Looking to understand time dilation

[Mike_Fontenot]

Tell me Mike, in your Twin Paradox scenario you proposed several posts back, how much tape gets played out?

And how does this relate to time dilation and length contraction?

In my given scenario, the traveler (he) is 20 years old when he does his turnaround. He will say the home twin (she) is 10 years old at that instant. But she will say that she was 40 years old when he turned around. They are both correct. That's the time dilation result.

Since he had been moving for 20 years of his time since he left home, he says he was 0.866*20 = 17.32 lightyears away from home at the turnaround. He would read 17.32 lightyears on HIS measuring tape when he turned around.

When the traveler was 20 years old (which was when he turned around), the home twin says that she was 40 years old. Since he had been moving for 40 years of her time since he left home, she says he was 0.866*40 = 34.64 lightyears from her at the turnaround. She would read 34.64 lightyears on HER measuring tape when he turned around.

Note that they EACH use the SAME velocity for their relative motion: 0.866c.

Those last three paragraphs are basically the length contraction result. She says he was TWICE as far away at the turnaround as he says he was. My previous post describes some potential pitfalls to avoid when using the length contraction result.

Mike Fontenot

 

[Mike_Fontenot]

I like the idea of the measuring tape as an instrument that measures accumulated distance similar to the way a clock measures accumulated elapsed time. I would add one small correction. The measuring tape reads the current separation according to the stay at home twin.
[...]

I think I FINALLY understand the point you were making, Yuiop. (And GHWellsJR was possibly thinking along the same lines).

That measuring tape that I described is indeed a very STRANGE tape.

The simplest way to understand the length contraction result (at the turnaround) in terms of a measuring tape, is for each twin to use a measuring tape that is stationary with respect to himself (or herself). That's the simplest, but it requires that each twin enlist the aid of another observer, who stationary in the given twin's frame, and who happens to be co-located with the other twin at the instant of the turnaround. That helper observer can (eventually) tell the given twin what the other twin's age was at the instant of co-location.

The measuring tape that I described previously is NOT such a tape: except for the reel itself, that tape is moving at velocity 0.866c with respect to the given twin (and is NOT moving with respect to the other twin).

Yet that tape gives the given twin the SAME reading, at the turnaround, as the tape that is stationary with respect to the given twin would give.

So the calibration marks on that moving tape WOULDN'T be correct, according to the given twin, if it were brought to a standstill with respect to the given twin ... its calibration marks WOULDN'T then match the marks on the permanently stationary tape. The moving tape is a tape that the given twin will find useful ONLY when it is moving at a velocity of 0.866c with respect to the given twin.

The moving tape is nevertheless a useful tape, because it gives him an IMMEDIATE answer to the question "How far away from me is my twin, currently?". He doesn't have to wait for any helper observer in his frame to tell him that answer.

But it's a very strange tape, because it has to be moving at 0.866c in order to be useful to him.

Of course, the EASIEST way the traveler can immediately determine his current distance from his home twin, is to not use any measuring tape at all, but rather, to simply multiply his velocity with respect to her by the amount of time that has elapsed for him since he left home.

Mike Fontenot

 

[Mike_Fontenot]

I decided that I needed to answer (at least for myself) two more questions about the traveler's STRANGE tape:

(1) If the traveler brings that strange tape to a standstill (with respect to himself), how will the markings on it compare with the markings on his own correct permanently stationary tape? (Recall that when his strange tape is moving at 0.866c (with respect to himself), the traveler will conclude (at each instant) that its marks have the same spacing as the marks on his permanently stationary tape).

(2) The traveler's strange tape is already stationary with respect to the home twin. How do the markings on that tape compare to the markings on her own correct permanently stationary tape?

If I've done the analysis correctly, here are the answers:

(1) If the traveler brings the strange tape to a standstill, its marks labeled as inches will be FARTHER APART than the inch marks on his correct permanently stationary tape, by a factor of gamma.

(2) The marks labeled as inches on the strange tape, will likewise be FARTHER APART than the inch marks on the home twin's correct permanently stationary tape, by that same factor gamma.

So the strange tape ISN'T the same as EITHER the home twins real tape, OR the traveler's real tape.

One more question: Is there a 12-step program I can join, for recovering "Physics Forum postaholics"?

Mike Fontenot

 

[ghwellsjr]

Mike, you need some direction:
1) Only use one frame of reference at a time.
2) Analyze what each observer measures locally and directly. No "remote measurements" allowed.
3) Assume each twin has an identical clock that flashes once an hour and they each count their own flashes and the other twin's flashes when they arrive. Keep track of flashes in transit.

If you do this correctly for multiple frames, you will discover that it doesn't matter which frame you use, each observer makes the same measurements but you have to keep in mind that you can see more than the observers can see.

You can call me George.

 

[yuiop]

I think I FINALLY understand the point you were making, Yuiop. (And GHWellsJR was possibly thinking along the same lines).

That measuring tape that I described is indeed a very STRANGE tape.

I think we are on the same page now. I just want to check by expanding on the numerical example you gave earlier, by introducing a space station at the turnaround point.

Earth to Space Station distance = 34.64 lightyears in Earth frame.
Velocity of traveller relative to Earth = 0.866c


It takes 40 years (Earth time) to travel from Earth to the Space Station (SS).
It takes 20 years (Traveller time) to travel from Earth to the SS.

The Earth and Space Station clocks are synchronised in the Earth frame.
When the traveler arrives at the SS he sees 40 years on the SS clock (and 20 years on his own).

The distance on the tape measure (one end fixed to the Earth) will read 34.64 lightyears when the traveler arrives at the SS. The distance according to the traveller using his velocity and the time he has been traveling is 0.866*20 = 17.32 lightyears and this disagrees with what his unspooling tape measure measures. If he had an odometer like that of a car that measures mileage, it would read the correct length contracted distance of 17.32 lightyears. If the traveller has a second tape measure that has the spool on the Earth and if he asks the Earth observer what the second tape measured at time t=10years Earth time then the Earth observer would say 17.32 lightyears, in agreement with what the traveller thinks the distance is when he arrives at the Space Station. The Earth observer on the other hand considers the time for the traveller to arrive at the SS to be 40 Earth years and if she looks at the second tape measure at time t=40 years Earth time, the tape will indicate 69.28 lightyears which is greater than the distance that either observer considers the Earth-SS distance to be, but it does agree with how far away from the Earth the traveller considers himself to be after 40 years have elapsed on the Earth. In short, the tape measure indicates the distance the observer with the fixed end of the tape considers the other observer to be but this information can only be obtained at a later time. The odometer device (rolling wheel rotations counter) or multiplying elapsed proper time by relative velocity gives an instantaneous local readout of distance traveled according to the traveller.

Agree?

 

[Dale]

But they are NOT completely equivalent. The fact the time and spatial coordinates have different signs in the metric is an immediate hint that they are not completely equivalent.

Yes, I agree.

 

[Mike_Fontenot]

[...]
The distance on the tape measure (one end fixed to the Earth) will read 34.64 lightyears when the traveler arrives at the SS.
[...]

No, that tape reads 17.32 lightyears at that instant. That tape is a STRANGE tape ... it is NOT a standard tape.

That tape is stationary with respect to the home twin, but if she compares the spacing of its "inch" marks, compared to the inch marks on her own standard stationary measuring tape, she will directly see that the "strange" tape's "inches" are longer than real inches, by a factor of gamma. She will regard it as a useless, incorrectly calibrated tape.

But the "inch" marks on the strange tape are exactly what is required, so that when the traveler reads that moving tape at the turnaround, it will (by design) give him the correct distance to his twin (according to HIM). The reading on that tape will be consistent with the traveler's own elementary distance calculations, using only (1) the elapsed time on his own watch during his outbound leg, and (2) his knowledge of his relative velocity, which together allow him to easily determine his current distance from his twin.

Since that tape is moving with respect to the traveler, its "inch" marks are length-contracted for him. The incorrect, too-large "inches" are designed so that, when they are contracted, they are exactly equal to one real inch, because of their motion. Obviously, that tape will be of no use to the traveler, if his (constant) velocity is anything OTHER than +-0.866c with respect to his twin.

It was your original comment that made me realize that the tape is NOT a standard tape. I hadn't originally realized that. Thanks.

Mike Fontenot

 

[granpa]

instead of a tape measure I like to use a
long line of stationary stations spaced 1 light sec apart and
each simultaneously sending out radio pulses at one sec intervals.

one can then create a second long line of stations' which are moving along with the moving twin.
from the traveling twins point of view these new stations' are spaced 1 light sec apart and
each is simultaneously sending out radio pulses at one sec intervals.

from the stationary twins point of view this new line of stations' is shrunk by a factor of gamma
and the time' between its pulses' is time dilated (increased) by a factor of gamma.

dont forget that the length of an object is the distance between front and back at one 'simultaneous' moment.

 

[ghwellsjr]

No, that tape reads 17.32 lightyears at that instant. That tape is a STRANGE tape ... it is NOT a standard tape.

That tape is stationary with respect to the home twin, but if she compares the spacing of its "inch" marks, compared to the inch marks on her own standard stationary measuring tape, she will directly see that the "strange" tape's "inches" are longer than real inches, by a factor of gamma. She will regard it as a useless, incorrectly calibrated tape.

But the "inch" marks on the strange tape are exactly what is required, so that when the traveler reads that moving tape at the turnaround, it will (by design) give him the correct distance to his twin (according to HIM). The reading on that tape will be consistent with the traveler's own elementary distance calculations, using only (1) the elapsed time on his own watch during his outbound leg, and (2) his knowledge of his relative velocity, which together allow him to easily determine his current distance from his twin.

Since that tape is moving with respect to the traveler, its "inch" marks are length-contracted for him. The incorrect, too-large "inches" are designed so that, when they are contracted, they are exactly equal to one real inch, because of their motion. Obviously, that tape will be of no use to the traveler, if his (constant) velocity is anything OTHER than +-0.866c with respect to his twin.

It was your original comment that made me realize that the tape is NOT a standard tape. I hadn't originally realized that. Thanks.

Mike Fontenot

Mike,

You are creating a new so-called paradox within the Twin Paradox but you are not even doing it correctly because your strange tape needs to be contracted, not stretched. But anyway, you are just demonstrating the ambiguity of accelerating a long object and how it can end up with different lengths depending on how the different parts of it are accelerated. This might be fun to analyze sometime, but it's not going to shed any light on understanding the Twin Paradox.

You had it right when you mentioned that the traveling twin can keep track of his distance traveled by multiplying his accumulated time by his speed, both of which he can measure.

Are you going to follow my suggested directions?

--George

 

[Mike_Fontenot]

[...]

Yeah, the simplest way to do a direct measurement of length is to use standard (Lorentz) spatial coordinates that are stationary with respect to the observer doing the measurements.

I haven't made any use of the "strange" tape before ... it just arose naturally in some of the previous discussion. Makes an interesting exercise, though.

BTW, I always find your posts to be very enlightening, and "spot-on".

Mike Fontenot

 

[Mike_Fontenot]

[...]
[...] you are just demonstrating the ambiguity of accelerating a long object [...]
[...]

There aren't any accelerations involved in the most recent discussions.

[...]
Are you going to follow my suggested directions?
[...]

No.

Mike Fontenot

 

[ghwellsjr]

There aren't any accelerations involved in the most recent discussions.

You don't think feeding out tapes that are stationary with respect to each twin on one end but attached to the other twin on the other end doesn't involve acceleration?

--George

 

[phyti]

Length contraction is just as physically real as time dilation is or else a light clock would keep a different time as it was rotated. (Remember MMX?)

-MMX with equal path lengths resulted in no interference. Kennedy-Thorndike
experiment using unequal path lengths resulted in no interference. Thus
interference is not a function of path length, therefore length contraction
is not an explanation.

 

[phyti]

No, I'm sorry but how can the 'speed of a body be relative to the speed of light which is 'c' for any and every FoR?

-The propagation speed of light in space is c, as verified by experiment. The
speed of light relative to an observer in an inertial frame is c by
definition. see paragraph 1 of Einstein's 1905 paper.
Consider the terms in gamma, where time dilation is a function of the ratio v/c.

Let me put this another way and try to shew you just what is bothering me. Maybe you can explain it to me...

A is a clock and it keeps 'regular' time - (however you want to define regular)- and it will accumulate time at the same rate for ever unless something happens to change it.

If we now introduce an observer B, and we don't meed to say where B is, he could be adjacent to A or he might be hundreds of Light Years away.

Let us say that B is traveling at velocity v relative to A, and therefore A is no longer stationary but traveling at v according to the observer B.

Now you say that B will observe A to be running slow, therefore it is running slow:

So we have clock A running along quite happily on its own at a steady rate but introduce an observer B at ANY distance from A and because they are moving with repect to one another A starts to run slow.

Can anyone explain cause and effect here??

-Begin with two identical clocks moving together at constant speed in space
relative to some random object (sun, moon, distant star, etc.). Clock1 moves
away from clock2 at constant speed. Ignoring the cmb or the 'fixed stars' as
a reference for determining 'absolute' motion, we don't know the initial
speed of the pair of clocks. Since the initial speed is unknown, the relative
speed cannot determine how much time dilation each clock experiences. The
relative speed could be divided between the two clocks in any number of ways,
a% for clock1 and (100-a)% for clock2. Assigning all the relative speed to
either clock, SR calculates the same time dilation for both, not because they
are moving at the same speed but because the theory is designed to produce
symmetrical results via the simultaneity definition. Relating this example to
the 'twins' problem within the framework of SR, both clocks are running
slower because both are moving, but it's not possible to determine if one
clock is running slower than the other. Only by reuniting them will any
difference in age be evident. In the simple case of the twins involving
constant speed, the one that moves to reunite is the youngest.

I would recommend researching the 'light clock' for understanding why a
moving clock runs slower, and the simultaneity definition.

 

[phyti]

If the space twin takes an extra clock with him and sets it to run gamma times his own clock rate, he will know what the Earth clock indicates at any time during his trip, without regard to delays and calculations.

 

[ghwellsjr]

Length contraction is just as physically real as time dilation is or else a light clock would keep a different time as it was rotated. (Remember MMX?)

-MMX with equal path lengths resulted in no interference. Kennedy-Thorndike
experiment using unequal path lengths resulted in no interference. Thus
interference is not a function of path length, therefore length contraction
is not an explanation.

Lorentz explained the null result of MMX by length contraction because he (and everyone else at the time) believed that MMX really was moving against an absolute stationary medium for light.

In Special Relativity, the same explanation is used for MMX when the frame of reference is not stationary with respect to MMX.

But beyond that point, are you disagreeing with my statement that length contraction is just as physically real as time dilation?

 

[ghwellsjr]

-The propagation speed of light in space is c, as verified by experiment. The
speed of light relative to an observer in an inertial frame is c by
definition. see paragraph 1 of Einstein's 1905 paper.

The round-trip speed of light experimentally measured by any inertial observer is c. This has nothing to do with any frame and nothing to do with Special Relativity, although Einstein affirms this fact in his 1905 paper.
The one-way speed of light in any inertial frame is arbitrarily defined to be c in Special Relativity, whether or not there is an observer at rest in that frame. For all other inertial observers (not at rest in that frame) the one-way speed of light is not c but the round-trip speed of light is c. These are very important distinctions pointed out by Einstein in his 1905 paper.

Consider the terms in gamma, where time dilation is a function of the ratio v/c.


-Begin with two identical clocks moving together at constant speed in space
relative to some random object (sun, moon, distant star, etc.). Clock1 moves
away from clock2 at constant speed. Ignoring the cmb or the 'fixed stars' as
a reference for determining 'absolute' motion, we don't know the initial
speed of the pair of clocks. Since the initial speed is unknown, the relative
speed cannot determine how much time dilation each clock experiences. The
relative speed could be divided between the two clocks in any number of ways,
a% for clock1 and (100-a)% for clock2. Assigning all the relative speed to
either clock, SR calculates the same time dilation for both, not because they
are moving at the same speed but because the theory is designed to produce
symmetrical results via the simultaneity definition. Relating this example to
the 'twins' problem within the framework of SR, both clocks are running
slower because both are moving, but it's not possible to determine if one
clock is running slower than the other. Only by reuniting them will any
difference in age be evident. In the simple case of the twins involving
constant speed, the one that moves to reunite is the youngest.

I would recommend researching the 'light clock' for understanding why a
moving clock runs slower, and the simultaneity definition.

The assigning of relative speeds between two clocks is not linear. You have to use the relativistic velocity formula. And it doesn't matter which frame of reference you view the two clocks or observers in, they view each other's time dilation as the same although from the point of view of the reference frame they can be different.

 

[Grimble]

-The propagation speed of light in space is c, as verified by experiment. The
speed of light relative to an observer in an inertial frame is c by
definition. see paragraph 1 of Einstein's 1905 paper.
Consider the terms in gamma, where time dilation is a function of the ratio v/c.


-Begin with two identical clocks moving together at constant speed in space
relative to some random object (sun, moon, distant star, etc.). Clock1 moves
away from clock2 at constant speed. Ignoring the cmb or the 'fixed stars' as
a reference for determining 'absolute' motion, we don't know the initial
speed of the pair of clocks. Since the initial speed is unknown, the relative
speed cannot determine how much time dilation each clock experiences. The
relative speed could be divided between the two clocks in any number of ways,
a% for clock1 and (100-a)% for clock2. Assigning all the relative speed to
either clock, SR calculates the same time dilation for both, not because they
are moving at the same speed but because the theory is designed to produce
symmetrical results via the simultaneity definition. Relating this example to
the 'twins' problem within the framework of SR, both clocks are running
slower because both are moving, but it's not possible to determine if one
clock is running slower than the other. Only by reuniting them will any
difference in age be evident. In the simple case of the twins involving
constant speed, the one that moves to reunite is the youngest.

I would recommend researching the 'light clock' for understanding why a
moving clock runs slower, and the simultaneity definition.

I'm sorry, but you've lost me here, not with what you've written here but with its relevance.
It would be helpful if you could address what I had described rather than going off at a tangent (a figurative one that is).

 

[ghwellsjr]

Grimble--

Some people here think I'm being a little too harsh on you because they think you picked up your false notions from some introductory textbook or other reference that states that there really is no age difference between the traveling twin and the home twin after they re-unite. I have asked you where you picked up that idea but you have not responded. Please tell us where you got that idea.

I am not trying to get you into trouble with anyone Mr. Wells(?) and your response to me is not a problem. As I have said the only 'introductory textbook' I have referenced is http://www.bartleby.com/173/" .

It is a very simple, straightforward and erudite text.

I have tried to shew what difficulties I have with SR as it is today by stating what I have understood and been shown in Einstein's writing.

I am, humbly, looking for explanation of the logic that obviously eludes me. If it is logical and complete then surely someone will be able to show it to me?

Again, many apologies for upsetting the applecart...

Sorry, in case I still haven't answered your question; That is the understanding I have had for many years. I left University in 1969 so you will excuse me, I hope if I cannot give you the titles and authors of the publications.

Einstein says that the traveling twin comes back younger. Why do you deny what he says?

It would be helpful if you could address what I had described rather than going off at a tangent (a figurative one that is).

And it would be helpful if you would answer my question. You claim that Einstein is your source of everything you know about relativity and you have given me a very good reference to one of his writings. Can you point to where in that document he says that a moving clock only apparently runs slower than a stationary clock and that when it stops moving, all the time it lost is suddenly regained?

If I can point you to another of Einstein's documents where he specifically says that a moving clock actually loses time even when it comes to rest at its starting point, will you give up your idea that time dilation is just an illusion?

 

[granpa]

instead of a tape measure I like to use a
long line of stationary stations spaced 1 light sec apart and
each simultaneously sending out radio pulses at one sec intervals.

one can then create a second long line of stations' which are moving along with the moving twin.
from the traveling twins point of view these new stations' are spaced 1 light sec apart and
each is simultaneously sending out radio pulses at one sec intervals.

from the stationary twins point of view this new line of stations' is shrunk by a factor of gamma
and the time' between its pulses' is time dilated (increased) by a factor of gamma.

dont forget that the length of an object is the distance between front and back at one 'simultaneous' moment.

from the traveling twins point of view
each individual clock on each individual stationary station is ticking at 1/gamma but
the cumulative time as told by each passing station (as it passes him) is accumulating at a rate of 1*gamma
(because in his frame those clocks are not synchronized)

when the traveling twin halts at the last clock then
those same clocks will be synchronized from his point of view and
they all agree with the last clock

 

[Grimble]

Hi Grimble, there are a few problems with this chart.

1) In spacetime diagrams the time axis is traditionally vertical and the space axis is horizontal. It is ok to switch it around, but then you should label it.

2) For AA (the clock going horizontally from A and staying on A) coordinate time will match proper time.

3) For BB coordinate time will match proper time.

4) For AB proper time will be slower than coordinate time by a factor of 0.8, meaning that on the line where coordinate time is 5 the proper time for AB is 4, not 5.

5) Similarly for BA.

6) You have not shown any lines of simultaneity for AB or BA, only for AA and BB.

The relationship of times as experienced by two moving FoR

http://img440.imageshack.us/img440/5448/specialrelativitydiagra.jpg

Uploaded with ImageShack.us

Ok, Let me deal with the points you have raised.

1). Accepted but this diagram will show both Length Contraction or Time dilation depending on which units are used.

2). yes, for AA coordinate = Proper = Blue figures.

3). Yes - once again shown in blue.

4). With reference to time Einstein based his formula on the ticking of the clock.

So 1 tick = i unit of proper time on A's axis which is equal to 6.25 (Proper dimensioned) units along AB so 5 proper units = 6.25 coordinate units. (the red label saying proper units is misleading here)

5). And the same for BA.

6). The only points of simultaneity here are the 'ticks' of the one physical clock here which is at rest in frame A and seen to be moving From frame B. We can see here that the moment the clock ticks is seen as 1 unit of time by A but as 1.25 units of time by B.

What other plane of simultaneity is there?

 

[Grimble]

Grimble--

Some people here think I'm being a little too harsh on you because they think you picked up your false notions from some introductory textbook or other reference that states that there really is no age difference between the traveling twin and the home twin after they re-unite. I have asked you where you picked up that idea but you have not responded. Please tell us where you got that idea.

I have calculated that. Einstein does not specify when the twin returns whether he is still travelling, at rest or whether that makes a difference, but looking at the LT equations brings me to the conclusion that it does make a difference.

I am not saying I am right, I am asking you to SHOW me how I am wrong.

Do you never make your own calculations to check how it all fits together? To check your understanding?

Please don't be annoyed I am trying to learn...

 

[Dale]

1). Accepted but this diagram will show both Length Contraction or Time dilation depending on which units are used.

Even so, it is best to label your axes when you are going against the usual convention.

2). yes, for AA coordinate = Proper = Blue figures.

3). Yes - once again shown in blue.

OK, you need to label things more clearly. You have 8 lines of numbers in various colors and locations, but only 2 different labels, which are repeated. So if the blue row along the AA line is the coordinate time what is the pink row along AA which is labeled coordinate time?

4). With reference to time Einstein based his formula on the ticking of the clock.

So 1 tick = i unit of proper time on A's axis which is equal to 6.25 (Proper dimensioned) units along AB so 5 proper units = 6.25 coordinate units. (the red label saying proper units is misleading here)

5). And the same for BA.

Use the Lorentz transform. The time dilation formula is a simplification of the Lorentz transform which beginners misuse all the time. Avoid it always, it will not save you time, and it will lead you into mistakes as it has here.

6). The only points of simultaneity here are the 'ticks' of the one physical clock here which is at rest in frame A and seen to be moving From frame B. We can see here that the moment the clock ticks is seen as 1 unit of time by A but as 1.25 units of time by B.

What other plane of simultaneity is there?

Your answer here doesn't make a lot of sense. There are 4 clocks, not just one. Also, A and B appear to indicate two different locations in one frame, not two different frames, so I don't know what you mean by frame B.

Remember simultaneity is relative, so the moving clocks AB and BA will judge different pairs of events to be simultaneous than the stationary clocks AA and BB. The dotted and dashed lines connect events that are judged to be simultaneous in the stationary reference frame. Again, use the Lorentz transform to determine which events are considered simultaneous in each frame.

 

[Grimble]

Einstein says that the traveling twin comes back younger. Why do you deny what he says?

I do not deny what he says at all:

Let us now consider a seconds-clock which is permanently situated at the origin (x' = 0) of K'. t' = 0 and t' = 1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks:
t = 0
and

As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but

seconds, i.e. a somewhat larger time. As a consequence of its motion the clock goes more slowly than when at rest. Here also the velocity c plays the part of an unattainable limiting velocity.

 

[Grimble]

From what I can see you've failed to understand the most basic concepts in his paper. My advice would be to approach this in a manner that's more conducive to actually learning the material. Far from upsetting the applecart, you're just being a little difficult, but far from shocking for someone new to the material. You're asking for the logic behind the papers?... they PRESENT their logic, which is new and then explains itself. Subsequent decades have only supported the conclusions presented there, so you might want to take in some more material designed to introduce someone to the basic concept of what "RELATIVE" means.

Your arguments ignore something which has been experimentally verified with cesium clocks in airplanes, and many times since. The logic is this: Einstein is presenting you with a new kind of universe from your previous Newtonian view... beyond that, what do you want, a burning bush to reassure you of SR's current standing?!

Einsteins paper presents no difficulties in understanding his logic that says that we must put aside the assumptions that science has made and clung to for so long.
That we should understand that time (and distance) is seen to pass at different rates depending on where and under what conditions it is measured.

I have questions that arise from this yet the only help it seems that I can get in understanding what are misconceptions or misunderstandings are such as you have stated here!

Everyone is very happy to say the logic is easy to understand yet no one is willing to say more than "... they [Einstein's own papers] PRESENT their logic, which is new and then explains itself."

I have spent hours going over these papers and their logic seems simple and straightforward, yet some of the conclusions and derivations from them do not and I am trying to determine why this is.

Just read threads such as this and the conclusion leaps out at you that none of the experts here even agree as to what it is and how you see it.

 

[Dale]

Just read threads such as this and the conclusion leaps out at you that none of the experts here even agree as to what it is and how you see it.

How do you come to that conclusion?

 

[Grimble]

All I want to see is the logic of how this can be... Science should never be a matter of faith surely.

Excuse for not replying in full to your calculations the reason being that although I am an unknowledgeable lover of mathematics I only find the formulae and comncepts of any beauty, numerical callculations detract from this beauty. I am also very lazy.

Sort of summing up your statements, yes, all measurements or observervations are observer dependent. That is, dependent on the observers state of motion, not on relative position, which is allowed for by taking into account signal travel times.

In regards to measurements being taken, it is my belief that the only clocks and rulers that you can reliably use are those at rest with respect to you. Almost anything that you can say about lengths or times in a frame moving relative to you is arrived at from measurements in your reference frame with your rods and your clocks. [/QUOTE]

Exactly that is what I am trying to show in this thought experiment!

Certain things, such as proper time and the intrerval, are the same for all observers and so are not frame or speed dependent, and these things are at the heart of the matter being as it were entities in themselves.

As for being a matter of faith, once you agree to accept the hypotheses as a working basis, which you are of course not obliged to, then whole of SR follows from logic, and we must insist that everyone buys into fundamental logical priciples.

The concepts are fairly simple, but they are not intuitive, that ids usually the problem.

Matheinste.

Yet that IS the problem that I have! Everyone says in essence what you have said there

... then whole of SR follows from logic, ...

And logic is exactly what I am applying BUT when I want to check that my understanding, my logic is correct everyone obfuscates!

No one will answer the points I make - such as that thought experiment, by showing me where I am wrong, they take single points and go off at tangents which have nothing to do with what I am querying.

Let me say here that I do not have a problem with Einstein's SR which I see as a simple beautiful, logical and elegant discovery. But I see no consistency in the different worlds and logics that others want to build upon it.

 

[ghwellsjr]

Grimble--

Some people here think I'm being a little too harsh on you because they think you picked up your false notions from some introductory textbook or other reference that states that there really is no age difference between the traveling twin and the home twin after they re-unite. I have asked you where you picked up that idea but you have not responded. Please tell us where you got that idea.

I have calculated that. Einstein does not specify when the twin returns whether he is still travelling, at rest or whether that makes a difference, but looking at the LT equations brings me to the conclusion that it does make a difference.

I am not saying I am right, I am asking you to SHOW me how I am wrong.

Do you never make your own calculations to check how it all fits together? To check your understanding?

Please don't be annoyed I am trying to learn...

Grimble, there is no expert here, in fact no novice here that I am aware of except you, that does not believe the traveling twin comes back younger.

I invite you to look at Einstein's 1905 paper which is the origin of the Twin Paradox. You can read this just before the end of the 4th section:

"From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B 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^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

"It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.

"If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: 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 has remained at rest the traveled clock on its arrival at A will be 1/2 t*v^2/c^2 second slow."

Notice in the first paragraph the words "remained at B". That means the traveling twin has stopped at a remote location (with respect to A). He uses the term "lags behind the other".

He then goes on in the second paragraph to explain that the same thing is true if B and A are at the same location, which is the more familiar explanation of the Twin Paradox.

Finally in the third paragraph, he gives a formula for a clock that travels in a circle from a starting point and returns to the starting point and very clearly says, the traveled clock will be slow.

So you can no longer use the excuse that you have learned from Einstein that it makes a difference whether the traveling twin has stopped or not, or that it makes any difference in the reading on his clock whether he has stopped or not.

You have to understand that a traveling clock returning to its starting point will have "lost" time. This is a fact of nature having nothing to do with Special Relativity or any other theory. Theories are our way of analyzing, explaining, and predicting what nature does but they don't cause nature to behave in any particular way. If you refuse to accept the fact that a traveling clock ends up losing time, then there is no point in trying to understand a theory that supports that fact because you will suspect the theory. That is what has been going on in this thread with you since the beginning.

This understanding is not a matter of doing any computation. It is a matter of doing experiments and, no, I have not repeated any of the experiments. But I do fully understand the computations and have done them myself and explained in very simple terms that does not even involve relativity how to see that the traveling twin comes back younger even though both always see the other one's clock as running slower than their own by the same amount.

 

[Grimble]

-The propagation speed of light in space is c, as verified by experiment. The
speed of light relative to an observer in an inertial frame is c by
definition. see paragraph 1 of Einstein's 1905 paper.
Consider the terms in gamma, where time dilation is a function of the ratio v/c.[

I'm sorry, are you saying "yes, you agree with me"?

-Begin with two identical clocks moving together at constant speed in space
relative to some random object (sun, moon, distant star, etc.).

You have me there. What has SR got to do with some sort of absolute motion ("... relevant to some random object ..." is still absolute, the random is what object defines the absolute reference)

Clock1 moves away from clock2 at constant speed. Ignoring the cmb

the what? - spelling mistake?

or the 'fixed stars' as a reference for determining 'absolute' motion, we don't know the initial speed of the pair of clocks.

They don't need any reference to anything there is no reason whatsoever why they should be considered to be travelling, is there?

Since the initial speed is unknown, the relative speed cannot determine how much time dilation each clock experiences. The relative speed could be divided between the two clocks in any number of ways, a% for clock1 and (100-a)% for clock2.

?

Assigning all the relative speed to either clock, SR calculates the same time dilation for both, not because they are moving at the same speed but because the theory is designed to produce symmetrical results via the simultaneity definition.

Or Relativity, as we call it.

Relating this example to the 'twins' problem within the framework of SR, both clocks are running slower because both are moving,[/quote] yes ... ?

but it's not possible to determine if one clock is running slower than the other.

except that their relative speed is (and has to be - they are relative to each other after all) and so according to LT the time dilation MUST be identical!

Only by reuniting them will any difference in age be evident. In the simple case of the twins involving
constant speed, the one that moves to reunite is the youngest.

Could you please explain what you mean by all this?

I would recommend researching the 'light clock' for understanding why a
moving clock runs slower, and the simultaneity definition.

I must apologise if you think that I have any difficulty with the light clock - one of the first things I did in my study of SR was to derive the LT equations from the light clock. But one must be careful to use the correct units in the light clock depending on where you are observing it from.

One question you might like to answer for me?

Einstein wrote:

Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event.

So why do we measure time in different FoRs using THE SAME SCALE?

Surely, using Einstein's own equation:

We should not assume, given the above quote, that the units of time in the different FoRs have the identical durations?

Is he perhaps saying that time t in the rest frame has the SAME DURATION as (=) the time

?

 

[Grimble]

Even so, it is best to label your axes when you are going against the usual convention.

OK, you need to label things more clearly. You have 8 lines of numbers in various colors and locations, but only 2 different labels, which are repeated. So if the blue row along the AA line is the coordinate time what is the pink row along AA which is labeled coordinate time?

Use the Lorentz transform. The time dilation formula is a simplification of the Lorentz transform which beginners misuse all the time. Avoid it always, it will not save you time, and it will lead you into mistakes as it has here.

Your answer here doesn't make a lot of sense. There are 4 clocks, not just one. Also, A and B appear to indicate two different locations in one frame, not two different frames, so I don't know what you mean by frame B.

Remember simultaneity is relative, so the moving clocks AB and BA will judge different pairs of events to be simultaneous than the stationary clocks AA and BB. The dotted and dashed lines connect events that are judged to be simultaneous in the stationary reference frame. Again, use the Lorentz transform to determine which events are considered simultaneous in each frame.

I will redraw it in separate stages there is oo much to explain when it is all together, sorry

 

[sylas]

the what? - spelling mistake?

cmb is cosmic background radiation; a useful reference against which velocities can be measured.

One question you might like to answer for me?

Einstein wrote:

So why do we measure time in different FoRs using THE SAME SCALE?

Surely, using Einstein's own equation:

We should not assume, given the above quote, that the units of time in the different FoRs have the identical durations?

You can't just add the durations from two frames, because that is assuming and absolute notion of simultaneity. This is invariably behind the confusions people have with respect to so-called paradoxes of relativity.

Cheers -- sylas

 

[Grimble]

How do you come to that conclusion?

By reading all the posts on this thread and seeing how many times one corrects another

 

[sylas]

By reading all the posts on this thread and seeing how many times one corrects another

Um... a much more sensible conclusion would be that not everyone contributing is an expert, rather than "none of the experts agree".

It can be good practice for a non-expert to try and identify the errors being made in so-called paradoxes or contradictions. The experts will generally provide helpful corrections, and I think THIS is what you are seeing.

Cheers -- sylas

 

[Dale]

By reading all the posts on this thread and seeing how many times one corrects another

Can you cite an example on where one expert corrected another expert? I think they are not as common as you think. The experts will be the ones with a Science Advisor label (minimally) or a Mentor label (preferably).

 

[ghwellsjr]

One question you might like to answer for me?

Einstein wrote:

Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event.

So why do we measure time in different FoRs using THE SAME SCALE?

Surely, using Einstein's own equation:

We should not assume, given the above quote, that the units of time in the different FoRs have the identical durations?

Is he perhaps saying that time t in the rest frame has the SAME DURATION as (=) the time

?

I can't believe you are sincere in presenting these kinds of questions. Can you cite your reference to Einstein's quote? I'm sure we will see that he is in the process of developing an argument and I'm sure you could figure it out if you would just read the rest of his argument instead of trying to convince all the rest of us that you alone understand Einstein and his theories.