Inertial Frames distinguished by proper times

[yogi]

Ich - I think you are trying to imply things I have not said. If you are saying that two separated clocks (say A and B) at rest wrt each other, and A is given a brief acceleration toward B, and travels most of the distance at constant velocity v, but just before reaching B, B is accelerated in the same direction to a velocity v, so now both A and B are moving together at velocity v relative to some object in the original frame of reference. Then A and B now run at the same rate because they are in the same frame (I hope we agree on that) but they do not read the same time - When B pulls alongside A, the time on the A clock will be less than the time on the B clock (do we agree on this)?

In fact, it is not necessary to wait until A is near B. For example, if A is at the origin of the X axis, and B is at X = 100 miles and both blast off in the direction of the + X axis at the same time with identical accelerations for identical time periods (their integrating accelerometers are set to cut off at the same velocity) Both will have reached a velocity v in the direction of the + X axis, so do you think there will be any difference in the reading on A's clock relative to B's clock now that they are traveling together in the same frame but still equally separated?

 

[Ich]

§1: Maybe I managed to misunderstand your claim. I agree with you that A will read less time.
What I had in mind was the following setup: A and B start at the same point, A being accelerated. A should "tick at a slower rate" from then. If you then accelerate B to match velocity with A, A and B should tick at the same rate. So if you bring A and B together, you would expect A to read less. SR says B will read less.

§2: We better don´t use accelerating frames. One has to be extremely careful with the setup and the calculations. For example, in your setup the distance between A and B would increase, and A will indeed read less time as he was at the bottom of a "gravity well". I think your §1 is enough to decide whether your view is consistent with SR´s.

 

[yogi]

§1: Maybe I managed to misunderstand your claim. I agree with you that A will read less time.
What I had in mind was the following setup: A and B start at the same point, A being accelerated. A should "tick at a slower rate" from then. If you then accelerate B to match velocity with A, A and B should tick at the same rate. So if you bring A and B together, you would expect A to read less. SR says B will read less.

After B accelerates to the same speed as A (same direction) then A and B will run at the same rate (I guess we agree on that). They are separated in space, but both are in the same frame at rest wrt each other (I assume we agree on that) My question is: "How do you propose to bring them together?" What if they are not brought together - each simply interrogates the other with radio signals? If they are not brought together, which clock will read more time?

 

[Ich]

We agree on both points.

My question is: "How do you propose to bring them together?" What if they are not brought together - each simply interrogates the other with radio signals? If they are not brought together, which clock will read more time?

To compare clocks, you either interrogate them with radio signals (same procedure as Einstein suggested in his paper) or you bring them together by slow transport (v<<c). It is a feature of SR that both procedures will give the same result: A will read more time.

 

[yogi]

Ich: That is interesting - let's see - we could actually do an orbiting version of what you suggest - let's launch a GPS satellite clock that has been corrected for height but not velocity - the two clocks are in sync in the Earth frame before launch - A is accelerated into orbit and flys for one year, during which time A will run slower than the B clock on earth. One year later we launch the B clock into an identical orbit (after correcting for altitude) and now A and B are side-by-side, so both are running at the same rate, but there will be a difference in the lapsed time accumulated on the A clock and the B clock during the one year that passed between the two launches. Are you saying the A clock will read more time, or are you saying the analogy is flawed?

I am assuming in the linear case that you proposed, the conclusion that A will read more time was arrived at using the methodology adopted by Einstein (1918) and Born to explain the twin thing. If so, I will comment upon that.

 

[Ich]

No, you could not do an orbiting version of this. I can explain later why not and how SR explains the effect in circular motion.
But for now I strongly suggest that we stop complicating things until we got the basics right.
I don´t know the methodology of Einstein and Born. It is simply the old simultaneity thing: Until A and B join frames, each one is equally right (or wrong) to say that the other´s clock is ticking slower. When B suddenly accelerates, A did not change his view of things. That means, B is still (nearly) at the same position, and his clock (nearly) shows the same time as before the acceleration. So the result that B shows less time still holds.
But B´s notion of simultaneity changed drastically. "Now" A´s clock is ahead of his, and this won´t change if the clocks are brought together.
WARNING: the following may be unintelligible. Ignore it, if you can´t make sense of it. It´s kind of a metaphor, but not too far from truth.
There is nothing important happening with A´s and B´s clock at this time. It is more like B suddenly recognized that his time was "flowing in the wrong direction" all the time. But which direction is the right one is decided only when you decide in which frame you want to compare clocks. If it would have been B´s frame, A´s time would have been flowing in the wrong direction.

 

[yogi]

Until A and B join frames, each one is equally right (or wrong) to say that the other´s clock is ticking slower. When B suddenly accelerates, A did not change his view of things. That means, B is still (nearly) at the same position, and his clock (nearly) shows the same time as before the acceleration. .

If I correctly picture what you painted - A and B are synchronized at the origin of an x-y coordinate system and A first accelerates to a velocity v along the positive x-axis - the A clock will run slower as long as this condition persists - for example, A could travel for a long time at v = 0.5c relative to B and would wind up with less accumulated time when arriving at Altair (we assume Altair is at rest relative to the origin of the coordinate system). Now just before A reaches Altair, B quickly accelerates to 0.5c wrt the coordinate axis 0,0 in the same direction (along the + axis toward Altair) - and you say that B's clock shows the same (nearly) time as it did before B commenced accelerating (OK agreed). So during A's long journey 1) the A clock either ran slower than the B clock, or 2) the spatial distance D between the origin and Altare is contracted from A's point of view so his clock only recorded a time L/v. (where L is the contracted length). Either way, before B accelerates, do you agree that A's clock will have recorded less time as A nears Altair (not yet decelerating) than clocks at rest with respect to the origin of the coordinate system where B has remained at rest? If so, then I do not understand how you arrive at a result that predicts B will show a Real (not apparent) lesser time than A after B completes his short duration of acceleration.

 

[Ich]

If I correctly picture what you painted - A and B are synchronized at the origin of an x-y coordinate system and A first accelerates to a velocity v along the positive x-axis - the A clock will run slower as long as this condition persists - for example, A could travel for a long time at v = 0.5c relative to B and would wind up with less accumulated time when arriving at Altair (we assume Altair is at rest relative to the origin of the coordinate system).

No, you start again mixing "observer" and "observer´s rest frame". It may sound like nitpicking, but it is crucial to understand the difference:
A will read less time than a clock positioned at Altair which is synchronized with B in their common IF. It will not read unambiguously less than B´s clock, because B himself is not at Altair, and comparing times at different positions is a very special thing in relativity.

Now just before A reaches Altair, B quickly accelerates to 0.5c wrt the coordinate axis 0,0 in the same direction (along the + axis toward Altair) - and you say that B's clock shows the same (nearly) time as it did before B commenced accelerating (OK agreed).

Yes.

So during A's long journey 1) the A clock either ran slower than the B clock, or 2) the spatial distance D between the origin and Altare is contracted from A's point of view so his clock only recorded a time L/v. (where L is the contracted length). Either way, before B accelerates, do you agree that A's clock will have recorded less time as A nears Altair (not yet decelerating) than clocks at rest with respect to the origin of the coordinate system where B has remained at rest?

Again, don´t compare clocks at different positions. That´s where all the trouble comes from. I agree that A will read less time than a clock synchronized with B when A passes it, eg the Altair clock when A is at Altair.

If so, then I do not understand how you arrive at a result that predicts B will show a Real (not apparent) lesser time than A after B completes his short duration of acceleration.

And again, B is not at Altair. A will read more time than a clock at Altair which is synchronized with B after B´s acceleration. B´s "simultaneity plane" or however you call it shifted during acceleration.
If you then bring A and B slowly together, the result stays the same: B will Really read less than A.

 

[yogi]

All the clocks in the coordinate frame can be synchronized prior to A's initial acceleration - So if we add a clock at Altair (call it D) it can be synced with A and B and will read the same as A and B before any accelerations have taken place. When A is accelerated, B and D are still in sync (reading the same time and logging time at the same rate).

1) When A arrives at D, A will read less than D. Do we agree on this?
I think from your post above, the answer must be yes

2) If Yes, do you agree that B and D still read the same time (prior to B's acceleration). If not, how did they get out of sync?

3) If yes to no 2 above, do you agree that A reads less than B immediately prior to B's acceleration?

- probably not - but in any case you conclude that after B's acceleration B will read less than A...and that is where we part company

Invariably the analysis of these interesting problems skips from actual real times (local times or proper, whatever you want to label them) logged by a clock to an apparent observation that typically involves a rapid shift in the slope of the plane of simultaneity

...so in the distant inertial system of A, the time on B's clock has rapidly changed as viewed by A - actually B's clock would have to lose a lot of time (run backwards) during a short period of acceleration - because prior to the acceleration, B clock should read the same as D clock - but physically the B clock cannot run backwords just to accommodate the book-keeping. While some folks are comfortable with such abstractions, I am not. To me the interest in these problems is in finding an explanation that is consistent with a physical reality.

 

[Hurkyl]

so in the distant inertial system of A ... because prior to the acceleration, B clock should read the same as D clock

The B clock and the D clock have never read the same time, as measured by any inertial reference frame in which A is stationary for its journey.

To me the interest in these problems is in finding an explanation that is consistent with a physical reality.

Coordinate charts are not physical reality. It is not inconsistent with reality for things to run backwards according to a coordinate chart. (Though technically such a thing is a generalization of a coordinate chart)

 

[yogi]

The B clock and the D clock have never read the same time, as measured by any inertial reference frame in which A is stationary for its journey.

In the quote you left out most of what I was saying - so it comes out wrong - B clock reads the same as D clock in the BD frame prior to B's acceleration. I did not say B and D are to be read in the A frame - B and D were synchronized initially and they stay in sync until B is accelerated. A reads less than D when A arrives at D, B reads the same as D prior to B's accelertion (B and D are still at rest in the same frame).

The real time lost on A's clock during the journey relative to B (analogous to the stay-at-home twin) is not going to be altered by B accelerating after the game is over- B can accelerate to v and immediately decelerate to 0 and wind up back in the BD frame only slightly displaced from the 0,0 origin - this will not influence any clock involved except to a small degree B - it doesn't wipe out the years of time lost by the A clock.

 

[Hurkyl]

I did not say B and D are to be read in the A frame

Then I am clueless as to what anything in your last paragraph of #59 means.

 

[Ich]

We are always encoutering the same problem: You compare times at different locations and treat the results as if they were invariant physical realities. This is not consistent with SR:

So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.

Example:

3) If yes to no 2 above, do you agree that A reads less than B immediately prior to B's acceleration?
- probably not - but in any case you conclude that after B's acceleration B will read less than A...and that is where we part company

You compare the reading of two clocks. One is defined adequately: B`s reading at the time and place where he starts accelerating. The value is observer-independent as it represents a proper time.
The other reading is not defined sufficiently: When do you read A´s time? Obviously "at the same time". And what does "at the same time" mean? Prior to B´s acceleration it means, for example, "when A passes D". After B´s acceleration it may mean "when A passes the next star behind Altair" (we used no numbers until now, so that will do). Of course A will read then more time.
But what happened? Did B "run backwards in time"? Did A "jump forwards in time"? No. We simply compared different things and came to different results.

Invariably the analysis of these interesting problems skips from actual real times (local times or proper, whatever you want to label them) logged by a clock to an apparent observation that typically involves a rapid shift in the slope of the plane of simultaneity

All the readings I mentioned are, of course, readings of proper time. Actual real times. Invariant. So where do all those "apparent" time shifts and all this come from?
You ask "which clock reads less?", and think implicitly that this question must have one invariant answer. That´s where you and SR part company.
I encourage you: try to find a paradox in my setup without comparing times at different locations. Use as many clocks as you like, sync them in which frame you like, and compare the readings of any two clocks which are at the same position.
The result will confirm that B reads less than A when they are brought together. It will contradict your claim that A reads less due to initial acceleration. In fact, it is completely irrelevant wheter A accelerated or not.

 

[yogi]

"All the readings I mentioned are, of course, readings of proper time. Actual real times. Invariant. So where do all those "apparent" time shifts and all this come from?
You ask "which clock reads less?", and think implicitly that this question must have one invariant answer. That´s where you and SR part company.
I encourage you: try to find a paradox in my setup without comparing times at different locations."

The local reading on a clock can be compared with the local reading on another clock which is displaced (separated) as long as they are not in relative motion. There is no mystery - until B accelerates, B and D read the same (they are at rest in the same frame). When A reaches D, D and A can each communicate to the other what their own local clock reads. If v = 0.5c, gamma = 0.866, so if D clock reads 100 years at the time of A's arrival, A clock will read 86.6 years. And since B has remained in sync with D all during the time of A's flight, B will also read 100 years, as will every other clock that has remained at rest wrt B and D. if you don't agree with this, tell me what you think B clock will read at the time A arrives at D.

B of course, does not have access to the information that A has arrived - nonetheless, B will know when to accelerate by the reading on his own clock - that is, since he knows v and he knows the distance to D as measured in the rest frame, he will know that he should accelerate when his own clock reads 100 years.

 

[yogi]

Then I am clueless as to what anything in your last paragraph of #59 means.

Sorry - to clarify - A and D can each read the other as A passes D. A does not directly read B at this time - the time on B is established (determined) by the prior synchronization of B and D. This holds good until B accelerates.

 

[yogi]

Here is a simple experiment to demonstrates that the first clock to be accelerated records less time - create two pions in the lab - immediately accelerate them to 0.999c and observe that on average they travel 15 meters which corresponds to a life extension of 20 times their at rest half life. Now create two pions A and B and immediately accelerate A to 0.999c. After pion A has traveled one meter, accelerate pion B to 0.999c Which one decays last (or to put it another way, which travels the farthest before decaying)? Obviously A.

 

[Ich]

if you don't agree with this, tell me what you think B clock will read at the time A arrives at D.

That´s easy:
B reads
-50 years for an ultrafast observer traveling B->D
-74.7 years for A
-100 years for D
-150 years for an ultrafast observer traveling D->B
-any value between those for other observers.
always assuming that B does not accelerate after 100 years.
And, as if it couldn´t be worse, check this out:
Clock E, comoving and synchronized with A but 43.3 LY behind (so that it meets B at the same time A meets D), proves without doubt that at the time A meets D, B reads 74,7 years - less than A.
"Proves without doubt" means that, following your argumentation, you are of course allowed to compare A and E as they are in the same frame:

The local reading on a clock can be compared with the local reading on another clock which is displaced (separated) as long as they are not in relative motion. There is no mystery - until B accelerates, B and D read the same (they are at rest in the same frame).

Do you agree that there must be something wrong with your view?

 

[yogi]

Something is wrong with someone's view. Clocks do not run backwards.

 

[Ich]

Something is wrong with someone's view. Clocks do not run backwards.

Very helpful, thanks.
I´ll return to the guided discussion style.
1.

B reads
-50 years for an ultrafast observer traveling B->D
-74.7 years for A
-100 years for D
-150 years for an ultrafast observer traveling D->B
-any value between those for other observers.

Do you agree that this is what SR says about B´s reading at the time A meets D?

Somehow you seem to have the notion that of all those values, 100 years is the "real" one, because B and D are at rest wrt each other.

2.

Clock E, comoving and synchronized with A but 43.3 LY behind (so that it meets B at the same time A meets D), proves without doubt that "at the time" A meets D, B reads 74,7 years - less than A.

E and A are at rest wrt each other. Do you agree that
a) SR says that B reads less than E "at the time" A meets D and
b) the statement "B reads less than A" is just as real as the statement "A reads less than B", because both are backed by clocks at rest wrt each other?

3. Do you agree that, as we get to contradictory statements, the notion of "absolute simultaneity" is for the circular file, even if both observers are at rest? (that´s why I quoted Einstein. He is very clear about this)

 

[yogi]

If you try to judge time on a distant clock in relative motion wrt the frame in which the meaurment is being made, you get apparent readings. Real time on a clock is proper time - proper time is local time - that read by an observer at rest wrt the clock. At the start of the experiment, A,B and D are all at rest and set to zero. All real times read on any of these clocks thereafter must be positive.

You posed this problem of a second high speed particle starting out at a later time than a first particle. I claim the local time on the clock that started last will be greater than the one that started first - you say otherwise.

In post #66 I suggested you think about it in terms of the physics of local time. Let's do it again - create two pions A and B at the same point - each has an average lifetime in the lab of 0.02 usec. Accelerate A immediately to 0.99c. Wait 0.01 and then accelerate B to 0.99c. Which one do you think will travel the longer distance?

 

[Ich]

Real time on a clock is proper time - proper time is local time - that read by an observer at rest wrt the clock.

"proper time is local time" is ok with me.
"that read by an observer at rest wrt the clock" is wrong and not compatible with SR.
What do you mean with "local"? 50 ly?

It might appear possible to overcome all the difficulties attending the definition of "time'' by substituting "the position of the small hand of my watch'' for "time.'' And in fact such a definition is satisfactory when we are concerned with defining a time exclusively for the place where the watch is located; but it is no longer satisfactory when we have to connect in time series of events occurring at different places, or--what comes to the same thing--to evaluate the times of events occurring at places remote from the watch.

You don´t have to be at rest with the clock to read it. You have to be near it.
This is about the 6th time I tell you that you can´t compare times at different locations unambiguously. Even if they are at rest. My post #67 shows that explicitly.
It makes no sense to continue the discussion until we clear this point.
Do you agree or not?
If not, address the points in #67.

 

[yogi]

Also from the same paper:


If at the point A of space there is a clock, an observer at A can determine the time values of events in the immediate proximity of A by finding the positions of the hands which are simultaneous with these events. If there is at the point B of space another clock in all respects resembling the one at A, it is possible for an observer at B to determine the time values of events in the immediate neighbourhood of B. But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an ``A time'' and a ``B time.'' We have not defined a common ``time'' for A and B, for the latter cannot be defined at all unless we establish by definition that the ``time'' required by light to travel from A to B equals the ``time'' it requires to travel from B to A. Let a ray of light start at the ``A time'' from A towards B, let it at the ``B time'' be reflected at B in the direction of A, and arrive again at A at the ``A time'' .


We assume that this definition of synchronism is free from contradictions, and possible for any number of points; and that the following relations are universally valid:--

If the clock at B synchronizes with the clock at A, the clock at A synchronizes with the clock at B.
If the clock at A synchronizes with the clock at B and also with the clock at C, the clocks at B and C also synchronize with each other.
Thus with the help of certain imaginary physical experiments we have settled what is to be understood by synchronous stationary clocks located at different places, and have evidently obtained a definition of ``simultaneous,'' or ``synchronous,'' and of ``time.'' The ``time'' of an event is that which is given simultaneously with the event by a stationary clock located at the place of the event, this clock being synchronous, and indeed synchronous for all time determinations, with a specified stationary clock.

 

[Ich]

Yes, that´s the definition of coordinate time. Einstein shows how you can establish an inertial frame with position and time coordinates.
Note that he says:
We "have evidently obtained a definition of 'simultaneous' "
Now add what I posted before:

So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.

And you find out why he said "a definition" and not "the definition".
There are as many definitions of simultaneity as there are reference frames. They all are different.
So you have to conclude that when you say in our example "D reads more than A when they meet", that´s an "universal truth", because everyone will agree on that.
But when you say "B reads the same as D all the time", that is no universal truth. It is (by the above definition) true in the B-D frame, but obviously wrong in the A-E frame. And the A-E frame is just another inertial frame, just as valid as B-D.
So when you compare A to B (at the time A meets D) you get different results, depending on the frame.
Do we agree on that?

 

[yogi]

I am comparing D and B only in the BD frame - I do not say that B reads the same as D after B is put in motion - where did you get that idea - after B is put in motion, the longer he travels the more out of sync B will be with D. But when B is first put in motion - the local time on his clock will not have changed much from what it read before he (B) accelerated

I missed something - what is the AE frame? What post did you introduce E?

Anyway, to continue

B and D each read 100 years when A arrives at D, and A reads 86.6 years. All readings are made when the guy holding the clock looks at it. A and D know when to look at their clocks because A see's D and yells ahoy. B knows that D's clock reads the same as his and he knows that A's trip will take 100 years in the BD frame - so the observer holding the B clock reconds A should arrive at D when his own clock (B) reads 100 years.

So are we in agreement as to the local readings before B accelerates?

 

[pervect]

We've been over this a lot. Here's my suggestion - if you draw a space-time diagram with jpicedt http://jpicedt.sourceforge.net/ or some other tool, and draw the wordline of every clock you are interested in on the diagram, there is at least some chance that you will receive enlightenment.

If not, if you go through all the trouble of drawing the diagram, I'll be willing to take the time to compute as best I can, what any specific clock on said space-time diagram will read, if you give the beginning point, the ending point, and the path that the clock takes on the diagram.

Since we may have clocks that accelerate on this diagram, I'd need to konw either a specific acceleration, or some general guideline like "in the limit of high acceleration" if you don't need that level of detail. (The high-acceleration limit is a lot easier to compute, please use it if you don't need the detailed effect of finite accelerations). The start and ending velocity would specify the world-line for the high-acceleration limit, the start and end velocity plus the proper acceleration would be one way of specifying the world-line for a finite acceleration case.

Note that I am going to insist that if you want to know the interval reading of a clock, that you specify its path in space-time, and the two points which mark the beginning and ending of the interval. The points can be specified by the receipt of specific light signals (you need to draw the appropriate light-beam on the diagram).

I'm assuming that all of this occurs in flat-space time (no massive bodies).

 

[Ich]

Yes, spacetime diagrams could help.
And, yogi, what would help even more: read my posts before answering.
In #67, I introduced clock E to show you where you run into contradictions. All my posts since then rely on that.
Besides that, we agree that B reads 100 years when he starts accelerating. That´s how we defined it.

 

[Ich]

pervect, I think you could help even w/o diagrams.

I spent the last ~20 posts explaining why SR gives a certain result, but now it occurs to me that yogi still does not believe the result at all.
Maybe he will believe you.

Gedankenexperiment:
Twins A and B are at the same position.
A accelerates quickly to v=+0.6c (0.6 is easier to calculate than 0.5).
After 100 years (proper time), B also accelerates to v=+0.6c.
Then, A and B are brought slowly together.

Yogis claim: A will be younger than B, because he accelerated first.

 

[pervect]

Nah, yogi and I have been over this ground before. (He's been over this ground with other people like robphy, too.)

Drawing the space-time diagrams has the potential to help yogi a lot more than it does me.

If yogi finds the motivation to draw the diagrams, I'll find the motivation to do the calculations. If you want to try answering his questions without the diagrams, go ahead and try.

But I'll predict that that won't work. The exercise of formulating the problem in geometric terms (as a specific line on the diagram) is probably essential for yogi to understand the point that we are trying to make.

[add]
My prediction, based on past experience, is that the problem that you will find in trying to answer the questions without a diagram is that the questions themselves will become "slippery" and/ or contain hidden assumptions.

 

[Hurkyl]

B and D each read 100 years when A arrives at D, and A reads 86.6 years. All readings are made when the guy holding the clock looks at it. A and D know when to look at their clocks because A see's D and yells ahoy. B knows that D's clock reads the same as his and he knows that A's trip will take 100 years in the BD frame - so the observer holding the B clock reconds A should arrive at D when his own clock (B) reads 100 years.

So are we in agreement as to the local readings before B accelerates?

I'm not sure -- are you claiming all of these are local readings? I object to:

B ... read 100 years when A arrives at D

B knows that D's clock reads the same as his

so the observer holding the B clock reconds A should arrive at D when his own clock (B) reads 100 years.
​

none of these are local readings. The only have meaning through the construction of the mathematical abstraction called the "BD inertial reference frame". But relative to that frame, I agree.

(At least I assume that you meant all three of those statements relative to the BD inertial reference frame -- it's rather irritating that you habitually omit such qualifications)

 

[yogi]

Hurkyl - this was not my problem - Ich introduced it - so I interpreted it to mean what I have said in post 74 - D is taken as the at rest frame since D does not accelerate at any part of the experiment. A, B and D are all at rest initially in the D frame and synchronized - in the first part of the problem A and B are together and separated from D - so the analysis is straightforward - A quickly accelerates to a velocity v and maintains this velocity for the rest of the journey - this is right out of Einsteins 1905 paper Part 4 - Einstein didn't draw any spacetime diagrams, they are not necessary - When A arrives at D, A clock will read less than D. B and D are stilll at rest in the D frame - so B must read the same as D.

Or if you insist, draw A and B at the X-T origin and D at some distance X_d along the positive X axis - initialy B and D both move vertically parallel to the T axis to the time 100 years (their world lines are vertical). The plane of simultaneity of B and D is still parallel to the X axis - A's world line starts at the origin and slopes upwardly to intersect the event D = 100 years with a space coordinate X_d. A's time will be less than D's time by the gamma factor. and since B and D have not moved since being synchronized, B will also read 100 years. All times are read by the guy accompanying the watch (proper times only) - What A reads for the B watch when A has reached D is an apparent time -

 

[yogi]

Ich - your post 67 - I recall now - I still don't understand the need to introduce all kinds of ultrafast observers to answer the question of what does B clock read at instant before it is accelerated - proper time is not dependent upon the distance or velocity of a relativly moving observer. Apparent times are misleading - and while it is true you can make measurments in a relatively moving frame and arrive at correct results if you do the proper book-keeping - the measurments themselves are not physical reality - e.g., this is the difficulty with attempting to explain the twin paradox by saying the clock of the stay at home twin appears to advance very fast when the traveling twin turns around - the reality of the situation is otherwise - clocks don't suddenly increase their readings because some other clock executes an acceleration. Einstein attempted to explain the twin paradox in 1918 by introducing a pseudo G field - if you get the correct answer - maybe it doesn't make any difference.

 

[Hurkyl]

so I interpreted it to mean what I have said in post 74 - D is taken as the at rest frame since D does not accelerate at any part of the experiment.

That is the point I wanted to make. Those three measurements I quoted are not local. They only make sense relative to a coordinate chart, so they are all "apparent", according to how you seem to be using the word.

 

[Ich]

Ich - your post 67 - I recall now - I still don't understand the need to introduce all kinds of ultrafast observers to answer the question of what does B clock read at instant before it is accelerated - proper time is not dependent upon the distance or velocity of a relativly moving observer.

But you did not ask about the instant before B is accelerated. You asked:

tell me what you think B clock will read at the time A arrives at D.

Do you really think that is the same?

And to make some progress:

Please answer these questions:
1. What time is the event "B accelerates" (t=100,x=0) in the A frame? (use v=.6, then gamma=1.25)

2. There is a difference between this time and B´s reading of 100 years. Will this difference remain unchaged if A and B then move slowly to meet each other?

3. Do we agree now?

 

[yogi]

Question 1 - if A stops when he reaches D, then B,A and D are all in the same frame (prior to B accelerating). When the owner of the A clock reads the A clock, it will read less than the reading made by the owner of the D clock when the D clock owner reads the D clock at the spacetime event marking the arrival of A.

B and D have remained at rest in the same frame and each will read the clock he owns as 100 years. D owner knows when to read D clock because he sees A arrive, A owner knows when to read A clock because he sees D. B owner checks the time on the B clock and makes sure he does not accelerate until his clock reads 100 years.



So the B and D clocks are running in sync until B accelerates, The A clock is now running at the same rate as B and D, but it is out of phase in the amount (100 years)/1.25 The world line of A is now vertical in the coordinate system of D, as are the world lines of B and D. Prior to pulling up to a stop at D, A would judge B clock to be running behind. ... But that is an apparent observation - analogous to the outbound twin mistakenly observing the stay at home twin to be aging more slowly by assuming the traveling frame to be at rest - but my whole purpose in starting this discussion is to avoid this type of analysis - as I have said - you get the correct answer if you translate the observations propertly - but it does not reveal anything about the physics - and that is what interest me.

Hurkyl and Ich ...What i am saying is that the times are local - the intent of this thread was to pose the question of whether clocks can be differentiated in relativly moving reference frames - In other words, does real time dilation involve local changes in the quantities that determine proper times.

As I stated in a previous post, you can do an experiment with lab generated pions and instantly accelerate some of the them (A pions) and delay the acceleration of some (B pions) and measure the average distance traveled by the A and B pions If any of you who wish to bet on the B pions traveling further (as per Ich) I will cover any and all takers.

Unfortunately, This thread got side-tracked and I am leaving for vacation w/o my computer.

Yogi