Acceleration is not important in the twin paradox

[Delta1212]

Alright, just in case I'm actually getting confused, because I do like to double check myself to make sure I'm not screwing up.

 

xyzt: Going off of your muon example. If the distance is measured to be 2 ly and, for the sake of simplicity the velocity is measured to be 13/15c (both from Earth's frame of reference), can you tell me:

 

The time the muon takes to travel that path in its own frame, the time that the Earth measures the muon to take in Earth's frame, the proper time of the muon, and the proper time of Earth, and explain how to find each? I realize some of those values will be the same, I just have a feeling this will be easier to understand if I can compare some actual numbers and see what they represent.

 

Edit: Like I said, I'm new to this, so if I'm making a fundamental mistake somewhere, I really need to figure out where any confusion of mine rests, and that will be easier if I can see the math done correctly than if I keep attempting it the way I've understood it which, if wrong, won't help me any.

Edited May 8, 2013 by Delta1212

[xyzt]

I don't need the basic stuff on time dilation.

The calculations are very simple, in A frame B and C meet at [math]t=\frac{D}{v_B+v_C}[/math]

The total elapsed proper time in frame A for:

 

1. B to reach the BC event is [math]\tau_B=\int_0^t{\sqrt{1-v_B^2}dt}=\frac{D}{v_B+v_C}\sqrt{1-v_B^2}[/math]

2. C to reach the BC event is [math]\tau_C=\int_0^t{\sqrt{1-v_c^2}dt}=\frac{D}{v_B+v_C}\sqrt{1-v_c^2}[/math]

 

Very simple.

After the encounter, two things happen:

- C "dumps" its accumulated proper time and gets its clock reset to [math]\tau_B=\frac{D}{v_B+v_C}\sqrt{1-v_B^2}[/math]

 

- C continues on towards A and arrives in A at the coordinate time [math]t'=\frac{D}{v_C}[/math] accumulating an additional:

 

 

 

[math]\tau'_C=(\frac{D}{v_C}-\frac{D}{v_B+v_C}) \sqrt{1-v_C^2}[/math]

 

So, the total time shown on C's clock when it arrives in A is:

 

 

 

[math]\tau^"_C=\tau_C+\tau'_C=(\frac{D}{v_C}-\frac{D}{v_B+v_C}) \sqrt{1-v_C^2}+\frac{D}{v_B+v_C} \sqrt{1-v_B^2}[/math]

 

Obviously, four things are true:

 

1. [math]\tau^"_C \ne \tau_A[/math]

 

2. A "jump" in C's clock has been inserted by the OP at the point where it met B via artificially resetting it to B's clock value. The effect is nothing but a sleigh of hand , fully equivalent to instantaneously accelerating from [math]v_C[/math] to [math]v_B[/math] retroactively (speak about unphysical stuff happening here!) at the point where C meets B.

 

 

3. Absent the sleigh of hand with resetting the clock at the BC event, the C clock will show:

 

[math]\tau^"_C=\frac{D\sqrt{1-v_C^2}}{v} [/math]

 

since C measures a length contracted distance to A.

 

 

4. for the trivial case whre [math]v_C=v_B=v[/math]

 

[math]\tau^"_C=\frac{D}{v} \sqrt{1-v^2}[/math]

 

i.e. the exercise reduces to the "instantaneous turnaround" for B, i.e. the case where B turns around with infinite acceleration.

 

So, the claim made in the title of the thread is bogus, you cannot get a twin "paradox" in the absence of acceleration (or, as in the case of this thread, a silly sleigh of hand that the naive would not notice).

Edited May 8, 2013 by xyzt

[Delta1212]

After the encounter, two things happen:

- C "dumps" its accumulated proper time and gets its clock reset to [math]\tau_B=\frac{D}{v_B+v_C}\sqrt{1-v_B^2}[/math]

 

- C continues on towards A and arrives in A at the coordinate time [math]t'=\frac{D}{v_C}[/math] accumulating an additional:

 

 

 

[math]\tau'_C=(\frac{D}{v_C}-\frac{D}{v_B+v_C}) \sqrt{1-v_C^2}[/math]

 

So, the total time shown on C's clock when it arrives in A is:

 

 

 

[math]\tau"_C=\tau_C+\tau'_C=(\frac{D}{v_C}-\frac{D}{v_B+v_C}) \sqrt{1-v_C^2}+\frac{D}{v_B+v_C} \sqrt{1-v_B^2}[/math]

 

Obviously, two things are true:

 

1. [math]\tau"_C \ne \tau_A[/math]

 

2. A "jump" in C's clock has been inserted at the point where it met B via artificially resetting it to B's clock value. The effect is nothing but a sleigh of hand , fully equivalent to instantaneously accelerating from [math]v_C[/math] to [math]v_B[/math] at the point where C meets B.

Wouldn't it be equivalent to accelerating from [math]v_B[/math] to [math]v_C[/math] since the first time measurement is made at [math]v_B[/math] and the second at [math]v_C[/math]? Or is that just semantics? Honest question.

 

For the sake of comparison, what would [math]\tau''_C[/math] be if C did not reset it's clock on passing B?

[xyzt]

Wouldn't it be equivalent to accelerating from [math]v_B[/math] to [math]v_C[/math] since the first time measurement is made at [math]v_B[/math] and the second at [math]v_C[/math]? Or is that just semantics? Honest question.

 

For the sake of comparison, what would [math]\tau''_C[/math] be if C did not reset it's clock on passing B?

I really don't feel like helping you just to see you turning around and claiming that you found "mistakes" in my posts. I don't like hypocrites, you have al the formulas, you can do the calculations yourself. You can stop giving me negatives, I don't really care about them.

Edited May 8, 2013 by xyzt

[Delta1212]

 

I really don't feel like helping you just to see you turning around and claiming that you found "mistakes" in my posts. I don't like hypocrites, you have al the formulas, you can do the calculations yourself. You can stop giving me negatives, I don't really care about them.

I don't keep giving you negatives. I rarely give people anything, actually.

[md65536]

Proper time is the time read by a clock between two events that the clock is present at. Everyone agrees on what a clock measures in its own frame. This is why proper time is frame invariant.

 

If the Earth measures a muon traveling to Earth from distance d (as measured by Earth) at velocity v (as measured by the Earth), then d/v * sqrt(1-(v/c)^2) will give you the proper time of the muon as it travels that path in the frame of the muon. But d/v will be the time that the Earth measures the muon taking in Earth's frame. It will, however, agree that the muon experiences d/v *sqrt(1-(v/c)^2) time during that period because that is the muon's proper time, which is agreed upon in all frames.

 

Proper time being frame invariant does not mean that every frame measures the same time between two events. It just means they all agree on what each other measures.

I can see what you're doing here, and I agree. I can't see what xyzt is doing, by always referring to proper times without defining them, or explaining what events they span.

 

 

The formula d/v *sqrt(1-(v/c)^2) calculates the proper time of a muon passing through events "enter atmosphere" and "hit Earth" where d is the (rest) distance between those two events. Setting t=d/v, then T_u = t*sqrt(1-(v/c)^2).

 

From the muon's perspective, setting t to the local time that the muon experiences passing through those two events, then the proper time on Earth of two events that are simultaneous ACCORDING TO THE MUON is also T_A = t*sqrt(1-(v/c)^2) using a different value of t than before. However this is fairly meaningless. Time t is not equal to d/v, because the distance (height of atmosphere) is length contracted according to the muon. At very high v, the height of the atmosphere is small for a muon, and the proper time of crossing it is small, and the time that elapses on Earth is tiny, and on Earth the tiny proper time between those events is meaningless, because the events don't describe anything relevant to the Earth observer. They're events on Earth that are simultaneous according to another observer (the muon) with the muon entering the atmosphere, and it hitting the Earth, respectively---the latter event can be agreed on by all.

 

I think that xyzt's confusion is in calculating "proper times" using remote events, thus not bothering to define events using any specific simultaneity (assuming any will do?), thus calculating different "proper times" depending on whose clock is used for the calculations, and THEN ignoring the problems of simultaneity by saying "proper time is invariant", as if the proper time at A between any two events must be the same as the proper time at A between any other 2 events, which is wrong. However, I can only guess that that is the error, because no details, no specific events are given. The events are always implied and they keep changing in xyzt's counter examples.

Edited May 8, 2013 by md65536

[Delta1212]

I really don't feel like helping you just to see you turning around and claiming that you found "mistakes" in my posts. I don't like hypocrites, you have al the formulas, you can do the calculations yourself. You can stop giving me negatives, I don't really care about them.

I pointed out something that seems like a mistake to me. It is possible that I am wrong, in which case, I need to understand why. If I see something contradictory in the explanation, I will point it out because it means that either there is a mistake in the explanation or I haven't understood it properly. Either way, there needs to be a correction for us to be on the same page.

 

I apologize if you don't like me trying to poke holes in what you're saying, and I do understand that, assuming you are totally correct, this can be frustrating. If I don't try to poke holes, though, I'm not going to figure out how it works. I don't mean to offend. You suggested I learn to follow the math, and I'm making a genuine effort to do that. I believe I understand what proper time and coordinate time are, but if I do, then you made a mistake, so if you're correct, then I'm obviously misinterpreting something.

 

I asked for specific numbers because if I can compare actual examples of proper time, I'll have a better grasp on where my understanding appears to differ from yours so that it can be corrected.

 

It's entirely possible that the fault lies with me, but I'm having trouble figuring out exactly where I'm going wrong, and simply saying that proper time is frame invariant doesn't seem to be helping because I thought I already understood that.

[xyzt]

 

 

I think that xyzt's confusion is in calculating "proper times" using remote events, thus not bothering to define events using any specific simultaneity (assuming any will do?), thus calculating different "proper times" depending on whose clock is used for the calculations, and THEN ignoring the problems of simultaneity by saying "proper time is invariant", as if the proper time at A between any two events must be the same as the proper time at A between any other 2 events, which is wrong. However, I can only guess that that is the error, because no details, no specific events are given. The events are always implied and they keep changing in xyzt's counter examples.

No, I am not doing any of the imagined mistakes you are accusing me of. Proof is, that if I dumb down the setup for the trivial cases, I recover the results obtained in the simplistic scenarios. The claim in your thread title is false, you will need to learn to live with it, no matter how you twist and turn.

[Delta1212]

xyzt: Can I ask some questions about the equations in the post where you did all the math? It's possible I'm misinterpreting how you're using some of the variables and/or not understanding how things are being plugged into the formulae. If that's the case, it would explain why I don't think your math adds up, because then I'm equating values that aren't meant to be the same. You don't have to actually do any math, I just want to know where a few things came from.

[xyzt]

Ok, if you're unfamiliar with spacetime diagrams then I'm sure this is just gonna look like a gigantic mess, but I'll do my best to be as descriptive as possible:

 

I adapted a diagram I had already made for Michel. The velocity is v = 0.6. If I used the velocity we've been using 13/15c the angles would be too small to work well.

 

 

The green and black arrowed line marked earth is earth's world line. The line marked [math]X_A[/math] is earth's spatial axis. The green dotted lines and green numbers are earth's coordinate system. The bold red line marked [math]\tau_B[/math] is B's world line. It bumps into C at the event marked BC. The bold red line marked [math]X_B[/math] is B's spatial axis. The red dotted lines and red numbers are B's coordinate system.

 

We are solving A's proper time (earth's proper time) for half of the experiment from B's perspective. Half the experiment is from event AB to event H. So, we know B's proper time and we know A's velocity relative to B, and that's it. That is all we know, and we cannot assume that B moves while A is stationary. That is cheating. We are solving from B's perspective. B's coordinate system.

 

We know the length of the red line from the event AB to BC in the red coordinate system. It is 5 years. Using the time dilation equation we can solve the length of the black line from the event AB to the event P in the green coordinate system.

 

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

 

[math]\tau_A = 5 \sqrt{1-0.6^2}[/math]

 

[math]\tau_A = 4[/math]

 

AB -> P in the green coordinate system should be 4 years. The diagram verifies this.

 

The red line that joins P and BC is the line of simultaneity for B when B bumps into C. The green line that joins H and BC is the line of simultaneity for A when B bumps into C. The length of P -> H has yet to be accounted for and that's what I used the following time Lorentz transformation for:

 

[math]t = \frac{t' + vx'}{\sqrt{1-v^2}}[/math]

 

It will find the length of the black line P2 -> H2 in the green coordinate system if you set t'=0 and x' = 3. We are looking for the black line P -> H which is the same length as P2 -> H2. Setting t'=0 and [math]x' = \tau_B \cdot v[/math] (which always equals the correct x') we get:

 

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

 

[math]\tau_A = 5 \frac{0.6^2}{\sqrt{1-0.6^2}}[/math]

 

[math]\tau_A = 2.25[/math]

 

The black line P -> H should therefore be 2.25 years in the green coordinate system. Checking the diagram verifies.

 

Add the two together

 

[math]\tau_A = 4 + 2.25[/math]

 

[math]\tau_A = 6.25[/math]

 

and one gets the proper time of A for half the trip. In other words, the length of the black line AB -> H is 6.25 years in the green coordinate system.

 

That is why I summed those two terms when solving from B's perspective. One finds the length of AB -> P and the other finds the length of PH. If one is familiar enough with the Lorentz transforms they should be able to do the above in their head (or with a small sketch) and come up with the equations needed.

 

enough said, I think.

 

 

 

---

 

 

 

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.

Edited May 8, 2013 by xyzt

[md65536]

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.

But 5 > 3.75/6, so you're saying that according to A, B's clock ticks faster? Nevermind, I can't read...

 

 

Okay, I think I agree with this math. Earlier the equation was [math]\tau_A=\frac{d \sqrt{1-(v/c)^2}}{v}[/math], incorrectly calculated for A instead of B, which is what I disagreed with.

Edited May 8, 2013 by md65536

[Delta1212]

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.

Ok, so A cannot measure a proper time between B leaving and B reaching C, because A is not present when B reaches C. A's clock will read one time (edited to add: in frame A) as being simultaneous with B reaching C, but an observer in a different frame will read a different time on A's clock as being simultaneous with B reaching C. Therefore the time that elapses on A's clock between B leaving and reaching C is not frame invariant and is not proper time, but coordinate time.

 

Did I get that right?

Edited May 8, 2013 by Delta1212

[xyzt]

Ok, so A cannot measure a proper time between B leaving and B reaching C, because A is not present when B reaches C. A's clock will read one time as being simultaneous with B reaching C, but an observer in a different frame will read a different time on A's clock as being simultaneous with B reaching C. Therefore the time that elapses on A's clock between B leaving and reaching C is not frame invariant and is not proper time, but coordinate time.

 

Did I get that right?

Yes, finally.

 

But 5 > 3.75/6, so you're saying that according to A, B's clock ticks faster? Nevermind, I can't read...

Give yourself a negative for the brainfart.

[Delta1212]

Yes, finally.

Ok, I have another question then. Would every frame agree upon the time at which A believes B to have arrived at C according to A's frame.

 

In other words, if A sets an alarm to go off at the time on A's clock that A believes is simultaneous with B reaching C, would observers in every frame agree on the time that A's clock read when the alarm went off, but disagree that it was simultaneous with B reaching C?

 

I'm hoping that was clear enough to understand.

[xyzt]

Ok, I have another question then. Would every frame agree upon the time at which A believes B to have arrived at C according to A's frame.

 

In other words, if A sets an alarm to go off at the time on A's clock that A believes is simultaneous with B reaching C, would observers in every frame agree on the time that A's clock read when the alarm went off, but disagree that it was simultaneous with B reaching C?

 

I'm hoping that was clear enough to understand.

Why is it relevant for the exercise? The OP has been proven FALSE. This is the end of the line.

[md65536]

No, I am not doing any of the imagined mistakes you are accusing me of. Proof is, that if I dumb down the setup for the trivial cases, I recover the results obtained in the simplistic scenarios. The claim in your thread title is false, you will need to learn to live with it, no matter how you twist and turn.

Alright I think that with your corrected equations we all might be coming to agreement?

 

I'll incorporate info from the 189 replies, and change the claim made in the title of the thread. As-is it doesn't make sense, because the twin paradox by name involves just 2 clocks, and one clock must not be inertial thus acceleration is important. So instead, a better title might be:

 

The time dilation measured in the twin paradox, in the case where the traveling twin's path can be separated into several independent inertial sections, can also be measured using a set of independent inertial clocks, one for each section, assuming that the clock postulate holds.

 

 

Or... to stick to the point:

 

Acceleration is not required to physically measure the time dilation effect demonstrated in the twin paradox.

Edited May 8, 2013 by md65536

[Delta1212]

Why is it relevant for the exercise? The OP has been proven FALSE. This is the end of the line.

I'm trying to figure out why what I did disagrees with what you did. I'm pretty sure that it's a result of mixing up proper time with coordinate time, but I'm looking to determine exactly where the error is.

 

If everyone agrees on the time that A believes is simultaneous with B meeting C, even if they all disagree that it is actually simultaneous, then you could get a measurement of the proper time that A measures between B leaving and the moment on A's clock that A believes is simultaneous with B's passing of C, but this would not be equal to the proper time on B's clock for the elapsed time of the journey. Right?

[xyzt]

Alright I think that with your corrected equations we all might be coming to agreement?

 

I'll incorporate info from the 189 replies, and change the claim made in the title of the thread. As is it doesn't make sense, because the twin paradox by name involves just 2 clocks, and one clock must not be inertial thus acceleration is important. So instead, a better title might be:

 

The time dilation measured in the twin paradox, in the case where the traveling twin's path can be separated into several independent inertial sections, can also be measured using a set of independent inertial clocks, one for each section, assuming that the clock postulate holds.

There is no "corrected equations", the correct equations have been there since post 73.

How about you simply admit that your claim in the title of the thread is FALSE?

While you are at it, feel free to reverse al the negs you have been giving me. Time to close this thread, it really promovates anti-mainstream ideas.

Edited May 8, 2013 by xyzt

[md65536]

There is no "corrected equations", the correct equations have been there since post 73.

How about you simply admit that your claim in the title of the thread is FALSE?

While you are at it, feel free to reverse al the negs you have been giving me. Time to close this thread, it really promovates anti-mainstream ideas.

Post #73 defines a proper time at A according to F_C, which is different from the times at A used in more recent posts. The equations of #73 might be correct equations for the wrong events... events which do not correspond to events described in a twin paradox nor the experiment described in this thread.

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.

 

[...]

 

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

This is the incorrect part. This calculation would work for tau_B, as in post #185 (which is what I considered "corrected"). Or it would work for tau_A if d was a rest length in F_C's frame.

 

If everyone agrees on the time that A believes is simultaneous with B meeting C, even if they all disagree that it is actually simultaneous, then you could get a measurement of the proper time that A measures between B leaving and the moment on A's clock that A believes is simultaneous with B's passing of C, but this would not be equal to the proper time on B's clock for the elapsed time of the journey. Right?

Right!

[xyzt]

Post #73 defines a proper time at A according to F_C, which is different from the times at A used in more recent posts. The equations of #73 might be correct equations for the wrong events... events which do not correspond to events described in a twin paradox nor the experiment described in this thread.This is the incorrect part. This calculation would work for tau_B, as in post #185 (which is what I considered "corrected"). Or it would work for tau_A if d was a rest length in F_C's frame.

 

Right!

[math]\tau_A[/math] is not necessary in proving that your conjecture is wrong. Your conceptuual error is identically exposed in both posts 73 and 185.

Edited May 8, 2013 by xyzt

[md65536]

[math]\tau_A[/math] is not necessary in proving that your conjecture is wrong. Your conceptuual error is identically exposed in both posts 73 and 185.

Post #185 agrees with the twin paradox, and the experiment described in post #1. There is no mention of acceleration in post #185.

[xyzt]

Post #185 agrees with the twin paradox, and the experiment described in post #1. There is no mention of acceleration in post #185.

post 185 exposes your sleigh of hand, the setup in post 1 just hides the acceleration by skipping proper elapsed time on the C clock (time has a discontinuity at event BC, introduced by your scenario).

post 73 already exposed the sleigh of hand, so we could have saved more than 100 posts if you admitted to the inner error

Edited May 8, 2013 by xyzt

[Delta1212]

post 185 exposes your sleigh of hand, the setup in post 1 just hides the acceleration by skipping proper elapsed time on the C clock (time has a discontinuity at event BC, introduced by your scenario).

post 73 already exposed the sleigh of hand, so we could have saved more than 100 posts if you admitted to the inner error

The point of the experiment isn't to measure C's time. Changing C's time doesn't yield a false measurement of C's time at the end, because no one thinks that that is the time as measured by C since the beginning of the experiment...

[md65536]

(time has a discontinuity at event BC, introduced by your scenario).

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.

Edited May 8, 2013 by md65536

[xyzt]

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. 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, and using a start time for C that is simultaneous, to A, with AB. Thus B and C age 1 year---proper time---before meeting, and each ages 1 year after.

 

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.

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?