Twin Paradox (by acceleration)

[Jonnyb42]

Ok so, supposedly you have two twins in [seperate] spaceships, one approaches the speed of light going away from his twin, then decelerates until going the same high speed but opposite direction, decelerates again to stop next to his twin buddy. The twin that did not move (relative to the very initial reference frame) is older than the twin who approached speed of light.

By symmetry, this time dilation MUST have occurred during the acceleration of the twin that moved because any time not during the acceleration, each saw the other moving at near the speed of light.

The twins start off at the same age (approximately... ) and end up with very different ages. Therefor the time dilation must have occurred during the only asymmetry, which is when one twin felt an 'inertial drag' so to speak. (and the other twin never felt this)

Why is the time dilation a function of velocity then? Shouldn't it have to do with acceleration?

 

[JesseM]

I always find it helpful to think in terms of the analogy to 2D geometry. If you have two paths through a 2D plane that cross at two points A and B, the straight-line path between A and B will always have a greater distance than a path that has a bend in it. Suppose the bent path consists of two straight segments at different angles joined by a short curved portion, like the letter V but with a slightly rounded bottom. Would you say that "by symmetry", the excess distance of the bent path (as compared to the straight path) must have occurred during the short non-straight segment? Obviously this isn't the case, since the curved portion can be made very short, and the sum of the lengths of the two straight segments of the bent path can be greater than the length of the straight-line path between A and B.

The analogy between "proper time along paths through spacetime" and "distance along paths through space" can actually be made fairly precise mathematically, see this post for some details...

 

[yuiop]

Ok so, supposedly you have two twins in [seperate] spaceships, one approaches the speed of light going away from his twin, then decelerates until going the same high speed but opposite direction, decelerates again to stop next to his twin buddy. The twin that did not move (relative to the very initial reference frame) is older than the twin who approached speed of light.

By symmetry, this time dilation MUST have occurred during the acceleration of the twin that moved because any time not during the acceleration, each saw the other moving at near the speed of light.

The twins start off at the same age (approximately... ) and end up with very different ages. Therefor the time dilation must have occurred during the only asymmetry, which is when one twin felt an 'inertial drag' so to speak. (and the other twin never felt this)

Why is the time dilation a function of velocity then? Shouldn't it have to do with acceleration?

Consider this triplet example. Triplet A stays home. Triplet B and triplet C press their launch buttons together and accelerate equally to 0.9c. After a while B presses his return home button which automatically slows down the rocket and then boosts it back towards A in a controlled manner. Sometime later C presses his return home button and goes through exactly the same acceleration pattern and heads back home. Both go through exactly the same deceleration pattern to come to a stop at A. Now B and C have gone through identical acceleration patterns but C has aged less than B. Therefore you can not put the difference in ageing between B and C down to acceleration.

The only way you can make the acceleration idea work, is to say that time dilation is a function of acceleration AND distance. In other words a clock with proper acceleration of 1 g at 1 light year from you time dilates differently to a clock with proper acceleration of 1g but two light years from you. The difficulty with this idea is how does the clock know how far it is from you? Having said that, Einstein made a similar analogy, saying that when the rocket accelerates it is equivalent to a gravitational field springing up and the further away the rocket is, the deeper the rocket is in the effective gravitational well. On the other hand, putting time dilation down to path lengths through spacetime works very nicely (So listen to what JesseM is telling you and look at the post he linked to ).

 

[Passionflower]

By symmetry, this time dilation MUST have occurred during the acceleration of the twin that moved because any time not during the acceleration, each saw the other moving at near the speed of light.

You are mixing up two things.

The difference in proper acceleration between the twins sets the relative clock rates. However the difference in total accumulated time depends on the duration of such a state.

 

[matheinste]

You are mixing up two things.

The difference in proper acceleration between the twins sets the relative clock rates. However the difference in total accumulated time depends on the duration of such a state.

In the standard twin paradox, no matter how you dress it up, the SR explanation of differential ageing depends on the spacetime path of the twins. If you accept the clock hypothesis then acceleration is only indirectly responsible as it is required to complete the traveling twins path. The inertial twin accumulates the most proper time. Some people insist that acceleration is somehow directly responsible, but I share the general view that it is only indirectly responsible. Be that as it may, path length is the key.

Matheinste.

 

[Jonnyb42]

What would happen if the traveling twin was never inertial, and accelerated the whole time, then decelerated and came back accelerated to halfway, then decelerated the rest of the way until at rest back at starting point with other twin?

 

[matheinste]

What would happen if the traveling twin was never inertial, and accelerated the whole time, then decelerated and came back accelerated to halfway, then decelerated the rest of the way until at rest back at starting point with other twin?

Path length determines proper time. In the above scenario the traveller passes through an infinite number of co-moving inertial frames and this fact is represented by the path length being the integral of the path traveled instead of, in the siplest case, the sum of two inertial paths.

Matheinste.

 

[JesseM]

What would happen if the traveling twin was never inertial, and accelerated the whole time, then decelerated and came back accelerated to halfway, then decelerated the rest of the way until at rest back at starting point with other twin?

The constant velocity path between two points in spacetime is always the one with the maximum elapsed aging (proper time), just like the straight-line path between two points in space is always the one with the minimum length. So, in this case the accelerating twin still ages less than the inertial one.

 

[starthaus]

What would happen if the traveling twin was never inertial, and accelerated the whole time, then decelerated and came back accelerated to halfway, then decelerated the rest of the way until at rest back at starting point with other twin?

This is the realistic case, it is treated very well here

 

[yuiop]

What would happen if the traveling twin was never inertial, and accelerated the whole time, then decelerated and came back accelerated to halfway, then decelerated the rest of the way until at rest back at starting point with other twin?

You could break the path up into short segments and calculate the average velocity for each segment and calculate the time dilation for each segment using that average velocity of that segment and add them all up. The more segments you break the path up into, the more accurate the aproximation becomes and this is roughly what is being done, when the integral of the path is taken as suggested by Matheinst. Essentially the time dilation is a function of instantaneous velocity at any given moment and for continuous acceleration you have to sum up all the infinitesimal velocities along the path.

 

[Aaron_Shaw]

Is it not simply the case that the speed difference causes time dilation, but acceleration to reach those speeds causes the dilation to be symmetrical because it shifts simultaneity between the two frames. Then when the moving frame, which we know is the moving one because it's accelerating, decelerates to a stop after completing the journey then the simultaneity is shifted back so that the two frames now experience events truly simultaneously according to a third observer. This shifting of simultaneity back exposes the fact that (in the context of a complete round trip) the moving frame has actually experienced the time dilation ?

 

[Demystifier]

Why is the time dilation a function of velocity then? Shouldn't it have to do with acceleration?

Good question!
The answer is presented here
http://xxx.lanl.gov/abs/physics/0004024 [Found.Phys.Lett. 13 (2000) 595]

 

[matheinste]

Is it not simply the case that the speed difference causes time dilation, but acceleration to reach those speeds causes the dilation to be symmetrical because it shifts simultaneity between the two frames. Then when the moving frame, which we know is the moving one because it's accelerating, decelerates to a stop after completing the journey then the simultaneity is shifted back so that the two frames now experience events truly simultaneously according to a third observer. This shifting of simultaneity back exposes the fact that (in the context of a complete round trip) the moving frame has actually experienced the time dilation ?

Time dilation is reciprocal.

Again all this may be true (or not) but it is just window dressing. Proper time difference due to path length difference is all you need to know or calculate.

Matheinste.

 

[curiousphoton]

I always find it helpful to think in terms of the analogy to 2D geometry. If you have two paths through a 2D plane that cross at two points A and B, the straight-line path between A and B will always have a greater distance than a path that has a bend in it.

Speaking in terms of 2-D geometry, wouldn't the straight-line path between two points, A and B, always have a smaller distance than a bent path between A and B.

 

[Fredrik]

Speaking in terms of 2-D geometry, wouldn't the straight-line path between two points, A and B, always have a smaller distance than a bent path between A and B.

I assume that we're talking about world lines in 1+1-dimensional spacetime. In that case, the statement you're looking for is that the straight line has the longest proper time. (Essentially it's because it's calculated by adding up contributions of the form $$\sqrt{dt^2-dx^2}$$ along the world line, and dx=0 for the straight path).

 

[JesseM]

Speaking in terms of 2-D geometry, wouldn't the straight-line path between two points, A and B, always have a smaller distance than a bent path between A and B.

Yes, note what I said in post #8:

The constant velocity path between two points in spacetime is always the one with the maximum elapsed aging (proper time), just like the straight-line path between two points in space is always the one with the minimum length.

This difference is, as Fredrik noted, because the formula for proper time along paths through spacetime contains a minus where the pythagorean formula for distance along paths through space contains a plus. I talked more about this in post #114 on this thread.

 

[Mike_Fontenot]

During the constant-speed legs of the trip, BOTH twins conclude that the other twin is ageing slower. But when the trip is over, they both agree that the stay-at-home twin is older. How is that possible?

It's possible because, during the turnaround, the traveler will conclude that the home twin quickly ages, with very little ageing of the traveler. The home twin concludes that neither of them ages much during the turnaround. When you add up all these segments of ageing, you get the result that the home twin is older (and both twins exactly agree on that).

Years ago, I derived a simple equation (called the "CADO" equation) that explicitly gives the ageing of the home twin during accelerations by the traveler (according to the traveler). The equation is especially easy to use for idealized traveling twin problems with instantaneous speed changes. But it also works for finite accelerations. I've got a detailed example with +-1g accelerations on my webpage:

http://home.comcast.net/~mlfasf

And I've published a paper giving the derivation of the CADO equation:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

 

[Mike_Fontenot]

Here's a brief description of my "CADO" equation:
____________________________________________________

Years ago, I derived a simple equation that relates the current ages of the twins, ACCORDING TO EACH TWIN. Over the years, I have found it to be very useful. To save writing, I write "the current age of a distant object", where the "distant object" is the stay-at-home twin, as the "CADO". The CADO has a value for each age t of the traveling twin, written CADO(t). The traveler and the stay-at-home twin come to DIFFERENT conclusions about CADO(t), at any given age t of the traveler. Denote the traveler's conclusion as CADO_T(t), and the stay-at-home twin's conclusion as CADO_H(t). (And in both cases, remember that CADO(t) is the age of the home twin, and t is the age of the traveler).

My simple equation says that

CADO_T(t) = CADO_H(t) - L*v/(c*c),

where

L is their current distance apart, in lightyears,
according to the home twin,

and

v is their current relative speed, in lightyears/year,
according to the home twin. v is positive
when the twins are moving apart.

(Although the dependence is not shown explicitely in the above equation, the quantities L and v are themselves functions of t, the age of the traveler).

The factor (c*c) has value 1 for these units, and is needed only to make the dimensionality correct.

The equation explicitly shows how the home twin's age will change abruptly (according to the traveler, not the home twin), whenever the relative speed changes abruptly.

For example, suppose the home twin believes that she is 40 when the traveler is 20, immediately before he turns around. So CADO_H(20-) = 40. (Denote his age immediately before the turnaround as t = 20-, and immediately after the turnaround as t = 20+.)

Suppose they are 30 ly apart (according to the home twin), and that their relative speed is +0.9 ly/y (i.e., 0.9c), when the traveler's age is 20-. Then the traveler will conclude that the home twin is

CADO_T(20-) = 40 - 0.9*30 = 13

years old immediately before his turnaround. Immediately after his turnaround (assumed here to occur in zero time), their relative speed is -0.9 ly/y. The home twin concludes that their distance apart doesn't change during the turnaround: it's still 30 ly. And the home twin concludes that neither of them ages during the turnaround, so that CADO_H(20+) is still 40.

But according to the traveler,

CADO_T(20+) = 40 - (-0.9)*30 = 67,

so he concludes that his twin ages 54 years during his instantaneous turnaround.

The equation works for arbitrary accelerations, not just the idealized instantaneous speed change assumed above. I've got an example with +-1g accelerations on my web page:

http://home.comcast.net/~mlfasf

The derivation of the equation is given in my paper

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

Mike Fontenot

 

[starthaus]

Good question!
The answer is presented here
http://xxx.lanl.gov/abs/physics/0004024 [Found.Phys.Lett. 13 (2000) 595]

This is an excellent paper. Should be made part of the textbooks dealing with the "paradox".

 

[Passionflower]

During the constant-speed legs of the trip, BOTH twins conclude that the other twin is ageing slower. But when the trip is over, they both agree that the stay-at-home twin is older. How is that possible?

You are mixing up two things.

It is true that when both twins are in relative motion it is not determinable whose clock is going slower. But just because something is not determinable does not imply that there are no real physical things happening.

When two locations A and B are 1 light year removed and a traveler, who undergoes a proper acceleration away from A, goes near light speed from A to B then it is true that during travel it cannot be determined which clock A or B is running slower, however after the traveler reached B it is easy to establish which clock in fact went slower. All B has to do is to send a message to A how much his chronometer has advanced during his trip from A to B. From this information is it easy for A to conclude that it was in fact the traveler's clock that ran slower.

No paradox.

 

[matheinste]

You are mixing up two things.

It is true that when both twins are in relative motion it is not determinable whose clock is going slower. But just because something is not determinable does not imply that there are no real physical things happening.

When two locations A and B are 1 light year removed and a traveler, who undergoes a proper acceleration away from A, goes near light speed from A to B then it is true that during travel it cannot be determined which clock A or B is running slower, however after the traveler reached B it is easy to establish which clock in fact went slower. All B has to do is to send a message to A how much his chronometer has advanced during his trip from A to B. From this information is it easy for A to conclude that it was in fact the traveler's clock that ran slower.

No paradox.

The wording of the description of the scenario is a bit ambiguous and so perhaps I am misunderstanding what you are saying.

Matheinste.

 

[Demystifier]

This is an excellent paper. Should be made part of the textbooks dealing with the "paradox".

I am glad to see that someone thinks so.

 

[starthaus]

I am glad to see that someone thinks so.

You are the "go-to" man when it comes to accelerated frames. I have also used extensively your other excellent paper, the one dealing with rotating frames.

 

[Jonnyb42]

I have also used extensively your other excellent paper, the one dealing with rotating frames.

Hey whoa what's going on here? It seems to me like Demystifier could save my thread https://www.physicsforums.com/showthread.php?t=404650".

Demystifier, I have a headache over this problem which to me is unsolved yet, could you help me out over there if you feel like it?

 

[yuiop]

Here is a thought experiment similar to the one given by Passionflower with an added twist, that gives a perhaps non-intuitive result. A and B are initially adjacent and at rest wrt each other. B accelerates to 0.99c in a couple of minutes (ignore the practical difficulties) and then cruises for aproximately one year according to A, who up to now has not undergone any acceleration. Now instead of B coming to rest in A's frame, A does a rapid burst of acceleration (lasting a few minutes) in B's direction and comes to rest in B's frame, but they are still far apart.

The slightly surpising result, if I have worked it out correctly (and I might not of, because I am a bit tired) is that A (who has a spent a year lazing around while B hurtles off at 0.99c relative to him) has aged less than B. It seems that the twin that ages the <edit> least </edit>, is the one that does the short final burst of acceleration to come to rest in the other twin's rest frame.

 

[matheinste]

Here is a thought experiment similar to the one given by Passionflower with an added twist, that gives a perhaps non-intuitive result. A and B are initially adjacent and at rest wrt each other. B accelerates to 0.99c in a couple of minutes (ignore the practical difficulties) and then cruises for aproximately one year according to A, who up to now has not undergone any acceleration. Now instead of B coming to rest in A's frame, A does a rapid burst of acceleration (lasting a few minutes) in B's direction and comes to rest in B's frame, but they are still far apart.

The slightly surpising result, if I have worked it out correctly (and I might not of, because I am a bit tired) is that A (who has a spent a year lazing around while B hurtles off at 0.99c relative to him) has aged less than B. It seems that the twin that ages the most, is the one that does the short final burst of acceleration to come to rest in the other twin's rest frame.

As I'm sure you know, a spacetime diagram shows exactly what is going on. Without giving it too much thought I would imagine that you can, by using appropriate figures, set up a similar scenario where they age by the same amount.

As usual its just about path lengths. Spacetime diagrams let you visualize what is going on and calculation confirms it.

Matheinste.

 

[Demystifier]

Hey whoa what's going on here? It seems to me like Demystifier could save my thread https://www.physicsforums.com/showthread.php?t=404650".

Demystifier, I have a headache over this problem which to me is unsolved yet, could you help me out over there if you feel like it?

But that is really easy. It is only Newtonian mechanics, no Einstein relativity is involved. From their point of view, one is rotating around the other (e.g. Moon around Earth or vice versa). They miss each other because the direction of force does not need to coincide with the direction of velocity. I do not see what is your problem with that.

 

[Demystifier]

You are the "go-to" man when it comes to accelerated frames. I have also used extensively your other excellent paper, the one dealing with rotating frames.

Thanks! Have you also published something on that stuff? (arXiv preprints also count as published.)

 

[starthaus]

Thanks! Have you also published something on that stuff? (arXiv preprints also count as published.)

No, but I have used extensivelyin order to derive new results. I consider your two papers on the subject the "nec plus ultra" :-)

 

[Demystifier]

No, but I have used extensivelyin order to derive new results. I consider your two papers on the subject the "nec plus ultra" :-)

In that case, perhaps two additional ones may also be helpful:
http://xxx.lanl.gov/abs/physics/9810017 [Am.J.Phys. 67 (1999) 1007]
http://xxx.lanl.gov/abs/gr-qc/9909035

 

[starthaus]

In that case, perhaps two additional ones may also be helpful:
http://xxx.lanl.gov/abs/physics/9810017 [Am.J.Phys. 67 (1999) 1007]
http://xxx.lanl.gov/abs/gr-qc/9909035

Thank you :-)

 

[Jonnyb42]

But that is really easy. It is only Newtonian mechanics, no Einstein relativity is involved. From their point of view, one is rotating around the other (e.g. Moon around Earth or vice versa). They miss each other because the direction of force does not need to coincide with the direction of velocity. I do not see what is your problem with that.

Yeah but, the coordinate system you are talking about does not rotate with the particles. The frame is centered around the center and both particles lie on the same axis, the x-axis for example. In that system, what explains the particles not colliding? I know you might say it is because the frame of reference is non-inertial, but what truly makes it non-inertial? Only a third observer would have to say that the reference frame is non-inertial.

 

[Demystifier]

I know you might say it is because the frame of reference is non-inertial, but what truly makes it non-inertial? Only a third observer would have to say that the reference frame is non-inertial.

That is not correct. Being in a non-inertial frame is an absolute statement. If you are in a non-inertial frame for one observer, then you are in a non-inertial frame for ANY observer. In particular, you can determine experimentally whether you are accelerating or not. Velocity is relative, but acceleration is absolute.

Now you will probably ask: But if I accelerate, then I accelerate with respect to what? In Newtonian mechanics it was not completely clear. For example, Mach thought that acceleration is defined with respect to (all) distant stars in the Universe. But in the relativity theory, this question can be answered: You accelerate with respect to the metric field (metric tensor).

 

[JesseM]

That is not correct. Being in a non-inertial frame is an absolute statement. If you are in a non-inertial frame for one observer, then you are in a non-inertial frame for ANY observer. In particular, you can determine experimentally whether you are accelerating or not. Velocity is relative, but acceleration is absolute.

To put it simply, inertial observers feel weightless, non-inertial observers feel G-forces (which can be measured with an accelerometer)

 

[Mike_Fontenot]

[...] It seems that the twin that ages the most, is the one that does the short final burst of acceleration to come to rest in the other twin's rest frame.

No, it's the opposite. The twin who does the accelerating, WHEN THEIR SEPARATION IS NON-ZERO, will be the younger, after they are again stationary with respect to one another.

The initial acceleration by B (when their separation is zero) has no effect: you can reformate the problem as two unrelated newborns, who are moving at a constant speed with respect to one another, and who happened to be momentarily adjacent at the instant they are both born, with neither one accelerating then.

As the two twins initially move apart (say, at a relative speed of 0.866c, to give a simple value of gamma = 2), they EACH will (correctly) conclude that the other is ageing more slowly (by a factor of 2). They are BOTH correct. Neither can adopt any other conclusion, without contradicting their own elementary measurements.

So, given the complete symmetry of this initial phase, you can actually consider B to be "the home twin", and A is "the traveler". If you do that, you can apply the CADO equation that I gave earlier, to get their two conclusions about their corresponding ages at various instants.

Suppose that the traveler (A) is 10 years old at the instant that he accelerates (call it point P), and they were both zero years old when they were co-located at birth.

The home twin (B) concludes that she is 20 years old when the traveler accelerates at point P, so she concludes that their separation is 20 (0.866) = 17.32 lightyears.

The CADO equation then says that CADO_T immediately before the acceleration is

CADO_T = CADO_H - L*v/(c*c)

or

CADO_T = 20 - (17.32)(0.866) = 20 - 15 = 5 years old.

This is B's age right before the acceleration by A, according to A. So, right before A accelerates, B says she is 20, but A says she is 5. (And they both agree that A is 10 then).

Immediately after the acceleration, the CADO equation says

CADO_T = 20 - (17.32)(0) = 20 years old.

So now, the twins AGREE on their relative ages (as they always will whenever their relative speed is zero). They both agree that when A is 10 (immediately AFTER his acceleration), B is 20.

And thereafter, they both agree that their rates of ageing are equal, and that B is always 10 years older than A.