Pure twin paradox...

[AntigenX]

We know what is twin paradox is, and also why it is not a paradox, but my question is, "if we can not define absolute motion, how can we decide that the twin which went to space will age less, and not the home one?"

That is to say that, when peter at home sees paul's clock moving slow (in space ship), similarly, paul should see peter's clock to be slowed down. So for paul, peter should be the younger! How is it that when paul comes, he is the one who is younger?

for example, let's say (arbitrarily) that, the absolute speed of earth (which is of course not measurable) is 70,000 km/s. the speed of paul's spaceship is 30,000 km/s. and they are moving in opposite direction. So, if we talk about the absolute speed, peter should be younger, and if we talk about relative speed, both should be younger for each other. What is it that lets us decide, that each time, paul will be the one who aged less?

[Doc Al]

We know what is twin paradox is, and also why it is not a paradox, but my question is, "if we can not define absolute motion, how can we decide that the twin which went to space will age less, and not the home one?"
One twin accelerates, the other doesn't; that breaks the symmetry. The accelerating twin experiences less proper time.

[AntigenX]

One twin accelerates, the other doesn't; that breaks the symmetry. The accelerating twin experiences less proper time.
Oh yes, but if the accelerating twin looks back, he will think, the earth is accelerating with respect to him. Or acceleration is not relative, while velocity is?

[Doc Al]

Acceleration is not relative.

[AntigenX]

How come acceleration is not relative?
Does that mean that when paul will look back, he will find that earth is in uniform relative motion with some velocity v with respect to him (and not accelerated with respect to him)? This is highly improbable.

[JinChang]

The twin on the spaceship is the one who is accelerating away from the inertial frame of reference (Earth) and will undergo such effects associated with the act of acceleration. Barring the effects of gravity and the motion of Earth around the Sun, the twin on Earth maintains the original inertial frame of reference. That's how I understood it.

[AntigenX]

The twin on the spaceship is the one who is accelerating away from the inertial frame of reference (Earth) and will undergo such effects associated with the act of acceleration. Barring the effects of gravity and the motion of Earth around the Sun, the twin on Earth maintains the original inertial frame of reference. That's how I understood it.

I understood more or less the same way.
But my question is, removing everything else, other than the earth and paul's spaceship (i.e. keeping only two frames of reference, earth and the spaceship), If paul is accelerated, will not he believe that the earth is accelerated as well with respect to him. accepted that he can measure his own acceleration without the reference to the earth (with an accelerometer), yet when he would look-back, he would notice the earth to be accelerating at the same rate, and hence, the time dilation for paul due to his own acceleration as observed by peter, will be applicable to peter as well when paul observes him.

[russ_watters]

If paul is accelerated, will not he believe that the earth is accelerated as well with respect to him.
Since he knows he is accelerating (after all, he's firing his engines), he will consider the earth stationary wrt him.

[jtbell]

And the twin on the earth doesn't "feel" anything happening to himself when his twin fires the spaceship's engines.

[AntigenX]

Since he knows he is accelerating (after all, he's firing his engines), he will consider the earth stationary wrt him.

"He's firing his engines" is not an answer! In such a case, If two space ships are going in the same direction (no relative velocity), and if they don't consider any reference to the other spaceship, or the earth, because they know that their engines are working, they will believe their clocks are slower? But I think that was just a friendly comment and not explanation.

And the twin on the earth doesn't "feel" anything happening to himself when his twin fires the spaceship's engines.
In relativity, we are concerned about what one observer feels (or observes) about other, and not what one feels about himself! The time for paul in his frame is going at a uniform speed for him. It's what peter's observation that paul's clock is slow. And thus, if it is the acceleration of paul, which let's him slow down his time, absolutely irrespective of peter, then that doesn't seem relativity to me.

Consider paul is in the ship, but he doesn't drive the ship, and just observes the relative motion, not visually, but by some radio signal strength or sort of thing. Now, before the engine is fired, some heavenly body strikes earth and accelerates to same amount, as would have been case if the engine would have been fired. Will he not feel that the engine has been fired? Now in such a situation, whose time will be slow? Paul's or Peter's?

[HallsofIvy]

Force equals mass times acceleration. The twin who is accelerating will fill a force during acceleration. If it were the earth that were accelerating, the twin on earth would feel an additional force.

This is all in "special relativity". In "general relativity", an "external" force can be interpreted as additonal gravity rather than acceleration. However, since a higher gravitational force results in slowing clocks, the result is the same.

[robphy]

Consider paul is in the ship, but he doesn't drive the ship, and just observes the relative motion, not visually, but by some radio signal strength or sort of thing.

(This comment accentuates those of russ_watters and jtbell.)

Consider a block (an ice cube) on a frictionless (icy) surface.
This setup is in the ship of each observer.

For the inertial observer, that block remains in the same position on the table throughout the experiment.

For the observer that accelerates (or is accelerated by the "driver"), that block is not in the same position.

[MeJennifer]

" 'He's firing his engines' is not an answer!"

"In relativity, we are concerned about what one observer feels (or observes) about other, and not what one feels about himself! "

"that doesn't seem relativity to me."

You are not listening AntigenX, people here are trying to help you with your problems in understanding relativity, but with comments like the ones above I fear not for long.

[AntigenX]

You are not listening AntigenX, people here are trying to help you with your problems in understanding relativity, but with comments like the ones above I fear not for long.
You are right, pardon me for being so arrogant and what not...

I have read comments of everybody, but I think we are missing the point here.

Simply put, If acceleration, which is absolute, and doesn't require any external frame of reference for its measurement (i.e. is self evident in the observer's own frame), then how come paul's time slow for peter, when it isn't relative to peter at all? I consider the gravity point of view, but how would we give explanation of this so called paradox in SR?

[Mentz114]

AntigenX:

you appear to mixing two different things here. The twin case applies to a situation where the the twins are together in one place, then one goes off on a round trip, and they compare clocks when the traveller returns. In this case the twin who did the most accelerating will have fewer ticks on his clock.

The other situation is what they will get if they measure each others clock rates while the travelling twin is in progress. In this case, each one will measure the others clock as being slower.

This contrasts with what they will see if they observe one another with telescopes, which is another case.

[AntigenX]

you appear to mixing two different things here. The twin case applies to a situation where the the twins are together in one place, then one goes off on a round trip, and they compare clocks when the traveller returns. In this case the twin who did the most accelerating will have fewer ticks on his clock.

The other situation is what they will get if they measure each others clock rates while the travelling twin is in progress. In this case, each one will measure the others clock as being slower.

This contrasts with what they will see if they observe one another with telescopes, which is another case.

May be I'm not putting the situation correctly. I am not mixing two situations though. As referred in my last post, If acceleration, which is absolute (and not relative) dilates paul's time, without any reference to peter, how should it matter to peter at all. As is our argument formulated, during the whole journey, paul is bound to accelerate at least 4 times, during which, his clock runs slower. My point is, slower with respect to which clock? Because, as noted above, acceleration is self evident in paul's frame for paul, and is not relative to peter.
If we say, with respect to peter's clock, then the time dilation in a situation of paul's acceleration is relative to peter (i.e. is not absolute), and hence, the same time dilation should be evident for peter as well, as observed from paul's point of view.

[DaleSpam]

My recommendation is to forget the acceleration. It only serves to break the symmetry and does not cause time dilation by itself. Instead, use the spacetime interval which allows you to extend the analysis to arbitrarily accelerating twins.

for example, let's say (arbitrarily) that, the absolute speed of earth (which is of course not measurable) is 70,000 km/s. the speed of paul's spaceship is 30,000 km/s. and they are moving in opposite direction. So, if we talk about the absolute speed, peter should be younger, and if we talk about relative speed, both should be younger for each other. What is it that lets us decide, that each time, paul will be the one who aged less?Your scenario here is fine, but you didn't actually do the math, so your conclusion is wrong.

In frame A (for "absolute") ve is .00023 c and vp is initially -.00010 c. The two start together at spacetime coordinates (0,0). If the outbound portion of the trip takes 1 year in frame A then at the end of the year the earth will be at spacetime coordinates (1,.00023) and paul will be at spacetime coordinates (1,-.00010) where both are measured in lightyears. At this point paul turns around and heads back to the earth where they re-unite at spacetime coordinates (2,.00046).

The spacetime interval for the earth's clock is
|(1,.00023)-(0,0)| + |(2,.00046)-(1,.00023)| = 1.9999999471 years.

The spacetime interval for paul's clock is
|(1,-.0001)-(0,0)| + |(2,.00046)-(1,-.0001)| = 1.9999998382 years.

So paul's clock reads less than the earth's clock by 3.4 seconds.

The point is that you can pick any arbitrary frame and reach the same conclusion that paul aged less. We don't have to "decide", we just have to follow the same rules in any inertial frame.

[Janus]

May be I'm not putting the situation correctly. I am not mixing two situations though. As referred in my last post, If acceleration, which is absolute (and not relative) dilates paul's time, without any reference to peter, how should it matter to peter at all. As is our argument formulated, during the whole journey, paul is bound to accelerate at least 4 times, during which, his clock runs slower. My point is, slower with respect to which clock? Because, as noted above, acceleration is self evident in paul's frame for paul, and is not relative to peter.
If we say, with respect to peter's clock, then the time dilation in a situation of paul's acceleration is relative to peter (i.e. is not absolute), and hence, the same time dilation should be evident for peter as well, as observed from paul's point of view.

Paul's acceleration does not cause his time dilation, rather it effects what he measures is happening to Peter's clock. You have to use different rules when dealing with the measurements of an accelerating observer. What Paul determines is happening to Peter's while he (Paul) is accelerating is determined by three things: The magnitude of the acceleration, The direction of the acceleration, and the distance between Peter and Paul as measured by Paul along a line parallel to the direction of the acceleration.

When Paul is accelerating away from Peter he measures Peter's clock as running slow, The greater the distance between the two, the more pronounced the slowing of Peter's clock. This is in addition to any slowing due to the relative velocity between them.

When Paul is accelerating towards Peter, he measures Peter's clock as running fast, minus any slowing due to relative velocity.Again, this becomes more pronounced as the distance between them is increased. Note that accelerating "towards" doesn't always means "moving towards". Braking to a stop while moving away counts as acceleration towards.

Thus Paul measures the following as happening to Peter and his clock.

As Paul accelerates away from Peter he sees Peter's clock run slow both due to the increasing velocity difference and his acceleration away. Since at this time the distance between the two is not large, the acceleration does not add much.

Once Paul reaches the desired cruising speed, he quits accelerating and coasts. During this time Peter's clock will run slow at a fixed rate.

Paul reaches the turn around point and begins to brake and then accelerate back towards Peter. During this period, he determines that Peter's clock speeds up due to his acceleration towards Peter. Since at this point, the distance between the two is its greatest, this speeding up of Peter's clock is enough to not only overcome any slowing due to relative velocity but actually causes Peter's clock to jump ahead of Paul's clock.

Paul begins the coasting leg of the return trip. He again sees Peter's clock as running slow. However Peter's clock starts out as ahead.

Peter and Paul reunite. The slowing of Peter's clock during the return trip is not enough to completely offset the time it gained during the turn around phase, and Paul finds that he and his clock has accumulated less time than Peter's.

[DocZaius]

For anyone having trouble understanding why the acceleration is crucial, take a look at this link. If you don't understand it (I didn't the first time through), then slow down and read it very carefully. I've found skimming through these sorts of things is just a waste of time.

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

[Mentz114]

Drawing a space-time diagram makes it very clear. Curvy world-lines represent accelerated frames, and they are longer than straight lines ( inertial frames ) joining any two points. The relative clock speeds are related to the lengths of the worldlines between the points of coincidence. It is immediately obvious why (how?) accelerations slow clocks.

[MeJennifer]

Drawing a space-time diagram makes it very clear. Curvy world-lines represent accelerated frames, and they are longer than straight lines ( inertial frames ) joining any two points. The relative clock speeds are related to the lengths of the worldlines between the points of coincidence. It is immediately obvious why (how?) accelerations slow clocks.
In spacetime the curved lines are shorter not longer than straight lines between two events. The length of the lines represent the total elapsed time. It is not correct to say that the clock on the curved line went slower, it went with the same rate as the clock on the straight line, it is just that the path was shorter thus less time was elapsed.

[MeJennifer]

Simply put, If acceleration, which is absolute, and doesn't require any external frame of reference for its measurement (i.e. is self evident in the observer's own frame), then how come paul's time slow for peter, when it isn't relative to peter at all? I consider the gravity point of view, but how would we give explanation of this so called paradox in SR?
Note that while velocity is relative in special relativity acceleration is relative under general relativity.

[neopolitan]

I have said it before, but it may not hurt to say it again.

There is really no twin paradox. It is a false paradox due to an imperfect framing of the scenario.

Acceleration is not really the answer, although the acceleration does give you an indication of symmetry breaking.

Think of it this way: the two twins are knocked out, and then they are revived just as they pass each other such that neither knows which of them could be considered to be stationary (it is actually neither of them in absolute terms). They both have identical clocks which are set in motion as they pass each other. When they reach a certain separation, they are both knocked out again and their clocks are stopped. They are allowed to keep moving relative to each other, then they undergo relative deceleration until their relative velocity is negative - so that they now approach each other at the same speed as they separated before.

When they reach the same separation as they had when they were knocked out, they are revived again and their clocks restarted. They are knocked out for a third time when they reach each other and their clocks are stopped for a second time. The experiment stops when they are collocated and stationary with respect to each other. They can now compare clocks, which only functioned during inertial portions of their voyages.

Which clock will read less? In the scenario as stated, both clocks should read the same. There is nothing to distinguish one twin from the other and, most importantly, the relative distance travelled is expressed as a separation from each other.

This is where there is a difference - in the standard expression of the "twin paradox", one twin travels to a location which is stationary with respect to the other twin then turns around and comes home again.

Say that you modify my scenario slightly:

When they reach a certain separation, they are both knocked out again and their clocks are stopped.
becomes
When one twin reaches a distant point which is at rest relative to the other twin (say at a distance of L), they are both knocked out again and their clocks are stopped.

Now you can introduce Lorentz transformations to see that if the relative distance travelled according to the twin "at rest" is 2L then, according to the other twin the relative distance travelled is:

2L' = ((L-vt)+(L+vt)) / sqrt(1-v^2/c^2) = 2L / sqrt(1-v^2/c^2)

This requires some interpretation, since the twin "at rest" doesn't go anywhere relative to the turning point. However, according to the other twin, the twin "at rest" does move, along with the turning point. 2L' is this perceived travelling distance, a greater distance than "actually" travelled. The time needed to travel this greater distance is greater given that the relative velocity is common to both, which is why the twin "at rest" will have a clock reading which is greater than the "travelling" twin's.

The acceleration itself has absolutely no effect. What has an effect is choosing a turning point which is at rest relative to one of the twins and it this which causes the symmetry break - not the acceleration.

cheers,

neopolitan

[neopolitan]

For anyone having trouble understanding why the acceleration is crucial, take a look at this link. If you don't understand it (I didn't the first time through), then slow down and read it very carefully. I've found skimming through these sorts of things is just a waste of time.

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

There is a problem here DocZaius.

Consider this, Joe and Jane do what I discussed in the previous post with a slight twist.

Jane travels to a spot which is a distance L from Joe. She travels inertially from Joe to this spot, starting her clock as she passes Joe (telling him to start his clock) and stopping it as she passes the reference spot. At that time she sends a message to Joe to stop his clock.

Then she turns around, travels back again, past the reference spot again, past Joe again and turns around a second time, passes Joe a third time, restarts her clock and tells Joe to restart his clock. She stops her clock again when she passes the reference spot for the third time and sends a message back to Joe to stop his clock.

When they join up again, they compare clocks.

Which will read the most?

We can work it out using relative distance travelled. According to Joe, Jane travelled from Joe to distant spot twice, a distance of 2L. According to Jane, the whole length Joe-distant spot moved past her twice. Because that length was in relative motion, it was contracted - so 2L'=2L . sqrt (1 - v^2/c^2).

According to Joe, the time taken for Jane to travel Joe -> distant spot was 2L/v.

According to Jane, the time take for the length Joe-distant spot to move past here was
2L'/v.

Therefore Joe's clock will read more.

No need for deceleration or acceleration. No need for comparison of world lines.

Again, the deceleration/acceleration is a useful indicator and the world lines are interesting for explaining what goes on ... but these are symptoms, not causes.

Here's a little mind bender:

Let's say Jane travels until Joe sends a message to stop and turn around.

Jane will take longer to get back to Joe because the message takes time to catch up to her. So if Joe sends the message at time t, the message has to travel a distance of vt plus whatever distance Jane travels in that time (and so on).

Then instead say that Jane travels till she gets sick of it (or lets the universe pass her by, if you prefer), then sends a message to Joe that she is turning around (or telling the universe to change relative direction).

Jane will get back to Joe quicker than expected, because the message takes time to get to him, time during which Jane is already on her way home.

But what will their clocks read, relative to each other? Will Jane be younger in both instances?

(To make it easier, try thinking of it from another perspective, what if Joe was in charge? If he decides that he wants Jane back, he can change the direction that the universe is moving relative to Jane (including himself) immediately. If Jane wants to return to Joe, she has to send him a message asking him to be so kind as to change the direction the universe is moving.)

The answer highlights that acceleration is not the culprit here.

cheers,

neopolitan

[Doc Al]

There is a problem here DocZaius.

Consider this, Joe and Jane do what I discussed in the previous post with a slight twist.

Jane travels to a spot which is a distance L from Joe. She travels inertially from Joe to this spot, starting her clock as she passes Joe (telling him to start his clock) and stopping it as she passes the reference spot. At that time she sends a message to Joe to stop his clock.

Then she turns around, travels back again, past the reference spot again, past Joe again and turns around a second time, passes Joe a third time, restarts her clock and tells Joe to restart his clock. She stops her clock again when she passes the reference spot for the third time and sends a message back to Joe to stop his clock.

When they join up again, they compare clocks.

Which will read the most?

We can work it out using relative distance travelled. According to Joe, Jane travelled from Joe to distant spot twice, a distance of 2L. According to Jane, the whole length Joe-distant spot moved past her twice. Because that length was in relative motion, it was contracted - so 2L'=2L . sqrt (1 - v^2/c^2).

According to Joe, the time taken for Jane to travel Joe -> distant spot was 2L/v.

According to Jane, the time take for the length Joe-distant spot to move past here was
2L'/v.

Therefore Joe's clock will read more.

No need for deceleration or acceleration. No need for comparison of world lines.

Again, the deceleration/acceleration is a useful indicator and the world lines are interesting for explaining what goes on ... but these are symptoms, not causes.
I fail to see how this example has much to do with the standard "twin paradox" scenario.

Also, don't forget to include the time it takes for Jane's signal to Joe to reach Joe.

[Mentz114]

In my earlier post I should have said 'proper length' instead of 'length'. This essay has a straightforward treatment of the the twin effect in terms of proper lengths.

http://astro.physics.sc.edu/selfpacedunits/Unit56.html

[Fimbulfamb]

Mentz114, I'm relatively (no pun intended) new to the theory of relativity, how come acceleration is not relative? If the traveling twin is the frame of reference, Earth is accelerating away from the rocket. Time, as well is space, is relative, therefor ms^{-2} should be relative. None of the above posts have cleared this up for me.

[paw]

...how come acceleration is not relative? If the traveling twin is the frame of reference, Earth is accelerating away from the rocket.

For two observers in relative uniform, that is inertial, motion neither one can claim to be the one moving. There is no way to determine if the relative motion can be attributed to one or the other.

The situation is different for acceleration. A simple accelerometer is all that's needed to determine which observer accelerated. So even though the travelling twin might try to claim it was the stationary twin (Earth) that accelerated the accelerometer can prove this to be false.

[Hurkyl]

Which clock will read less? In the scenario as stated, both clocks should read the same. There is nothing to distinguish one twin from the other
Incorrect -- you have not provided enough information to make any conclusion about the relative time on the clocks.

Furthermore, the twins (and their trajectories) are clearly distinguishable; it's just that you have not given me the necessary information.

[neopolitan]

Incorrect -- you have not provided enough information to make any conclusion about the relative time on the clocks.

Furthermore, the twins (and their trajectories) are clearly distinguishable; it's just that you have not given me the necessary information.

Sort of. I wondered if anyone would pick up on this or whether I would have to explain it.

I implied something that gives you the information to make the conclusion. You probably overlooked it.

When they reach a certain separation, they are both knocked out again and their clocks are stopped.

This implies that they reach a certain separation they are knocked out, and their clocks are stopped - simultaneously. This places a huge restriction on the scenario, since the twins could really only be knocked out simultaneously relative to a third observer. Once you apply this restriction and set your scenario up so that they twins are indeed knocked out simultaneously relative to the third observer, then you will find that the clocks read the same.

The twins will not be knocked our simultaneously relative to each other and their clocks will not be stopped simultaneously relative to each other.

This unstated assumption of a third observer determining what is simultaneous is equivalent to the unstated assumption in the twins' paradox, namely that one twin travels between the a "stationary" twin and a distant point which is at rest relative to the stationary twin.

My point is that acceleration is not the answer. It is merely an indicator.

cheers,

neopolitan

[neopolitan]

I fail to see how this example has much to do with the standard "twin paradox" scenario.

Also, don't forget to include the time it takes for Jane's signal to Joe to reach Joe.

The standard "twin paradox" is poorly stated, and that is why it appears to be a paradox (note Hurkyl's complaint in an earlier post). What I am showing here is that the same result can be achieved with no acceleration, which is often what is brought in to explain the broken symmetry.

And I didn't forget the time it takes for Jane's signal to reach Joe.

Firstly, I specifically mentioned a spot which was anchored to Joe, it is a distance of L from Joe. That means that the intellectual effort required to subtract the signal's travel time is trivial. You can do it, I can do it and I assumed that Joe could do it. This is another one of these unstated assumptions.

Secondly, Joe doesn't even need a clock. We tend to always talk in terms of two clocks, because it may seem simpler, but in my scenario Joe's clock was entirely redundant. Jane moves between Joe and the spot which is at a distance of L at a velocity of v. Again, we are talking about a trivial intellectual effort. According to Joe, it should take Jane a period of L/v to travel from where he is to the spot at a distance of L. He doesn't need a clock to work that out. Once the signal's travel time is factored in, his totally redundant clock will confirm that this calculation is correct.

cheers,

neopolitan

[Doc Al]

The standard "twin paradox" is poorly stated, and that is why it appears to be a paradox (note Hurkyl's complaint in an earlier post).
The standard twin paradox seems clear and unambiguously stated to me. I trust you realize that "paradox" is used in the sense of "A seemingly contradictory statement that may nonetheless be true". No one really thinks that the example of the twins represents a true contradiction within relativity.
What I am showing here is that the same result can be achieved with no acceleration, which is often what is brought in to explain the broken symmetry.
I fail to see why you think that your example, with its arbitrary stopping and starting of clocks, has anything to do with the "twin paradox" or is of any special interest on its own.

And I didn't forget the time it takes for Jane's signal to reach Joe.

Firstly, I specifically mentioned a spot which was anchored to Joe, it is a distance of L from Joe. That means that the intellectual effort required to subtract the signal's travel time is trivial. You can do it, I can do it and I assumed that Joe could do it. This is another one of these unstated assumptions.
Looks like you forgot it to me! :wink: (All of these calculations are trivial.)

Secondly, Joe doesn't even need a clock. We tend to always talk in terms of two clocks, because it may seem simpler, but in my scenario Joe's clock was entirely redundant. Jane moves between Joe and the spot which is at a distance of L at a velocity of v. Again, we are talking about a trivial intellectual effort. According to Joe, it should take Jane a period of L/v to travel from where he is to the spot at a distance of L. He doesn't need a clock to work that out. Once the signal's travel time is factored in, his totally redundant clock will confirm that this calculation is correct.
So? (I agree that you are talking about a trivial intellectual effort.)

[Mentz114]

neopolitan;
This unstated assumption of a third observer determining what is simultaneous is equivalent to the unstated assumption in the twins' paradox, namely that one twin travels between the "stationary" twin and a distant point which is at rest relative to the stationary twin.
(my emphasis).
The point where the twins meet up can be just some coordinates. It cannot be said to be at rest or moving relative to either twin.

My point is that acceleration is not the answer. It is merely an indicator.
Relative velocities arise through different acceleration histories. So in that sense acceleration is a cause.

But proper lengths and the invariance of the proper length is the best way to understand it. I urge you to read the link in my earlier post.

[neopolitan]

So? (I agree that you are talking about a trivial intellectual effort.)

Context, Doc Al.

I was responding to DocZaius' claim that acceleration is crucial. It isn't.

What is crucial is that in the twins' paradox it is tacitly assumed that the travelling twin travels to a location which has a fixed distance from the stay-at-home twin (for example one twin travels to Alpha Centauri - inaccurately stated, Alpha Centauri is a fixed distance from us here on Earth).

It doesn't actually have to be any special location, as Mentz rightly points out. While he may also be right in that acceleration leads to relative velocities and hence in that fashion acceleration leads to time dilation, where he is wrong is that acceleration doesn't lead to the symmetry break per se.

I explained in another thread already - here (http://www.physicsforums.com/showpost.php?p=1728168&postcount=14). Even if the "travelling" twin described in the linked post doesn't undergo any acceleration, it will still be that "travelling" twin who will experience less time elapsed. (And yes I am aware of the real world difficulties associated with decelerating and accelerating an extremely long rod, that's why we make use of mind experiments.)

cheers

neopolitan

[Mentz114]

neopolitan;
What is crucial is that in the twins' paradox it is tacitly assumed that the travelling twin travels to a location which has a fixed distance from the stay-at-home twin
This statement does not make sense to me. The distance you mention does not exist until the second coincidence. So how can it be in any way be fixed or not-fixed ?

I bow out of this thread with a final injunction to read a good essay on SR ( and forget about simultaneity !).

[Doc Al]

Context, Doc Al.

I was responding to DocZaius' claim that acceleration is crucial. It isn't.

What is crucial is that in the twins' paradox it is tacitly assumed that the travelling twin travels to a location which has a fixed distance from the stay-at-home twin (for example one twin travels to Alpha Centauri - inaccurately stated, Alpha Centauri is a fixed distance from us here on Earth).
No, that's not relevant.

It doesn't actually have to be any special location, as Mentz rightly points out. While he may also be right in that acceleration leads to relative velocities and hence in that fashion acceleration leads to time dilation, where he is wrong is that acceleration doesn't lead to the symmetry break per se.
Wrong again.

I explained in another thread already - here (http://www.physicsforums.com/showpost.php?p=1728168&postcount=14). Even if the "travelling" twin described in the linked post doesn't undergo any acceleration, it will still be that "travelling" twin who will experience less time elapsed. (And yes I am aware of the real world difficulties associated with decelerating and accelerating an extremely long rod, that's why we make use of mind experiments.)
Your "explanation" in the other thread is incorrect. For the twins to reunite, one must have accelerated. And it's the one who accelerates, regardless of whether he drags a rod along or not, that experiences less proper time.

[Hurkyl]

Sort of. I wondered if anyone would pick up on this or whether I would have to explain it.

I implied something that gives you the information to make the conclusion. You probably overlooked it.
I'm pretty sure that, if you choose any two positive numbers for the readings on the clocks, I can devise a scenario where Alice's clock and Bob's clock have those readings (in that order), and is consistent with the provided information.

For example, I think one main conceptual omission you've having is that Alice and Bob need not have the same velocity relative to your third observer.

(I will confess to having misread the scenario the first time -- otherwise my initial objection would simply have been that your scenario is ill-defined, because their 'separation' is not an absolute notion)

[Mettra]

Mentz114, I'm relatively (no pun intended) new to the theory of relativity, how come acceleration is not relative? If the traveling twin is the frame of reference, Earth is accelerating away from the rocket. Time, as well is space, is relative, therefor ms^{-2} should be relative. None of the above posts have cleared this up for me.

Instead of thinking of it like that, think of it for the moment from this angle:

When astronauts go up into space using a rocket, they feel a very large acceleration during the trip. Do we on earth feel that acceleration (forget Newton's third law forces for now, some people on the ground near the takeoff site will feel some affects of the rocket - but that's not what I'm talking about)? See, in this case, we have considered the earth to be the reference frame of choice (so we consider it at rest).

Now let's take that example again, taking the earth to be our at rest reference frame (except this time, forget the rotational motion). The astronauts are in space, but are still going, still accelerating until they get to Mars. Do we on the earth feel the acceleration that these astronauts feel? The answer is: no we don't. Acceleration is special because you can feel it. Where there is acceleration, there is a force.

When you ride in a car along the highway at constant velocity (if there is not much air resistance) you can travel at a constant velocity. Here you feel no forces, since velocity is 'free' (Newton's first law). If the driver suddenly nailed the gas, you would be pushed by a very real force into your seat. If the driver suddenly nailed the breaks, you would pushed again by a very real force into the windshield. The rest of the world does not feel the acceleration that you feel.

So in summary, if acceleration were relative, then different observers might disagree on the results of an experiment (they can disagree on measurements, not results). You could juggle forces around by picking different frames of reference.

There are deeper, more mathematical explanations of this, but an intuitive grasp is probably more useful. Anything beyond either of these is getting to be a philosophical question.

[MeJennifer]

So in summary, if acceleration were relative, then different observers might disagree on the results of an experiment (they can disagree on measurements, not results). You could juggle forces around by picking different frames of reference.

There are deeper, more mathematical explanations of this, but an intuitive grasp is probably more useful. Anything beyond either of these is getting to be a philosophical question.
As I wrote before acceleration is relative under general relativity but not under special relativity.

[DaleSpam]

As I wrote before acceleration is relative under general relativity but not under special relativity.Although coordinate acceleration is relative under GR, physical acceleration* is not relative. Physical acceleration can be unambiguously detected with an accelerometer in both theories.

*there may be better terms than coordinate and physical acceleration

[MeJennifer]

Although coordinate acceleration is relative under GR, physical acceleration* is not relative. Physical acceleration can be unambiguously detected with an accelerometer in both theories.

*there may be better terms than coordinate and physical acceleration
I disagree, physical acceleration is relative under GR. In fact under GR all forms of motion are relative.

[robphy]

Although coordinate acceleration is relative under GR, physical acceleration* is not relative. Physical acceleration can be unambiguously detected with an accelerometer in both theories.

*there may be better terms than coordinate and physical acceleration

I disagree, physical acceleration is relative under GR. In fact under GR all forms of motion are relative.

It seems DaleSpam is referring to "4-acceleration", aka "worldline curvature", u^a\nabla_a u^b which measures the deviation of the worldline from being a geodesic.

MeJennifer, what is your definition of "[physical] acceleration"?

[A.T.]

I disagree, physical acceleration is relative under GR. In fact under GR all forms of motion are relative.
Proper acceleration (what an accelerometer measures) is absolute. Just like proper time (what a clock measures).

[MeJennifer]

Proper acceleration (what an accelerometer measures) is absolute.
What an accelerometer measures could be the result of acceleration or gravity it cannot be determined by the measurement which one it is. Similarly as in special relativity where it cannot be determined which object is at rest and which object is moving we can only say that they are in relative motion.

[A.T.]

What an accelerometer measures could be the result of acceleration or gravity it cannot be determined by the measurement which one it is.

No, it is always the result of acceleration. In GR an object held up by a force against a gravitational field is accelerated, just like a rocket in open space firing its engines.

But the question was: "Is it absolute?" And since every observer will agree what the accelerometer measures, it is absolute.

[DaleSpam]

Proper acceleration (what an accelerometer measures) is absolute. Just like proper time (what a clock measures).Thanks, I knew there was a better term than "physical acceleration" but I just couldn't think of it.

If proper acceleration is not absolute then GR is an easily falsified theory.

[MeJennifer]

If proper acceleration is not absolute then GR is an easily falsified theory.
Really? How is it falsified?

No, it is always the result of acceleration. In GR an object held up by a force against a gravitational field is accelerated, just like a rocket in open space firing its engines.
Think how an accelerometer can indicate acceleration in a strong gravitational field due to tidal forces.

In other words the state of acceleration in a rod can in principle be caused by gravitation, hence acceleration is relative.

[DaleSpam]

Really? How is it falsified?If a theory claimed that proper acceleration were relative then accelerometer readings would be frame variant, which would lead to logical inconsistencies like an accelerometer reading 0 and g at the same time.

In general, all experimentally measureable outcomes must be invariant, not relative. Hence, although coordinate time is frame variant, proper time is absolute. Similarly with coordinate acceleration which is relative and proper acceleration which is absolute.

[neopolitan]

... although coordinate time is frame variant, proper time is absolute.

Proper time. Hm, if we have proper time, can we have "proper simultaneity"? Since we can take a single event as our reference, it can be the big bang event or it could be the flash of a light which conceptually radiates photons throughout the universe, can we then not work out proper time for each location in space and therefore have "proper space"?

Probably there are limitations to the application of "proper time" :)

cheers,

neopolitan

[DaleSpam]

if we have proper time, can we have "proper simultaneity"? I am 90% sure that you are going to misinterpret this, but the short answer is yes.

If two clocks, A and B, at rest wrt each other are synchronized by some arbitrary synchronization convention in their rest frame and pass through events, Ea and Eb, then all observers will be able to determine wether or not the clocks read the same or different. Iff they read the same then the events were simultaneous in their frame and by their synchronization convention.

Note, there is still no physical significance to the simultaneity because if the events are spacelike separated in one frame they will be spacelike separated in all frames and thus not causally connected.

Since we can take a single event as our referenceA single event does not define a reference frame in any sense. There is nothing "proper" about it. What direction does a single point define?

[neopolitan]

A single event does not define a reference frame in any sense. There is nothing "proper" about it. What direction does a single point define?

I didn't say the event was a reference frame, did I?

You seem to be talking about "proper synchronisation" not about "proper simultaneity".

People seem to hung up about clocks and rods, which are supposed to be representative. Can you not point to a location on a spacetime diagram drawn from the perspective of the frame in which you are at rest and say "according to me, this is now"? Or are you limited to point to clocks? Same with a distance, can you not point to a location and say "this is a distance of L from me"? Or do you have to have a rod?

Take one event and allow photons to radiate out from it in all directions. If you have "proper time", is the distance from the reference event traveled by a photon in that "proper time" not "proper separation"? Are not all the 4-space locations associated with the photons which are causally linked to the reference event (since the event spawned all the photons we are discussing) properly simultaneous? (Note that I am not saying they are simultaneous with anything other than themselves. They may be, but we don't have the information necessary to make that determination.) Does that hypersurface of simultaneity spring into existance because the photons are released, or does it exist even if we don't release photons, since we can say these events constitute the 4-space locations which the photons would have reached iff and we released them?

Perhaps I am misinterpreting you :) Either way, I don't feel an overwhelming need to talk about clocks and synchronisation conventions.

cheers,

neopolitan

[DaleSpam]

Either way, I don't feel an overwhelming need to talk about clocks and synchronisation conventions.Me neither.

FYI, I have never seen the term "proper simultaneity" before your post; it doesn't have an established meaning. I was simply providing a possible meaning, but I won't try to argue for it since I see no value in it.

[Mentz114]

'Proper' means - 'as measured by an observer in his own frame'. There cannot be such a thing as proper simultaneity.

People seem to be hung up about clocks and rods, which are supposed to be representative.
You have absolutely no understanding of physics. It is about measurements, which are always done with clocks and rulers.

It is you who is 'hung up' on simultaneity.

[Bosemann]

For anyone having trouble understanding why the acceleration is crucial, take a look at this link. If you don't understand it (I didn't the first time through), then slow down and read it very carefully. I've found skimming through these sorts of things is just a waste of time.

Acceleration is not really the answer.
Think of it this way: one twin travel to a distance L and turn around, another twin travel to a distance L/2 and turn around. The twin travel longer will age less when they meet again.
Since they both experience acceleration, so acceleration is not the cause of age difference

[DocZaius]

Acceleration is not really the answer.
Think of it this way: one twin travel to a distance L and turn around, another twin travel to a distance L/2 and turn around. The twin travel longer will age less when they meet again.
Since they both experience acceleration, so acceleration is not the cause of age difference

I didn't say acceleration is the only cause of age difference. I said that acceleration was the crucial factor explaining why there was a break in symmetry in that particular scenario.

[Bosemann]

I didn't say acceleration is the only cause of age difference. I said that acceleration was the crucial factor explaining why there was a break in symmetry in that particular scenario.

Was there a break in symmetry in this particular scenario (both twins experience acceleration)?
If yes, was acceleration the crucial factor explaining why there was a break in symmetry?

[DocZaius]

Was there a break in symmetry in this particular scenario (both twins experience acceleration)?
If yes, was acceleration the crucial factor explaining why there was a break in symmetry?

In the scenario in which they both experience acceleration? Of course acceleration is not the crucial factor there.

[neopolitan]

'Proper' means - 'as measured by an observer in his own frame'. There cannot be such a thing as proper simultaneity.


You have absolutely no understanding of physics. It is about measurements, which are always done with clocks and rulers.

It is you who is 'hung up' on simultaneity.

I had to have a bit of a laugh at this. Is this a faith based thing for you Mentz? If so, I apologise for questioning your faith.

But as far as I am concerned physics is not just measuring things. It is more about understanding how things work, even simultaneity. And part of the process of improving understanding is to ask questions and challenge assumptions. When it comes to religious thought you just ask the experts and accept the doctrine. When it comes to science (including Physics) you are actively encouraged to challenge the theories - a theory after all must be falsifiable.

The use of ad hominem arguments in a physics discussion indicates that perhaps it is not (only) me who has no understanding of it. (Note I said "People seem ..." Your statement is far more categorical.)

Relax and be open to challenges to your assumptions. Defend your assumptions if are not able to challenge your assumptions yourself. But if you attack people who challenge your assumptions, then any wrong assumptions you have will never be addressed.

And that is not physics, is it?

cheers,

neopolitan

[Mentz114]

neopolitan,

are you saying that the statement 'physics is all about clocks and rulers' is an assumption, or an article of faith ?
If that's the case, you don't know your assumptions from your elbow.

Physical theories try to explain measurable things. Theories about non-measurable things are just hot-air.

Everything you've said above merely strengthens my point.

M

[granpa]

consider this similar question:
if observer a sees b as being length contracted then why does b see a as being length contracted instead of seeing a as being stretched out. clearly the answer is because of loss of simultaneity. couldnt the same be true for the twin paradox?

[Mentz114]

consider this similar question:
if observer a sees b as being length contracted then why does b see a as being length contracted instead of seeing a as being stretched out. clearly the answer is because of loss of simultaneity. couldnt the same be true for the twin paradox?

There is no contradiction in the fact that both observers see each others rulers contracted.

Loss of what simultaneity ? There are some events that the observers would agree were simultaneous, and others they would not agree on. It means nothing.

There is no twin paradox !

[granpa]

There is no contradiction in the fact that both observers see each others rulers contracted.


there is no contradiction if you take loss of simultaneity into account. the length of an object is the distance between the position of the front and back of the object at one simultaneous moment.

[Mentz114]

granpa, I can see where you're coming from. Another way of measuring the length would be to measure the time taken for the rod to pass a fixed point in your frame.

[neopolitan]

neopolitan,

are you saying that the statement 'physics is all about clocks and rulers' is an assumption, or an article of faith ?
If that's the case, you don't know your assumptions from your elbow.

Physical theories try to explain measurable things. Theories about non-measurable things are just hot-air.

Everything you've said above merely strengthens my point.

M

Mentz,

I admit to being pedantic here, but physics is about explaining "testable" things, measurements being the mechanism by which mosts tests are conducted. However, since physics theories must be (or at the very least ought to be) not only testable but falsifiable, then some tests won't require measurements at all. An example was Young's attempt to test the wave theory of light with the double-slit experiment. No clocks, no rulers. Either the distribution of light on the other side of the double slit would support the wave theory of light or falsify it. As it was it supported the theory (but did not prove it, since you can't "prove" a theory, just provide overwhelming support, until such time as someone clever comes along and finds a way to disprove it).

Anyway, my initial comment (which was not you but rather to DaleSpam) was about the fact that events have a temporal component irrespective of whether a clock is there or not. That temporal component can be deduced retrospectively, if you so wished, in terms of the frame in which you are at rest. Similarly, in the frame in which I am at rest, there is a spatial separation between me and a spot 3m distant from my desk about 2m above the floor, even if there is no rod between me and that spot and, to my eyes, nothing to distinguish that spot from any other spot in my room which is simularly unoccupied by people, furniture etc. However, at some point of time, that spot will be inhabited by an oxygen molecule, and I could talk about the distance between the exact centre of my skull and that oxygen molecule event despite the total and complete absence of rods.

To go further, there is no need for anything to be in that location for me to label it ... does there? Do I actually need to have a rod extending out from the centre of my skull to that spot to discuss it? Do I need to have a clock to meaningfully talk about an event 5 minutes from now? My not having a clock is not going to prevent it from happening ...

Perhaps this is all too difficult, along with the question I posed in the original offending post, to wit:

Take one event and allow photons to radiate out from it in all directions. If you have "proper time", is the distance from the reference event traveled by a photon in that "proper time" not "proper separation"? Are not all the 4-space locations associated with the photons which are causally linked to the reference event (since the event spawned all the photons we are discussing) properly simultaneous? (Note that I am not saying they are simultaneous with anything other than themselves. They may be, but we don't have the information necessary to make that determination.) Does that hypersurface of simultaneity spring into existance because the photons are released, or does it exist even if we don't release photons, since we can say these events constitute the 4-space locations which the photons would have reached iff and we released them?

To get back to "proper", this is from Wikipedia:

In relativity, proper time is time measured by a single clock between events that occur at the same place as the clock. It depends not only on the events but also on the motion of the clock between the events. An accelerated clock will measure a shorter proper time between two events than a non-accelerated (inertial) clock between the same events.

This means that "proper simultaneity" which I talked about doesn't work, since I thought about a range of locations - which means a break with a core feature of "proper time". DaleSpam was right to be leery of the term.

Note that I object to the wikipedia description to the extent that it is not the acceleration per se which causes the shorter proper time, merely that for one clock the events are in the same location (hence shorter proper time) and for another clock the events are in two different locations (hence longer "improper time").

You can prove this with mathematics and with mind experiments with clocks and rods, or at the very least you can disprove that it is acceleration which causes it. But since it is part of the popular mythology now, the idea that it is acceleration behind the phenomenon is pretty hard to shift. (Note that quite a few others, some with official status on the forums, have also stated that acceleration does not cause the shorter proper time per se.)

cheers,

neopolitan

[DaleSpam]

To go further, there is no need for anything to be in that location for me to label it ... does there? Do I actually need to have a rod extending out from the centre of my skull to that spot to discuss it? Do I need to have a clock to meaningfully talk about an event 5 minutes from now? My not having a clock is not going to prevent it from happening ...By "label it" I assume you mean "assign it a coordinate value". Whatever physical mechanism you use to assign the coordinates is equivalent to some system of clocks and rods.

The terms "clock" and "rod" in SR are thus just shorthand for "a physical method of measuring time" and "a physical method of measuring distance" respectively. In that sense if you do not use "clocks and rods" then your coordinate system is not measurable and therefore non-physical, hence Mentz's objection. I don't think that was your intention.

[neopolitan]

IPeople seem to hung up about clocks and rods, which are supposed to be representative.

This is from the original post which Mentz attacked.

I don't suggest taking away methods of measuring, or even the idea of measuring. All I am saying is that I can discuss a distance in my rest frame (the frame in which I am at rest) with or without positing an actual physical rod to measure it. Similarly, I can discuss an interval of time without having an actual clock to measure it.

In the same way as the "observers" don't have to be real observers, rods and clocks don't have to be real.

The difference is between my pointing to two events on my hypersurface of simultaneity and your discussing "two clocks, A and B, at rest wrt each other are synchronized by some arbitrary synchronization convention in their rest frame" (and then there is Mentz with his "forget about simultaneity, read a good essay on SR").

For me, the clocks being synchonised has little to do with simultaneity - since two clocks reading completely different times can in fact be simultaneous (like my watch and your watch, which are probably in different time zones, simultaneous, but reading a few hours apart - give or take). The same applies with rods, I can put two scratchs close to each end of a rod and write 1m between them. Then I can get a shorter rod, and do the same thing. It's not the rod. It is the distance that the rod represents that matters.

In reality, I should talk about changing the intervals between the ticks of a clock, not the time shown, to make it equivalent to using different length rods. Usually when we synchronise clocks or watches, we make sure they read the same, not check that they keep time with each other. Making sure clocks keep time is possibly more important, but the synchronisation conventions usually discussed on this forum are about making clocks at rest with respect to each other read the same - with an assumption that they will keep time once synchronised. But, this is getting bogged down in actual clocks again, rather than thinking about what the clocks represent.

I have no problem with what you have to say, DaleSpam and I don't think we have any substantial disagreement.

cheers,

neopolitan

[Mentz114]

neopolotan,

I'm sorry if my post seemed like an 'attack'. You obviously think about these things a lot, but in my opinion you have missed a vital thing. You say above,

But, this is getting bogged down in actual clocks again, rather than thinking about what the clocks represent.

Clocks don't 'represent' anything, and time in physics can only be defined as 'what is measured by a clock'. Just as distance can only be defined as 'what is measured by a ruler'.

Any other definitions are mostly philosophical. 'The End of Time' by Julian Barbour is a good read.

I don't have time to write anymore now.

M

[neopolitan]

I'm sorry if my post seemed like an 'attack'.

"You have absolutely no understanding of physics." If it was meant as helpful advice, I missed the subtlety. Sorry.

Clocks don't 'represent' anything, and time in physics can only be defined as 'what is measured by a clock'. Just as distance can only be defined as 'what is measured by a ruler'.

Any other definitions are mostly philosophical.

Perhaps I am getting philosophical here, but I would agree that "time in physics is 'that which could be measured by a clock'" - note I object to your use of "is". You are implying, perhaps unintentionally, that there was no time until someone invented a clock to measure it. And think we all agree that that is a bit silly.

Similarly with distance.

Time and space existed before clocks and rods. They existed before clocks and rods and will almost certainly exist long after all the clocks and rods have been burnt up or crushed in a passing black hole. Unless of course you believe in a god who has a supply of clocks and rods and will be handing them out in heaven to good physicists who know their assumptions from their elbow. :smile:

[Mentz114]

neopolitan,
You are implying, perhaps unintentionally, that there was no time until someone invented a clock to measure it. And think we all agree that that is a bit silly.
You're being silly. I did not imply any such thing.

There can be no physics without clocks and rulers. This is a physics forum.

You still don't understand what physics is about, in spite of all the time people have wasted trying to set you straight.

[Al68]

Hi all,

neopolitan, I brought up your point on this forum a couple of years ago, and got similar replies, but, like now, the obvious point was ignored. The "twins paradox" can be setup many ways, with or without acceleration. With or without anyone changing reference frames. Just have the traveling twin pass earth at speed, start clocks, and stop the clocks when he passes the distant star system (the earth twin will have to subtract light transit time from his clock reading, or have a clock at the destination) the traveling twin's clock will read less time. No acceleration, no change of reference frames. Same result. Common sense tells me that if we get the same result without acceleration, then acceleration isn't the reason for the result. This even seems too obvious to bother pointing out. Maybe I'm missing something.

And, like neopolitan says, there is an obvious asymmetry in the twins paradox that has nothing to do with acceleration. The turnaround point is defined to be a fixed distance (at rest with) one of the observers, but not the other. This is the key to the whole thing.

The traveling twin has less elapsed time because the distance traveled is smaller. Whichever twin measures the smaller distance between the events will also measure less elapsed time between the events. Can someone come up with a scenario where this is not true?

This is the only asymmetry I can see in the twins paradox that will not go away simply by slightly changing the scenario.

Al

[matheinste]

Hello A168.

In this never ending saga of the twins nonparadox how can the two be present at the start event and the meeting event and follow different spacetime paths without one or both accelerating. If they are not present at the two events there is no apparent paradox anyway because the whole idea is to show the "remarkable" and puzzling effect of the differential in ageing

In all the scenarios they must both be present at the two events to realistically compare their elapsed time ( ages ).

Also the one that follows the shortest spacetime path experiences the most elapsed time. An inertial, non accelerated, path is the shortest spacetime distance of all.

The resolution of his nonparadox is explained countless times in this forum and in most textbooks.

Matheinste.

[Doc Al]

And, like neopolitan says, there is an obvious asymmetry in the twins paradox that has nothing to do with acceleration. The turnaround point is defined to be a fixed distance (at rest with) one of the observers, but not the other. This is the key to the whole thing.

Nonsense. Not only is the turnaround point being "at rest" not the key, it's not even relevant or meaningful. The traveling twin can turn around at any point--it doesn't matter! And he must turnaround at some point, and that point has some specific distance from the other twin (different in each frame, of course).

[AntigenX]

The traveling twin has less elapsed time because the distance traveled is smaller. Whichever twin measures the smaller distance between the events will also measure less elapsed time between the events. Can someone come up with a scenario where this is not true?

This is the only asymmetry I can see in the twins paradox that will not go away simply by slightly changing the scenario.

Al

Hi Al,

You must account for acceleration, because, without acceleration, the traveling twin and the stationary twins are equivalent. We can not decide in SR which is moving and which is traveling. So, though the twin in spaceship is traveling, he is traveling wrt the stationary twin, and thus, he will also see the stationary twin's clock slow. In such a case, without acceleration (which breaks symmetry, I don't know how!), no twin can age less!

Though I don't quite understand the effect of acceleration, without acceleration no one can age less.

[matheinste]

Hello AntigenX.

The acceleration breaks the symmetry by making the travelling twin's spacetime interval longer than the non-accelerated twin's spacetime interval. The object ( twin ) with the longer spacetime interval accumulates less elapsed time and hence remains younger. Do not confuse spatial distance travelled with spacetime interval. They are very different things.

As said above a spacetime diagram makes this clear.

Matheinste.

[robphy]

Hello AntigenX.

The acceleration breaks the symmetry by making the travelling twin's spacetime interval longer than the non-accelerated twin's spacetime interval. The object ( twin ) with the longer spacetime interval accumulates less elapsed time and hence remains younger. Do not confuse spatial distance travelled with spacetime interval. They are very different things.

As said above a spacetime diagram makes this clear.

Matheinste.

I take issue with the portion of your comment that I highlighted above.
The acceleration DOES NOT MAKE the travelling twin's spacetime interval longer... it DOES INDICATE that a noninertial worldline is being used from A to B and such a worldline is necessarily shorter than an inertial worldline from A to B.
(In addition... "travelling twin's spacetime interval" [along his worldline from A to B] is SHORTER.)

As I wrote in an earlier post (in " "twins' paradox" and acceleration ") (http://www.physicsforums.com/showthread.php?p=1738289#post1738289),
which you quoted in your post (in "Impure twin's paradox") (http://www.physicsforums.com/showthread.php?p=1739679#post1739679),
"The symmetry break (between inertial and noninertial) is the "presence of an acceleration (worldline curvature) somewhere during the trip" for the noninertial observer. Neither are causes of the shorter-elapsed-proper-time from A to B... they are correlated with the shorter-elapsed-time because they indicate that a noninertial (i.e. nongeodesic) worldline was used to experience both A and B."

(One of the best papers on the Twin Paradox/Clock Paradox:
http://www.jstor.org/pss/2309916 (requires institutional access)
"The Clock Paradox in Relativity Theory", Alfred Schild, The American Mathematical Monthly, Vol. 66, No. 1. (Jan., 1959), pp.1-18.)

[granpa]

longer spacetime interval


thats a strange way of putting it. in Minkowski space the interval is defined as (d^2-T^2)^0.5
so its actually shorter. thats why i was so confused earlier. but i know what you mean though.


the length of the line segment that corresponds to the interval would be (d^2+T^2)^0.5. thats what you are referring to.

[AntigenX]

Hello AntigenX.

The acceleration breaks the symmetry by making the travelling twin's spacetime interval longer than the non-accelerated twin's spacetime interval. The object ( twin ) with the longer spacetime interval accumulates less elapsed time and hence remains younger.

Greetings matheinste,

Yes, I understand the "translated" meaning, but can't get it how? As I have asked earlier several times and also in this (http://www.physicsforums.com/showthread.php?t=236229) thread, as acceleration is not relative to the stationary twin (but absolute), how can it's effect be relative? I mean only one clock is slowed more and not other?

Do not confuse spatial distance travelled with spacetime interval. They are very different things.


Have I?


As said above a spacetime diagram makes this clear.

Matheinste.

Well, space time diagrams are good If you understand the things. I'm not clear about the things, so spacetime diagrams are just boring math for me. I don't want to get in to the habit of learning physics via math. Though others may not agree, this is just absurd for stupids like me:wink:.

Thanks for the suggestion anyways...

[DocZaius]

Well, space time diagrams are good If you understand the things. I'm not clear about the things, so spacetime diagrams are just boring math for me. I don't want to get in to the habit of learning physics via math. Though others may not agree, this is just absurd for stupids like me:wink:.

I think this quote is applicable:

“To those who do not know mathematics it is difficult to get across a real feeling as to the beauty, the deepest beauty, of nature … If you want to learn about nature, to appreciate nature, it is necessary to understand the language that she speaks in.”

–R. Feynman

[Doc Al]

Of the several ways to understand relativistic effects, the use of spacetime diagrams is probably the least mathematical (and most profound). Sure, it takes a bit of effort to figure out how to interpret them, but well worth it if you're serious. Verbal handwaving won't help.

[AntigenX]

I think this quote is applicable:

“To those who do not know mathematics it is difficult to get across a real feeling as to the beauty, the deepest beauty, of nature … If you want to learn about nature, to appreciate nature, it is necessary to understand the language that she speaks in.”

–R. Feynman

Oh yes, But I think you are getting me wrong. By calling myself stupid I don't propose to say I don't know math (neither you would have inferred it, I suppose), instead, It's just what einstein used to say "... equations and math are for book-keeping, not for learning or understanding...". But of course, It was einstein, and his thinking may be different like mine, yours and R. Feynman's.

[DocZaius]

Oh yes, But I think you are getting me wrong. By calling myself stupid I don't propose to say I don't know math (neither you would have inferred it, I suppose), instead, It's just what einstein used to say "... equations and math are for book-keeping, not for learning or understanding...". But of course, It was einstein, and his thinking may be different like mine, yours and R. Feynman's.

As you have said, I wasn't inferring you don't know math. My only point was that the application of mathematics to your study of nature is very useful and brings out much more meaning (in my opinion) than the lack of it.

I'm surprised Einstein said that equations and math are not for learning or understanding. They seem to be invaluable tools.

[AntigenX]

As you have said, I wasn't inferring you don't know math. My only point was that the application of mathematics to your study of nature is very useful and brings out much more meaning (in my opinion) than the lack of it.

I'm surprised Einstein said that equations and math are not for learning or understanding. They seem to be invaluable tools.

Don't worry about that, and it doesn't matter even if you consider me stupid, as long as you are teaching me something or trying to help me learn something.

Einstein said them in some of his lectures, where he was explaining the importance of clocks and rods and thought experiments and their philosophical and physical interpretations. Even he himself has never relied on spacetime diagrams but thought experiments and their interpretations were his tools of the trade (as far as I have read him)...

EDIT: He also said later that the theories of relativity (especially GR) have become so mathematical that he himself doesn't understand it (jokingly of course)...

[matheinste]

Hello robphy.

I apologise for my error.

granpa.

You are correct instaed of spacetime interval i should have said length of spacetime worldline.

Mateinste

[AntigenX]

Of the several ways to understand relativistic effects, the use of spacetime diagrams is probably the least mathematical (and most profound). Sure, it takes a bit of effort to figure out how to interpret them, but well worth it if you're serious. Verbal handwaving won't help.

I have never disregarded it, and several people told me about this as well. It's just I wish to try the other way. "Verbal handwaving" is doing a good job currently, and is my time tested method. I don't see any way to convince you that I am serious, but I wish only if you could believe me...

[Al68]

Hello A168.

In this never ending saga of the twins nonparadox how can the two be present at the start event and the meeting event and follow different spacetime paths without one or both accelerating. If they are not present at the two events there is no apparent paradox anyway because the whole idea is to show the "remarkable" and puzzling effect of the differential in ageing

In all the scenarios they must both be present at the two events to realistically compare their elapsed time ( ages ). No they don't. The earth twin could have an agent at the distant star system with a clock.

Also the one that follows the shortest spacetime path experiences the most elapsed time. An inertial, non accelerated, path is the shortest spacetime distance of all.

The resolution of his nonparadox is explained countless times in this forum and in most textbooks.
Yes. And not only are they unsatisfactory, they contradict each other.

Al

[Al68]

Nonsense. Not only is the turnaround point being "at rest" not the key, it's not even relevant or meaningful. The traveling twin can turn around at any point--it doesn't matter! And he must turnaround at some point, and that point has some specific distance from the other twin (different in each frame, of course).

Why does the traveling twin have to turn around? Just so we can have the novelty of the twins meeting again?

Al

[Al68]

Hi Al,

You must account for acceleration, because, without acceleration, the traveling twin and the stationary twins are equivalent. We can not decide in SR which is moving and which is traveling. So, though the twin in spaceship is traveling, he is traveling wrt the stationary twin, and thus, he will also see the stationary twin's clock slow. In such a case, without acceleration (which breaks symmetry, I don't know how!), no twin can age less!

Though I don't quite understand the effect of acceleration, without acceleration no one can age less.

Again, since we get the same result even if there is no acceleration involved, it seems to me that I don't have to account for acceleration. Even in the standard twins paradox, I ignore acceleration and get the same result.

The simple fact is that at a given speed, it will take less time to traverse a smaller distance. It will always take less time to traverse a smaller distance. Whichever twin measures the most distance between events will have the most elapsed time between those events.

This is true no matter how we set up the scenario, with or without acceleration, is it not?

Al

[matheinste]

Hello A168.

Quote:-

-----No they don't. The earth twin could have an agent at the distant star system with a clock.-----

That is not how the paradox is presented. The whole idea is to make it look like a contradicton. ( but of course the resolutions should not be contradictory )

Quote:-

---Why does the traveling twin have to turn around? Just so we can have the novelty of the twins meeting again?-----

For the same reason.

If you find the reasons for the age difference unsatisfactory i can't help you. If you find them contradictory have you considered that some of the answers may be wrong.

Matheinste.

[AntigenX]

Again, since we get the same result even if there is no acceleration involved, it seems to me that I don't have to account for acceleration. Even in the standard twins paradox, I ignore acceleration and get the same result.

What same result? that the traveling twin will age less? We don't get the same result. Because, as I pointed out earlier, we don't really have any means to say that traveling twin is really traveling!


The simple fact is that at a given speed, it will take less time to traverse a smaller distance. It will always take less time to traverse a smaller distance. Whichever twin measures the most distance between events will have the most elapsed time between those events.


How will you measure the speed of the traveling twin?


This is true no matter how we set up the scenario, with or without acceleration, is it not?

Al

No. Though I don't understand the effects of acceleration, without the acceleration the situation is perfectly symmetric. In such a case, both twins will observe the other twin to age less, and we have no reasons to prefer any one over the other.

[matheinste]

Hello AntigenX.

I believe some of your reasoning is inceorrect and contradict's the generally accepted answer as derived from the axioms of SR. Textbooks which give the resolution to the seeming paradox agree on the answer. There should be no problem

Matheinste.

[Mentz114]

AntigenX,

What same result? that the traveling twin will age less? We don't get the same result. Because, as I pointed out earlier, we don't really have any means to say that traveling twin is really traveling!

It doesn't matter what the situation is, the twin with the longest proper interval will age less. There's no paradox or difficulty. Just learn how to calculate the proper interval and all these cases can be worked out with the same recipe.

[AntigenX]

Hello AntigenX.

I believe some of your reasoning is inceorrect and contradict's the generally accepted answer as derived from the axioms of SR. Textbooks which give the resolution to the seeming paradox agree on the answer. There should be no problem

Matheinste.

Sorry for that. I thought twice before posting though!

Can you please tell me which are those points? I am asking this because, may be my english is not proper, and hence I am misinterpreted many times.

[matheinste]

Hello AntigenX

Your english is fine.

Quote:-

----Whichever twin measures the most distance between events will have the most elapsed time between those events.----

It is the other way around. The travelling twin ( the accelerated one in the case of the proposed paradox ) follows the longer spacetime path and so accumulates the lesser elapsed time.

Matheinste.

[AntigenX]

Hello AntigenX

Your english is fine.

Quote:-

----Whichever twin measures the most distance between events will have the most elapsed time between those events.----

It is the other way around. The travelling twin ( the accelerated one in the case of the proposed paradox ) follows the longer spacetime path and so accumulates the lesser elapsed time.

Matheinste.

Thanks Matheinste for the compliment:blushing:!!! But I never said that... In fact I contradicted that, and as you are saying, I was correct.

[matheinste]

Hello AntigenX.

Many apologies. I of course retract my statement about the incorrectness of your post

Matheinste.

[Al68]

Hello AntigenX

Your english is fine.

Quote:-

----Whichever twin measures the most distance between events will have the most elapsed time between those events.----

It is the other way around. The travelling twin ( the accelerated one in the case of the proposed paradox ) follows the longer spacetime path and so accumulates the lesser elapsed time.

Matheinste.

I think I had it right. In this case the traveling twin (with the longer spacetime path) measures a shorter distance between events, and a shorter elapsed time.

[Al68]

Hello A168.

Quote:-

-----No they don't. The earth twin could have an agent at the distant star system with a clock.-----

That is not how the paradox is presented. The whole idea is to make it look like a contradicton. ( but of course the resolutions should not be contradictory )

Quote:-

---Why does the traveling twin have to turn around? Just so we can have the novelty of the twins meeting again?-----

For the same reason.

If you find the reasons for the age difference unsatisfactory i can't help you. If you find them contradictory have you considered that some of the answers may be wrong.

Matheinste.

Hi Matheinste,

I think you said it. The whole point of the way the scenario is normally presented is for the novelty of making it look like a contradiction.

And I don't have a problem with the age difference. I get the same answer if I pretend the earth twin accelerated with everything else the same. Just pretend (for no reason) that the earth twin "felt" the turnaround and do the math. Ship's twin still ages less. Amazing. SR math will work exactly the same and show the twin who measures the shorter distance traveled to have the shorter elapsed time. Amazing again.

It is simple to have a scenario where neither twin accelerates during the test. just have the ship's twin stop his clock just before he turns around, and have a clock at the turnaround point (synched with earth) record the time of the same event. Then the twins can turnaround, fly in circles, or whatever, it won't change the clocks since they are stopped. And again, the twin who measures the shorter distance will have the least elapsed time.

How about a challenge for all: Come up with a scenario in which my claim is wrong. The claim is that whichever twin measures the shorter distance between two events will have less elapsed time between those events.

Al

[Hurkyl]

I think you said it. The whole point of the way the scenario is normally presented is for the novelty of making it look like a contradiction.
It's not simply a novelty -- it's for educational purposes. People really do make that mistake and other similar ones (even people that should know better), so its important to spend some time teaching students to identify the flaw, and demonstrating that it really is flawed.

It is simple to have a scenario where neither twin accelerates during the test. just have the ship's twin stop his clock just before he turns around
Stopping the clock during an experiment doesn't stop the experiment.

How about a challenge for all: Come up with a scenario in which my claim is wrong. The claim is that whichever twin measures the shorter distance between two events will have less elapsed time between those events.
State your claim precisely, please. Current problems include:
(1) Which events?
(2) If you mean the events where the twins separate and reunite, then they are timelike separated, and there is no intrinsic meaning to the 'distance' between them. (It would make sense to ask about the proper duration, but of course, everyone would measure the same value)
(3) If you mean to refer to coordinate-dependent quantities, then I think you are going to need to put some constraints on what coordinate charts each twin uses.

[matheinste]

Hello Al68

Quote:-

----I think I had it right. In this case the traveling twin (with the longer spacetime path) measures a shorter distance between events------

He experiences a shorter elapsed time but travels a longer spacetime path. Having less accumulated time does not mean he travels a shorter spactime path.

Matheinste.

[Hurkyl]

The travelling twin ( the accelerated one in the case of the proposed paradox ) follows the longer spacetime path
In this case the traveling twin (with the longer spacetime path)
The "spacetime path" travelled by an object1 is time-like; the notion of 'distance' doesn't make sense. The 'duration' of the path, however, is exactly what the observer's wristwatch measures.


1: A tardyonic object, at least. This doesn't apply to tachyons

[matheinste]

Hello Hurkyl

I don't think i used the word distance, except in quotes, for a spacetime path. I was just using the term spacetime path, perhaps inacurately, as a sort of measure of some sort of separation between events. If i gave the impression that i meant spatial distance this was unintended as of course what you say is correct.

Matheinste.

[Al68]

It's not simply a novelty -- it's for educational purposes. People really do make that mistake and other similar ones (even people that should know better), so its important to spend some time teaching students to identify the flaw, and demonstrating that it really is flawed.


Stopping the clock during an experiment doesn't stop the experiment.

No, but I meant that we could redefine the end of the experiment, too.


State your claim precisely, please. Current problems include:
(1) Which events?
(2) If you mean the events where the twins separate and reunite, then they are timelike separated, and there is no intrinsic meaning to the 'distance' between them. (It would make sense to ask about the proper duration, but of course, everyone would measure the same value)
(3) If you mean to refer to coordinate-dependent quantities, then I think you are going to need to put some constraints on what coordinate charts each twin uses.

1) u pick em
2) I mean cumulative distance traveled.
3) no constraints, as long as each twin measures everything properly.

Really, I would just like to see why people believe that acceleration is crucial, when the experiment could be presented without acceleration with the same result. By same result I mean that the twin who measured the shorter distance has less elapsed time, not that the twins reunite. For example, it has been presented as two separate trips with a third observer traveling from the distant star system to earth, nobody accelerates, and we just add up the two trips and get the same result. Or the experiment could end when the ship passes the turnaround point. With a third observer there with a clock. The twins don't have the novelty of reuniting, but we have the same result of two defined events, and one twin has less elapsed time between them than the other.

And I think it's worth mentioning that, in the normal twins paradox, that the twins' reunion doesn't really change anything. It's not like the laws of physics change because they reunite.

I don't dispute that the traveling twin will age less, but he will age less (have less elapsed time between events) during a one-way journey as well. And we don't need to reunite the twins to show this. Unless we consider the most important thing here is to have two twins look at each other and have one say "Gee, you're older than me, how did that happen?"

Sure, in the common example, the twin who accelerates does indeed age less, but is there any evidence (or logical deduction) that shows that this is the reason? SR certainly doesn't make such a claim. Using SR, we can ignore acceleration altogether and get the same result. We can even say that the ship never accelerated, and the earth (and distant star system) were moved by magic/God/Unknown reasons, and when the twins reunite, the one in the ship is younger according to SR. Yes, that's silly, I know. Just making a point.

And I have seen many posts saying I'm wrong, but none that say how, or provide any substantiation of the claim that acceleration is important as a general rule, not just in a specific scenario where the accelerated twin happens to age less.

That's what I'm asking for.

Al

[Al68]

He experiences a shorter elapsed time but travels a longer spacetime path. Having less accumulated time does not mean he travels a shorter spactime path.

I agree. I never said otherwise. I said: "Whichever twin measures the most distance between events will have the most elapsed time between those events."

Al

[Hurkyl]

Really, I would just like to see why people believe that acceleration is crucial
When people seek to resolve the twin paradox, they seek to point out a flaw in the logical argument in the twin paradox. If you are talking about something that isn't the twin paradox (e.g. any experiment involving twins that don't reunite), then that is something irrelevant.

I don't dispute that the traveling twin will age less, but he will age less (have less elapsed time between events) during a one-way journey as well.
In a one-way journey, it is impossible for both twins to be present at both events. And there is no intrinsic way to compare their ages.

[Hurkyl]

2) I mean cumulative distance traveled.
Events don't travel; they're events. This doesn't make sense.
Similarly, you mentioned 'elapsed time'; elapsed time of what?

3) no constraints, as long as each twin measures everything properly.
You do realize that, in any experimental setup, by choosing the appropriate coordinate chart, I can make either twin (properly!) compute any value I want for any coordinate-dependent quantity I want, right?

[matheinste]

Hello Al68

Quote:-

---- agree. I never said otherwise. I said: "Whichever twin measures the most distance between events will have the most elapsed time between those events."----

As Hurkyl pointed out to us both, distance is the wrong word. However your quote is still wrong. It should read " Whichever twin measures the most 'distance' between events will have the least elapsed time between those events"

Edit. Reading the latest post from Hurkyl #105 i t seems i have also been using the phrase elapsed time rather loosely but you will know wehat i mean

Matheinste.

[Hurkyl]

Here's a scenario: Alice and Bob are both sitting on Earth. They never leave each other's side. They consider the worldline of a rocket travelling from here to alpha centauri. I choose as my two events: "The rocket taking off" and "The rocket arriving". Alice decides to measure things relative to an Earth-centered inertial frame. Bob decides to measure things relative to an inertial frame where the rocket is (mostly) stationary. Bob measures the shorter coordinate distance between the two events (zero!), but also measures the longer coordinate time between those two events. And the aging of the twins seems entirely irrelevant to anything in the setup.

Same scenario, but this time Bob uses a rescaled version of the chart Alice uses: one that doubles both lengths and times. This time, Bob measures the longer coordinate distance between the two events, and the longer coordinate time between the events.


Here's another fun one. This time, Bob gets on the rocket, and uses the coordinate chart he originally used. This time, I will involve four events involved:
1. The rocket's departure
2. The rocket's arrival
3. Earth, at the time simultaneous with (2), as measured by Alice's chart
4. Earth, at the time simultaneous with (2), as measured by Bob's chart
Everyone can compute that:
(A) Alice's aging between (1) and (3) is more than Bob's aging between (1) and (2).
(B) Alice's aging between (1) and (4) is less than Bob's aging between (1) and (2).
This is a common one-way travel setup... how would you like to treat it?

[Al68]

The acceleration itself has absolutely no effect. What has an effect is choosing a turning point which is at rest relative to one of the twins and it this which causes the symmetry break - not the acceleration.

cheers,

neopolitan
neopolitan, you and I seem to be the only people around who see this obvious source of asymmetry in the twins paradox. The only asymmetry that cannot be eliminated simply by a minor change in the scenario. The only asymmetry that actually comes into play in the SR equations. The only asymmetry that can be used to show the ship's twin to age less just by doing the math and not making claims that are not even mentioned in SR.

Al

[Al68]

When people seek to resolve the twin paradox, they seek to point out a flaw in the logical argument in the twin paradox. If you are talking about something that isn't the twin paradox (e.g. any experiment involving twins that don't reunite), then that is something irrelevant.


In a one-way journey, it is impossible for both twins to be present at both events. And there is no intrinsic way to compare their ages.

Just have a third observer at the second event. Easy. And he's at rest with earth, so even clock synch is easy.

And an experiment involving twins who don't reunite is relevant to my point. Which is that a different experiment could yield a similar result even without acceleration involved.

Al

[Al68]

Events don't travel; they're events. This doesn't make sense.I never mentioned "traveling events".

Similarly, you mentioned 'elapsed time'; elapsed time of what?
a clock.


You do realize that, in any experimental setup, by choosing the appropriate coordinate chart, I can make either twin (properly!) compute any value I want for any coordinate-dependent quantity I want, right?

Yes, but I have no interest in getting that far off track.

Al

[Hurkyl]

Just have a third observer at the second event. Easy. And he's at rest with earth, so even clock synch is easy.
Being at the second event doesn't mean he's at rest with Earth. To wit, the spacebound twin is at the second event, and he's not at rest with Earth.

I assume your plan is to synchronize the wristwatches of the Earthbound twin and the third observer according to the Einstein convention, and rather than compare the the aging of the two twins (which is not a well-defined question), you intend to compare the aging of the spacebound twin with the difference between the third observer's age at the second event with the Earthbound twin's age at the first event.

But why do it that way? Why not have the third observer at rest with the spacebound twin, and arrange things so that the third observer's clock when he meets the Earthbound twin reads the same as the spacebound twin's clock when he arrives at the second event? Or why not use a pair of observers at rest with respect to each other, but not relative to either of the twins?


And an experiment involving twins who don't reunite is relevant to my point.
But your point is not the twin paradox -- you have no business criticizing resolutions of the twin paradox on the grounds that they are not relevant to your alternative scenario.

[MeJennifer]

The "spacetime path" travelled by an object1 is time-like; the notion of 'distance' doesn't make sense. The 'duration' of the path, however, is exactly what the observer's wristwatch measures.

I completely disagree. In spacetime there is a clear definition of the distance between two events. In spacetime there are no durations only distances between events.

[Hurkyl]

In spacetime there is a clear definition of the distance between two events.
Only for space-like separated events.

In spacetime there are no durations only distances between events.
There are durations for time-like separated events.

When (c \Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 is positive, the events are time-like separated, and the (proper) duration between the two events is (1/c) \sqrt{(c \Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2}. When it's negative, the events are space-like separated, and the (proper) distance between the two events is \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2 - (c \Delta t)^2}

[Al68]

Hello Al68

Quote:-

---- agree. I never said otherwise. I said: "Whichever twin measures the most distance between events will have the most elapsed time between those events."----

As Hurkyl pointed out to us both, distance is the wrong word.
I used the word distance so that someone wouldn't mistakingly think I meant spacetime path. I mean spacial distance.
However your quote is still wrong. It should read " Whichever twin measures the most 'distance' between events will have the least elapsed time between those events"

No, I'm pretty sure you've got it backward. In the twins paradox, the ship's twin measures less distance traveled than the earth twin, hence less time elapsed.

Al

[matheinste]

Hello Al68.

If i have got it back to front ( i am sure i have not ) no-one else has picked up on it.

Matheinste

[Al68]

Being at the second event doesn't mean he's at rest with Earth. To wit, the spacebound twin is at the second event, and he's not at rest with Earth.

OK. I meant an observer at rest with earth.


But your point is not the twin paradox -- you have no business criticizing resolutions of the twin paradox on the grounds that they are not relevant to your alternative scenario.

Huh? I think you missed my point completely. It is easy to construct a scenario like the twins paradox, but where there is no acceleration, yet with the same result. My point is not that the answer given in the common resolutions is wrong, but that the claim of acceleration being important is unsubstantiated. Because you could have the same result if there were no acceleration involved.

Al

[MeJennifer]

Only for space-like separated events.


There are durations for time-like separated events.

When (c \Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 is positive, the events are time-like separated, and the (proper) duration between the two events is (1/c) \sqrt{(c \Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2}. When it's negative, the events are space-like separated, and the (proper) distance between the two events is \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2 - (c \Delta t)^2}
The metric of spacetime defines the distance between two events. This metric is well defined (ignoring for the moment if this space is actually Hausdorff) for both flat (Minkowski) and curved (Lorentzian) spacetimes.

http://en.wikipedia.org/wiki/Metric_(mathematics)

[Al68]

Hello Al68.

If i have got it back to front ( i am sure i have not ) no-one else has picked up on it.

Matheinste

Well, your statement would have the earth twin younger than the ship's twin when they reunite.

Al

[Hurkyl]

but that the claim of acceleration being important is unsubstantiated.
Acceleration is important because it precisely demonstrates the flawed reasoning -- one twin accelerates, is not stationary in any coordinate chart, and thus the time dilation argument that the Earthbound twin should age less is invalid. If you do not bring up acceleration (or something equivalent to it), then you cannot invalidate the twin paradox.

[matheinste]

Hello Al68.

I think you are wrong. However i would rather let someone decide this for us rather than continue disagreeing.

It's 4AM here so i will be disappearg soon.

Matheinste.

[Hurkyl]

The metric of spacetime defines the distance between two events. This metric is well defined (ignoring for the moment if this space is actually Hausdorff) for both flat (Minkowski) and curved (Lorentzian) spacetimes.

http://en.wikipedia.org/wiki/Metric_(mathematics)
First off, you've cited the definition of the wrong word: see Metric tensor (http://en.wikipedia.org/wiki/Metric_tensor#Lorentzian_metrics_from_relativity). Secondly, see proper length (http://en.wikipedia.org/wiki/Proper_length) and proper time (http://en.wikipedia.org/wiki/Proper_time#In_special_relativity).

[MeJennifer]

Every serious book on relativity writes about distances between events in spacetime.

What I write seems to fall on deaf ears, so I won't respond anymore to this.

[Al68]

Hello Al68.

I think you are wrong. However i would rather let someone decide this for us rather than continue disagreeing.

It's 4AM here so i will be disappearg soon.

Matheinste.

I don't understand why there's a disagreement. In the twins paradox, the ship's twin measures the distance traveled to be less than that measured by the earth twin. If I'm right, that would mean that the ship's twin would age less. I thought we all agreed on which twin aged less. I thought the only disagreement was why.

Al

[matheinste]

Hello Al68.

I have no disagreement about which twin ages less. I only disagree about the spactime distance. Tomorrow i will consult my textbooks and then can hopefully quote some relevant passages.

Matheinste.

[Al68]

Hello Al68.

I have no disagreement about which twin ages less. I only disagree about the spactime distance. Tomorrow i will consult my textbooks and then can hopefully quote some relevant passages.

Matheinste.

Well, there's the problem, I was refering to spacial distance, not spacetime length.

Al