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In his 1905 paper Einstein introduced two coordinate systems which he called K and k. It seems that both systems are inertial because there are no external forces and they are moving at the constant relative speed v.
But I'm not certain, so are they both inertial? Is it correct that either one can be considered the 'stationary' system and the other one the 'moving' system?
George Jones
Jun30-09, 02:53 PM
In his 1905 paper Einstein introduced two coordinate systems which he called K and k. It seems that both systems are inertial because there are no external forces and they are moving at the constant relative speed v.
But I'm not certain, so are they both inertial? Is it correct that either one can be considered the 'stationary' system and the other one the 'moving' system?
Yes and yes.
So, with K( X,Y,Z,T) stationary the clocks of k( x,y,z,t) are moving in the + X direction at speed v. Putting X=vT in the transfforms leads to t=T/m, where m is the quantity Einstein calls beta. So t is less than T.
With k being stationary the clocks of K are moving in the -x direction at speed v. Putting x=-vt in the transforms leads to T=t/m, ie T is less than t.
Is there general agreement with this result? Is everyone OK with it?
HallsofIvy
Jun30-09, 03:24 PM
Yes, an observer in the K system would take the K system as stationary, K' as moving, and observe time in the K' system, t', as slower than t in the K system. An observer in the K' system would take the K' system as stationary, K as moving, and observe time in the K system, t, as slower than t' in the K' system.
Is there general agreement with this result? Is everyone OK with it
Yes.....It will have to do until a better theory is developed!
Thanks to you both. So can this example be generalized to say that ' Any analysis carried out with K stationary can also be carried out with k stationary, with suitable choice of k coordinates' ? ( In the example the -x coordinate was considered equivalent to the +X coordinate.)
To continue:
The 1905 paper adopted K as the stationary system for all its analysis. For the 'clock paradox' the result was t=T/m, the same as the slow clock.
Based on the above discussion, k can be chosen as stationary, and the 'clock paradox' calculated using the same concepts and math, resulting in T=t/m.
These results are compatible because the comparison is between the time of the stationary system and the time of the moving system 'as viewed from the stationary system'
The equivalence of K and k has apparently been overlooked, and consequently many writers have treated K as the only alternative, and have extensively elaborated this case.
OK?
To continue:
The 1905 paper adopted K as the stationary system for all its analysis. For the 'clock paradox' the result was t=T/m, the same as the slow clock.
Based on the above discussion, k can be chosen as stationary, and the 'clock paradox' calculated using the same concepts and math, resulting in T=t/m.
These results are compatible because the comparison is between the time of the stationary system and the time of the moving system 'as viewed from the stationary system'
The equivalence of K and k has apparently been overlooked, and consequently many writers have treated K as the only alternative, and have extensively elaborated this case.
OK?
What specific part are you referring to when you talk about the "clock paradox"? Is it this part from section 4 of the paper (http://www.fourmilab.ch/etexts/einstein/specrel/www/)?
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 travelled clock on its arrival at A will be (1/2)t*v^2/c^2 second slow.
Is so, note that there is an asymmetry in how the problem was described--he said that the two clocks were synchronized in the K frame up until the moment clock A was moved. This is crucial to understanding why clock A is the one that is behind when the two clocks meet. If he had instead said the clocks were still at rest in the K frame, but synchronized in the k frame, and then when clock A was accelerated it came to rest in the k frame while B continued to move towards it in the k frame (i.e. B continued to be at rest in the K frame), then in that case it would be clock B that was behind when they met.
Hello again, JesseM,
Yes, I refer to p.4 of '1905.
I agree with your description of the original 'clock paradox'. A and B start at rest in K, A moves and returns, and A ends up slow compared to B.
From your last sentence I gather that you recognize a valid alternative analysis of the clocks where B ends up slow compared to A.
Do you think that these two results are compatible, and that the results apply to the 'twin paradox?
Hello again, JesseM,
Yes, I refer to p.4 of '1905.
I agree with your description of the original 'clock paradox'. A and B start at rest in K, A moves and returns, and A ends up slow compared to B.
From your last sentence I gather that you recognize a valid alternative analysis of the clocks where B ends up slow compared to A.
It's not just an alternative "analysis" of the same physical situation though, it's an actually physically distinct scenario where the two clocks have initially been synchronized differently. In any given physical scenario involving two clocks that meet, there will be only one correct answer to what the clocks read at the moment they meet.
Do you think that these two results are compatible, and that the results apply to the 'twin paradox?
Again, the alternative physical scenario I mentioned doesn't contradict the fact that in the scenario as Einstein defined it, clock A will definitely be behind clock B, this is true regardless of what frame you analyze the scenario in. Likewise, in the twin paradox you will find that the inertial twin elapsed more time than the twin who accelerated regardless of which frame you use to analyze the voyage.
JesseM,
Please listen-
I have already said that I agree with Einsteins analysis.
What I am saying is that k has an equal right to consider himself to be stationary! Thus , if K sees k moving away, turning around, and returning, then k, considering himself to be stationary,sees K moving away, turning around and returning. Then by your description, k being stationary elapses more time than K who turns around. Einstein did not describe this situation.
Do you deny k the right to consider himself stationary?
JesseM,
Please listen-
I have already said that I agree with Einsteins analysis.
What I am saying is that k has an equal right to consider himself to be stationary! Thus , if K sees k moving away, turning around, and returning, then k, considering himself to be stationary,sees K moving away, turning around and returning. Then by your description, k being stationary elapses more time than K who turns around. Einstein did not describe this situation.
Do you deny k the right to consider himself stationary?
If k is not inertial, then k cannot be stationary, in any inertial frame. If k turns around, then k is not stationary, and there is no symmetry between k and K. But I don't think Einstein gave a single co-ordinate system for an observer that turned around and came back. You can't given an inertial co-ordinate system for such an observer.
JesseM,
Please listen-
I have already said that I agree with Einsteins analysis.
What I am saying is that k has an equal right to consider himself to be stationary! Thus , if K sees k moving away, turning around, and returning, then k, considering himself to be stationary,sees K moving away, turning around and returning. Then by your description, k being stationary elapses more time than K who turns around. Einstein did not describe this situation.
Do you deny k the right to consider himself stationary?
What do you mean "moving away, turning around, and returning"? That would suggest that k and K are not moving inertially relative to each other as Einstein defined them to be. In the clock scenario, I was assuming k was the inertial frame where A and B were initially moving at the same constant speed, then after A accelerated A was at rest in k while B continued to move at the same constant speed until it met with A and they compared clock readings. Are you thinking of k as a non inertial frame whose velocity in K's frame is not constant for all eternity? In this case k would not have an equal right to consider himself stationary and use the ordinary SR time dilation equation to calculate the time elapsed on moving clocks, because the SR time dilation equation only works in inertial frames. And remember, there is an absolute truth in SR about whether a given observer is moving inertially or non-inertially--a non-inertial observer will feel G-forces as he accelerates which can be measured by an accelerometer he carries alongside himself.
Lets attend carefully to what Einstein said as quoted be JesseM in his post no.8.
The first paragraph, 'From this...', is his statement of the 'slow clock' idea. His result is the approximation of the square root in the formula t=T/m, see post 3. I have identified clock B with K and clock a with k to agree with Einsteins definitions in his par. 3, where K is stationary and B doesn't move and k and clock A move. Both A and B are inertial.
In the second par. he says '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.'
By 'this result' he clearly means the 'slow clock' formula he just derived in the immediately preceding par. In addition he cites the 'slow clock' formula in the third par. regarding the continuously curved line.
Thus he is saying that the time lag experienced by a clock moving in a closed curve is the same as the lag experienced by a clock moving in a straight, inertial line. So he is totally ignoring any effects of accelerations, or any other effects of the curved or segmented path of the moving clock.
Since A and B are both inertial, either may consider himself to be stationary. The reasoning of the first par. is equally valid with A stationary and B moving, with the resulting time lag expressed by T=t/m.
Lets attend carefully to what Einstein said as quoted be JesseM in his post no.8.
The first paragraph, 'From this...', is his statement of the 'slow clock' idea. His result is the approximation of the square root in the formula t=T/m, see post 3. I have identified clock B with K and clock a with k to agree with Einsteins definitions in his par. 3, where K is stationary and B doesn't move and k and clock A move. Both A and B are inertial.
In the second par. he says '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.'
By 'this result' he clearly means the 'slow clock' formula he just derived in the immediately preceding par. In addition he cites the 'slow clock' formula in the third par. regarding the continuously curved line.
Thus he is saying that the time lag experienced by a clock moving in a closed curve is the same as the lag experienced by a clock moving in a straight, inertial line. So he is totally ignoring any effects of accelerations, or any other effects of the curved or segmented path of the moving clock.
Since A and B are both inertial, either may consider himself to be stationary. The reasoning of the first par. is equally valid with A stationary and B moving, with the resulting time lag expressed by T=t/m.
Good grief, no.
k is NOT the clock moving in polygons. It is a single inertial co-ordinate system, for the single leg moving from A to B. The "result that holds good" is that the clock moved from A to B ends up lagging behind the clocks at A and B which are synchronized in frame K.
Hence, when you continue to move the clock in a new direction but at the same speed, it will continue to lag further and further behind clocks that are at rest wrt A and synchronized with A. Hence, the end result is that the clock moving in a closed curve from A back to A again ends up by showing less time -- lagging behind -- the clock which remains at A throughout. THAT is what Einstein is saying if you play careful attention to his words; in terms of the twin paradox, the twin that moves in a curve is the one that ages less.
I've seen these kinds of discussions before; so be aware, I have no interest in arguing with you if you take the line that you are reading correctly and trying to explain Einstein. Einstein is quite clear and speaks for himself with admirable clarity... and you've misunderstood it. I can help explain if you are are genuinely uncertain.
Nothing in Einstein's words suggests that there's a frame "k" which applies for the clock all through its polygon or curved trip from A back to A again.
Cheers -- sylas
Lets attend carefully to what Einstein said as quoted be JesseM in his post no.8.
The first paragraph, 'From this...', is his statement of the 'slow clock' idea. His result is the approximation of the square root in the formula t=T/m, see post 3. I have identified clock B with K and clock a with k to agree with Einsteins definitions in his par. 3, where K is stationary and B doesn't move and k and clock A move. Both A and B are inertial.
In the second par. he says '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.'
By 'this result' he clearly means the 'slow clock' formula he just derived in the immediately preceding par. In addition he cites the 'slow clock' formula in the third par. regarding the continuously curved line.
But the "slow clock" formula is just meant to calculate the time elapsed on a clock which is moving in whatever inertial frame you're using. It is indeed true that if you pick different inertial frames K and k, and use them to analyze the time elapsed on a clock which moves on a polygonal line (an approximation to a continuous curve), you can use the same time dilation formula in each frame to calculate the time elapsed on each segment of the polygon--this would just be (difference in time coordinates between the beginning and end of the segment in your frame)*(time dilation factor in your frame)--and add them up to get the total time elapsed on the polygonal path, and both frames will end up predicting exactly the same time elapsed (if you like I could given an example of a simple polygonal path analyzed in different frames). But nowhere did Einstein mean to suggest that the same time dilation formula could be used in a non-inertial frame!
Thus he is saying that the time lag experienced by a clock moving in a closed curve is the same as the lag experienced by a clock moving in a straight, inertial line.
No, he certainly isn't! He's just saying you can break the polygonal path into segments and calculate the time elapsed on each segment the same way you would for the time between two events on the path of an inertial clock, then add up the segments to find the total time elapsed. But again, for each segment you must use an inertial frame to calculate the time elapsed (though it's not necessary to use the same inertial frame for different segments, since all frames make the same prediction for the time elapsed on a given segment).
Of course you are also free to take the worldline of the clock B that never changes its velocity and divide it into segments, calculating the time elapsed on each segment and adding them up to find the total time elapsed. But if you do this you must make sure that the end of each earlier segment corresponds to the same event as the beginning of the next segment. So you can't have one segment's end be "the event on clock B's worldline that's simultaneous with clock A accelerating in frame K", and the beginning of the next segment be "the event on clock B's worldline that's simultaneous with clock A accelerating in frame k", because according to the relativity of simultaneity these are two entirely different events which may leave an unaccounted-for section of B's worldline between the end of the first segment and the beginning of the second.
JesseM
Refering to your quote of Einstein in post 8:
In the first par. it states that the clock moved in a line from A to B lags behind B by the amount t v.v/2c.c. (c.c = c squared, etc.)
In the third par. it states that a clock moved in a closed curved line at constant velocity lags behind the statiionary clock by the amount t v.v/2 c.c.
These two formulas are identical. Therefore the conclusion is that the time lag experienced by a clock moving in a closed curve is the same as the lag experienced by a clock moving in a straight line.
One possible conclusion from this is that the curved path has no effect on the time lag. This is supported by Wikipedia-Twin Paradox. According to Langevin the acceleration ( of the curved path) is the cause of the asymmetry and not of the aging itself.
JesseM
Refering to your quote of Einstein in post 8:
In the first par. it states that the clock moved in a line from A to B lags behind B by the amount t v.v/2c.c. (c.c = c squared, etc.)
In the third par. it states that a clock moved in a closed curved line at constant velocity lags behind the statiionary clock by the amount t v.v/2 c.c.
Just a thought. You do understand the difference between "speed" and "velocity" don't you?
Velocity has both magnitude (= speed) and direction. A change in direction counts as a change of velocity and as acceleration. Something moving at constant speed around a closed curve is still accelerating and is not moving inertially. It would feel a "G-force" to prove the difference.
Apologies if you realise this already.
These two formulas are identical. Therefore the conclusion is that the time lag experienced by a clock moving in a closed curve is the same as the lag experienced by a clock moving in a straight line.This is true for two clocks both being measured by the same inertial observer. It's not true the other way round, if an inertial observer and an accelerating observer are both measuring the same clock. (Einstein doesn't consider accelerating observers in his paper.)
Thanks for your post, DrGreg
Yes I understand the difference, and Einstein surely did also. The appearnce of 'velocity' instead of 'speed' is likely due to the translation.
Since clock B is stationary for both the 'straight line' path of A and the 'closed curve' path, I assume you agree with my statement you quoted. Do you also agree with my last comment in post 17?
JesseM
Refering to your quote of Einstein in post 8:
In the first par. it states that the clock moved in a line from A to B lags behind B by the amount t v.v/2c.c. (c.c = c squared, etc.)
In the third par. it states that a clock moved in a closed curved line at constant velocity lags behind the statiionary clock by the amount t v.v/2 c.c.
The v in these formulas represents speed rather than velocity, and the formulas only work if you are using v to represent the speed as measured in an inertial frame. Do you understand that when he talks about motion in a closed curve at constant speed v, he is still talking about constant speed as measured in some inertial frame?
One possible conclusion from this is that the curved path has no effect on the time lag. This is supported by Wikipedia-Twin Paradox. According to Langevin the acceleration ( of the curved path) is the cause of the asymmetry and not of the aging itself.
Yes, that's correct. If you know a clock's speed as a function of time v(t) in some inertial frame, and you want to know how much time will elapse on the clock between two coordinate times t0 and t1 in that inertial frame (which could be the times of the clock departing from and returning to an inertial clock), then the formula is just a function of speed, not acceleration:
\int_{t_1}^{t_0} \sqrt{1 - v(t)^2/c^2} \, dt
For motion at constant speed v this just reduces to (t_1 - t_0)\sqrt{1 - v^2/c^2}, and if you subtract this from the time elapsed between the same events on a clock which is motionless in this frame, i.e. (t_1 - t_0) - (t_1 - t_0)\sqrt{1 - v^2/c^2}, you get a value for the time difference which is close to Einstein's approximation of \frac{tv^2}{2c^2} (he mentions that this is just an approximation in the paper).
JesseM; Thanks for your post. Isn't Einsteins clock B stationary and inertial while clock A performs it's circular motion?
JesseM; Thanks for your post. Isn't Einsteins clock B stationary and inertial while clock A performs it's circular motion?
Are we still talking about this 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 travelled clock on its arrival at A will be (1/2)t*v^2/c^2 second slow.
In the first paragraph it is A that is accelerated while B moves inertially (though A is only accelerated instantaneously and then moves at constant velocity relative to B), while in the third paragraph where he talks about continuous acceleration, it seems he switches to having A be the inertial clock while the other clock (not given a label) is accelerated in a closed curve with constant speed in A's frame.
The context of this thread is the Twin Paradox. As presented in Wikipedia, Einsteins clock paradox was considered to be an absolute result, and evidence of absolute motion. It was questioned on the basis that " the laws of physics should exhibit symmetry. Each twin sees the other as traveling, so each should see the other aging more slowly. How can an absolute effect result from a relative motion?" Investigations by Langevin and vonLaue supported and explained Einsteins result, but did not provide the desired symmetry.
Another viewpoint can be based on the observation that the time lag of the closed-curve clock is the same as the lag of a clock moving in a single straight line. Posts 3 and 4 of this thread show that straight line clocks have the desired symmetry, each sees the other as aging more slowly. Note how each clock is selected in turn to be the stationary/inertial clock.
In the early stages of the twins journey both are inertial, no idea of turnaround has been decided, and each can be considered stationary/inertial. In the Langevin/vonLaue analyses the stay home twin has been arbitrarily chosen to be inertial. The traveling twin can be correctly chosen to be inertial, and the same methods of analysis used by Lang/von applied to find the stay home twin to be less aged, thus obtaining the symmetry shown by the straight line clocks.
The context of this thread is the Twin Paradox. As presented in Wikipedia, Einsteins clock paradox was considered to be an absolute result, and evidence of absolute motion.
Absolute acceleration, not absolute "motion" in general. You can analyze the twin paradox from the perspective of any inertial frame you like, including ones where the inertial twin has a higher speed than the non-inertial one for part of the trip, and you'll still reach the same conclusion that the inertial twin has aged more when they reunite.
It was questioned on the basis that " the laws of physics should exhibit symmetry. Each twin sees the other as traveling, so each should see the other aging more slowly. How can an absolute effect result from a relative motion?" Investigations by Langevin and vonLaue supported and explained Einsteins result, but did not provide the desired symmetry.
Do you mean the desired asymmetry? To explain why the rhetorical argument above is wrong, you just have to note that acceleration is absolute since it causes the accelerating twin to feel G-forces, and the law that a moving clock ticks more slowly is only intended to apply to inertial reference frames. So the G-forces provide an asymmetry between the two twins, and that explains why the rhetorical argument that says their perspectives should be symmetrical (and that they should each predict the other will have aged less) is incorrect according to SR.
Another viewpoint can be based on the observation that the time lag of the closed-curve clock is the same as the lag of a clock moving in a single straight line. Posts 3 and 4 of this thread show that straight line clocks have the desired symmetry, each sees the other as aging more slowly.
Of course if you look at each clock's inertial rest frame after clock A has been accelerated, the two frames disagree about which clock ticks more slowly after the acceleration. But I already responded to this argument by pointing out that part of Einstein's setup was that the two clocks were synchronized in B's rest frame before A was accelerated. This means that if you look at the frame where A was at rest after acceleration, in this frame the two clocks were out-of-sync before A accelerated, with B significantly ahead of A; thus even though it's true that A was then ticking faster than B after the acceleration, you still get the prediction that B's reading will be ahead of A's when they meet, because B had a "head start" in this frame. If you are confused on this point or aren't convinced, I'd be happy to give you a simple numerical example where I analyze the situation from the perspective of both frames, and show that both frames agree that B will be ahead of A when they meet, by the same amount (again assuming the clocks were initially synchronized in B's rest frame prior to A accelerating).
In the early stages of the twins journey both are inertial, no idea of turnaround has been decided, and each can be considered stationary/inertial. In the Langevin/vonLaue analyses the stay home twin has been arbitrarily chosen to be inertial. The traveling twin can be correctly chosen to be inertial, and the same methods of analysis used by Lang/von applied to find the stay home twin to be less aged, thus obtaining the symmetry shown by the straight line clocks.
Of course, as long as both move inertially then in each one's frame the other has aged less. But in order for them to come together and reunite after they've been moving apart inertially, one of them has to accelerate, and whichever accelerated will be the one to have aged less when they reunite. For example, if we imagine attaching giant rockets to the Earth so it can catch up with a ship that has been moving away from the Earth inertially, then in that case it will be the Earth-twin who has aged less when they reunite, not the rocket-twin.
Mentz114
Jul17-09, 07:20 PM
The context of this thread is the Twin Paradox. As presented in Wikipedia, Einsteins clock paradox was considered to be an absolute result, and evidence of absolute motion. It was questioned on the basis that " the laws of physics should exhibit symmetry. Each twin sees the other as traveling, so each should see the other aging more slowly. How can an absolute effect result from a relative motion?" Investigations by Langevin and vonLaue supported and explained Einsteins result, but did not provide the desired symmetry.
Another viewpoint can be based on the observation that the time lag of the closed-curve clock is the same as the lag of a clock moving in a single straight line. Posts 3 and 4 of this thread show that straight line clocks have the desired symmetry, each sees the other as aging more slowly. Note how each clock is selected in turn to be the stationary/inertial clock.
In the early stages of the twins journey both are inertial, no idea of turnaround has been decided, and each can be considered stationary/inertial. In the Langevin/vonLaue analyses the stay home twin has been arbitrarily chosen to be inertial. The traveling twin can be correctly chosen to be inertial, and the same methods of analysis used by Lang/von applied to find the stay home twin to be less aged, thus obtaining the symmetry shown by the straight line clocks.
As Jesse has said, there's no evidence or argument for absolute motion in the twin 'paradox'. This idea there's some kind of symmetry between the frames is misleading. To paraphrase another poster in this forum (apologies to them, I can't find the post )-
"geodesic motion maximises proper time. The greater the deviations of the worldline from the geodesic, the smaller the proper length and so the elapsed time. "
There's only symmetry if both twins have identical elapsed times. It must be the continuing and incorrect use of the word 'paradox' that draws so many ignorant people to this topic, who obviously believe there is something inconsistent going on. There isn't.
Help! I don't know how to work the 'Quote' function.
JesseM and Mentz114- Do you or do you not agree with the first four posts of this thread? Do you understand the thought process involved in 'designation K as stationary' compared with 'designating k as stationary'?
JesseM and Mentz114- Do you or do you not agree with the first four posts of this thread? Do you understand the thought process involved in 'designation K as stationary' compared with 'designating k as stationary'?Are you using K and k to refer to two inertial frames as in Einstein's 1905 paper? or are you referring to the twins paradox where one is non-inertial, since it has applied force?
In the twins paradox, the ship's twin can certainly consider himself stationary, but not stationary in an inertial frame, since force is applied to the frame during the turnaround. He can consider himself stationary in an accelerated frame.
Einstein addressed the twins paradox considering the ship's twin to be stationary in a 1918 paper. Here's a link: http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_rela tivity
The end result of course is the same as the standard analysis, the ship's clock has less elapsed time at the reunion.
I don't know how to work the 'Quote' function.
Hit the "Quote" button underneath a post. :smile: This incorporates the quoted post, in its entirety, into your post, enclosed in [ quote ] and [ /quote ] tags. (The tags do not actually include any spaces; I added spaces here so the forum software wouldn't interpret them as actual quote tags.)
Very important: Please delete any material between the quote tags that is not directly relevant to your response! You can edit this material the same way that you edit your own material: click the mouse and hit the delete key, or whatever. Just don't delete the quote tags themselves.
JesseM and Mentz114- Do you or do you not agree with the first four posts of this thread? Do you understand the thought process involved in 'designation K as stationary' compared with 'designating k as stationary'?
As Al68 said, it depends on whether you're talking about inertial or non-inertial frames--the standard SR time dilation equation is only meant to be used in inertial ones. I agree that as long as K and k are inertial frames, you can treat either one as stationary, but as long as you use both frames to analyze the same physical scenario (including details about how the clocks were synchronized initially), then both frames will make the same prediction about what any two clocks read when they meet, in spite of the fact that they may disagree about which clock was ticking slower during any particular phase of the journey. Do you disagree that inertial frames will always reach the same conclusions about what clocks read when they meet as long as both frames are analyzing the same physical scenario?
Al68 JesseM
OOOHKaY? I'm really trying to find out whether there'a anything at all that we can agree on, so help out here.
I sense that you haven't read the first four posts of this thread. So I think these are yes/no questions.
Have you read these posts?
Is there any thing in them that you don't understand?
Al68 JesseM
OOOHKaY? I'm really trying to find out whether there'a anything at all that we can agree on, so help out here.
I sense that you haven't read the first four posts of this thread. So I think these are yes/no questions.
Have you read these posts?
Is there any thing in them that you don't understand?
I am not sure that you have ever really grasped the fact that "k" is a not a frame for the clock as it moves in polygons, or circles.
That error was clearest in msg #11, where you said
What I am saying is that k has an equal right to consider himself to be stationary! Thus , if K sees k moving away, turning around, and returning, then k, considering himself to be stationary,sees K moving away, turning around and returning.
The earlier posts in the thread are fine as worded, but only if you recognize that "k" refers to an inertial frame for an observer or clock in straight line unaccelerated motion. If, however, "what you were saying" in those earlier posts was predicated on k being somehow associated with the clock moving in polygons, or a closed circle, then the answers are all different. In fact, a clock that moves and returns does not have an "equal right" to consider itself stationary, and k is not a frame for motions in any kind of curve.
Cheers -- sylas
I sense that you haven't read the first four posts of this thread. So I think these are yes/no questions.
Have you read these posts?
Is there any thing in them that you don't understand?
Of course I read them, and I already pointed out my disagreement with your fourth post (post #7) in post #8 (specifically the part where you said T=t/m in the k frame, which I took to mean that you were claiming the k frame would predict the inertial clock B would be behind the accelerated clock A when they met). Again, if we use the physical situation as Einstein described it, with the two clocks being initially synchronized in the K frame where both clocks were at rest before A accelerated, then even if we analyze things from the perspective of the k frame where both clocks were moving and then A came to rest after it accelerated, this frame will still predict that A will be behind B when they meet, in spite of the fact that this frame predicts A was ticking faster than B after they accelerated. As I said, the reason is that the two clocks were not initially synchronized in this frame, B had a significant "head start" at the time A was accelerated and came to rest in this frame. Do you have trouble understanding this analysis, or do you disagree with it in any way?
Al68, Sylas: Thanks for your comments.
These threads diverge so quickly and far that I've tried to start at the very simplest, basic ideas of Einsteins relativity. In my first post I identified K and k as Einsteins original coordinates, he even calls them K and k. They are both inertial, as agreed by Mentor George Jones. The connection between K,k and the curved/ segmented path is so unclear that I don't want to try to connect them.
I am trying to find something that we can agree on as a starting place for our discussion. I propose that we put our blinders on and focus on the first four posts of this thread. I don't want to assume anything, and it would encourage me greatly to hear a 'yes' in agreement with these posts..( I almost fell off my chair when I read Jones 'yes and yes'!)
Al68, Sylas: Thanks for your comments.
These threads diverge so quickly and far that I've tried to start at the very simplest, basic ideas of Einsteins relativity. In my first post I identified K and k as Einsteins original coordinates, he even calls them K and k. They are both inertial, as agreed by Mentor George Jones. The connection between K,k and the curved/ segmented path is so unclear that I don't want to try to connect them.
I am trying to find something that we can agree on as a starting place for our discussion. I propose that we put our blinders on and focus on the first four posts of this thread. I don't want to assume anything, and it would encourage me greatly to hear a 'yes' in agreement with these posts..( I almost fell off my chair when I read Jones 'yes and yes'!)
George's reponse (yes, and yes) was to your message #1. That's quite unexceptional. I don't know why you found it surprising.
Your message #3 is poorly expressed. You refer to co-ordinates t,x,y,z and T,X,Y,Z; and show t=T/m. The m here is the gamma factor, and this is a description of the time dilation effect.
But think about it. t and T are not variables denoting a single value, but co-ordinates, which range over many values. When you write t = T/m, you are comparing a whole range of events that are simultaneous for one observer... but NOT the other. If you switch to the other frame, you get T = t/m. But this is involving a different notion of simultaneity.
Specifically. For observer in the k frame, the other clock is moving at v and reads T/m simultaneously with their own clock reading t.
Equivalently, for the observer in the K frame, the first clock is the one moving at v, and it reads t/m simulaneously with their own clock reading T.
By writing t=T/m, and then also T=t/m, you are not in fact writing two inconsistent equations. You are writing the relations for what one clock reads simultaneously with the other; for two different notions of simultaneity... depending on the frame.
This seems to be the difficulty everyone has with relativity. It nearly always comes down to failing to appreciate that simultaneity depends on the frame being used. Einstein's paper does it correctly. Using the Lorentz transformations gives you the correct answers.
Cheers -- sylas
In his 1905 paper Einstein introduced two coordinate systems which he called K and k. It seems that both systems are inertial because there are no external forces and they are moving at the constant relative speed v.
But I'm not certain, so are they both inertial? Is it correct that either one can be considered the 'stationary' system and the other one the 'moving' system?All correct.
So, with K( X,Y,Z,T) stationary the clocks of k( x,y,z,t) are moving in the + X direction at speed v. Putting X=vT in the transfforms leads to t=T/m, where m is the quantity Einstein calls beta. So t is less than T.To make this clear, I would spell out when x = 0, X = vT and t = T/m.
With k being stationary the clocks of K are moving in the -x direction at speed v. Putting x=-vt in the transforms leads to T=t/m, ie T is less than t.To make this clear, I would spell out when X = 0, x = -vt and T = t/m.
The bits in bold are important, and make it clear that the two paragraphs apply to two different situations. It's not possible for x and X to both be zero (except for the special case when t and T are both zero).
JM, I have no problems with your first post. As for the second, you say:
So, with K( X,Y,Z,T) stationary the clocks of k( x,y,z,t) are moving in the + X direction at speed v. Putting X=vT in the transfforms leads to t=T/m, where m is the quantity Einstein calls beta. So t is less than T.
With k being stationary the clocks of K are moving in the -x direction at speed v. Putting x=-vt in the transforms leads to T=t/m, ie T is less than t.
Is there general agreement with this result? Is everyone OK with it?
I agree with this, but just to be clear, you understand that you're talking about two different events here right? X=vT would be some event on the worldline of the clock at rest in k, so in t=T/m, t would be the time on this k-clock and T would be the time in the K frame of the event of the k-clock reading t. On the other hand, x=-vt would be some event on the worldline of the clock at rest in K, so in T=t/m, T would be the time on this K-clock and t would be the time in the k frame of the event of the K-clock reading T.
Sylas, DrGreg, JesseM:
Thank you for your responses, they are positive and encouraging. There is a lot to think about in them, I want to take some time to understand and prepare appropriate responnses. So stay tuned.
Sylas writes: "Your message #3 is poorly expressed."
My posts may be abbreviated, with details added as necessary.
And "...this is a description of the time dilation effect."
This particular expression of time dilation has the advantages of its direct derivation from the transforms, and its symmetry. For convenience we might refer to it as 'double slow clocks', with each half called a 'slow clock'.
But think about it. t and T are not variables denoting a single value, but co-ordinates, which range over many values. When you write t = T/m, you are comparing a whole range of events that are simultaneous for one observer... but NOT the other. If you switch to the other frame, you get T = t/m. But this is involving a different notion of simultaneity.
Specifically. For observer in the k frame, the other clock is moving at v and reads T/m simultaneously with their own clock reading t.
Equivalently, for the observer in the K frame, the first clock is the one moving at v, and it reads t/m simulaneously with their own clock reading T.
Cheers -- sylas
I agree with this analysis.
The basic statement of The Relativity of Simultanaity (ROS) is ' A number of events that are simultaneous in one coordinate system are not necessarily simultaneous in any other coordinate system.' That doesn't seem too difficult to understand. But the application of this principle to a particular situation is not so easy to see.
The book 'Special Relativity' by A. P. French includes good descriptions in the sections 'The Relativity of Simultaneity' p.74, and Relativity is Truly Relative, p.111. The use of ROS there provides unique insights into the behavior of clocks, and develops the same resulst as given in posts 3 and 4.
By writing t=T/m, and then also T=t/m, you are not in fact writing two inconsistent equations. You are writing the relations for what one clock reads simultaneously with the other; for two different notions of simultaneity... depending on the frame.
Cheers -- sylas
This idea seems to be related to the process of designating a coordinate/observer as being 'stationary', or 'the reference system'. This is important and related to the question of how many events there are. To be discussed.
All correct.
To make this clear, I would spell out when x = 0, X = vT and t = T/m.
To make this clear, I would spell out when X = 0, x = -vt and T = t/m.
The bits in bold are important, and make it clear that the two paragraphs apply to two different situations. It's not possible for x and X to both be zero (except for the special case when t and T are both zero).
Agreed. In his analysis of the slow clock Einstein specified that the expression X=vT describes the position of the clock at x=0 of k.
I agree with this, but just to be clear, you understand that you're talking about two different events here right? X=vT would be some event on the worldline of the clock at rest in k, so in t=T/m, t would be the time on this k-clock and T would be the time in the K frame of the event of the k-clock reading t. On the other hand, x=-vt would be some event on the worldline of the clock at rest in K, so in T=t/m, T would be the time on this K-clock and t would be the time in the k frame of the event of the K-clock reading T.
I see what JesseM is driving at here, and in the context of world lines I agree.
The description of the physical situation under posts 3/4 needs to be clarified. To begin, there is no known location of absolute rest, and only relative motions have meaning. So these posts describe only one physical event, namely the two coordinates/observers are separating at speed v. The two results T=t/m and t=T/m come from the process of designating one or the other as 'stationary', or equivalently taking the viewpoint of one observer or the other as each views the same event. Isn't the world line approach consistent with this idea?
Comments?
Slow Clocks.
The idea expressed in post 4,that each obsercer 'sees' the others clock running slow, has been considered berore, for example by G Builder, Phil. Sci., Apr. 1959, p. 135-144. He considers this idea paradoxical because, he says, the two results t=T/m and T=t/m are contradictories, i.e. if one is true the other must be false. On the other hand both results are considered here to be true. Explanation is foound in understanding of the meaning of the terms.
Builders analysis indicates that he views the calculations to apply directly to the time reading of the moving clock. He says that ...one of the clocks must be retarded relative to the other. But in deriving the time delay from the Lorentz trasnforms,Einstein clearly states that '...the time marked by the (moving) clock ([B]viewed in the stationary system[B])is slow...' Emphasis added. Builder has apparently ignored this qualifying phrase, as is commonly done even nowadays. But there is a sugnificant difference between 'the time on the clock' and 'the time perceived be the stationary viewer'. Thus each observer can correctly state that the others clock appears to run slow. An example sometimes used to illustrate this point is the difference between the actual height of a person standing at some distance away and the reduced height perceived by each of the facing observers.
Agreement on these ideas would remove a significant source of confusion for me, and I hope for others.
The Twins again
In the paper cited above,Builder considers the clock/twin paradox in the form of a thought exaperiment. In the nomenclature of this thread, he pictures a clock k, originally at the origin of inertial coordinates K , that moves at speed v along the X axis of K, stops, and returns to the origin of K. He considers several methods for calculating the time retardation when the clock k arrives back at the origin, and concludes that all methods give the same result, t =T/m. He also notes that the asymmetry of the predicted retardation is obviously consistent with the dynamical asymmetry involved in the experiment itself, which requires that k should be subjected to accelerations while K remained at rest. His analysis is of spcial intest because it is representative of the ideas pesented by many others.
continued next post
Twins, cont.
So, lets consider whether there are other thought experiments that can be develpped from Einsteins description of hes clock paradox, par. 4 of the 1905 paper. Note that the clock paradox involves only the clocks A and B, and that no connection is made between A,B and K,k. So we are on our own to make the connecrion. Builder has evidently chosen to identify the stationary B with coordinates K, and specifies the path of A as the back-and=forth motion of k along the X axis.
It is equally corect to identify the stationary B with coordinates k, and the path of A as a back-and-forth motion of K along the negative x axis. The analysis methods Builder uses are equally applicable to this second experiment, with the result T = t/m.
cont,
Twins, concluded
The presence of acceleration at the turn around may be part of Builders experiment,, but it is not part od Einseteins, who diesn't mention accelerations. Accelaerations are not necessary anyway because they don;t affect the calculated time retardation, alnd because designation of which coordinate is stationsary is enough to specify the experimental conditions.
For the reasons given in post 43 the twins journey can be considered as one event with two parts, namely 1. the twins separate, and 2 the twins re-unite. Builders experiment takes Ks perspetive, who 'sees' k recede and then retrn. The second experiment takes ks perspective, who 'sees' K recede and then return. Thus when the twins re-unite each 'sees' the others clock to be running slow.
So thats all that I meant to say in this thread. Is it 'QED' or 'back to the drawing board'?
What do y'all think?
I agree with this, but just to be clear, you understand that you're talking about two different events here right? .
Right. When K is at rest the expression X=vT identifies the clock at rest with K that is adjacent to the origin of the k system. So this event takes place at the origin of k. When k is at rest the expression x = -vt identifies the clock at rest with k that is located at the origin of the K system. So this event takes place at the origin of K. Since the origins are at different places the events are separate.
I believe that Silas and DrGreg were making the same point about there being two different events.
OK?
In the 1905 paper Einstein spent much time developing his ideas aboout time. Does his reasoning lead to the result that when any one clock at rest with K reads a specific time, such as cT= 10, then all the clocks at rest with K also read '10'?
In the 1905 paper Einstein spent much time developing his ideas aboout time. Does his reasoning lead to the result that when any one clock at rest with K reads a specific time, such as cT= 10, then all the clocks at rest with K also read '10'?
According to the definition of simultaneity in their inertial rest frame (and assuming they've been synchronized using the procedure Einstein gave in that paper), yes.
So does the result of post 49 apply also to coordinates k, namely 'when any one clock at rest with k reads a specific time , such as ct =8, then all clocks at rest with k also read '8''?
May I assume that the answer to post 51 is yes, with the qualifiers given in post 50? If so, can it be concluded that when K is at rest the relation t = T/m holds when comparing the clock at rest at the origin of K with the clock of k that is nearby?
The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?
The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?
It seems contradictory only if you haven't grasped what the relativity of simultaneity is about. To compare two clocks that are separated by a distance, you have to measure both clocks at the same time. Different observers disagree over what "at the same time" means, and there is no definition that everyone can agree on. So that's why one observer says k is slower than K and another observer says K is slower than k. They are both right according to their own definition of simultaneity.
Thanks for looking in DrGreg,
To compare two clocks that are separated by a distance, you have to measure both clocks at the same time.
I'm not sure what youre referring to here. In my post in both cases 1. and 2. the comparison is between the clock at the origin of K and the clock of k that is immediately next to the origin of K.
I dont think the results are contradictory but I read it stated in a reference as an absolute, and I'm trying to find out how y'all explain it.
JM
The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?Those statements are true, but the clocks cannot be local at both the beginning and end of a measurement unless the relative velocity is zero. It makes no sense to say that each clock is slower than the other over a distance of zero and an elapsed time of zero.
So each statement is true for any non-zero time period, meaning that the distance between the clocks must be non-zero either at the beginning or end of the time period.
Thanks for coming in Al68.
Those statements are true, but the clocks cannot be local at both the beginning and end of a measurement unless the relative velocity is zero. It makes no sense to say that each clock is slower than the other over a distance of zero and an elapsed time of zero.
So each statement is true for any non-zero time period, meaning that the distance between the clocks must be non-zero either at the beginning or end of the time period.
Agreed. The clocks are not together at the beginning of the time period.
The problem considered here is based on Einsteins slow clock calculation, as described in posts 3 and 4 of this thread. The clocks of both K and k are set to zero when the origins of each coincide. Afte a time T the origin of k has moved a distance vT along the X axis of K. At this time there will be a clock of k that is coincident with the clock at the origin of K. The question is how each observer, K and k, will view the comparative time of the two clocks. So are the statements you refer to still true? And how does one work out the answer?
DrGreg- Specific example-
It seems contradictory only if you haven't grasped what the relativity of simultaneity is about.
I agree that it is important to understand the Relativity of Simultaneity (ROS). But its not clear to me how the principle enters the problem under consideration. I have restated the problem in the just previous post. Would it be possible to describe the solution to this problem so we can all see how its done?
This is not a homework problem.
Thanks for looking in DrGreg,
I'm not sure what youre referring to here. In my post in both cases 1. and 2. the comparison is between the clock at the origin of K and the clock of k that is immediately next to the origin of K.
I dont think the results are contradictory but I read it stated in a reference as an absolute, and I'm trying to find out how y'all explain it.
JM
In order to talk about the rate a clock is ticking in a given frame, you must consider two points on that clock's worldline, and compare the time that elapsed on the clock between those points with the coordinate time between those points in your chosen frame. So even if the first point is the one where K and k are at a common location, the second point on either clock's worldline will have to be one where that clock has moved away from the other clock, and simultaneity issues come into play. For example, if both clocks read 0 when they are at the same position, and in K's frame the k clock reads 0.08 nanoseconds simultaneously with the K clock reading 0.1 nanosecond, then this means that in K's frame the k clock is slowed down by a factor of 0.8; but in k's frame the event of the k clock reading 0.08 nanoseconds is not simultaneous with the K clock reading 0.1 nanosecond, instead it is simultaneous with the K clock reading 0.064 nanoseconds, so in k's frame it is the K clock that is running slow by a factor of 0.8. These simultaneity disagreements are of central importance no matter how tiny the interval between the first reading and the second (even if it is considered to be infinitesimal, so you're talking about the instantaneous rate of ticking between some time t and t + dt).
Thanks for coming in Al68.
Agreed. The clocks are not together at the beginning of the time period.
The problem considered here is based on Einsteins slow clock calculation, as described in posts 3 and 4 of this thread. The clocks of both K and k are set to zero when the origins of each coincide. Afte a time T the origin of k has moved a distance vT along the X axis of K. At this time there will be a clock of k that is coincident with the clock at the origin of K. The question is how each observer, K and k, will view the comparative time of the two clocks. So are the statements you refer to still true? And how does one work out the answer?Yes both statements would be true. Each clock would run slow in the other frame. Notice that relativity of simultaneity now comes into play, so if we define an event (after the origin of K and k coincide) as clock A (at rest in K) reading t, then clock B (at rest in k) will read t2 (less than t) simultaneous (in K) with clock A reading t. But the events of clock A reading t and clock B reading t2 are not simultaneous in k. In k, clock A reading t is simultaneous with clock B reading t3 (more than t).
The situation would be similar if we defined an event prior to the clocks passing. Either way, we have two events, one in which the clocks are local, the other in which they are not.
I notice in your post 3 you say that the twins scenario can be considered to have only two parts: 1. the twins separate, and 2 the twins re-unite.
This is true, but the standard SR equations are not valid in the ship's non-inertial frame in that case, because they are only valid in inertial (unaccelerated) reference frames.
That's why the turnaround is usually treated as a separate event, because then the standard SR equations can be used between each events, since each twin is unaccelerated between the events.
I always recommend Einstein's own twins paradox resolution of 1918, not because it's better or simpler than the rest, but because it directly addresses the point of concern many people have, instead of "dodging" it. And he doesn't bother with the math, he just explains the physics of the situation. And he explains it using the non-inertial frame in which the ship is stationary the whole time.
Agreed. The clocks are not together at the beginning of the time period.
The problem considered here is based on Einsteins slow clock calculation, as described in posts 3 and 4 of this thread. The clocks of both K and k are set to zero when the origins of each coincide. Afte a time T the origin of k has moved a distance vT along the X axis of K. At this time there will be a clock of k that is coincident with the clock at the origin of K. The question is how each observer, K and k, will view the comparative time of the two clocks.
When two clocks coincide at the same point in time and space, all frames always agree on what their respective readings are...otherwise you could get genuinely different physical predictions, like if one clock was set to explode when it reached a different time and different frames disagreed about whether another clock would be next to it when that happened. So both frames agree that the clock at the origin of K is behind the clock of k that it's next to at the moment that the k-clock reads at time T (the clock at the origin of K will read T*\sqrt{1 - v^2/c^2} at this moment). However, in the K frame note that this does not imply the clock at the origin of K was ticking slower than that k-clock, since in the K frame that k-clock would have already shown a time later than zero at the moment the two origins coincided and the clocks at the origins both read zero (again, the relativity of simultaneity).
According to the definition of simultaneity in their inertial rest frame .
JesseM- What is the definition of simultaneity for clocks at rest in K?
I always recommend Einstein's own twins paradox resolution of 1918, not because it's better or simpler than the rest, but because it directly addresses the point of concern many people have, instead of "dodging" it. And he doesn't bother with the math, he just explains the physics of the situation. And he explains it using the non-inertial frame in which the ship is stationary the whole time.
Al68-Thanks for your suggestion. Can you give me a more exact reference to the paper of 1918?
Thanks.
Jm
DrGreg, JesseM, Al68- I appreciate your efforts to respond to my question. But I'm still having a problem putting it all in context. Can you suggest a tutorial on ROS? There seems to be agreement that each observer can see the others clock to be running slow, even when the clocks of K and k are co-located. But there is also a suggestion that when co-located both observers must see the same relation, i.e. if K sees k running slow then k must see K running fast, or did I misunderstand?
I hope for further discussion.
JM
JesseM- What is the definition of simultaneity for clocks at rest in K?
Simultaneity in each inertial frame is defined using the Einstein synchronization convention (http://en.wikipedia.org/wiki/Einstein_synchronisation), which is based on assuming that light moves at the same speed in both directions in that frame. The method Einstein originally gave was that if you have two clocks A and B at rest in the same inertial frame, and clock A sends a light signal to B when A reads T1, and B bounces back the signal when it hits it, then if A reads T2 when the bounced signal hits it, the two clocks are synchronized if B's time when the light hit it was exactly halfway between T1 and T2--i.e. B read a time of T1 + (T1 + T2)/2 when the light hit it. An equivalent procedure which I find a little more intuitive is to set off a flash at the exact midpoint of A and B, and set both clocks to read the same time at the moment the light from the flash reaches them.
With the latter procedure, it's not hard to see why different observer must disagree about simultaneity if they each assume light travels at c in all directions in their own frame. For example, suppose I am on a rocket passing you at high speed, and I want to synchronize clocks at the front and back of the rocket by setting off a flash at the midpoint of the rocket, and setting both clocks to the same time when the light reaches them. In your frame, the front of the rocket is moving away from the point where I set off the flash, while the back of the rocket is moving towards that point, so if you assume the light travels at the same speed in both directions in your frame, naturally you will conclude that the light must hit the back clock before the light hit the front clock, which means if both clocks read the same time when the light hits them, in your frame they must be out-of-sync.
It works out that if two clocks are synchronized and a distance L apart in their own rest frame, then in a frame where they are moving at velocity v (parallel to the axis between them), they will be out-of-sync by vL/c^2. So for example, suppose we have two clocks at rest in k, k0 and k1, which are synchronized and a distance of 20 light-seconds apart in the k frame, with k0 being at the origin and reading a time of 0 seconds at the moment it is next to the clock K0 at the origin of the K frame (which also reads 0 seconds at that moment). Suppose the clock K0 then moves towards k1 at 0.8c as seen in the k frame, so in the k frame it will take a time of 20 light-seconds/0.8c = 25 seconds to reach k1. So, at the time K0 reaches k1, k1 reads 25 seconds, while K0 has been slowed down by time dilation so it only reads 25*\sqrt{1 - 0.8c^2} = 25*0.6 = 15 seconds. But now if we re-analyze the same situation from the perspective of the K frame, things look different--because of length contraction, k0 and k1 are only 0.6*20 = 12 light-seconds apart in this frame, and because of the relativity of simultaneity they are also out-of-sync by vL/c^2 = 0.8c*20 l.s./c^2 = 16 seconds, with the clock at the rear (k1) being ahead of the leading clock (k0) by this amount. So in K's frame, at the moment K0 is next to k0 and both read a time of 0 seconds, k1 is 12 light-seconds away and already reads a time of 16 seconds. Since k1 is approaching K0 at 0.8c, it will take 12/0.8 = 15 seconds for k1 to reach K0 in this frame, meaning K0 will read 15 seconds at the moment k1 passes it, which is what we found before in the other frame. In this frame it is the k clocks that are slowed down by a factor of 0.6 relative to the K clocks, so during these 15 seconds, the clock k1 only ticked forward by 15*0.6 = 9 seconds. But since k1 already had a "head start" of 16 seconds when K0 was passing k0, this means that when k1 passes K0, k1 will read 16 + 9 = 25 seconds, which again matches what we found in the other frame, even though the two frames disagree about whether K0 or k1 was running slower between the event of K0 passing k0 and the event of K0 passing k1.
Can you suggest a tutorial on ROS?
This one is my favorite of the one's I've seen (http://www.pitt.edu/~jdnorton/Goodies/rel_of_sim/index.html), but if you google "relativity of simultaneity" you should be able to find others.
There seems to be agreement that each observer can see the others clock to be running slow, even when the clocks of K and k are co-located. But there is also a suggestion that when co-located both observers must see the same relation, i.e. if K sees k running slow then k must see K running fast, or did I misunderstand?
You misunderstood that part, observers in each frame always measure clocks in the other frame to be running slow (though they can sometimes see them running fast visually due to the Doppler effect--'measure' refers to how they assign time-coordinates to events on each clock's worldline). As I said in post #59, when talking about the "rate" a clock is ticking you always need at least two measurements at different times, even if the time-interval between these measurements is infinitesimal, so it's potentially misleading to talk about what they measure at the exact instant they are co-located.
JesseM, Thank you for the clear and complete analysis, and the reference.
At the moment I'm trying to reconcile the results with the results described in posts 3 and 4. Can I get some help on this?
Re SR dimensions.
In ordinary applications of the equation x = a y, where a is a non-dimensional constant, the units of x and y must agree. If y is in inches then x must be in inches, for example. Is this convention followed in SR also?
JesseM, Thank you for the clear and complete analysis, and the reference.
At the moment I'm trying to reconcile the results with the results described in posts 3 and 4. Can I get some help on this?
Do you mean posts #3 and #4 on the thread (post #4 being by HallsofIvy) or do you mean your own third and fourth posts?
Re SR dimensions.
In ordinary applications of the equation x = a y, where a is a non-dimensional constant, the units of x and y must agree. If y is in inches then x must be in inches, for example. Is this convention followed in SR also?
Yes.
Simultaneity in each inertial frame is defined using the Einstein synchronization convention (http://en.wikipedia.org/wiki/Einstein_synchronisation), which is based on assuming that light moves at the same speed in both directions in that frame. The method Einstein originally gave was that if you have two clocks A and B at rest in the same inertial frame, and clock A sends a light signal to B when A reads T1, and B bounces back the signal when it hits it, then if A reads T2 when the bounced signal hits it, the two clocks are synchronized if B's time when the light hit it was exactly halfway between T1 and T2--i.e. B read a time of T1 + (T1 + T2)/2 when the light hit it. An equivalent procedure which I find a little more intuitive is to set off a flash at the exact midpoint of A and B, and set both clocks to read the same time at the moment the light from the flash reaches them.
By the way, I made a little math error in this paragraph--in order to ensure that "B's time when the light hit it was exactly halfway between T1 and T2", B's time when the light hit it should be T1 + (T2 - T1)/2.
Do you mean posts #3 and #4 on the thread (post #4 being by HallsofIvy) or do you mean your own third and fourth posts?
Yes.
I meant 3 and 4, where 4 is by Halls.
Following up the 'dimensions' question: Since the clocks of K and k are all set to zero when the axes origins coincide, and time is measured in seconds for both K and k, doesn't that mean that the clocks of K and k are always in synch?
Does an observer ever see his own time dilated or his own dimensions contracted? Doesnt saying that K's time is dilated and his length shortened mean that the view taken is k's?
I meant 3 and 4, where 4 is by Halls.
OK, I already commented on post #3 in my own post #37, and I have no disagreement with what HallsofIvy says in post #4.
Following up the 'dimensions' question: Since the clocks of K and k are all set to zero when the axes origins coincide, and time is measured in seconds for both K and k, doesn't that mean that the clocks of K and k are always in synch?
No, that's where the relativity of simultaneity comes in--the two frames disagree about what distant clocks read "at the same time" that the two clocks at the origins are passing each other and both read 0. In the K frame, all the K clocks read 0 "at the same time" that the clocks at the origin are next to each other reading 0, but all the k clocks are out-of-sync with one another and only the one at the origin reads 0. In the k frame, the reverse is true--all the k clocks read 0 "at the same time" that the clocks at the origin are next to each other reading 0, but all the K clocks are out-of-sync with one another and only the one at the origin reads 0. It may help to take a look at the diagrams I drew up in this thread (http://www.physicsforums.com/showthread.php?t=59023), showing two rulers with clocks attached at regular intervals moving past one another at relativistic speeds, with all the clocks on a given ruler being synchronized in that ruler's rest frame but out-of-sync in the other ruler's rest frame.
Does an observer ever see his own time dilated or his own dimensions contracted?
No.
Doesnt saying that K's time is dilated and his length shortened mean that the view taken is k's?
Yes, K's time is dilated from the perspective of the k frame, in K's own rest frame his clocks are running normally.
Yes, K's time is dilated from the perspective of the k frame, in K's own rest frame his clocks are running normally.
I take it that your 'yes' indicates that you agree that saying 'Ks clocks are dilated' means that the analysis is taking ks viewpoint. In that case both of the calculations in your post 65 take ks viewpoint, and that is why you get the same answer for both, and why you dis agree with post 4 that says each observer sees the others clocks as slow.
No, that's where the relativity of simultaneity comes in--
What I'm saying is that two clocks that are set to zero at the same time and thereafter run at the same rate will always be in synch.
I don't see that ROS has any thing to do with it. I'm not suggesting that each obserrver has to look at the other for this result to be true.
It feels to me that this thread has dead-ended at ROS. Anyone else have any comments?
DrJM
The explanation for the result t = T/m is not found in the behavior of clocks, but in the 'Fundamental Principle of Relativity'.
Consider the segment of the x axis of k extending from x=0, where a light source is located, to x=l, where a detector is located. Let a light ray be emitted when the origins of K and k coincide. From ks viewpoint the light ray reaches l at t =l/c. From Ks viewpoint the segment is moving at speed v, so the light ray reaches x =l at T = l/( c - v). So T = t/ (1-v/c). Thus T and t are the coordinates that describe the same event, the arrival of the light ray at x = l, as seen by the respective observers K and k. As measured by clocks that are always in synch.
matheinste
Sep17-09, 08:02 AM
What I'm saying is that two clocks that are set to zero at the same time and thereafter run at the same rate will always be in synch.
I don't see that ROS has any thing to do with it. I'm not suggesting that each obserrver has to look at the other for this result to be true.
It feels to me that this thread has dead-ended at ROS. Anyone else have any comments?
DrJM
When you set the clocks to zero when the origins are colocated, you do not alter the rate at which time passes for either set of clocks. This difference in the rate of time passing, time dilation, is caused by the relative velocity and the result is that each observer observes others clocks to be running slow compared to his own, and so the others clocks appear desynchronized. Zeroing the clocks when the origins are colocated does not change any of this as the relative motion is still the same. It does not put both sets of clocks permanently in synch, it only sets the accumulated time to zero or, if you like, sets the datum to zero. As the other frames clocks pass by you they appear to you to be progressively more and more out of synch with your own. This is of course a reciprocal effect and applies to both observers.
Does that help? If not when I have a little more time i could perhaps give a clearer explanation. I am sure that JesseM and some other respondants have a better grasp of this than I do but, sometimes another wording helps.
Matheinste.
I take it that your 'yes' indicates that you agree that saying 'Ks clocks are dilated' means that the analysis is taking ks viewpoint.
Yes.
In that case both of the calculations in your post 65 take ks viewpoint, and that is why you get the same answer for both, and why you dis agree with post 4 that says each observer sees the others clocks as slow.
No, I analyzed the same situation from both frames in post 65. Read it again, in the first part I described things from the perspective of the k frame:
It works out that if two clocks are synchronized and a distance L apart in their own rest frame, then in a frame where they are moving at velocity v (parallel to the axis between them), they will be out-of-sync by vL/c^2. So for example, suppose we have two clocks at rest in k, k0 and k1, which are synchronized and a distance of 20 light-seconds apart in the k frame, with k0 being at the origin and reading a time of 0 seconds at the moment it is next to the clock K0 at the origin of the K frame (which also reads 0 seconds at that moment). Suppose the clock K0 then moves towards k1 at 0.8c as seen in the k frame, so in the k frame it will take a time of 20 light-seconds/0.8c = 25 seconds to reach k1. So, at the time K0 reaches k1, k1 reads 25 seconds, while K0 has been slowed down by time dilation so it only reads 25*\sqrt{1 - 0.8c^2} = 25*0.6 = 15 seconds.
Then I analyzed everything again from the perspective of the K frame:
But now if we re-analyze the same situation from the perspective of the K frame, things look different--because of length contraction, k0 and k1 are only 0.6*20 = 12 light-seconds apart in this frame, and because of the relativity of simultaneity they are also out-of-sync by vL/c^2 = 0.8c*20 l.s./c^2 = 16 seconds, with the clock at the rear (k1) being ahead of the leading clock (k0) by this amount. So in K's frame, at the moment K0 is next to k0 and both read a time of 0 seconds, k1 is 12 light-seconds away and already reads a time of 16 seconds. Since k1 is approaching K0 at 0.8c, it will take 12/0.8 = 15 seconds for k1 to reach K0 in this frame, meaning K0 will read 15 seconds at the moment k1 passes it, which is what we found before in the other frame. In this frame it is the k clocks that are slowed down by a factor of 0.6 relative to the K clocks, so during these 15 seconds, the clock k1 only ticked forward by 15*0.6 = 9 seconds. But since k1 already had a "head start" of 16 seconds when K0 was passing k0, this means that when k1 passes K0, k1 will read 16 + 9 = 25 seconds, which again matches what we found in the other frame, even though the two frames disagree about whether K0 or k1 was running slower between the event of K0 passing k0 and the event of K0 passing k1.
You can see that in this second section, it is k's clocks that are running slow ('In this frame it is the k clocks that are slowed down by a factor of 0.6 relative to the K clocks...'), the distance between k's clocks are shrunk ('because of length contraction, k0 and k1 are only 0.6*20 = 12 light-seconds apart in this frame'), and it is the k clocks that are out-of-sync ('because of the relativity of simultaneity they are also out-of-sync by vL/c^2 = 0.8c*20 l.s./c^2 = 16 seconds').
What I'm saying is that two clocks that are set to zero at the same time and thereafter run at the same rate will always be in synch.
What two clocks are you talking about? Are you just talking about the clocks at the origin? I thought you wanted to talk about other clocks at different positions in each frame, like the clock k1 in my example above which passes K0 at a different moment so we can compare the readings on K0 and k1. If you're just talking about the clocks at the origin, k0 and K0 in my example, then it's true that the relativity of simultaneity doesn't enter into it, since all frames agree they both read 0 at the same moment. However, all frames do not agree that they "thereafter run at the same rate", I don't understand where you got that assumption--in any frame where they have different velocities, they run at different rates due to time dilation. For example, in the K frame the k0 clock is running slow, and in the k frame the K0 clock is running slow.
The explanation for the result t = T/m is not found in the behavior of clocks, but in the 'Fundamental Principle of Relativity'.
All time coordinates in any frame are meant to stand for the readings on clocks. Einstein based his coordinate systems on the idea of a physical system of rulers and clocks at rest in a given frame, with the clocks synchronized according to his simultaneity convention; then the position and time coordinates of an event are determined by purely local readings on this system, seeing which marking on the ruler the event was next to as it happened, and what reading was showing on the clock attached to that marking as the event happened.
The equation t = T/m is meaningless unless you specify a particular event which is supposed to have time coordinate t as measured in one frame (i.e. measured using that frame's system of rulers and clocks), while it is supposed to have time coordinate T in the other frame. If T is the time coordinate in the K frame and t is the time coordinate in the k frame, and m is supposed to represent the gamma factor, then this equation t = T/m only works if you are talking about an event on the worldline of the clock k0. On the other hand, if you were talking about an event on the worldline of the clock K0, we would have T = t/m. This was the point I was making in post #37--do you disagree?
Consider the segment of the x axis of k extending from x=0, where a light source is located, to x=l, where a detector is located. Let a light ray be emitted when the origins of K and k coincide. From ks viewpoint the light ray reaches l at t =l/c.
Yes, that's true in the k frame.
From Ks viewpoint the segment is moving at speed v, so the light ray reaches x =l at T = l/( c - v). So T = t/ (1-v/c).
This is wrong. From K's viewpoint, the segment does not have length l, you're forgetting about length contraction--in the K frame the length of the segment is only l * \sqrt{1 - v^2/c^2}. So it will take a time of T = l * \sqrt{1 - v^2/c^2} / (c - v) for the light to reach the end of this segment in the K frame.
Thus T and t are the coordinates that describe the same event, the arrival of the light ray at x = l, as seen by the respective observers K and k. As measured by clocks that are always in synch.
Again, where do you get the idea that the clocks are always in sync? The clocks at the origin don't stay in sync in either frame because of time dilation. And in SR you wouldn't normally measure the time of the event of the light reaching the end of that segment using clocks at the origin, you need to measure it using different clocks in each frame which are actually at the position of the end of that segment when the light reaches it. Only by using local measurements can you avoid the issue of the delays between when an event actually happens and when the event is seen by the person who is noting the time on his own clock (although of course you can correct for transmission delays if you know the distance, but this will give exactly the same time-coordinates as following the standard procedure of using local measurements on an array of clocks synchronized using the Einstein clock synchronization convention).
Hit the "Quote" button underneath a post. :smile: This incorporates the quoted post, in its entirety, into your post, enclosed in [ quote ] and [ /quote ] tags. (The tags do not actually include any spaces; I added spaces here so the forum software wouldn't interpret them as actual quote tags.)
Thanks, JT, I got it. So, how do I work the multi qoute button?
25*\sqrt{1 - 0.8c^2}[/tex] = 25*0.6 = 15 seconds. But now if we re-analyze the same situation from the perspective of the K frame, things look different--because of length contraction, k0 and k1 are only 0.6*20 = 12 light-seconds apart in this frame, and because of the relativity of simultaneity they are also out-of-sync by vL/c^2 = 0.8c*20 l.s./c^2 = 16 seconds, with the clock at the rear (k1) being ahead of the leading clock (k0) by this amount. So in K's frame, at the moment K0 is next to k0 and both read a time of 0 seconds, k1 is 12 light-seconds away and already reads a time of 16 seconds. Since k1 is approaching K0 at 0.8c, it will take 12/0.8 = 15 seconds for k1 to reach K0 in this frame, meaning K0 will read 15 seconds at the moment k1 passes it, which is what we found before in the other frame. In this frame it is the k clocks that are slowed down by a factor of 0.6 relative to the K clocks, so during these 15 seconds, the clock k1 only ticked forward by 15*0.6 = 9 seconds. But since k1 already had a "head start" of 16 seconds when K0 was passing k0, this means that when k1 passes K0, k1 will read 16 + 9 = 25 seconds, which again matches what we found in the other frame, even though the two frames disagree about whether K0 or k1 was running slower between the event of K0 passing k0 and the event of K0 passing k1.
1. Where do you get the expressions for out-of-synch, time dilation, and length contraction?If not from Lorentz transforms, then where?
2. How do you know how to manipulate these expressions, regarding sign and magnitude? The reference you gave tells about history but not methods.
3. Is your analysis intended to adress the situation presented in post 3? If so why is your answer different ? Note that post 3 is based on the Lorentz transforms, and that T and t refer to total elapsed time since the origins coincided so two points in time are compared. If not, what situation are you addressing?
4. If you agree with sylas, post 35. that 'Using the Lorentz transformations gives you the correct answers.' then what is the point of using the more-difficult ROS procedure?
how do I work the multi qoute button?
Hit the MULTI QUOTE button on each post that you want to quote from. This highlights the button, but doesn't do anything else immediately. Then hit the QUOTE button on one of those posts. All of the selected posts will be quoted (separately) in your response. Delete unnecessary text and intersperse your comments as usual.
1. Where do you get the expressions for out-of-synch, time dilation, and length contraction? [...]
(just as an example)
Thanks for prodding me to figure out how to do it (had to do a Google search on "vbulletin mulitiquote"). When I've needed to do something like this before, I copied and pasted text, and inserted quote tags by hand for stuff from posts other than the one I used the QUOTE button on.
Lets review some posts related to 'slow clocks' and see what conclusion result. This thread started with Einsteins coordinate frames K and k:
In his 1905 paper Einstein introduced two coordinate systems which he called K and k. It seems that both systems are inertial because there are no external forces and they are moving at the constant relative speed v.
But I'm not certain, so are they both inertial? Is it correct that either one can be considered the 'stationary' system and the other one the 'moving' system?
and the response was:
Yes and yes..
On that basis clock rates were calculated for both frames:
So, with K( X,Y,Z,T) stationary the clocks of k( x,y,z,t) are moving in the + X direction at speed v. Putting X=vT in the transfforms leads to t=T/m, where m is the quantity Einstein calls beta. So t is less than T.
With k being stationary the clocks of K are moving in the -x direction at speed v. Putting x=-vt in the transforms leads to T=t/m, ie T is less than t.
Is there general agreement with this result? Is everyone OK with it?
and the results supported:
Yes, an observer in the K system would take the K system as stationary, K' as moving, and observe time in the K' system, t', as slower than t in the K system. An observer in the K' system would take the K' system as stationary, K as moving, and observe time in the K system, t, as slower than t' in the K' system..
Thus, it can be concluded that each observer K or k, when taken to be at rest, observes the other observers clocks to be slow, as viewed from is own rest position.
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For clarity post 3 can be explained further.
The coordinates of K and k are related by the Lorentz transforms, derived in Einsteins 1905 paper
(1) x = m( X – vT), and ct = m( cT –vX/c). These equations can be inverted to obtain
(2) X = m( x + vt), and cT = m( ct + vx/c). For y,z = Y,Z = 0,0.
The meaning of these equations is that when K observes an event occurring at X,T, then k observes the same event occurring at x, t. Thus each observer is assigning his own coordinates to the event.
These equations can work in either direction. Arbitrary values of X, T can be entered on the right of eqn.( 1) and the corresponding values of x,t determined. This process is referred to as ‘taking the point of view of K’, or ‘taking K to be at rest’. Or similarly, arbitrary values of x,t can be assumed and the corresponding values of X, T determined from eqn. (2). This process is referred to as ‘taking the point of view of k’, or ‘taking k to be at rest’.
These equations have a ‘reciprocal’ property. Specific values of X, T entered into eqn (1) produce specific values of x,t. For example when X, cT = 20, 25 and v =0.8c, eqn (1) produces x,t = 0,15. Entering these values into eqn(2) produces the original values of X, T, namely 20, 25. This process is not the same as ‘taking k to be at rest’ as describe above, because the values of x,t are not arbitrary, but are specifically obtained from eqn (1).
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There is more, but I think I will post this and follow with another post.
Comments continued from previous post.
When it was pointed out in Post 37 that post 3 refers to two different events, one occurring at the origin of K and the other at the origin of k, the synchronization procedure of Einstein ( and ROS)was recalled as described in Post 65. This background leads to the following reasoning.
Post 49:
In the 1905 paper Einstein spent much time developing his ideas aboout time. Does his reasoning lead to the result that when any one clock at rest with K reads a specific time, such as cT= 10, then all the clocks at rest with K also read '10'?
Post 50
According to the definition of simultaneity in their inertial rest frame (and assuming they've been synchronized using the procedure Einstein gave in that paper), yes.
Post 51
So does the result of post 49 apply also to coordinates k, namely 'when any one clock at rest with k reads a specific time , such as ct =8, then all clocks at rest with k also read '8''?
Post 52
May I assume that the answer to post 51 is yes, with the qualifiers given in post 50? If so, can it be concluded that when K is at rest the relation t = T/m holds when comparing the clock at rest at the origin of K with the clock of k that is nearby?
Post 53
The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?
As stated in Post 53 another way, the conclusion is that no matter which two clocks are considered, one of K and one of k, each observer views the other observers clock to be slow. This is a significant result of this discussion, because the conconclusion is not always clear from the the literature. It is a unique property of Special Relativity, embodying , in a way, the Relativity Principle that the laws of physecs must be the same whether referred to either of two inertial coordinate frames. It's also a 'mind bender' when you try to reason it our using 'common sense'.
The result is supported from another viewpoint as follows:
Post 54
It seems contradictory only if you haven't grasped what the relativity of simultaneity is about. To compare two clocks that are separated by a distance, you have to measure both clocks at the same time. Different observers disagree over what "at the same time" means, and there is no definition that everyone can agree on. So that's why one observer says k is slower than K and another observer says K is slower than k. They are both right according to their own definition of simultaneity.
The above presents my understanding of the 'slow clockx' phenomenon. Any comments?
Comments continued from previous post.
When it was pointed out in Post 37 that post 3 refers to two different events, one occurring at the origin of K and the other at the origin of k, the synchronization procedure of Einstein ( and ROS)was recalled as described in Post 65. This background leads to the following reasoning.
Post 49:
In the 1905 paper Einstein spent much time developing his ideas aboout time. Does his reasoning lead to the result that when any one clock at rest with K reads a specific time, such as cT= 10, then all the clocks at rest with K also read '10'?
Post 50
According to the definition of simultaneity in their inertial rest frame (and assuming they've been synchronized using the procedure Einstein gave in that paper), yes.
Post 51
So does the result of post 49 apply also to coordinates k, namely 'when any one clock at rest with k reads a specific time , such as ct =8, then all clocks at rest with k also read '8''?
Post 52
May I assume that the answer to post 51 is yes, with the qualifiers given in post 50? If so, can it be concluded that when K is at rest the relation t = T/m holds when comparing the clock at rest at the origin of K with the clock of k that is nearby?
Post 53
The point of this is the following:
1, When K is at rest the c lock of k that is near to the origin of K is slow compared to the clock at the origin of K, and
2. When k is at rest the clock at the origin of K is slow compared with the clock of k that is near the origin of K.
The question is whether this result is explainable and accepted using SR, or whether the two results are contradictory and impossible, as some writers suggest?
As stated in Post 53 another way, the conclusion is that no matter which two clocks are considered, one of K and one of k, each observer views the other observers clock to be slow. This is a significant result of this discussion, because the conconclusion is not always clear from the the literature. It is a unique property of Special Relativity, embodying , in a way, the Relativity Principle that the laws of physecs must be the same whether referred to either of two inertial coordinate frames. It's also a 'mind bender' when you try to reason it our using 'common sense'.
The result is supported from another viewpoint as follows:
Post 54
It seems contradictory only if you haven't grasped what the relativity of simultaneity is about. To compare two clocks that are separated by a distance, you have to measure both clocks at the same time. Different observers disagree over what "at the same time" means, and there is no definition that everyone can agree on. So that's why one observer says k is slower than K and another observer says K is slower than k. They are both right according to their own definition of simultaneity.
The above presents my understanding of the 'slow clockx' phenomenon. Any comments?
I want to thank all those who gave their time and effort to contribute to this thread. Their comments help me stretch and sharpen my ideas.
I hope that the lack of comments on my last two posts indicates that there is no important disagreement with the ideas stated there. It would help me to feel that some generally agreed conclusion has been reached.
I note that there have been over 4000 views of this thread, so there must be some interest. Would some of you viewers give your opinions of the thread, did you learn anything, would you like to see more on these subjects? It takes time and effort to carry on these conversations so an encouraging word would help.
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