Acceleration is not important in the twin paradox

[md65536]

the reason that you keep getting unphysical answers

I'm getting physical answers and they're consistent, and they're consistent with SR. Feel free to check them, I may have made additional mistakes. But please stick to the original or equivalent experiments instead of changing it.

[xyzt]

I'm getting physical answers and they're consistent, and they're consistent with SR. Feel free to check them, I may have made additional mistakes. But please stick to the original or equivalent experiments instead of changing it.

 

 

This is the whole point, in the BOGUS scenario that you describe in the first post, the clocks DO exhibit a jump via the fact that B assigns its time to C. You need to follow the argument, it is pretty basic.

[md65536]

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

It's been pointed out by others that the experiment described in post #1 is physically possible. Please stop trolling this thread, you're misleading people.

[xyzt]

It's been pointed out by others that the experiment described in post #1 is physically possible. Please stop trolling this thread, you're misleading people.

It is physically possible but it produces BOGUS results. Try following the counter-argument instead of pretending that it isn't there. The math is pretty elementary.

Edited May 6, 2013 by xyzt

[md65536]

This is the whole point, in the BOGUS scenario that you describe in the first post, the clocks DO exhibit a jump via the fact that B assigns its time to C. You need to follow the argument, it is pretty basic.

Hrm, setting a clock does not constitute a jump in time.

Proper time is the time elapsed between 2 events, and in all the variations C is measuring the proper time between BC and AC. Setting the clock in the original experiment is equivalent to starting a measurement of proper time.

 

To speak only in terms of proper time, the equivalent experiment description in post #3 is clearer. The original experiment has C measuring a proper time of 1 between time 1 and 2 on its clock... I can see how this is confusing, the modifications to the experiment---where no clock sync happens---clear this up.

Edited May 6, 2013 by md65536

[xyzt]

Hrm, setting a clock does not constitute a jump in time.

Proper time is the time elapsed between 2 events, and in all the variations C is measuring the proper time between BC and AC. Setting the clock in the original experiment is equivalent to starting a measurement of proper time.

 

To speak only in terms of proper time, the equivalent experiment description in post #3 is clearer...

LOL, the C clock shows one time when it arrives in P and a DIFFERENT time when it leaves P.

I agree with you, this thread belongs in Speculation. It belonged from the very first post.

[md65536]

It is physically possible but it produces BOGUS results. Try following the counter-argument instead of pretending that it isn't there. The math is pretty elementary.

You described a different experiment. You've modified it to be symmetrical from frame F's perspective, and then started C's clock according to F instead of actually measuring the events described in the original experiment.

[xyzt]

You described a different experiment. You've modified it to be symmetrical from frame F's perspective, and then started C's clock according to F instead of actually measuring the events described in the original experiment.

Nope, it is the same "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."

 

The only difference is that you set the speed of C to be equal to the speed of B. According to my explanation, this makes B totally irrelevant (I told you this several times already). What I did is to generalize to "u \ne v" and to show the absurdities your scenario leads to. Instead of rushing your answers in trying to defend your position, how about you read the counter-argument I wrote. It dissects your scenario with math, not with empty words. It is hard to write any math on this forum, this is why people don't post any, they just talk arond the subject.

Edited May 6, 2013 by xyzt

[Delta1212]

 

It is physically possible but it produces BOGUS results. Try following the counter-argument instead of pretending that it isn't there. The math is pretty elementary.

No one is claiming that the time on Clock C at the end of the experiment is the time that Clock C has experienced since the beginning of the experiment. You are arguing against a strawman of the experiment in the OP.

[xyzt]

No one is claiming that the time on Clock C at the end of the experiment is the time that Clock C has experienced since the beginning of the experiment. You are arguing against a strawman of the experiment in the OP.

That is false, the time on C is CHANGED to the time on B when the two meet. Pay attention before you post such mistakes.

 

"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."

Edited May 6, 2013 by xyzt

[md65536]

"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."

Yes, I was sloppy there. The last sentence should say either, "they'll record the same discrepancy on their clocks", or "they'll record the same difference in proper time between AC and BC+AC".

 

The "proper time" from AB to BC to AC is the sum of the proper times measured by 2 different clocks in my experiment, which I'm saying is equivalent to the proper time of a single clock passing through the same events.

 

However, it is a mistake of mine to call it a proper time, because proper time applies to a single clock. I should have called it a sum of proper times.

That is false, the time on C is CHANGED to the time on B when the two meet. Pay attention before you post such mistakes.

How do you know that? The time on C's clock before meeting B is never mentioned (it's actually irrelevant). How do you know it didn't start 1 unit of time (B's proper time between AB and BC used in examples) earlier, and found that it already had the same time as B when it meets?

 

AND if it did that, does that change its measurement of the time between BC and AC?

Edited May 6, 2013 by md65536

[xyzt]

I'm getting physical answers and they're consistent, and they're consistent with SR. Feel free to check them, I may have made additional mistakes. But please stick to the original or equivalent experiments instead of changing it.

 

Yes, I was sloppy there.

Good, you are making some progress in admitting you are wrong.

The last sentence should say either, "they'll record the same discrepancy on their clocks", or "they'll record the same difference in proper time between AC and BC+AC".

 

The "proper time" from AB to BC to AC is the sum of the proper times measured by 2 different clocks in my experiment, which I'm saying is equivalent to the proper time of a single clock passing through the same events.

 

However, it is a mistake of mine to call it a proper time, because proper time applies to a single clock. I should have called it a sum of proper times.

The mistake runs much deeper than that: in your naive setup, B is made totally irrelevant because it has the same speed as C, B doesn't play any role, A and C move inertially towards each other and record the same proper time, nothing to do with the "twins paradox" . So, your argument in the OP falls apart. If, on the other hand B and C have different speeds, your argument falls apart for a different reason, C clock has been advanced (or retarded) by a non-null jump when it encountered B. So, your "experiment" is bogus.

 

How do you know that? The time on C's clock before meeting B is never mentioned (it's actually irrelevant). How do you know it didn't start 1 unit of time (B's proper time between AB and BC used in examples) earlier, and found that it already had the same time as B when it meets?

 

AND if it did that, does that change its measurement of the time between BC and AC?

I know becuase I took the trouble to write the exact mathematical description instead of piling words. This is how I know.

Edited May 6, 2013 by xyzt

[md65536]

I know becuase I took the trouble to write the exact mathematical description instead of piling words. This is how I know.

I see what you're saying now. If we universally call event AB as the "start" of the experiment and AC the end, then over the experiment...

 

According to symmetrical observer F, A ages 4 years while C ages 4 years. This is your argument?

 

According to A, A ages 4 years, while C ages 2 years. According to A, the aging of C over the experiment is the same as the aging of B from AB to BC, plus the aging of C from BC to AC. This is (sort of) my argument.

 

According to C, A ages 4 years, while C ages 8 years.

 

I think we are in agreement.

Edited May 6, 2013 by md65536

[xyzt]

I see what you're saying now. If we universally call event AB as the "start" of the experiment and AC the end, then over the experiment...

 

According to symmetrical observer F, A ages 4 years while C ages 4 years. This is your argument?

 

According to A, A ages 4 years, while C ages 2 years. According to A, the aging of C over the experiment is the same as the aging of B from AB to BC, plus the aging of C from BC to AC. This is my argument.

 

According to C, A ages 4 years, while C ages 8 years.

 

I think we are in agreement.

You are still wrong, proper time does not depend on the observer, it is frame invariant. So you cannot have:

 

"According to A, A ages 4 years, while C ages 2 years.

According to C, A ages 4 years, while C ages 8 years.."

 

I cannot ever agree with the stuff you just wrote, basic SR says you are wrong and you continue along the path of being wrong.

Edited May 6, 2013 by xyzt

[md65536]

You are still wrong, proper time does not depend on the observer, it is frame invariant. So you cannot have:

 

"According to A, A ages 4 years, while C ages 2 years.

According to C, A ages 4 years, while C ages 8 years.."

 

I cannot ever agree with the stuff you just wrote, basic SR says you are wrong and you continue along the path of being wrong.

I am still right.

Observer C never passes through event AB, it starts the experiment at a distance and with velocity relative to A. Thus the time at C that is simultaneous with event AB does depend on observer.

 

 

You're right though... all observers agree A's proper time between AB and AC is 4 units, B's proper time between AB and BC is 1 unit, and C's proper time between BC and AC is 1 unit, where gamma=2 in these examples.

 

And all agree that C's proper time between its start of your experiment as determined by F, is 4 units.

Edited May 6, 2013 by md65536

[xyzt]

I am still right.

Observer C never passes through event AB, it starts the experiment at a distance and with velocity relative to A. Thus the time at C that is simultaneous with event AB does depend on observer.

Of course you are right, even when you are totally wrong, even when you demonstrate that you don't know basics.

Edited May 6, 2013 by xyzt

[Delta1212]

Of course you are right, even when you are totally wrong, even when you demonstrate that you don't know basics.

Except you keep saying that he's wrong, and then saying things that are frequently, well, wrong.

[xyzt]

Except you keep saying that he's wrong, and then saying things that are frequently, well, wrong.

You need to learn how to follow the math.

It appears that md is starting to understand where he went wrong. Sometime, in the future, maybe you will understand as well.

Edited May 6, 2013 by xyzt

[Delta1212]

You need to learn how to follow the math.

It appears that md is starting to understand where he went wrong. Sometime, in the future, maybe you will understand as well.

Quick question as I'm actually trying to go through the math at your suggestion, I'm sure it was just a typo.

 

"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)"

 

I'm finding this difficult to parse properly because there are two open parentheses that aren't closed, and I don't want to make an assumption about what is being grouped in case I'm wrong and introduce an error that wasn't actually there.

 

Also, technically, the time shown on C when it reaches A would be tau_BP+tau_PA as tau_CP was, in your words, "dumped" but you appear to be using the correct variables so I assume you saying that C dumped its time for B's was meant to convey that tau_CP now equals tau_BP.

 

Edit: Just went and did the addition myself and figured out where the parentheses should have gone.

Edited May 6, 2013 by Delta1212

[Iggy]

Of course you are right, even when you are totally wrong, even when you demonstrate that you don't know basics.

 

He is right. Post 88 is correct, and post 90 correctly explains the basics for why it is right.

 

The statement in quotes:

 

You are still wrong, proper time does not depend on the observer, it is frame invariant. So you cannot have:

 

"According to A, A ages 4 years, while C ages 2 years.

According to C, A ages 4 years, while C ages 8 years.."

 

I cannot ever agree with the stuff you just wrote, basic SR says you are wrong and you continue along the path of being wrong.

 

is 100% accurate. The invariance of proper time doesn't conflict with it, and the relativity of simultaneity is the thing that makes it correct.

[Delta1212]

Let me try using some actual numbers to see if I can get my head around this. For the sake of simplicity, we'll say A is at rest, observes B and C as each traveling at 0.8c, and measures each leg as taking 5 years.

proper time for A would be

10*sqrt(1-(0/c)^2)

which is, obviously, 10 years.

A sees B travel at 0.8c for 5 years, so B will be measured to have traveled 4 ly. B's proper time will be:

5*sqrt(1-(0.8c/c)^2) = 5 * sqrt(0.36) = 3 years

From the meeting point of B and C, C sets its clock to read 3 years. A then observes C traveling for 5 years at 0.8c covering a distance of 4 ly. C's proper time during the second leg of the trip is

5*sqrt(1-(0.8c/c)^2) = 5 * sqrt(0.36) = 3 years

which is added on to the 3 years it was set to at the meeting of B and C.

3 years + 3 years = 6 years.

6 years < 10 years.

I've never done the math for relativity myself before and I'm only passingly familiar with the equations I just used, so I'm quite sure I must have gotten something wrong. Can you point out where I made my error so that I can correct it?

[Iggy]

Let me try using some actual numbers to see if I can get my head around this. For the sake of simplicity, we'll say A is at rest, observes B and C as each traveling at 0.8c, and measures each leg as taking 5 years.

Would it make more sense to use 13/15c, to directly compare the numbers with the original problem?

edit:

for the velocity I mean, and where [math]\tau_A[/math] is 4 yrs

Edited May 6, 2013 by Iggy

[Delta1212]

 

Would it make more sense to use 13/15c, to directly compare the numbers with the original problem?

edit:

for the velocity I mean, and where [math]\tau_A[/math] is 4 yrs

It would, but I didn't know that was the velocity until after I'd already done my own math because that was literally my first attempt at doing any kind of math for relativity, and I wasn't familiar enough with what I'm assuming must be rather commonly used velocities and their associated gammas to recognize that 13/15c is the velocity in that problem.

 

I arbitrarily picked 0.8c and was kind of surprised to wind up with a whole number of years, to be honest. I'll give it a try for the sake of practice, though.

[xyzt]

 

He is right. Post 88 is correct, and post 90 correctly explains the basics for why it is right.

 

The statement in quotes:

 

 

is 100% accurate. The invariance of proper time doesn't conflict with it, and the relativity of simultaneity is the thing that makes it correct.

You have the math to work with, the language of physics is math, "literary composition" doesn't cut it.

This problem is judged from one frame only , so it has nothing to do with RoS. Perhaps if you stopped making sweeping statements and you started writing math, you could be more believable. For the time being, not.

 

 

Let me try using some actual numbers to see if I can get my head around this. For the sake of simplicity, we'll say A is at rest, observes B and C as each traveling at 0.8c, and measures each leg as taking 5 years.

proper time for A would be

10*sqrt(1-(0/c)^2)

which is, obviously, 10 years.

A sees B travel at 0.8c for 5 years, so B will be measured to have traveled 4 ly. B's proper time will be:

5*sqrt(1-(0.8c/c)^2) = 5 * sqrt(0.36) = 3 years

From the meeting point of B and C, C sets its clock to read 3 years. A then observes C traveling for 5 years at 0.8c covering a distance of 4 ly. C's proper time during the second leg of the trip is

5*sqrt(1-(0.8c/c)^2) = 5 * sqrt(0.36) = 3 years

which is added on to the 3 years it was set to at the meeting of B and C.

3 years + 3 years = 6 years.

6 years < 10 years.

I've never done the math for relativity myself before and I'm only passingly familiar with the equations I just used, so I'm quite sure I must have gotten something wrong. Can you point out where I made my error so that I can correct it?

If B anC have the same speed, B is irrelevant to the problem, the paradox reduces to the form where B turns around instantaneously, with infinite acceleration. This is the fourth time when I point this out. No matter how you three guys twist and turn, the paradox is explained only in the presence of acceleration.

Edited May 6, 2013 by xyzt

[Iggy]

You have the math to work with, the language of physics is math, "literary composition" doesn't cut it.

This problem is judged from one frame only , so it has nothing to do with RoS....

 

Am I to understand that you think this refers to one frame?...

 

According to A, A ages 4 years, while C ages 2 years.

 

According to C, A ages 4 years, while C ages 8 years..

 

You do understand that "according to A" is a different frame from "according to C", and that their simultaneity is relative?

 

I think this may be the source of the problem.