Twin "paradox" depends on timing, not proper acceleration -- NEVERMIND, I must be mistaken

[md65536]

Edit: I'm double-checking the math and I think I've made an error somewhere here...
----

Here's an outline of a thought experiment that shows that the effects of the twin paradox are not determined by who experiences proper acceleration.


Start with a basic twin setup:
Observer E stays on Earth,
observer R is on a rocket that passes Earth at time 0, moves at a velocity of [math]v[/math] for a proper time of t, then turns around and returns at a rate of [math]-v[/math] for a time of t.
R then has aged 2t while E has aged gamma*(2t), where gamma is the Lorentz factor.

Only R experiences proper acceleration.

Now modify the experiment so that R doesn't turn around.
Instead, R keeps moving inertially for the full 2*t, but at time t it sees E instantaneously accelerate and approach at a relative closing velocity of [math]-v[/math]. E catches up to R at time 2t.

In this version, only E has experienced proper acceleration during the experiment, but it still ages at a rate of gamma relative to R. What allows this is that the proper times of the two phases of the experiment aren't equal to each other for the accelerating twin, as it is in the usual twin setup.



So, I haven't actually analysed this to show it's true. Is it intuitive enough to be certainly true?

 

Edit: So I decided I'd better do the math and it's not working out, I think I'm forgetting about relative simultaneity or I'm using the wrong velocity... :S

----------------

 

I think I Gerry'd this up completely.

Where's my (many) error???

 

I think I forgot to consider that in the second experiment, in E's frame the "last clock tick before instant acceleration" and "first clock tick after accel" are essentially simultaneous, but they're not simultaneous in R's frame.

 

If R remains at rest I don't think there's any way to make E's clock tick relatively faster just with velocity, so I think the hypothesis is WRONG.

 

-----------------

 

So I think the main mistake is thinking that if R sees E change its relative velocity halfway (for R) through the experiment, that E would take the same amount of time (in R's frame) to appear to catch up as it did to recede. In reality, E would be seen to accelerate after very little Earth time has passed, and it would appear to catch up in I think the same amount of E time (not Earth time, now!) appears to pass, which would happen in very very little R time. So R would see E aging less than itself in this experiment.

 

Mythbusted!

 

Sorry if anyone feels they wasted their time

Edited March 22, 2013 by md65536

[michel123456]

Edit: I'm double-checking the math and I think I've made an error somewhere here...

----

 

Here's an outline of a thought experiment that shows that the effects of the twin paradox are not determined by who experiences proper acceleration.

 

 

Start with a basic twin setup:

Observer E stays on Earth,

observer R is on a rocket that passes Earth at time 0, moves at a velocity of [math]v[/math] for a proper time of t, then turns around and returns at a rate of [math]-v[/math] for a time of t.

R then has aged 2t while E has aged gamma*(2t), where gamma is the Lorentz factor.

 

Only R experiences proper acceleration.

 

Now modify the experiment so that R doesn't turn around.

Instead, R keeps moving inertially for the full 2*t, but at time t it sees E instantaneously accelerate and approach at a relative closing velocity of [math]-v[/math]. E catches up to R at time 2t.

 

In this version, only E has experienced proper acceleration during the experiment, but it still ages at a rate of gamma relative to R. What allows this is that the proper times of the two phases of the experiment aren't equal to each other for the accelerating twin, as it is in the usual twin setup.

 

 

 

So, I haven't actually analysed this to show it's true. Is it intuitive enough to be certainly true?

 

Edit: So I decided I'd better do the math and it's not working out, I think I'm forgetting about relative simultaneity or I'm using the wrong velocity... :S

 

----------------

 

I think I Gerry'd this up completely.

Where's my (many) error???

 

I think I forgot to consider that in the second experiment, in E's frame the "last clock tick before instant acceleration" and "first clock tick after accel" are essentially simultaneous, but they're not simultaneous in R's frame.

 

If R remains at rest I don't think there's any way to make E's clock tick relatively faster just with velocity, so I think the hypothesis is WRONG.

 

 

-----------------

 

So I think the main mistake is thinking that if R sees E change its relative velocity halfway (for R) through the experiment, that E would take the same amount of time (in R's frame) to appear to catch up as it did to recede. In reality, E would be seen to accelerate after very little Earth time has passed, and it would appear to catch up in I think the same amount of E time (not Earth time, now!) appears to pass, which would happen in very very little R time. So R would see E aging less than itself in this experiment.

 

Mythbusted!

 

Sorry if anyone feels they wasted their time

It's a mess.

But a good idea. I am sure the correct result must show that both E and R have aged the same time. There is no reason to search for a paradox.

 

You said:

Now modify the experiment so that R doesn't turn around.

 

That's my point too.

In the twin "paradox" all the maths I have seen so far say nothing about the u-turn.

 

The result of the maths show what happen when the observer E on Earth is separated from R by a huge distance.

 

In this case there is no paradox: each one will observe the other with a younger age.

 

The "paradox" arises when someone ply the thought experiment in 2 like a sheet of paper, making E and R coincide in space.

IMHO the plying of the sheet of paper (read the "u-turn") must have some mathematical implications.

Edited March 22, 2013 by michel123456

[md65536]

I am sure the correct result must show that both E and R have aged the same time.

[...]

The result of the maths show what happen when the observer E on Earth is separated from R by a huge distance.

 

In this case there is no paradox: each one will observe the other with a younger age.

 

The "paradox" arises when someone ply the thought experiment in 2 like a sheet of paper, making E and R coincide in space.

IMHO the plying of the sheet of paper (read the "u-turn") must have some mathematical implications.

It turns out (as should have been expected) that if R remains at rest it ages LESS than E does.

The way that I was getting R to age more, and it would be the same problem with getting E and R to age the same amount, is to try to force it so that "E spends t seconds moving away from R at a velocity of v, and then spends 14t seconds (for example when gamma=2) moving toward R at a velocity of -v" which is not possible in Euclidean geometry.

 

 

The separation of a huge distance is important, and I'll try to explain why. The u-turn itself is not that important. For example, in the case where only R is accelerating: suppose instead of a u-turn that R decided to make a huge arc around E so that the turn was more of a triangle, or even a semicircle. That would take some time, and there could be relativistic effects to calculate, but it would not change the physics of separation or return trips. Or, say that R got 100 LY away, and then accelerated a thousand times back and forth in 1 second. There are a thousand HUGE changes in relative simultaneity with E, but they cancel each other out and in the end this addition only adds 1second*gamma to E's aging in the whole equation.

 

The equations depend on v and t (and d)... what's important is how long an observer spends traveling at v. But v*t = d, so I might be tempted to say "Really it's all about distance!" And yes, distance is important because you can't spend a long time traveling at high velocity without an accumulation of distance. But why it's WRONG to think "it's all about distance" is that there are different ways to get to the same distance... different paths to take and different velocities, and so different proper times or even two paths with the same proper time at R but that will have different relative aging at E. If you know d and t at every step, you can work it out (as v can also be determined). If you know a and t at every step, you can work it out.

 

But how you accelerate can be neglected. If you can accelerate from v to -v in 1 second, or if you can accelerate in one millionth of a second, the difference in aging of the other observer will differ by less than gamma*1 second. If you must take a thousand years to do a u-turn, you can take a snapshot of the experiment at the start of the u-turn, and a thousand years later, and (assuming the u-turn starts and ends at the same point) you can remove the aging of R and of E during the u-turn and it will work out the same as if there was an instantaneous u-turn. The effects of acceleration only matter during the u-turn, so you can choose to neglect that time with negligible effect on the experiment.

Edited March 22, 2013 by md65536

[michel123456]

That is insane.

 

Take E and R again.

make R go away as you did.

Then after a year, make E go away.

 

R and E are 1 year from each other.

 

Then make R stop and wait for the arrival of E.

 

When they meet, is there any difference in age? I think not. There is no paradox.

 

The velocities, the accelerations do not matter: when they meet together they MUST observe the same thing, the same age. Any other result is wrong.

 

I think the misunderstanding is caused by the direction of the velocity vector. When someone goes away the vector is positive, when he comes back the vector is negative. Even if speed remains the same.

 

Isn't that the difference between velocity and speed?

Edited March 22, 2013 by michel123456

[swansont]

The velocities, the accelerations do not matter: when they meet together they MUST observe the same thing, the same age. Any other result is wrong.

 

No, if the speeds are not the same you don't expect the time to be the same. Even for a constant distance the time dilation is still a function of v.

 

 

But how you accelerate can be neglected.

 

If the time spent accelerating is small compared to the time of the trip. If it's not, then it becomes part of the path you have to evaluate.

 

In many ways, time dilation might be more easily understood as frequency dilation — the fractional frequency of any clock changes by gamma, which depends on speed. To find the phase (time) shift you integrate over the time of the trip. The final result thus depends on both speed and elapsed time, which is why it can be seen to depend on distance.

 

That the effect is integrated over the trip is key no matter how you approach it.

[michel123456]

No, if the speeds are not the same you don't expect the time to be the same. Even for a constant distance the time dilation is still a function of v.

Even under Newtonian dynamics, when the speed is lower, you take more time to make the travel. The point is that when they meet, they should observe the same thing. After all, it is fundamental condition of relativity, all observers must agree on what they observe, even when they are together at the same time and the same place.

 

i believe people are too much concerned with the twin paradox and find paradoxes everywhere.

 

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I mean, if the 2 observers are together at the beginning and at the end, it means that their average velocity is the same. Otherwise they don't meet.

Edited March 22, 2013 by michel123456

[ACG52]

After all, it is fundamental condition of relativity, all observers must agree on what they observe

 

Observers in different frames will NOT agree on what they observe. That's what relativity tells you.

[md65536]

The point is that when they meet, they should observe the same thing. After all, it is fundamental condition of relativity, all observers must agree on what they observe, even when they are together at the same time and the same place.

"Everyone must agree" means everyone agrees that when R and E come together, E has aged 4 years and R has aged 2, as an example. It doesn't mean that everyone aged the same amount.

 

Essentially it means, "Everyone may measure things differently, but everyone agrees on what anyone else measures." All the different measurements agree.

 

Twin paradox is focused on because it's a very simple situation with a clear classically unintuitive effect.

Take E and R again.

make R go away as you did.

Then after a year, make E go away.

 

R and E are 1 year from each other.

 

Then make R stop and wait for the arrival of E.

 

When they meet, is there any difference in age? I think not. There is no paradox.

 

The velocities, the accelerations do not matter: when they meet together they MUST observe the same thing, the same age. Any other result is wrong.

Oh wait... I reread this and I see what you're saying.

Yes, in this version when they meet they'll have aged the same amount.

In this version both E and R make symmetrical movements, according to an observer on Earth for example, just at different times. In this version both observers experience the same proper acceleration.

 

Edit: But the velocities DO matter. In this version they age the same amount specifically because you made them travel at the same velocities for the same proper times.

 

Also, the essential "paradox" is still there: With time dilation, each twin measures that the other is aging slower while moving, so how can they end up at the same age? The answer is "because of length contraction and relative simultaneity".

Edited March 22, 2013 by md65536

[swansont]

Even under Newtonian dynamics, when the speed is lower, you take more time to make the travel.

 

But the physics involved is known to be wrong, so this doesn't matter.

 

I mean, if the 2 observers are together at the beginning and at the end, it means that their average velocity is the same. Otherwise they don't meet.

 

Time dilation is not a (linear) function of average speed.

[michel123456]

Observers in different frames will NOT agree on what they observe. That's what relativity tells you.

we are not talking about observers in different FOR. when they meet together they are in the same FOR.

 

You know what the paradox is about, no?

 

the paradox is NOT that the one has another age with the other.

the paradox is that both should observe the other having aged less.

[ACG52]

the paradox is that both should observe the other having aged less.

 

 

But to compare their ages they must be brought together in the same FOR. In order for that to happen, one twin, the traveling twin, must undergo non inertial motion, and it's that which breaks the symmetry between them. that's why there's no paradox.

[md65536]

But to compare their ages they must be brought together in the same FOR. In order for that to happen, one twin, the traveling twin, must undergo non inertial motion, and it's that which breaks the symmetry between them. that's why there's no paradox.

No, they just need to be brought together. If they pass and touch, even if they have relative velocity, everyone will agree on the "touch event"; everyone will agree on the two observer's proper times (age) of the event.

 

Edit: There is only one *essential* change of inertial frame in the twin paradox, and that's the one that happens when they're separated.

 

Anyway, I think we're all now talking about slightly different setups. Not all of us are talking about the standard twin paradox. I think we all agree there's no physical paradox here.

Edited March 22, 2013 by md65536

[Delta1212]

It's less a "Twin paradox" than a "Twin counter-intuitive implication of relativity" but that's not as snappy a title.

[DimaMazin]

Clock,which travels relatively of massive bodies,is slower.Only wrong theory can have paradox.

Edited April 2, 2013 by DimaMazin

[Iggy]

So I think the main mistake is thinking that if R sees E change its relative velocity halfway (for R) through the experiment, that E would take the same amount of time (in R's frame) to appear to catch up as it did to recede. In reality, E would be seen to accelerate after very little Earth time has passed, and it would appear to catch up in I think the same amount of E time (not Earth time, now!) appears to pass, which would happen in very very little R time. So R would see E aging less than itself in this experiment.

 

Have you considered the shift in the plane of simultaneity?

 

When E jumps to catch up to R, the clocks on R (from E's perspective) skip ahead. They wouldn't see them skip ahead, but they would calculate while they're on earth that the clock on R at the present instant is one thing, then after they jump to catch up they would calculate that the clock on R at the present instant is suddenly quite a bit further along.

 

The shift in simultaneity would work just like the normal twin paradox. Let me quote wiki or something because I don't think I'm explaining well at all...

 

 

Resolution of the paradox in special relativity

 

<snip>

 

There are indeed not two but three relevant inertial frames: the one in which the Earth-based twin remains at rest, the one in which the traveling twin is at rest on his outward trip, and the one in which he is at rest on his way home. It is during the acceleration at the U-turn that the traveling twin switches frames. That is when he must adjust his calculated age relative to the Earth-based twin.

 

In special relativity there is no concept of absolute present. The present from the point of view of a given observer is defined as the set of events that are simultaneous for that observer. The notion of simultaneity depends on the frame of reference (see relativity of simultaneity), so switching between frames requires an adjustment in the definition of the present. If one imagines a present as a (three-dimensional) simultaneity plane in Minkowski space, then switching frames results in changing the inclination of the plane.

 

 

In the spacetime diagram on the right, drawn for the reference frame of the Earth-based twin, that twin's world line coincides with the vertical axis (his position is constant in space, moving only in time). On the first leg of the trip, the second twin moves to the right (black sloped line); and on the second leg, back to the left. Blue lines show the planes of simultaneity for the traveling twin during the first leg of the journey; red lines, during the second leg. Just before turnaround, the traveling twin calculates the age of the Earth-based twin by measuring the interval along the vertical axis from the origin to the upper blue line. Just after turnaround, if he recalculates, he'll measure the interval from the origin to the lower red line. In a sense, during the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the Earth-based twin. The traveling twin reckons that there has been a jump discontinuity in the age of the Earth-based twin.

 

When E jumps to catch up to R in your version you'd get the same thing. E's plane of simultaneity would instantly sweep over a large part of R's world line.

 

So, you couldn't... if you're calculating the proper time of R from E's perspective you can't just say that R first has one relative velocity then another and add the two time dilated results together. You also have to add a third bit of time corresponding to the shift in simultaneity.

Edited April 2, 2013 by Iggy

[md65536]

So, you couldn't... if you're calculating the proper time of R from E's perspective you can't just say that R first has one relative velocity then another and add the two time dilated results together. You also have to add a third bit of time corresponding to the shift in simultaneity.

Yes, I certainly didn't consider all of the details.

 

When I first wrote this, it was because I was wondering what would happen if you basically put an entire twin paradox experiment in a box and then accelerated the box to counteract one of the observer's acceleration. Then I drew some conclusions and posted without thinking it through fully.

 

I think in the most basic case, you simply couldn't accelerate everything in a box at the same time according to all observers, and the results of the experiment would be different than without accelerating the box.

 

Perhaps frame-dragging could be exploited to accelerate an experiment in a box and get identical results to the original experiment??? But then anyone experiencing proper acceleration in the original experiment would experience it in the modified experiment too.

[Iggy]

Yes, I certainly didn't consider all of the details.

 

I read your OP closer and noticed... It looks like we were on the same page when you said "I forgot to consider that in the second experiment, in E's frame the 'last clock tick before instant acceleration' and "first clock tick after accel' are essentially simultaneous, but they're not simultaneous in R's frame."

 

You had it. I was playing catch up

 

When I first wrote this, it was because I was wondering what would happen if you basically put an entire twin paradox experiment in a box and then accelerated the box to counteract one of the observer's acceleration. Then I drew some conclusions and posted without thinking it through fully.

 

I think in the most basic case, you simply couldn't accelerate everything in a box at the same time according to all observers, and the results of the experiment would be different than without accelerating the box.

 

Perhaps frame-dragging could be exploited to accelerate an experiment in a box and get identical results to the original experiment??? But then anyone experiencing proper acceleration in the original experiment would experience it in the modified experiment too.

 

I think you're right. If the box doesn't drag everything inside it then everything would maybe have a different coordinate acceleration from the box's perspective but their proper acceleration would be the same.

 

 

The frame dragging... yeah... Einstein did consider the twin paradox in those terms... Here it is. The twin paradox part starts down near the diagram. If I remember right, it was translated from German and written as a dialog between two people, so hard to follow, but it is quite near what you're talking about.

[md65536]

I think you're right. If the box doesn't drag everything inside it then everything would maybe have a different coordinate acceleration from the box's perspective but their proper acceleration would be the same.

You could rig it so that the box does drag everything. For example, if you're using rockets for acceleration, you could attach additional rockets and time them to implement the action of the box. (Then, another mistake I made is to consider moving the box using Newtonian physics, but consider the contents using SR.)

Einstein did consider the twin paradox in those terms... Here it is.

Thanks for the link. It's interesting to see how Einstein is able to think about all of this in such abstract ways, which I've never been able to do. I suppose that comes from understanding the the maths and their meaning.

[md65536]

Just because I have OCD, here's how I think that the twin 'E' can leave at velocity 0.866c, and appear to return at velocity -0.866c, and trick twin 'R' (who remains at rest) into thinking that E has aged twice as much as R (the original hypothesis of the thread):

 

After a time of 1 year at R, E has appeared to age 0.268 years, and appears to have reached a rest (R's frame) distance of 0.232 LY. If at that moment E disappears (turned off its signal), and an imposter E' at a rest distance of 3.232 LY appears, timed to seem simultaneous to R, then E' can travel at velocity -0.866c for a time of 3.732 years, appearing to age that much while R ages 1 year. So E and the imposter E' are seen aging a total of 4 years while R ages 2.

 

 

R has been tricked, as I have, by not realizing that E' had to travel so much more on the inbound trip than E traveled on the outbound trip. The trick might work if R is unable to distinguish between the two different rest distances, and is dum.

Edited April 4, 2013 by md65536

[phyti]

In the simple 'twin pardox' triangle, where the twins (or clocks) separate,

the one that returns has the least amount of accumulated time.

 

 

[LaurieAG]

Here's what happens to the paradox when you forget about relative to the speed of light in a vacuum timers and use solid state timers.

 

The attached image shows two digital counters that would show the same count regardless of how fast the glass plate was spinning. Each count is created by connecting an electric current between A and B that increments the equidistant counters, the wires run under the glass plate and each wire is permanently connected to the central spindle by frictionless brushes, or whatever, on different sides of the plate.

 

Can anybody prove how the timers can be different despite their relative speed difference.

[LaurieAG]

Additionally if you send one signal for each rotation of the glass plate any difference between the calculated electron speeds going one way and the other will cancel each other out (i.e. same as MM) and the timer counts will remain the same.

 

This was originally in response to a similar paradox with regards to a relativistic train that would only stop if optical signals were sent from the front and the back and both signals arrived at the center simultaneously. Move up in scale and the timer on the glass plate is in your relativistic (stationary) train on a huge circular (moving) track. This is similar to using galactic year scales (i.e. 1 galactic rotation) for non relativistic comparisons of cosmological observations of photons emitted from sources rotating around consistent galactic centers.