. In this case, all inertial frames will agree that the clock whirling around ticks less time after a full revolution than the clock I am holding (which is at rest in an inertial frame), so we can simply say that the whirling clock "must go more slowly" without referring to which inertial frame we're talking about, just as Einstein said the same thing about the clock at the equator. But the fact that the whirling clock is moving in a circle is critical here--if instead we were comparing a clock held by me with a clock moving in a straight line at constant velocity relative to me, then different frames would disagree about which clock is going more slowly, so we couldn't make a statement like that without specifying which frame we're talking about. Do you agree that the two situations are different in this sense?

As for GPS, no, the satellite clocks will not behave like they are in an inertial frame--for example, if a GPS clock is moving at velocity v(t) in the instantaneous rest frame of a clock on the Earth's surface at time t as measured by that clock, you can't find the elapsed time on the GPS clock by integrating \sqrt{1 - v(t)^2/c^2}.

Jesse - With regard to your last statement in post 91 - in the example of A and B, after the brief impulse in getting A up to speed to move toward B, both A and B are in inertial frames - yet one continues to age more than the other.

Not in any frame-independent sense, no. B ages more than A in B's rest frame, and A ages more than B in A's rest frame. The situation is completely symmetrical, as I tried to show with the addition of objects C and D to this example at the end of post #87.

 

[yogi]

Jesse - it can't be symmetrical. In the linear straight path case - Its only symmetrical in the sense that either A could have taken off toward B (in which case A will be younger when they are compared at arrival) or B takes off toward A in which case B is younger when the clocks meet. Once you decide upon which one takes off, things are not symmetrical - even though A, once up to speed, travels in an inertial frame. This is the unresolved mystery of SR...but Einstein clearly tells you that the two clocks will be out of sync when the meet - and yet he disregards any influence of acceleration. If it is A that moves toward B, all during that journey the A clock will running at a reduced rate - because when they meet A will have accumulated less time as mesured by the B clock - and the longer the distance betwen A and B at the outset, the greater the age difference. And that is the puzzle of the Special Theory - it obviously isn't a puzzle to You just just like the Trinity isn't a worry to Catholics - but it is to me and many others ---this aspect of relativity is at best - unexplained. You can't have it both ways...as we have both said, once A is up to speed, A's inertial frame should be as good as B's (there is no apparent reason why A's clock should continue to run slow after the acceleration period has ended, and there is no reason why it should run slow during the initial takeoff).

Given that all inertial frames are guaranteed by the constitition to be created equal, maybe, like people, some are more equal than others

 

[gonzo]

These seem strange things to assert, Yogi. I don't really understand what you mean in a few places.

For example, what do you mean by "takes off"? Is this another way of saying "changes inertial frames"?

You seem to be trying to make it more complicated than it is, and it is very odd to invoke religion here. There is nothing unresolved or confusing here as far as I understand it.

Are you talking about A and B starting in sync and then one of them changing inertial frames? Have you tried drawing the world lines? That often clears these things up. The other thing to look at is assumptions you make about simultaneity, that is often another point of artificial confusion.

Time is measured differently in different inertial frames. Simultaneity is different. I don't really see the issue or the need to compare it to religous belief.

 

[JesseM]

Jesse - it can't be symmetrical.

What can't be symmetrical? Look over the example I gave in post #87 with the addition of D (who moves at constant velocity towards B, and who is traveling alongside A after A accelerates) and C (who is a constant distance d from D in D's rest frame, until C passes B, at which point it changes velocity so it is at rest relative to B):

To show more clearly how reciprocal it is, imagine that when A and B are at rest relative to each other, A is x meters to the right of B in their rest frame. Meanwhile, suppose that as D heads left towards A, D also has another object C which is at rest relative to D, and which is x meters to the left of D in their rest frame. At the moment D reaches A's location, A instantaneously accelerates so it's now at rest relative to D; at the moment C reaches B's location, C instantaneously accelerates so it's now at rest relative to B. Also, until C and A accelerate, A is synchronized with B in the AB rest frame and C is synchronized with D in the CD rest frame. Now, in terms of how D and A's times relate when they meet, and how C and B's times relate when they meet, you have two choices:

1. When D meets A, their clocks read the same time, and read the same time thereafter. This means C's clock will not read the same time as B's when they meet.
2. When C meets B, their clocks read the same time, and read the same time thereafter. This means A's clock will not read the same time as D's when they meet.

You can't have both, but do you agree there's no reason to prefer one over the other, since if you pick #1 the situation in B's frame looks exactly how the situation in D's frame looks if you pick #2, and vice versa? (this also means that if you pick #1, and when all four meet A and D's clocks read time x while C's reads y and B's reads z, then if you pick #2, you know that when all four meet C and B's clocks will read time x while A's reads y and D's reads z--once again, totally reciprocal).

Do you agree with all the statements in my last paragraph? If not, please specify which ones you think are wrong.

Once you decide upon which one takes off, things are not symmetrical - even though A, once up to speed, travels in an inertial frame.

Sure they're symmetrical. Suppose you synchronize A such that, in D's frame, when A and B both read exactly the same time at the moment that A started accelerating and traveling alongside D. In this case, when A and B meet, B will read less time. It's only if you choose to synchronize A and B such that they read the same time in B's frame at the moment A accelerates that A will read less time than B when they meet. Once again, it all comes down to a completely arbitrary choice of synchronization, there is no "real" truth about whether A or B has aged less since the moment A accelerated.

 

[yogi]

gonzo - I was being facetious - doesn't really have anything to do with religeon or the US constitution. When I say takes off - that implies a change from sitting at rest in the AB frame and the start of A's motion toward B. When A takes off, A goes from zero velocity to v (granted an acceleration and a change of frame). A then continues toward B for a long time at v. When he gets to B his clock reads less than B.

The situation is very simple - except it doesn't follow. During the major part of his trip A moves at constant velocity - so he is in an inertial frame. His clock should not run at a different rate than B's clock since two identical mechanical devices cannot run at different speeds w/o a mechanical reason. The reason is absent. It has nothing to do with observations in other frames - there are none - only a beginning event (take-off) and an ending event (arrival) These two events exist at the same spatial point in both frames - but the arrival event corresponds to different time in each frame.