Acceleration is not important in the twin paradox

[Iggy]

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

 

It doesn't count. It counts for nothing.

 

Until you can say why and how and give equations and explain their why and how... your empty biased support means absolutely nothing. It is a waste of your time to write and a waste of our time to read.

 

Less than nothing.

[xyzt]

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 ?

 

It is clear that he would not admit to error, even if you clearly outlined it in post 2 and I exposed the sleigh of hand in the scenario in both posts 73 and 177

Edited May 9, 2013 by xyzt

[michel123456]

It doesn't count. It counts for nothing.

 

Until you can say why and how and give equations and explain their why and how... your empty biased support means absolutely nothing. It is a waste of your time to write and a waste of our time to read.

 

Less than nothing.

So you are not surprised that after so many posts full of equations the argument is still not resolved.

 

You don't need math to solve this argument. Relativity is based on the principle that there is no preferential inertial F.O.R.

When you describe a situation where all observers are inertial ones there should exist no paradox. It is a simple way to CHECK the maths. If you obtain a discrepancy, your maths are wrong.

 

Md's argument is that one can construct an experiment where all observers are inertial ones and at the end obtain a discrepancy. He should look for the error. It is disappointing that he refuses to search for it. Worse, when showed where the "trick" is, he proceeds and invents another one. He behaves exactly like all those who present a free energy device.

And you are supporting him.

[md65536]

Md's argument is that one can construct an experiment where all observers are inertial ones and at the end obtain a discrepancy. He should look for the error. It is disappointing that he refuses to search for it. Worse, when showed where the "trick" is, he proceeds and invents another one. He behaves exactly like all those who present a free energy device.

Oh brother. I HAVE searched for it. I can't find it. I've been told that it's "hidden". So someone show it to me! But not with nonsense, hand-waving, or even with math if it backs up what I'm claiming.

 

But Iggy's right, opinion---anybody's---counts for nothing, it has to be backed up.

[xyzt]

Oh brother. I HAVE searched for it. I can't find it. I've been told that it's "hidden". So someone show it to me! But not with nonsense, hand-waving, or even with math if it backs up what I'm claiming.

 

But Iggy's right, opinion---anybody's---counts for nothing, it has to be backed up.

Read posts 73 and 177. They expose where your sleigh of hand is.

Post 2, by Markus exposes a different , more fundamental , flaw in your post 1.

I suggest that you read all these posts.

Edited May 9, 2013 by xyzt

[md65536]

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.

Are you sure about that?

 

The "paradox" of the twin paradox is simply time dilation.

Time dilation occurs between any 2 relatively moving clocks, even ones that are each inertial. It's a reciprocal relationship, so they (or anyone) can't agree that one actually ticks slower than the other unless they have a way to universally establish it (enter the same inertial frame and compare at agreed upon events, or come together at a single event) which the twin paradox experiment accomplishes using a turn-around, with acceleration.

 

Two clocks on 2 different paths between the same 2 events will generally measure a difference in proper time. Both paths can't be inertial if they're different, in flat spacetime. ALL I'M SAYING is that a single clock measures the same proper time along a given path, as would 2 clocks with the path split into 2 parts with each part measured separately by one of the two clocks. This does not disagree with SR, in fact SR assumes that it is true! Any two clocks, side-by-side in an inertial frame, will both tick at one second per second. Contrary to xyzt's belief, no clock jumps in proper time.

 

Your "obvious and well established fact" is neither obvious nor established nor a fact, but if you have a reference to back it up I'll check it out... I could be wrong.

Edited May 9, 2013 by md65536

[xyzt]

Are you sure about that?

 

The "paradox" of the twin paradox is simply time dilation.

 

No, it isn't, this is a common mistake. The twins paradox is about elapsed proper time, not about time dilation.

[xyzt]

Contrary to xyzt's belief, no clock jumps in proper time.

The jumps are explained in post 177, they are due to your sleigh of hand in resttingC to the time of B at the BC encounter. This has been explained multiple times to you, the mathematical explanation is in post 177 (a repeat of post 73).

Edited May 9, 2013 by xyzt

[Iggy]

So you are not surprised that after so many posts full of equations the argument is still not resolved.

 

I actually think it was solved on the first page of posts. If one considers a coordinate transformation to be acceleration then acceleration is key to the twin paradox. If not then not.

 

Subsequent argument (mine included) has mostly missed the mark.

 

You don't need math to solve this argument. Relativity is based on the principle that there is no preferential inertial F.O.R.

When you describe a situation where all observers are inertial ones there should exist no paradox. It is a simple way to CHECK the maths. If you obtain a discrepancy, your maths are wrong.

 

I agree. Well said. There is, indeed, no discrepancy. No paradox. Post 130 shows, the math checks out. Nothing untoward here.

 

Md's argument is that one can construct an experiment where all observers are inertial ones and at the end obtain a discrepancy.

 

It's not a discrepancy. It is an expected difference in proper time. Consistent with the principle of extremal aging.

 

 

He should look for the error.

 

There is no error. I think you may be taking some people in this thread too seriously.

 

It is disappointing that he refuses to search for it.

 

Nothing to search for. All correct results have been rightly expected.

 

Worse, when showed where the "trick" is, he proceeds and invents another one.

 

Nothing invented.

 

He behaves exactly like all those who present a free energy device.

 

No miracle cures here.

 

And you are supporting him.

 

No, I support relativity. I assure you, the second Md says anything inconsistent with relativity, I will pounce on him most harshly. You'll see fangs and blood, I swear it!

[Delta1212]

I actually think it was solved on the first page of posts. If one considers a coordinate transformation to be acceleration then acceleration is key to the twin paradox. If not then not.

 

Subsequent argument (mine included) has mostly missed the mark.

Yes, basically.

[Markus Hanke]

The "paradox" of the twin paradox is simply time dilation.

 

You see, this is where you are being imprecise, and hence run into problems.

The twin paradox is not about time dilation as a result of Lorentz transformations between inertial frames. In fact, it is the exact opposite - it is about disagreements between clocks which cannot be related by Lorentz transformations. In other words, it is about relations between frames which are not symmetric. You are correct that there is time dilation involved, but, due to the equivalence principle, it is gravitational time dilation that we are dealing with, not Lorentz time dilation. This is actually made clear in my own post #2 - inertial frames ( as used in SR ) all have constant metric tensors associated with them; the very constancy of the metric tensor in Minkowski space-time explicitly rules out any disagreements between clock readings. In the twin paradox however, the metric tensor in the frame of the travelling twin is not constant; locally, this is equivalent to the presence of a gravitational field. That is the reason why his clock disagrees with the one of the stay-behind twin. You cannot replicate that result with any arrangement of purely inertial frames that you can possible dream up.

 

Here's a pretty good and intuitive overview without complicated maths - have a read through it :

 

http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm

 

Edited May 10, 2013 by Markus Hanke

[md65536]

Here's a pretty good and intuitive overview without complicated maths - have a read through it :

 

http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm

It seems to be three vs three in this thread. I'll wait before replying to your counter argument, hoping that an expert will weigh in, because I don't think I have any credibility left.

 

What do you make of this line from the conclusion in your link: "It is important to point out, however, that appealing to General Relativity is not necessary to resolve the paradox, as demonstrated above."?

[xyzt]

It seems to be three vs three in this thread. I'll wait before replying to your counter argument, hoping that an expert will weigh in, because I don't think I have any credibility left.

 

What do you make of this line from the conclusion in your link: "It is important to point out, however, that appealing to General Relativity is not necessary to resolve the paradox, as demonstrated above."?

Science is not a democracy, it is not settled by vote. Out of curiosity, who do you think is on your side?

[StringJunky]

Science is not a democracy, it is not settled by vote. Out of curiosity, who do you think is on your side?

 

I don't think MD is saying that, he's just saying there seems to be two sets of polarised views.

[Iggy]

You see, this is where you are being imprecise, and hence run into problems.

The twin paradox is not about time dilation as a result of Lorentz transformations between inertial frames. In fact, it is the exact opposite - it is about disagreements between clocks which cannot be related by Lorentz transformations. In other words, it is about relations between frames which are not symmetric.

 

No one is saying it's symmetric, but it certainly is possible to solve the traditional twin paradox with special relativity. The though experiment Md has given has been solved with the Lorentz transforms in this thread, so I don't quite understand your argument.

 

You are correct that there is time dilation involved, but, due to the equivalence principle, it is gravitational time dilation that we are dealing with, not Lorentz time dilation.

 

The usual equivalence principle way to solve the twin paradox is to consider both time dilation from gravitational potential and velocity. Both contribute. Here, for example, is Einstein:

 

 

It should be kept in mind that in the left and in the right section exactly the same proceedings are described, it is just that the description on the left relates to the coordinate system K, the description on the right relates to the coordinate system K'. According to both descriptions the clock U2 is running a certain amount behind clock U1 at the end of the observed process.

 

When relating to the coordinate system K' the behaviour explains itself as follows: During the partial processes 2 and 4 the clock U1, going at a velocity v, runs indeed at a slower pace than the resting clock U2. However, this is more than compensated by a faster pace of U1 during partial process 3. According to the general theory of relativity, a clock will go faster the higher the gravitational potential of the location where it is located, and during partial process 3 U2 happens to be located at a higher gravitational potential than U1. The calculation shows that this speeding ahead constitutes exactly twice as much as the lagging behind during the partial processes 2 and 4. This consideration completely clears up the paradox that you brought up [that each clock would be running behind the other at the end]...

-Einstein 1918

 

So I couldn't agree that we're only dealing with gravitational and not 'Lorentz'. But, it seems like your point is to say that the traditional twin paradox has to have coordinate acceleration, and I don't think anyone would argue with that.

 

This is actually made clear in my own post #2 - inertial frames ( as used in SR ) all have constant metric tensors associated with them; the very constancy of the metric tensor in Minkowski space-time explicitly rules out any disagreements between clock readings. In the twin paradox however, the metric tensor in the frame of the travelling twin is not constant; locally, this is equivalent to the presence of a gravitational field. That is the reason why his clock disagrees with the one of the stay-behind twin. You cannot replicate that result with any arrangement of purely inertial frames that you can possible dream up.

 

Here's a pretty good and intuitive overview without complicated maths - have a read through it :

 

http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm

 

 

I firstly disagree that observers in flat spacetime cannot have disagreements between clock readings. The time measured between two events depends on the velocity relative to a world line intersecting the events.

 

But, I don't think that is important. I believe you and Md are arguing different things. Md is saying that proper acceleration is not the cause of the results of the paradox. I think you could agree that it is the non-linear nature of the metric that causes the time dilation to be non-symmetric, and not any inertial force felt by the clock.

 

For example, two clocks in orbit at different altitudes are both inertial but have non-symmetric time dilation. You could also do a version of the twin paradox where the traveling twin feels no acceleration at the turnaround, for example, the way the Apollo 13 astronauts used a free return trajectory around the moon. The clocks on board felt no proper acceleration as the ship turned around, but the results are consistent.

 

I think it is most correct to say that it is the metric that causes both inertial forces and time dilation. Not that acceleration causes non-symmetric time dilation. I would hope you and Md could agree on that.

[Markus Hanke]

For example, two clocks in orbit at different altitudes are both inertial

 

You mean approximately inertial locally. Globally the entire setup is not inertial, because the clocks are at points of different gravitational potentials.

 

I think it is most correct to say that it is the metric that causes both inertial forces and time dilation.

 

Yes, I can agree to this statement, even though I am not certain what you mean by "inertial forces". That is also the reason why I kept mentioning the metric tensor in my explanations.

 

What do you make of this line from the conclusion in your link: "It is important to point out, however, that appealing to General Relativity is not necessary to resolve the paradox, as demonstrated above."?

 

What they mean is that the "paradox" arises only from the erroneous assumption that both frames are inertial; if that was the case then there shouldn't be any differences in clock readings. Recognizing that one of the frames is not in fact inertial immediately shows that there really isn't a paradox in the first place, because the disagreement in readings is a physical necessity due to the fact that the non-intertial frame "carries" a non-constant metric tensor.

 

This is pretty much what the last two sentences in the link say.

[Iggy]

You mean approximately inertial locally. Globally the entire setup is not inertial, because the clocks are at points of different gravitational potentials.

 

 

Yes, I can agree to this statement, even though I am not certain what you mean by "inertial forces". That is also the reason why I kept mentioning the metric tensor in my explanations.

 

I think we're in good agreement. I mean pseudo forces, or what wikipedia annoyingly calls fictitious forces

 

We can have non-symmetrical time dilation when the clocks don't feel proper acceleration is all I mean.

 

edit: let me have another run at that. I'm not the best at articulating...

 

I think it is most correct to say that it is the metric that causes both inertial forces and time dilation.

 

So, for example, if you're standing on a planet it is the non-linear nature of spacetime that causes both the proper acceleration that you feel as you stand there, and the time dilation between your feet and your head. It is more accurate to say that curved spacetime is the cause of both than to say acceleration is the cause of time dilation.

 

ok

Edited May 11, 2013 by Iggy

[Markus Hanke]

So, for example, if you're standing on a planet it is the non-linear nature of spacetime that causes both the proper acceleration that you feel as you stand there, and the time dilation between your feet and your head. It is more accurate to say that curved spacetime is the cause of both than to say acceleration is the cause of time dilation.

 

ok

 

Yeah, I am happy with that

Given that, I think it is quite clear that in the scenario of three clocks, all of which are in uniform relative motion with respect to one another, you will never find a "twin paradox". That is really all I was trying to point out. For me, it is most clearly obvious in the proper time integral for each observer.

[md65536]

Now that everyone seems to be in agreement, perhaps someone can help me figure out the error, or "hidden trick" in my original premise. Please tell me the first statement here that you disagree with:

 

1) In a twin paradox using constant outbound v = sqrt(3)/2 ~= 0.866, "instant" turnaround, and inbound v = -v, where the traveling twin B ages 1 year proper time on each of the two legs of the trip, inertial twin A will age 4 years between B's departure and B's return (gamma = 2).

 

2) This works regardless of the method of travel of B. If B is on an outbound train, and then "jumps" onto a different inbound train, instantly accelerating and experiencing infinite proper acceleration during its frame switch, this properly implements the twin paradox experiment described in (1).

 

3) A clock on the train will tick at the same rate as B's clock while B is at rest on that train. So, a clock on the outbound train will age one year while B is on it, and a clock on the inbound train will age one year while B is on it.

 

4) Before and after B jumps trains, it will have a different relative simultaneity relative to A. Before, B calculates that A has aged 0.5 years since leaving, and after, B calculates that A has aged 3.5 years since leaving. In negligible proper time of B, B has calculated a change of 3 years of A's proper time.

 

5) The relative time depends on distance and velocity. So while B is at rest on the first train, B and the train agree on the relative simultaneity with respect to A, ie. just before the trains pass, both agree that A has aged 0.5 years. Just after the trains pass, and B is at rest on the inbound train, B and the inbound train agree that A has aged 3.5 years since B's leaving. To see this is true, according to (3) both B and train will age 1 year until arriving at A, both will calculate that A ages 0.5 years during that time, and both will agree on A's age (4 years) when they arrive.

 

6) The relative simultaneity of events at A as measured by the trains is the same whether or not B is on the train. If B misses the jump, the outbound train still ages 1 year before passing the inbound train, and the inbound train still ages 1 year before arriving at A, and A has still aged 4 years between trains, even though B may be stuck on the outbound train or splatted somewhere in space.

 

7) If B stays at rest on the outbound train, no object accelerates at the passing of the outbound and inbound train.

 

8) Any observer can calculate the relative simultaneities of the various objects, and all observers agree on the proper time or aging of the trains and of A and B between any of the events they pass.

 

9) Therefore the effects of the twin paradox are evident in objects that follow a part of the same path as the traveling twin, and the full effect is evident in a set of objects that cumulatively follow the full path of the traveling twin, even if no single object undergoes acceleration.

 

10) Acceleration is still important to implement the twin paradox experiment in flat spacetime using only one clock per twin.

Edited May 11, 2013 by md65536

[xyzt]

Now that everyone seems to be in agreement, perhaps someone can help me figure out the error, or "hidden trick" in my original premise.

post 177 explains your sleigh of hand quite clearly.

Edited May 11, 2013 by xyzt

[Delta1212]

 

Yeah, I am happy with that

Given that, I think it is quite clear that in the scenario of three clocks, all of which are in uniform relative motion with respect to one another, you will never find a "twin paradox". That is really all I was trying to point out. For me, it is most clearly obvious in the proper time integral for each observer.

You would not find a twin paradox in the sense that no clocks will measure a different elapsed time between two events at which both clocks were present (since, seeing as they're all in uniform motion, no two clocks will ever be present at the same two events).

 

Would you agree that you could set up the clocks to measure the elapsed time of the two paths taken in the twin paradox, one inertial and one non-inertial, without accelerating any of the clocks? The final elapsed time for the non-inertial path made up of the paths of two of the clocks would represent the elapsed time of a hypothetical accelerated twin, but would not represent the elapsed time measured by any one of the clocks in the experiment.

[Phi for All]

post 177 explains your sleigh of hand quite clearly.

 

!

Moderator Note

It doesn't to the people you're discussing this with, and they've told you so. Please, think of another way to explain this, and break up the communication log jam.

 

And to all involved, thanks for the increased attention to civility.

[Iggy]

Yeah, I am happy with that

 

Sweet!

 

Given that, I think it is quite clear that in the scenario of three clocks, all of which are in uniform relative motion with respect to one another, you will never find a "twin paradox". That is really all I was trying to point out. For me, it is most clearly obvious in the proper time integral for each observer.

 

I'm positive that doesn't follow. It follows with two observers, but not in the confines of the thought experiment Md has given with three clocks.

 

I have to think you're making the same point you made in your second post in the thread:

 

If there is only uniform relative motion between the three reference frames, then the frames are symmetric, and the clocks can't disagree, because you can exchange any two of them via Lorentz transformations without effecting the physical outcome.

 

Clocks in uniform relative motion do disagree, so let's try this with a different approach to see if we can clear it up.

 

If an alien ship were passing though our solar system inertially at close to c relative to earth would you agree that it could cover the distance between the earth and the sun in one second (according to the clocks on the ship)?

Edited May 11, 2013 by Iggy

[xyzt]

!

Moderator Note

It doesn't to the people you're discussing this with, and they've told you so. Please, think of another way to explain this, and break up the communication log jam.

 

And to all involved, thanks for the increased attention to civility.

It is very simple , really. If you reset the C clock to the value of the B clock, there is a jump in elapsed time because at the event BC , when the clocks met, they have accumulated different amounts of proper time. This is what post 177 demonstrates mathematically. Post 73 showed the same calculations (not in Latex).

Edited May 11, 2013 by xyzt

[DimaMazin]

It is very somple , really. If you reset the C clock to the value of the B clock, there is a jump in elapsed time because at the event BC , when the clocks met, they has accumulated different amounts of proper time. This is what post 177 demonstrates mathematically. Post 73 showed the same calculations (not in Latex).

Calculations can't be scientifically proved without experimental proof. Indications of clocks don't jump. Just speed of moving of arrows can jump.And arrows don't came back.