I Does the "space twin" benefit from length contraction?

[PeroK]

I don't disagree with any of these details, but my simple question should have a simple answer I believe. Me and Romer synchronize our clocks to zero before the acceleration and agree after the acceleration that we are still attached to the same spacecraft but disagree as to the time for acceleration. Our delta V is manisfestly the same yet our times for it differ...does that not require the acceleration we measure (eg with our buttocks) to be different?

First, for an acceleration phase of months, say, the difference in time is only $h/c$ even for a final velocity close to $c$.

This is the same margin for error as ignoring length contraction in the IRF.

To resolve this, you would need to analyse the problem exactly. For example, in the IRF view, the ship could stay the same length - i.e. its proper length would increase and then you wouldn't have exactly a rigidly accelerating frame.

 

[PeroK]

PS it would be important to analyse coming out of the acceleration phase. If they both stop accelerating when their respective clocks reach a certain time, then what does that look like in the IRF?

 

[PAllen]

I don't disagree with any of these details, but my simple question should have a simple answer I believe. Me and Romer synchronize our clocks to zero before the acceleration and agree after the acceleration that we are still attached to the same spacecraft but disagree as to the time for acceleration. Our delta V is manisfestly the same yet our times for it differ...does that not require the acceleration we measure (eg with our buttocks) to be different?

That’s where the notion of rigidity comes into play. If the ship is taken to be rigid, then an accelerometer at the front will measure lower acceleration than the rear. The amount is extremely small, a lot less than the change in acceleration from front to back if the ship were sitting on a planet with the same acceleration at the bottom.

 

[PAllen]

Here are two scenarios:

1) In an initial inertial frame, front and back begin accelerating at the same time by the same amount. Then, per the same frame, deceleration begins at the same time, by the same amount until front and back are both at rest again. These will be simultaneous for this frame. Assume acceleration for a month. Then, the back will see the front move slowly away, to many times the initial separation. Then, before the back starts decelerating, they will see the front moving towards them. When the back finally stops decelerating, the original length will be restored. Both rocket clocks will show the exact same time, much, much less than clocks in the original frame.

2) In an inertial frame, back starts accelerating, accelerates for a month per the initial frame, then decelerates to stop. The motion of the front is governed by the rule that it always looks the same distance away, per the back. To achieve this, the front accelerates and decelerates less. This is a frame independent statement. Per the initial frame, the distance between them will shrink to a tiny fraction of the initial distance, then grow back. The front clock will have accumulated more time than the back, but this difference will be very small compared to the how much either is behind the initial frame clocks.

[edit: A couple of corrections to case 2, above. During deceleration, for born rigid motion, what was the front will have decelerate faster than the back. And, without doing the math, I don’t know how the clocks will end up, except they will be extremely close, because the front will be ticking slower than back, per the back, during deceleration. Also, from the original frame, both front and back will have covered the exact same distance, but with different velocity profile, so I see no obvious argument as to exactly how they end up comparing to each other.]

 

[PeroK]

That’s where the notion of rigidity comes into play. If the ship is taken to be rigid, then an accelerometer at the front will measure lower acceleration than the rear. The amount is extremely small, a lot less than the change in acceleration from front to back if the ship were sitting on a planet with the same acceleration at the bottom.

... looking at my notes, for a rigidly accelerating ship,we have a proper acceleration of:

$\frac{a}{1 + ah/c^2}$

And, we have a proper time of:

$\tau = (1 + ah/c^2)\tau_0$

Hence, as you say, the proper acceleration varies inversely with proper time along the ship.

 

[hutchphd]

Thank you all is now copacetic in my relativity brains...I appreciate your help.

 

[nomadreid]

This thread has lots of very helpful analyses, but despite having read the thread and Baez’s explanation, watched Hewitt’s videos, followed the calculations and so forth, there is one point which intrudes upon my getting a good intuition of the asymmetry. I will start with the geometric viewpoint, that between two points in spacetime there is only one straight line, so that an asymmetry is inevitable. On one side one says that acceleration is not important, on the other side one emphasizes that it is the turn-around, the switching IRF’s, that is important. But isn’t the turn-around an acceleration? So the traveler must assume that she is the one turning around, not the earth, which she can only do by noticing a force. No? So isn't the classic explanation that it is the acceleration which makes the difference still valid, even if one adds the cautionary note that one doesn't need GR, only SR, to handle the difference that the acceleration makes?

Putting the question another way: the only way that the traveler knows that she is turning around is either to experience a force or acceleration (the popular explanation that the thread quickly discards) or to assume that the homebody’s IRF is the one with the straight line: but if the Traveler does not assume this and ignores the force (or acceleration), then what is to stop her from assuming that hers is the straight line and the homebody’s the one with the crook?

To make the question in the spirit of the diagrams in Baez’s post (rotated 90 degrees,), one could compare, say, the graphs
{ Trav: y=|x|; Home: the line from y=(-5,5) to(5,5) }
and
{Trav: the line from (-5,0) to (5,0); home: y=5-|x|. }
The only way to find out (before one gets back to Earth to compare clocks) which one of the two twins in on the straight line is by noticing the acceleration, why is it said that the acceleration is not the important factor?

 

[Nugatory]

then what is to stop her from assuming that hers is the straight line and the homebody’s the one with the crook?

The presence or absence of a crook is a physical and mathematical fact about a path through spacetime.

It is true that you cannot force an object to follow a path with a crook in it without subjecting the object to acceleration, but we can measure the proper time along a path without forcing an object to follow that path (transfer the reading on the outbound clock to an inbound clock as they pass one another, similar to the way that we avoid acceleration at the beginning and end of the round trip), and these measurements will confirm the presence of the crook.

 

[Herman Trivilino]

So isn't the classic explanation that it is the acceleration which makes the difference still valid, even if one adds the cautionary note that one doesn't need GR, only SR, to handle the difference that the acceleration makes?

Perhaps parts of what you're saying are correct, but the part of it that I quoted is not correct. But the thing is, it's the straight-line paths through spacetime that are the cause of the difference in twin ages. If all the traveling twin did after departure is slow down, turn around, and return to the reunion, the difference is twin ages would be smaller than if the traveling twin moved away at a steady speed before executing the very same turn-around to make it back to the reunion.

In other words, you can have the very same amount of acceleration in each case with very different outcomes. So it cannot be that the acceleration accounts for the difference in ages.

 

[PeroK]

In other words, you can have the very same amount of acceleration in each case with very different outcomes. So it cannot be that the acceleration accounts for the difference in ages.

And, you can have very much the same differential ageing with different acceleration profiles. If we insist on acceleration, then we get the same result regardless of whether there are two, three or four acceleration phases. Which is the same result as when we simply switch reference frames by transferring clock readings.

The small differences in these cases can be explained by the different velocity profiles during the various acceleration phases.

So, it's really the word "cause" that is the issue. The acceleration cannot be seen as a direct cause of differential ageing.

 

[laymanB]

@nomadreid , as you can see from reading through this thread and the reply posts above, acceleration is not needed to explain the differential aging. The important thing to remember is that any proper analysis will come up with the same results. With that being said, personally I think some ways are simpler than others to analyze the problem, based on how it is presented. With the standard twin paradox problem setup, the easiest way for me to analyze it is to say that the space twin is the only one of the two twins to feel an acceleration, therefore we know that his worldline is not a straight line in flat spacetime between the two events. And because a straight worldline between two events in flat spacetime has the longest proper time, ergo; the Earth twin will have aged more. But if you present the solution like that then people come away thinking that the acceleration is necessary to explain the differential aging.

Don't worry if you still don't have a rock solid intuition about the asymmetry, you are in good company. I have read through many forums where people seem to know GR very well, claim that it requires acceleration and GR, work out the equations, and still come to the faulty conclusion that the problem is symmetric without someone experiencing acceleration.

 

[Herman Trivilino]

@nomadreidAnd because a straight worldline between two events in flat spacetime has the longest proper time, ergo; the Earth twin will have aged more. But if you present the solution like that then people come away thinking that the acceleration is necessary to explain the differential aging.

The acceleration is necessary, and it does explain the asymmetry, but it is not the cause of the difference in the ages of the twins.

If you're hit by a drunk driver your presence at the collision is necessary to explain the collision. But your presence is not the cause of the collision.

 

[laymanB]

The acceleration is necessary, and it does explain the asymmetry, but it is not the cause of the difference in the ages of the twins.

If you're hit by a drunk driver your presence at the collision is necessary to explain the collision. But your presence is not the cause of the collision.

Good point. In the standard twin paradox setup acceleration is a necessary condition for the twins to reunite. Acceleration in this setup is a necessary but not sufficient condition for differential aging.

I should have been more precise: In any generalized setup of the twin paradox, with or without acceleration, acceleration is not a necessary condition to explain the asymmetry of the problem.

 

[hutchphd]

So, it's really the word "cause" that is the issue. The acceleration cannot be seen as a direct cause of differential ageing.

I think the semantics get very slippery here. If we lock Homer and Romer in the usual opaque elevators at birth then the only measurable difference in their experiences is the acceleration. So will they agree that that is causal? I also note that Romer (or Homer!) needs only the measured time course of his acceleration to completely characterize his path..

In other words, you can have the very same amount of acceleration in each case with very different outcomes. So it cannot be that the acceleration accounts for the difference in ages.

The fact that the total amount of acceleration is the same does not mean the acceleration is exactly the same. If I know only the measured acceleration profiles (and the initial conditions), I can characterize each twin completely and uniquely determine outcome. So you can't say "cannot be causal". Does this make it a "cause"?? "direct cause?? It is manifestly sufficient, yes?

 

[PeroK]

I think the semantics get very slippery here. If we lock Homer and Romer in the usual opaque elevators at birth then the only measurable difference in their experiences is the acceleration. So will they agree that that is causal? I also note that Romer (or Homer!) needs only the measured time course of his acceleration to completely characterize his path..

The fact that the total amount of acceleration is the same does not mean the acceleration is exactly the same. If I know only the measured acceleration profiles (and the initial conditions), I can characterize each twin completely and uniquely determine outcome. So you can't say "cannot be causal". Does this make it a "cause"?? "direct cause?? It is manifestly sufficient, yes?

I don't disagree with any of that. When I was learning SR, I quite quickly gave up all talk of twins and various other devices that seemed to me only to cloud the issue. I started to think in terms of clocks. Clocks can be reset and used to measure proper time along a worldline without worrying about their experience from birth!

For example, in this case you could have two stop-clocks on board Rover's ship, one of which you pause for the acceleration phases. That would give you the proper time for the entire trip on one clock and the proper time for the inertial phases only on the second clock.

That then gets to the heart of the matter, where thinking about twins is too inflexible.

 

[Herman Trivilino]

If we lock Homer and Romer in the usual opaque elevators at birth then the only measurable difference in their experiences is the acceleration.

They can also measure the amount of time they spend not accelerating.

I also note that Romer (or Homer!) needs only the measured time course of his acceleration to completely characterize his path..

They also need the amounts of time they spend not accelerating.

 

[hutchphd]

They can also measure the amount of time they spend not accelerating.
They also need the amounts of time they spend not accelerating.

Measurement of the acceleration certainly includes zero as a possible measurement...(semantic issue).

 

[hutchphd]

I don't disagree with any of that. When I was learning SR, I quite quickly gave up all talk of twins and various other devices that seemed to me only to cloud the issue. I started to think in terms of clocks. Clocks can be reset and used to measure proper time along a worldline without worrying about their experience from birth!

For example, in this case you could have two stop-clocks on board Rover's ship, one of which you pause for the acceleration phases. That would give you the proper time for the entire trip on one clock and the proper time for the inertial phases only on the second clock.

That then gets to the heart of the matter, where thinking about twins is too inflexible.

For me the opposite is true. In particular thinking about Rover seeing the dephasing of the clocks in his frame as he decelerates to turn around is what makes the whole thing make sense to me. Thinking only about proper time is too inflexible!

 

[Nugatory]

Rover seeing the dephasing of the clocks in his frame as he decelerates to turn around

But now how do you explain the acceleration-free but otherwise equivalent case in which Rover turns around by executing a tight hyperbolic orbit around some mass conveniently placed at the the turnaround point?

 

[PeroK]

For me the opposite is true. In particular thinking about Rover seeing the dephasing of the clocks in his frame as he decelerates to turn around is what makes the whole thing make sense to me. Thinking only about proper time is too inflexible!

The proper time along a worldline is the physics. SR is about the nature of space and time. You don't gain anything by considering a pair of twins as your clocks other than confusion and obfuscation - which this thread proves.

Considering clocks allows you to see the physics, rather than getting bogged down in misconceptions.

Finally, you know, SR and GR are far more than simply the twin paradox. Your thinking about SR in terms of acceleration cannot be extended to GR, where you must focus on the geometry of spacetime intervals.

There is, thankfully, no popular extension of the twin paradox to GR.

 

[hutchphd]

But now how do you explain the acceleration-free but otherwise equivalent case in which Rover turns around by executing a tight hyperbolic orbit around some mass conveniently placed at the the turnaround point?

If we limit ourselves to the "full" paradox sequence: birth, traveling interval, meet again, there is no acceleration--free case is there?? If rover is unaware of the existence of the mass he will be surprised that he has returned to his twin but the initial and final perceived acceleration will still be necessary (but no longer sufficient) to achieve this path. I believe there is no nontrivial closed (complete) trajectory that does not require a perceived acceleration.for Romer. (?)

 

[jbriggs444]

If we limit ourselves to the "full" paradox sequence: birth, traveling interval, meet again, there is no acceleration--free case is there?? If rover is unaware of the existence of the mass he will be surprised that he has returned to his twin but the initial and final perceived acceleration will still be necessary (but no longer sufficient) to achieve this path. I believe there is no nontrivial closed (complete) trajectory that does not require a perceived acceleration.for Romer. (?)

There is no requirement that the twins begin or end at rest with respect to one another. That dispenses with the need for an initial or a final acceleration.

However, inclusion of a mass and a hyperbolic swing-by makes this a problem for general relativity. It goes beyond the flat space-time background normally assumed for the classical twin paradox.

 

[hutchphd]

The proper time along a worldline is the physics. SR is about the nature of space and time. You don't gain anything by considering a pair of twins as your clocks other than confusion and obfuscation - which this thread proves.

For me the point is to imagine that I am Romer. This is easier than imagining I am a clock!

 

[jbriggs444]

For me the point is to imagine that I am Romer. This is easier than imagining I am a clock!

So put a watch on your wrist. That's easier than measuring your hair length, waistline or secondary sexual characteristics.

 

[hutchphd]

So put a watch on your wrist. That's easier than measuring your hair length, waistline or secondary sexual characteristics.

And of course it would be silly to try to imagine riding along with, say, an electromagnetic wave...what a useless enterprise!

 

[jbriggs444]

And of course it would be silly to try to imagine riding along with, say, an electromagnetic wave...what a useless enterprise!

Yes. An electromagnetic wave would out-pace you. So yes, that would be silly unless you contrived to have it bounce back and forth in some kind of light clock.

[Sarcasm really does not work well in an electronic medium. Please try to say what you mean simply and directly]

 

[PeroK]

And of course it would be silly to try to imagine riding along with, say, an electromagnetic wave...what a useless enterprise!

Actually, here is my alternative twin paradox. Owing to a) the boredom of the space flight; b) the lack of fresh air and exercise; c) the deleterious effects of cosmic radiation: Rover actually looks and feels much older than Homer upon his return. Thus confirming that Homer is the "younger" twin.

 

[hutchphd]

Yes. An electromagnetic wave would out-pace you. So yes, that would be silly unless you contrived to have it bounce back and forth in some kind of light clock.

[Sarcasm really does not work well in an electronic medium. Please try to say what you mean simply and directly]

The point is that measurement means something very specific and it involves "being there". Quantum mechanics indicates this in a surprising way. I have always found that reducing problems to real human beings making simple real measurements eliminates the chance of making some very subtle errors. This is my method...you are free to have your own but not to disparage mine.
The light beam comment was of course an homage to the 16 yr old Albert wrestling with Maxwell's equations

 

[jbriggs444]

The point is that measurement means something very specific and it involves "being there".

The point is that in order to measure time, one needs a clock. The human body is a remarkably poor one.

 

[nitsuj]

When I was learning SR, I quite quickly gave up all talk of twins and various other devices that seemed to me only to cloud the issue. I started to think in terms of clocks. Clocks can be reset and used to measure proper time along a worldline without worrying about their experience from birth!

For example, in this case you could have two stop-clocks on board Rover's ship, one of which you pause for the acceleration phases. That would give you the proper time for the entire trip on one clock and the proper time for the inertial phases only on the second clock.

Well said!