Acceleration is not important in the twin paradox

[md65536]

So, you continue to persist in the lie you have created? Even after it has been shown how your scenario is nothing but a manipulation of C's clock at the encounter with B?

Yyyyyyyyup. I've modified my experiment to suit your counter-arguments, without changing the relevant conditions or the outcome. There is no discontinuity of time for clock C at event BC.

 

There is no discontinuity in proper time according to any clock in the experiment.

 

There is no discontinuity of proper time experienced by any clock ever, within the domain of SR.

Edited May 8, 2013 by md65536

[Iggy]

Fascinating thread. going too fast for the time i can spend here.

From post #156 on page 8

If t'=0 then I suppose x=t'v=0 also.

 

No, t'=0 doesn't necessitate that x' is zero. x' can be any value in the lorentz transforms regardless of the value of t'.

 

---

 

 

What you did is to work out the trivial case where C's speed wrt A is equal to B's speed wrt A.

 

The math I did proves the general case. The diagram I did only illustrates a specific case for elucidation. If you want C's speed to be different then simply rotate the blue line. Nothing else has to change.

[xyzt]

Yyyyyyyyup. I've modified my experiment to suit your counter-arguments, without changing the relevant conditions or the outcome. There is no discontinuity of time for clock C at event BC.

 

There is no discontinuity in proper time according to any clock in the experiment.

 

There is no discontinuity of proper time experienced by any clock ever, within the domain of SR.

In other words, I proved you wrong and you surreptitiously edited the scenario? What post post did you post edit?

 

I'll try this again!

 

 

What's important in the twin paradox, which results in the asymmetry between the twins, is that one of the twins remains in an inertial frame while the other uses primarily two different inertial frames. It is tempting then to think that a mechanical switch between the frames somehow "causes" the relativistic effects---and further that the only way to switch frames is to accelerate---but this is not true.

 

This can be shown by running the experiment with 3 moving clocks, none of which needs to accelerate during the experiment.

 

Start with 2 passing clocks, A and B, which are each set to zero at passing. Let B travel some distance at velocity v, where it passes clock C traveling in the opposite direction at the same speed, and have C set its clock to match B's as they pass. When C and A meet they'll record the same difference in proper time as if the experiment was run with clock B instantly turning around as it passes clock C.

 

What this shows is that the effect of time dilation does not depend on physically turning around, it only depends on what happens while in the various inertial frames.

 

 

If B and C pass at a distance of 0, their meeting can be considered a single event that is simultaneous according to all observers. However at that moment, their different frames are important (not any mechanical effect of switching anything between their frames) as they have very different measures of simultaneity relative to A. C will have measured A's aging as greater than B has measured A's aging, even though they are at the same place at the moment of their passing.

 

 

 

So, to fix the ideas, the stuff in post 1, the one that gives the name to the thread has been proven to be false.

[md65536]

So, to fix the ideas, the stuff in post 1, the one that gives the name to the thread has been proven to be false.

Everything you quoted looks good to me.

As mentioned before the thread title could be replaced.

It would be useful to include a statement that the equivalence assumes the clock postulate.

[Iggy]

---

The fact that it should be d1 and d2 is actually the entire basis of my argument.

Good argument then.

If you go back and look at xyzt's post that I'm taking this from, d and v aren't generic variables for distance and velocity. He specifically defines them as d being the distance travelled by C as measured by A and v as the velocity of C as measured by A.

This:

[math]\tau_C = \frac{d}{v}\sqrt{1-v^2}[/math]

would give the right answer for [math]\tau_C[/math] if d were measured by A. v, by the way, is measured the same by A and C. They get the same value for that. The above equation assumes that C is moving and A is at rest. C's clock runs slow, in other words.

He then does tau_A = d/v *sqrt(1-(v/c)^2) from frame C. d here should be d(2).

That wouldn't work for D or D2 (I know I implied that it might in my last post but when talking about specific events, it won't). The equation would be:

[math]\tau_A = \frac{D_C}{v \sqrt{1-v^2}}[/math]

That equation would work. It, again, assumes that C is moving and A is at rest.

 

d/(u+v)(sqrt(1-(u/c)^2-sqrt(1-(v/c)^2)+d/v*sqrt(1-(v/c)^2) which defines the proper time experienced by B on the first leg with C on the second leg, where all of the d's are d(1) as if d(1) and d(2) are the same variable. He then uses the result to explain why the whole thing falls apart, when he was actually using a single variable to define to separate distances, which resulted in him subtracting the proper time experienced by C over the course of the experiment while thinking he was subtracting the proper time of A over the course of the experiment.

When I used the same variables that he did, I derived that contradiction because it's wrong.

Please go back and check his math to see whether I'm way off base here. It's in post #73.

I wasn't trying to do that. I was pointing out that you did do that.

 

I don't quite follow that. We know the experiment works. xyzt has admitted as much, just saying that the acceleration is hidden. If a certain method doesn't work then it is the method.

 

[xyzt]

Everything you quoted looks good to me.

As mentioned before the thread title could be replaced.

It would be useful to include a statement that the equivalence assumes the clock postulate.

Why not do the honorable thing and admit that the claim in the title is false?

[Iggy]

A quick proof that my approach obtains the same exact results as the one using the Minkowski diagrams above (but it is much more general).

 

The amount of coordinate time needed for B to reach BC as calculated in frame A is:

 

[math]T=\frac{x_{BC}}{v_B}=\frac{3.75}{0.6}[/math]

 

The amount of proper time elapsed on B's clock is:

 

[math]\tau_B=\int_0^T{\sqrt{1-v_B^2}dt}=T*\sqrt{1-v_B^2}=0.8 \frac{3.75}{0.6}=5[/math]

 

....which is exactly the number that can be read from the Minkowski diagram.

 

I found the proper time of A given the proper time of B assuming B is at rest. You just found the proper time of B given A.

 

That is not "the exact same results". It isn't even the same problem. I'll say again, I know you can find the proper time of the traveling twin. That's all you've done in this thread, and we're all very aware of how to do it.

Edited May 8, 2013 by Iggy

[md65536]

In other words, I proved you wrong and you surreptitiously edited the scenario? What post post did you post edit?

The experiment in #201 is exactly the same as the experiment in #1, with some additional unimportant details given in each.

 

The clock C in #1 is set at BC, but setting a clock does not affect proper time. C still ages 2 years at a rate of one year per yer whether or not it sets its clock midway. Setting a clock doesn't make it jump in age.

[xyzt]

I found the proper time of A given the proper time of B assuming B is at rest. You just found the proper time of B given A.

 

That is not "the exact same results". It isn't even the same problem. I'll say again, I know you can find the proper time of the traveling twin. That's all you've done in this thread, and we're all very aware of how to do it.

I did a lot more than that, I have proven that the claim in the title is false.

 

Not according to SR.

 

 

Using A's frame of reference and the original values, A ages 4 years between AB and AC, while B ages 2 years, and C ages 2 years. The B and C times are coordinate times, or a proper time for B between AB and B_end where B_end is simultaneous with AC according to A, and a proper time for C between C_start and AC where C_start is simultaneous with AB according to A.

 

Both B and C age at the same rate according to A, because v and -v give the same value of gamma. B and C meet halfway through the experiment according to A (or B or C) using these newly defined events. Thus B and C age 1 year---proper time---before meeting, and each ages 1 year after (using these events).

 

In the original post I was ignoring as irrelevant the aging of C before BC, and the aging of B after, however these can be calculated consistently with everything else. It is irrelevant because all I need to know is the proper time for B between AB and BC, and the proper time for C between BC and AC, which can be compared to the proper time for A between AB and AC. Each describes a path from AB to AC, which can meaningfully be compared.

 

No hidden acceleration.

So, all you did now is that you made B and C to have the same speed, thus they meet midway, so you just dumbed down the scenario such that it would fit your claim. But, as I already pointed out, when you claim:

 

"Both B and C age at the same rate according to A, because v and -v give the same value of gamma. B and C meet halfway through the experiment"

 

you simply reduced the scenario to the case where B has no role whatsoever. We have been over this several times already.

 

I found the proper time of A given the proper time of B assuming B is at rest. You just found the proper time of B given A.

 

That is not "the exact same results". It isn't even the same problem. I'll say again, I know you can find the proper time of the traveling twin. That's all you've done in this thread, and we're all very aware of how to do it.

So, I did the reverse, I found the proper time for B knowing the data from A. The point I made is that the two methods, yours and mine are equivalent. Yours is much more complicated <shrug>

Edited May 8, 2013 by xyzt

[md65536]

Why not do the honorable thing and admit that the claim in the title is false?

I admit the claim in the title is false.

 

So, all you did now is that you made B and C to have the same speed, thus they meet midway, so you just dumbed down the scenario such that it would fit your claim.

You're right, the experiment would be equivalent to a corresponding twin paradox even with different speeds.

[xyzt]

I admit the claim in the title is false.

Good, finally.

Please change the title to: "Acceleration plays a key role in the twin paradox"

[hypervalent_iodine]

!

Moderator Note

xyzt,

 

I'm issuing this as your final warning. You are to stop with the hostile/bullying approach to discussions and stop insulting people or you will be suspended from posting.

[Iggy]

So, I did the reverse, I found the proper time for B knowing the data from A. The point I made is that the two methods, yours and mine are equivalent. Yours is much more complicated <shrug>

 

I was proving that the velocity of B and C could be anything shy of c and you would get the same results as the traditional twin paradox. The easiest way to do that was to solve from the perspective of B and C. Solving from their perspective is more complicated. That is probably why you couldn't do it. I am truly sorry about that, but it's no reason for you to be belligerent.

[md65536]

Good, finally.

Please change the title to: "Acceleration plays a key role in the twin paradox"

I can't change it but I prefer my previously suggested: "Acceleration is not required to physically measure the time dilation effect demonstrated in the twin paradox."

[Iggy]

I admit the claim in the title is false.

 

You're right, the experiment would be equivalent to a corresponding twin paradox even with different speeds.

 

How so? A lot has happened in the thread since I was here last.

 

---

 

 

I can't change it but I prefer my previously suggested: "Acceleration is not required to physically measure the time dilation effect demonstrated in the twin paradox."

 

Oh, I do like that better.

[xyzt]

!

Moderator Note

xyzt,

 

I'm issuing this as your final warning. You are to stop with the hostile/bullying approach to discussions and stop insulting people or you will be suspended from posting.

 

 

I am sorry, I did not mean to insult. I just have a very low tolerance to cheating and lying. I have proven my point at post 73, yet , it took another 140 posts to conclude that the claims in the OP are indeed false. I hope that I will not encounter another thread like this, and, if I do, I will stay away or simply report it , as I have successfully done with the two threads that you moved to "Speculation".

I can't change it but I prefer my previously suggested: "Acceleration is not required to physically measure the time dilation effect demonstrated in the twin paradox."

This is not correct, the correct statement is : "Acceleration IS required to physically measure the time dilation effect demonstrated in the twin paradox."

How so? A lot has happened in the thread since I was here last.

---

 

 

 

Oh, I do like that better.

This is not correct, the correct statement is : "Acceleration IS required to physically measure the time dilation effect demonstrated in the twin paradox." . This was known since my post 73.

I was proving that the velocity of B and C could be anything shy of c and you would get the same results as the traditional twin paradox. The easiest way to do that was to solve from the perspective of B and C. Solving from their perspective is more complicated. That is probably why you couldn't do it. I am truly sorry about that, but it's no reason for you to be belligerent.

I have long proved it, post 73.

Edited May 9, 2013 by xyzt

[Iggy]

 

I was proving that the velocity of B and C could be anything shy of c and you would get the same results as the traditional twin paradox...

I have longed proved it, post 73.

 

No, you specifically said exactly the opposite which is why I went through the trouble of doing the math for the general case. The thread has a record. You said it here:

 

 

 

xyzt, would you say this statement is accurate:

 

Adding the proper time measured by Clock B between events AB and BC to the proper time measured by Clock C between events BC and AC will yield a number that is equivalent to the proper time experienced by an observer that travels from event AB to event BC and then instantly accelerates to travel back to event AC?

OK, I understand your notation. The above is:

1. true only if B and C have the same speed wrt A

2. false if B and C have different speeds wrt A

 

I showed this in my writeup.

 

Funny how post 73 has since proved opposite claims

Edited May 8, 2013 by Iggy

[xyzt]

 

No, you specifically said exactly the opposite which is why I went through the trouble of doing the math for the general case. The thread has a record. You said it here:

 

 

 

Funny how post 73 has since proved opposite claims

I don't think you understand what post 73 shows, it shows that there is a skip in the C clock at the BC event.

The skip is explained again, in Latex at post 185. Did you read the math at post 73? There is no such wording as you claim in post 73, there is just math.

Edited May 8, 2013 by xyzt

[Iggy]

 

I was proving that the velocity of B and C could be anything shy of c and you would get the same results as the traditional twin paradox...

 

I have longed proved it, post 73.

There is no such wording as you claim in post 73

 

I accept your retraction.

Edited May 9, 2013 by Iggy

[xyzt]

 

I accept your retraction.

It means your claim is false, there is no wording you claim in my post 73. There is no "retraction", you just made a false claim.

Edited May 9, 2013 by xyzt

[Iggy]

In means your claim is false, there is no wording you claim in my post 73. There is no "retraction", you just made a false claim.

 

I don't know what that means. I actually don't know what "in means" means.

 

Whatever. I'm going to stop talking to you now.

 

I encourage everyone else in this thread whom you've discouraged to do the same.

[md65536]

In response to "You're right, the experiment would be equivalent to a corresponding twin paradox even with different speeds.":

How so? A lot has happened in the thread since I was here last.

In the experiment I have B and C traveling at the same speed relative to A, only for the sake of simplicity, but it would work at any speeds. A traveling astronaut would age less by a predictable amount even if the return trip was faster or slower.

 

(Note: In post #1 I say B "passes clock C traveling in the opposite direction at the same speed" but I don't say for whom it is the "same". In the twin paradox it's A, and that's probably clear because if it was interpreted to mean the same speed according to B, any speed at all suffices! It would just have to be fast enough that A and C eventually meet: v_BC would have to be greater than v_AB... if it was interpreted to mean "B passes C at the same relative speed that it passed A" then A and C would be at rest. But as long as A and C meet, the twin paradox effect is still demonstrated.

 

But actually on second thought nevermind this note!, by saying B is traveling at v it is clearly referring to the frame where B is moving, that is A's.

)

I don't think you understand what post 73 shows, it shows that there is a skip in the C clock at the BC event.

The skip is explained again, in Latex at post 185. Did you read the math at post 73? There is no such wording as you claim in post 73, there is just math.

Clocks don't skip. What is the theoretical basis for predicting such a thing???

 

Is there a skip in the A clock at the AB event? Observer A resets its clock at the AB event, does that bother you?

 

In the experiment that I described, I have 3 inertial clocks each passing the other 2 clocks in 2 respective events. Why do you think that event BC is special?

Edited May 9, 2013 by md65536

[michel123456]

Event BC is special because there is a frame jumping.

 

A similar frame jumping happens at AB and at AC.

 

I'd suggest to open a new thread about frame jumping. There is no need to consider the particular case of a turning point: any frame jumping implies weird results that can be solved only by considering frame jumping as an extreme case of acceleration. IMHO of course.

 

And IMHO xyzt is right against everybody. If my opinion counts...

[Markus Hanke]

I am surprised that this thread is still going. It is an obvious and well established fact that, if only uniform relative motion in Minkowski space-time is involved, it is impossible to get any disagreements between clock readings. The twin paradox cannot exist under such circumstances, no matter what specific scenario is conjured up.

 

Do you really need the maths again ?

[xyzt]

 

Clocks don't skip. What is the theoretical basis for predicting such a thing???

 

Is there a skip in the A clock at the AB event? Observer A resets its clock at the AB event, does that bother you?

 

In the experiment that I described, I have 3 inertial clocks each passing the other 2 clocks in 2 respective events. Why do you think that event BC is special?

But in the scenario you contrived, they DO. Look at post 177 (or 73). At BC you made C dump its accumulated time and pick up the time shown by B. This results into a step, jump i, discontinuity in C's reading. This is the sleigh of hand that I have been referring to.

Edited May 9, 2013 by xyzt