Are Both Coordinate Systems in Special Relativity Considered Inertial?

[JM]

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]

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.

 

[JM]

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]

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.

 

[Naty1]

Is there general agreement with this result? Is everyone OK with it

Yes...It will have to do until a better theory is developed!

 

[JM]

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.)

 

[JM]

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?

 

[JesseM]

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?

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)t*v^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

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

If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be (1/2)t*v^2/c^2 second slow.

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.

 

[JM]

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?

 

[JesseM]

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.

 

[JM]

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?

 

[sylas]

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]

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.

 

[JM]

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.

 

[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.

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

 

[JesseM]

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.

 

[JM]

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.

 

[DrGreg]

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 realize 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.)

 

[JM]

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]

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).

 

[JM]

JesseM; Thanks for your post. Isn't Einsteins clock B stationary and inertial while clock A performs it's circular motion?

 

[JesseM]

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 traveled 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.

 

[JM]

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.

 

[JesseM]

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]

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.

 

[JM]

Help! I don't know how to work the 'Quote' function.

 

[JM]

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'?

 

[Al68]

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_relativity

The end result of course is the same as the standard analysis, the ship's clock has less elapsed time at the reunion.

 

[jtbell]

I don't know how to work the 'Quote' function.

Hit the "Quote" button underneath a post. 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]

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?