Acceleration is not important in the twin paradox

[Iggy]

B simply transfers to C its reading. At the time they meet, B and C show the same reading. This makes B totally irrelevant to the exercise, since C simply accumulates proper time as if B didn't exist in the exercise. So, as shown above, when A and C meet, \tau_A=\tau_C.

 

Unless they performed an Einstein synchronization before the start of the experiment [math]\tau_A[/math] would not equal [math]\tau_C[/math]. That is an ad hoc way of explaining how they could be forced to get the same results at the end of the experiment by doing something that has nothing to do with the experiment at all.

 

By the way, you should maybe research "Einstein synchronization" because you just said "you can only compare co-located clocks"

 

Do you still maintain that "Everyone will agree on what time the two passing clocks read at the moment they pass each other." is false?

[xyzt]

Unless they performed an Einstein synchronization before the start of the experiment [math]\tau_A[/math] would not equal [math]\tau_C[/math]. That is an ad hoc way of explaining how they could be forced to get the same results at the end of the experiment by doing something that has nothing to do with the experiment at all.

They better did, otherwise their clocks can be offset by an arbitrary and unknown amount rendering the whole exercise meaningless. Please don't talk down to me, I understand what Einstein synch is.

Edited May 5, 2013 by xyzt

[michel123456]

That is false, you can only compare co-located clocks, i.e. A and C. When A and C are next to each other you can compare them. and they will show the same exact time since the scenario has been dumbed down by taking away the acceleration.

 

B simply transfers to C its reading. At the time they meet, B and C show the same reading. This makes B totally irrelevant to the exercise, since C simply accumulates proper time as if B didn't exist in the exercise. So, as shown above, when A and C meet, \tau_A=\tau_C.

Yes, You are correct.

By transfering from B to C the exact same reading, there is no gap of time between B & C. For the gap, see here and here down the page.

[md65536]

No, in the absence of acceleration there is no asymmetry:

 

1. A measures \tau_A=Integral{\sqrt{1-(v/c)^2}dt}

2. C measures \tau_C=Integral{\sqrt{1-(v/c)^2}dt}

 

The interval of integration for above are the same, the speeds are the same so, \tau_A=\tau_C.

 

You are confusing total elapsed time with ticking rate.

Can you describe what you think A and C are measuring proper time from? What event(s) coincide with the start of their elapsed times?

[Iggy]

They better did, otherwise their clocks can be offset by an arbitrary and unknown amount rendering the whole exercise meaningless. Please don't talk down to me, I understand what Einstein synch is.

 

No, Md's thought experiment works perfectly fine if all three clocks read something different at the start of the experiment. The results are independent of that factor, just as the results are independent of that factor in the traditional twin paradox. If one person leaves earth age 20 and returns age 21 and the person staying on earth ages from 40 to 60 then the twin paradox is tested true. Your condition is unreasonable.

 

I'm not trying to talk down to you. I am demonstrating frustration which I think is an appropriate response to evasiveness.

 

Do you still maintain that "Everyone will agree on what time the two passing clocks read at the moment they pass each other." is false?... because it was extremely off putting the way you talked down to Md telling him how false that statement is.

Edited May 5, 2013 by Iggy

[md65536]

There's a simple way to figure out what's going on in this version of the paradox: Run it in parallel with the typical twin paradox.

 

In the twin:

A and B separate,

B turns around,

A and B meet.

 

In this version:

A and B separate,

B passes C going in the opposite direction,

A and C meet.

 

So let B instantly turn around and return with C.

Do you agree that B and C could make that journey "side by side"?

Do you agree that B and C would age the same amount during that return trip?

[xyzt]

Each of A and C are measuring proper time since passing B.

If A and C are observed symmetrically (or any other way), A passes B earlier than C passes B, and measures a greater proper time.

 

The velocity of B relative to A is 0.866c after passing (gamma=2). The velocity of B relative to C is about 0.990c (gamma=7).

Let's try this: There is a frame in which A and C start simultaneously and move wrt each other (with +v/2, respectively -v/2). In that frame B and C start simultaneously and B moves towards C with +v.

B and C travel the same amount of proper time (since they meet somewhere, closer to C since B has double the speed) and B transfers its reading to C (which is irrelevant since their proper time readings are the same when they meet).

C continues to travel towards A and A travels towards C and they meet half-way. A and C total elapsed proper times are the same. B was just a red herring all along.

 

So let B instantly turn around

 

 

Err, "B instantly turn around" means infinite acceleration.

Edited May 5, 2013 by xyzt

[md65536]

If the results do rely on B then I don't fully understand Md's assertion that information transferred from B is unimportant, or that A, B, and C don't have to know about each other.

I say this because you can modify the experiment to not include such details as specific information transfers. If you're using the same variables (v, distances, etc) then the exact same theory applies, and gives the exact same results.

 

 

Not to derail the fighting, but here's an example where information from C to A is completely taken out of the picture, and yet the timing remains the same:

 

Using the same values, suppose that there is some ship at rest relative to A, and that it sends a probe to A once every second at v=0.866c relative to the ship, and that the probe takes 1 year of proper time to arrive at A. Now suppose A observes a meteor (B) pass A at v=0.866c, and that it collides with the ship, destroying it (and C, the probe that would be launched at event BC, is destroyed!). The result: A stops receiving probes 4 years after the meteor passes, THE SAME aging it would experience if it was a twin following the path of the meteor and the missing destroyed probe.

 

So what's the point of this? The probes that were already launched before the meteor destroyed the ship clearly do not receive any information from the meteor. However, each of the probes makes the same journey at the same speed, and they all experience it the same, they all age the same amount, they all record the same proper time of the journey. They do this REGARDLESS of any "memory" of past events ie. information from B. Here, we have accurately measured the time that the destroyed probe would have taken to arrive at A, without it even existing. All the timing is exactly what would be measured in a twin experiment.

Let's try this: There is a frame in which A and C start simultaneously and move wrt each other (with +v/2, respectively -v/2). In that frame B and C start simultaneously and B moves towards C with +v.

You're not getting it. A and C are not recording proper time since the start of the experiment. They're recording proper time since their individual passings of B. It says so in like the first or second line of the description of the experiment. The experiment is SET UP specifically to time the proper time since events with B. I don't see the point of choosing alternative times to start, and then describing a completely different experiment, and then making universal conclusions based on that.

 

Anyway, in your version, A and C are starting separated and with a relative velocity. You've chosen a symmetrical observer F to decide a single "starting time", but you should be aware that those start times are not simultaneous for general observer, and C starting at that time is meaningless for everyone else involved (even C, who in your experiment does not measure starting at the same time as A), and that F is not mentioned in the original experiment and is not relevant at all.

 

Err, "B instantly turn around" means infinite acceleration.

Yyyyup! Agreed.

 

Would you say that C is affected by whether B just passes by, or turns around to follow C? If so, how?

Edited May 5, 2013 by md65536

[xyzt]

I say this because you can modify the experiment to not include such details as specific information transfers. If you're using the same variables (v, distances, etc) then the exact same theory applies, and gives the exact same results.

 

 

You're not getting it. A and C are not recording proper time since the start of the experiment. They're recording proper time since their individual passings of B. It says so in like the first or second line of the description of the experiment. The experiment is SET UP specifically to time the proper time since events with B. I don't see the point of choosing alternative times to start, and then describing a completely different experiment, and then making universal conclusions based on that.

If you take this stance, I hope that you realize that what you are describing is a hidden acceleration: it is exactly as if C traveled at v up to the point it meets with B and at v/2 after that. So, your scenario simply "hides" acceleration. You can't get a difference in proper times UNLESS you have asymmetry. You can't have assymmetry UNLESS you create it via accelerating one twin.

Edited May 5, 2013 by xyzt

[michel123456]

You all: I'd like to know

 

1. In a regular twin experiment, where the turnaround is made instantaneously, what is the observation of B a second before the turn and a second after the turn?

 

2. In Md65536 's experiment, what is the observation of C one second after the passing of info from B?

Edited May 5, 2013 by michel123456

[md65536]

If you take this stance, I hope that you realize that what you are describing is a hidden acceleration: it is exactly as if C traveled at v up to the point it meets with B and at v/2 after that. So, your scenario simply "hides" acceleration. You can't get a difference in proper times UNLESS you have asymmetry. You can't have assymmetry UNLESS you create it via accelerating one twin.

What happens to C before meeting B is irrelevant. I've described it in a case with no acceleration.

 

 

 

The only way I could even come close to agreeing with you on this is that the experiment "without acceleration" DOES require that all observers have a specific velocity, and something back in the history of time would have had to set up these initial conditions. However the point stands, that the theory and calculations of SR as used in the twin paradox do not change depending on past history of the observers.

[xyzt]

What happens to C before meeting B is irrelevant. I've described it in a case with no acceleration.

 

This is of course, false. Your scenario simply hides the acceleration by attempting to hide the fact that , by transfering the B proper time to C you made it look as if C traveled at two different speeds:

-v up to the time of encountering B

-v/2 after the time of encountering B

 

It is just a sleigh of hand, easily discoverable. No matter how you twist and turn you can only generate an asymmetry via acceleration. You can think of clever ways of covering the acceleration but the acceleration is there.

[Delta1212]

If you take this stance, I hope that you realize that what you are describing is a hidden acceleration: it is exactly as if C traveled at v up to the point it meets with B and at v/2 after that. So, your scenario simply "hides" acceleration. You can't get a difference in proper times UNLESS you have asymmetry. You can't have assymmetry UNLESS you create it via accelerating one twin.

Yes, the results are the same you would get if you had an accelerated frame, but that's the entire point. You get the same results without any of the clocks physically undergoing acceleration. Changing the frame in which the measurement is being done partway through the measurement is all that is required. You don't need to use the same device throughout, however, and so no device needs to switch frames itself.

 

Saying that switching between measurement devices in different frames is a "hidden" acceleration because the result is the same as if you used a single device that accelerated demonstrates a fundamental lack of comprehension of the experiment because the fact that the effect is identical is what md is trying to illustrate.

 

You're literally arguing over whether acceleration means a change in velocity of a physical object or a change in frames used to make a measurement, only you don't appear to realize that this is what you're arguing.

[xyzt]

Yes, the results are the same you would get if you had an accelerated frame, but that's the entire point. You get the same results without any of the clocks physically undergoing acceleration.

The point is that the clocks in the scenario do undergo acceleration. It is just that there is a sleigh of hand that tries to cover the presence of acceleration, as explained.

[Delta1212]

This is of course, false. Your scenario simply hides the acceleration by attempting to hide the fact that , by transfering the B proper time to C you made it look as if C traveled at two different speeds:

-v up to the time of encountering B

-v/2 after the time of encountering B

 

It is just a sleigh of hand, easily discoverable. No matter how you twist and turn you can only generate an asymmetry via acceleration. You can think of clever ways of covering the acceleration but the acceleration is there.

The final time registered by C's clock is not supposed to represent the time experienced by C. It's supposed to represent the time experienced along the path travelled by B up until the collocation of B and C and then C from that point until the collocation of A and C.

 

This is the same path travelled by the accelerated twin in the twin experiment, but nothing is undergoing acceleration, unless you define acceleration as the act of switching between frames, with or without anything actually changing velocity to accomplish the switch.

 

The point is that the clocks in the scenario do undergo acceleration. It is just that there is a sleigh of hand that tries to cover the presence of acceleration, as explained.

At no point is any force applied to any of the clocks, nor do they undergo a change in velocity (obviously). The path made up of the joint paths of B from A to C and C from B to A is measured in different frames, however, which results in the same effect as if a single clock travelled the entire path and accelerated at point BC.

[md65536]

Yes, the results are the same you would get if you had an accelerated frame, but that's the entire point. You get the same results without any of the clocks physically undergoing acceleration. Changing the frame in which the measurement is being done partway through the measurement is all that is required. You don't need to use the same device throughout, however, and so no device needs to switch frames itself.

 

Saying that switching between measurement devices in different frames is a "hidden" acceleration because the result is the same as if you used a single device that accelerated demonstrates a fundamental lack of comprehension of the experiment because the fact that the effect is identical is what md is trying to illustrate.

 

You're literally arguing over whether acceleration means a change in velocity of a physical object or a change in frames used to make a measurement, only you don't appear to realize that this is what you're arguing.

Thank you! This pretty much sums up everything I wanted to say with this thread.

The point is that the clocks in the scenario do undergo acceleration. It is just that there is a sleigh of hand that tries to cover the presence of acceleration, as explained.

No clock experiences proper acceleration (in this experiment).

 

You can always add coordinate acceleration through a choice of observer, but that has no bearing on what is physically happening to the observers.

Edited May 5, 2013 by md65536

[xyzt]

The final time registered by C's clock is not supposed to represent the time experienced by C. It's supposed to represent the time experienced along the path travelled by B up until the collocation of B and C and then C from that point until the collocation of A and C.

 

This is the same path travelled by the accelerated twin in the twin experiment, but nothing is undergoing acceleration, unless you define acceleration as the act of switching between frames, with or without anything actually changing velocity to accomplish the switch.

 

 

At no point is any force applied to any of the clocks, nor do they undergo a change in velocity (obviously). The path made up of the joint paths of B from A to C and C from B to A is measured in different frames, however, which results in the same effect as if a single clock travelled the entire path and accelerated at point BC.

I pointed out the frame-jumping several pages ago. No matter what sleigh of hand you use in creating the asymmetry, it is still a sleigh of hand that once exposed makes the whole scenario fall apart like a house of cards.

So your choices are :

 

-frame jumping

-hidden acceleration

 

take your pick.

Edited May 5, 2013 by xyzt

[md65536]

You might like this article: Clock Postulate.

That certainly is an important assumption in the twin paradox. I've always taken it for granted as true without considering it.

 

My claim here that the proper time measured in the non-accelerated, 3-observer experiment is equivalent to a twin paradox with instantaneous acceleration, implicitly assumes that the clock postulate is true. However, all twin paradox calculations that use only the Lorentz transformation, assume that it is true. Basically, the traveling twin experiences:

 

1) Travel a length-contracted (relative to other twin) distance for some proper time,

2) Instantaneously turn around, changing its proper time by some amount

3) Travel back for the same proper time (same speed) as in (1)

 

The clock postulate assumes that the change in proper time in step (2) is zero. As the link notes, the postulate is not proven, though "it has been verified experimentally up to extraordinarily high accelerations, as much as 1018 g in fact".

 

The twin paradox can be calculated using only the Lorentz transformation, only if the clock postulate holds. And if it doesn't hold, the effect would be an unknown, unless someone can measure it or theorize a calculation.

 

The 3-observer variation without acceleration would not be affected by whether or not the clock postulate holds.

1. In a regular twin experiment, where the turnaround is made instantaneously, what is the observation of B a second before the turn and a second after the turn?

"A second" depends on which clock you're using, but assuming you mean proper time, then...

Assuming the clock postulate holds, B ages 2 seconds between those two events.

Edited May 5, 2013 by md65536

[michel123456]

(...)

Assuming the clock postulate holds, B ages 2 seconds between those two events.

No, what I ment is that at the turning point, B looks back at the Earth and observes that the Earth is 6 LY away. C also looks at the Earth from the same point and at the same moment and sees the Earth is indeed 6 LY away.

With the only difference that B sees the Earth in 2000 (say) and C sees the Earth in 2004!(say).

And it is not a question of clocks anymore, they observe different things! Which makes no sense.

 

The only way to make it somehow sensible is to change the straight path lines (in a spacetime diagram) with smooth curves, which mean "acceleration".

[Delta1212]

No, what I ment is that at the turning point, B looks back at the Earth and observes that the Earth is 6 LY away. C also looks at the Earth from the same point and at the same moment and sees the Earth is indeed 6 LY away.

With the only difference that B sees the Earth in 2000 (say) and C sees the Earth in 2004!(say).

And it is not a question of clocks anymore, they observe different things! Which makes no sense.

 

The only way to make it somehow sensible is to change the straight path lines (in a spacetime diagram) with smooth curves, which mean "acceleration".

B and C have different velocities, so they see the same Earth (let's say 2002) but they measure the distance as being different.

[md65536]

No, what I ment is that at the turning point, B looks back at the Earth and observes that the Earth is 6 LY away. C also looks at the Earth from the same point and at the same moment and sees the Earth is indeed 6 LY away.

With the only difference that B sees the Earth in 2000 (say) and C sees the Earth in 2004!(say).

And it is not a question of clocks anymore, they observe different things! Which makes no sense.

 

The only way to make it somehow sensible is to change the straight path lines (in a spacetime diagram) with smooth curves, which mean "acceleration".

Are you talking common sense or Lorentz transformation? It makes sense and is consistent in terms of the Lorentz transformation.

 

I wrote what B and C would measure of A at their meeting using the original values, previously:

To elaborate further on where the asymmetry might be "seen", suppose that in the example observer A (clock one) ages 4 years before meeting C, while B (clock two) ages 1 year before meeting C, and C (clock 3) ages 1 year from there until meeting A (gamma=2).

 

Then, when B and C meet, B can say "I have aged 1 year, while A has aged 0.5 years," but C will say "I agree B has aged 1 year, but A has aged 3.5 years since B left it."

 

The difference in their measurements is the same as the difference in measurements before and after a twin's instantaneous turnaround.

Note that this is what they "measure" by accounting for the delay of light. What they see is different, and that is to say, that uh, B and C will actually agree on how A looks at the instant of their meeting. To see why this is so, imagine that A sends a signal to B and C that arrives at the moment of their passing. Since they pass at the same location, these signals are sent to the same location, and take the same time to arrive according to A, so they represent a single instant of how A looks to B and C when they pass.

 

In this example, both B and C agree A appears to have aged 0.268 years since B passed it (via relativistic Doppler factor).

Edited May 5, 2013 by md65536

[michel123456]

Are you talking common sense or Lorentz transformation? It makes sense and is consistent in terms of the Lorentz transformation.

 

I wrote what B and C would measure of A at their meeting using the original values, previously:

 

 

md65536, on 04 May 2013 - 19:52, said:

To elaborate further on where the asymmetry might be "seen", suppose that in the example observer A (clock one) ages 4 years before meeting C, while B (clock two) ages 1 year before meeting C, and C (clock 3) ages 1 year from there until meeting A (gamma=2).

 

Then, when B and C meet, B can say "I have aged 1 year, while A has aged 0.5 years," but C will say "I agree B has aged 1 year, but A has aged 3.5 years since B left it."

 

The difference in their measurements is the same as the difference in measurements before and after a twin's instantaneous turnaround.

 

 

 

 

Extracting the bold part from the quote, you stated:

Then, when B and C meet, B can say "I have aged 1 year, while A has aged 0.5 years," but C will say "I agree B has aged 1 year, but A has aged 3.5 years since B left it."

 

My comment is:

It is not what the say, it is what they see, what they observe. During the experiment all observers are constantly in reach of sight from each other. It is not a situation where observers close their eyes during the travel and open it when they meet someone. Observation and measurement are the same all along the travel.

IOW B sees A in June 2000 and C sees A in June 2003. from the same point, from the same instant.

 

Note that this is what they "measure" by accounting for the delay of light. What they see is different, and that is to say, that uh, B and C will actually agree on how A looks at the instant of their meeting. To see why this is so, imagine that A sends a signal to B and C that arrives at the moment of their passing. Since they pass at the same location, these signals are sent to the same location, and take the same time to arrive according to A, so they represent a single instant of how A looks to B and C when they pass.

 

In this example, both B and C agree A appears to have aged 0.268 years since B passed it (via relativistic Doppler factor).

 

 

 

 

 

To me that is not correct: what they observe accounts for the delay of light, exactly as what they measure accounts for the delay of light. Observation and measurement coincide.

Edited May 6, 2013 by michel123456

[xyzt]

Are you talking common sense or Lorentz transformation? It makes sense and is consistent in terms of the Lorentz transformation.

 

I wrote what B and C would measure of A at their meeting using the original values, previously:Note that this is what they "measure" by accounting for the delay of light. What they see is different, and that is to say, that uh, B and C will actually agree on how A looks at the instant of their meeting. To see why this is so, imagine that A sends a signal to B and C that arrives at the moment of their passing. Since they pass at the same location, these signals are sent to the same location, and take the same time to arrive according to A, so they represent a single instant of how A looks to B and C when they pass.

 

In this example, both B and C agree A appears to have aged 0.268 years since B passed it (via relativistic Doppler factor).

 

 

the reason that you keep getting unphysical answers is that you do not realize that the exercise, as set up, hides acceleration. Not only that it hides acceleration but, in certain instances, it produces contradictory results. At the root of all the problems is the "jump" in elapsed proper time on C's clock. This forum is horrible for math, so you will have to bear with what is to follow.

 

In the description, we have A and B at one end of a segment of length d (as measured by A, at rest wrt the segment) and C at the other end. In frame F_A, B and C start simultaneously from the ends of the segment with speeds u and v, respectively.

 

They meet at t=d/(u+v) as measured by A in point P.

 

At that point, the proper times shown on their clocks are:

 

tau_BP=d sqrt(1-(u/c)^2)/(u+v)

tau_CP=d sqrt(1-(v/c)^2)/(u+v)

 

C "dumps" its time and "assumes" the time shown by B.

C continues towards A accumulating an additional:

tau_PA=sqrt(1-(v/c)^2)(d/v-d/(u+v))

 

So, the total time shown by C when it meets A is:

 

tau_CA=tau_CP+tau_PA=d/(u+v)(sqrt(1-(u/c)^2-sqrt(1-(v/c)^2)+d/v*sqrt(1-(v/c)^2)

 

On the other hand, the proper time accumulated by A can be calculated by a frame comoving with C, F_C as :

 

tau_A=d/v*sqrt(1-(v/c)^2)

 

(all frames agree on proper time so we might as well use F_C)

 

It means that the difference between the times showed by C and A is:

 

tau_CA-tau_A=d/(u+v)(sqrt(1-(u/c)^2-sqrt(1-(v/c)^2)

 

We have three possibilities;

 

1. u=v .

This is the case usually shown in textbooks. B plays no role (at the P "exchange" B and C show the same elapsed time). So:

 

tau_CA=tau_A (there is no difference in total elapsed time , so the whole "paradox" is NONEXISTENT)

 

2. u>v

 

tau_CA< tau_A

 

3. u<v

 

tau_CA>tau_A

 

case 3 is particularly disturbing because it CONTRADICTS the spirit of the "twin paradox", the "traveling twin" shows MORE elapsed proper time. The reason why the results are BOGUS (sorry for the nonscientific term but it applies) is that the whole setup is BOGUS, C's clock is forced to "jump" (forward or backward, depending on u<v, u>v) at point P. A BOGUS scenario results into BOGUS results. So, the OP was false from the get go.

Edited May 6, 2013 by xyzt

[md65536]

To me that is not correct: what they observe accounts for the delay of light, exactly as what they measure accounts for the delay of light. Observation and measurement coincide.

I double-checked and I think it's correct.

 

One way to describe the discrepancy in MEASUREMENT while observing the same thing, is to consider the differing distances that the light signal travels from AB to BC. Remember each observer measures light traveling at a speed of c, and if you accept that, then relativity of simultaneity has these unintuitive effects.

 

Using gamma=2, v=.866, proper time T from AB to BC = 1 year (measured by B)...

 

For A, the rest distance between itself and BC (imagine a marker in space at that location, or consider that events occur in all frames, they don't have an inertial frame and can be considered at rest relative to anyone) is the length-contracted distance traveled by B, converted to A's frame: v*T*gamma = 1.732 LY. Observer A ages 2 years while B ages the one, and t = 2-1.732 = 0.268, the age A is when it must send a signal to reach event BC.

 

For B: If A sent a signal at time t, the time at B is t*gamma = 0.536, and the light spends the rest of proper time T incoming, ie. T-t*gamma = 0.464 years. For B, the signal travels 0.464 LY, during which A ages 0.464/gamma = 0.232. Add t, the age when A sent the signal, and you get that A aged 0.5 years when B reaches C, in agreement with the Lorentz transformation.

 

For C: Observer C is closing on B extremely quickly. Using composition of velocities, its relative velocity is 0.990c, Lorentz factor gammaB=7. While B ages 1 year, C ages 7 before they pass. Taking 7 years to reach BC, it measures that A ages 7/gamma (it is also moving at -v relative to A) = 3.5 years. If A sends its signal at 0.268 years old, it will age 3.5-0.268 = 3.232 years while the light signal approaches BC, during which C ages 3.232*gamma = 6.464 years.

 

So to recap: Due to relativity of simultaneity, and light traveling the same relative speed according to all the observers, A observes that the signal from AB to BC travels 1.732 LY, B measures 0.464 LY, and C measures 6.464 years.

 

This is possible because AB and BC are not simultaneous for anyone, so they're not measuring the same distance.

At the root of all the problems is the "jump" in elapsed proper time on C's clock.

Clocks do not experience jumps in proper time.

The jump is not in the original experiment description, nor any of my variations. I'm curious what physical effect you think causes a jump in proper time, but I think it would belong in Speculations. Have you read Iggy's link, http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html ? The clock postulate says that acceleration does not have an effect on proper time. Are you suggesting that the postulate (and SR and GR along with it) is wrong?

Edited May 6, 2013 by md65536

[xyzt]

 

Clocks do not experience jumps in proper time.

The jump is not in the original experiment description, nor any of my variations. I'm curious what physical effect you think causes a jump in proper time, but I think it would belong in Speculations

 

 

This is the whole point, in the BOGUS scenario that you describe in the first post, they DO.

Edited May 6, 2013 by swansont
fix quote tag