B Twin paradox explained for laymen

  • #51
PeterDonis said:
The stationary twin's pseudo-gravitational time dilation in the traveling twin's rest frame is due to the stationary twin's position (at much higher "altitude" than the traveling twin). But if you are going to call that an influence of path curvature, manifesting as pseudo-gravity, on time dilation, it is not direct, only indirect. So in both cases (traveling twin in stationary twin's rest frame, or stationary twin in traveling twin's rest frame), the path curvature of the traveling twin's worldline does affect time dilation, but only indirectly.
I think, the influence of pseudo-gravitational potential-difference on time dilation is directly. But that's no problem, because time dilation (tickrate-ratios at a certain instance of time) is frame-dependent. Only the end-result (age-difference when meeting the 2nd time) is absolute. The influence of pseudo-gravitational potential-differences on that is only indirectly. The age difference depends, amoung others, on the pseudo-gravitational time-dilation, integrated over the turnaround-time.
 
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  • #52
Sagittarius A-Star said:
I think, the influence of pseudo-gravitational potential-difference on time dilation is directly.

"Pseudo-gravitational potential difference" is not the same as "path curvature". The latter is the invariant, and, as I said, it affects time dilation indirectly, and does so in both frames. So there is no need to point to any frame-dependent quantity like "pseudo-gravitational field" as a cause of time dilation; the invariant, path curvature, works fine as a cause in any frame.

Sagittarius A-Star said:
time dilation (tickrate-ratios at a certain instance of time) is frame-dependent

Under this definition of "time dilation" (which I have no issue with, but not everyone is careful about using the term only with this definition), it doesn't need to have a cause, because frame-dependent things don't need causes. Only invariant things, like the age difference of the twins when they meet again, need to have causes.
 
  • #53
PeterDonis said:
Under this definition of "time dilation" (which I have no issue with, but not everyone is careful about using the term only with this definition)
That's right. For example in the English Wikipedia they define time dilation as a difference in elapsed time, measured by two clocks (due to velocity or gravitational potential difference), and in the German Wikipedia that processes in a physical system tick slower relative to the observer, if the system is moving relative to the observer.
 
  • #54
From a geometric viewpoint,
time-dilation compares
the measurer's t-component of the 4-velocity of the measurer [which is 1, since 4-velocities are unit vectors]
with
the measurer's t-component of the 4-velocity of an astronaut [which is what the measurer measures as the elapsed time between two ticks (marked by events) of the astronaut's clock].

These are equal only for the Galilean spacetime used in PHY 101: measurer measures \Delta t = 1.
In special relativity, measurer measures \Delta t \geq 1 (since \gamma=\cosh\theta \geq 1).. this is time dilation.
In Euclidean geometry (by analogy), \Delta t \leq 1 (since \cos\phi \leq 1.
See the diagrams below.

From this viewpoint, the key idea is that:
the general situation is that \Delta t \ne 1
and that the Galilean case is the exceptional case [not typical case].

It seems to me that the root of time-dilation is that the real spacetime we live in isn't Galilean...
but is pseudoriemannian with a Minkowski signature [for dilation].
Once this is realized, there's no need to include gravity or acceleration etc...
unless you want to calculate specific elapsed times.
Including them probably clouds the real issue above.

Our low-relative-speed lifestyles have led to our "common sense" Galilean notions of time.
Using https://www.desmos.com/calculator/wm9jmrqnw2
for Special Relativity (E=1), Galilean (E=0), and Euclidean (E=-1)...
1595131113048.png
\quad
1595131138335.png
\quad
1595131173199.png


The bottom line is to calculate the elapsed time along different worldlines from event O to event Z.
The general case is that they are different.

1595131787842.png
 
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  • #55
robphy said:
but is pseudoriemannian with a Minkowski signature [for dilation].
Once this is realized, there's no need to include gravity or acceleration etc...
unless you want to calculate specific elapsed times.
Including them probably clouds the real issue above.
That's all fine, if you teach relativity to physics students. But the OP asks for a laymen explanation of the typical "twin paradox" scenario and states wrongly, that a scenario without real gravity would be symmetrically.

I think, for laymen an explanation with 4-vectors and 4D-spacetime may be too difficult to understand. An explanation comparing the empirical viewpoint of both twins maybe didactically better for an explanation to laymen.

And the asymmetry, related to tick-rates, can be explaind by the fact, that in the inertial frame of the "stationary" twin the tick-rate of the other twin's wristwatch can be calculated as only velocity-dependent effect, while in the accelerated frame of the "travelling" twin also pseudo-gravitational potential difference influences the tick-rate of the other twin's wristwatch.

The OP came to the wrong idea, that real gravitation is needed to break the symmetry of the typical "twin paradox" scenario. I can understand why, because at gravitational time-delation, both twins agree on, whose wristwatch ticks slower and whose faster. But in reality in this case, the symmetry breaking of tick-rates does not come from real gravitation, but from pseudo-gravitational potential-difference in one of the rest frames. So in some sense, the OP was close to this real explanation of the needed asymmetry.
 
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  • #56
Sagittarius A-Star said:
because at gravitational time-delation, both twins agree on, whose wristwatch ticks slower and whose faster.
However, the amount by which the ages differ is a function of the amount of time the traveller spends coasting and is independent of the duration and magnitude of the pseudo-gravitational time dilation at turnaround.
 
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  • #57
Nugatory said:
However, the amount by which the ages differ is a function of the amount of time the traveller spends coasting and is independent of the duration and magnitude of the pseudo-gravitational time dilation at turnaround.
That's right, because the pseudo-gravitational time dilation relates to the fast tick-rate of the stationary twin's wrist watch while the turnaround und not to the tick-rate of the traveling twin's wrist watch, which stays "normal" in his frame. And the integral of the pseudo-gravitationally time dilated tick-rate over the turnaround-time is pre-defined in the scenarion, as the acceleration of the traveller's frame shall change his velocity in the stationary twin's frame from +v to -v in a certain approximate distance.
 
  • #58
Sagittarius A-Star said:
That's right, because the pseudo-gravitational time dilation relates to the fast tick-rate of the stationary twin's wrist watch while the turnaround und not to the tick-rate of the traveling twin's wrist watch, which stays "normal" in his frame. And the integral of the pseudo-gravitationally time dilated tick-rate over the turnaround-time is pre-defined in the scenarion, as the acceleration of the traveller's frame shall change his velocity in the stationary twin's frame from +v to -v in a certain approximate distance.
Suppose the traveller takes two stopwatches on the journey. The first watch stays on for the whole journey (including the several acceleration phases). The second watch is switched on only during the inertial phases of the motion, and is stopped during the acceleration phases.

If we assume that the acceleration phases are short (they can, theoretically, be made arbitrarily short), then the two watches show approximately the same time on return the Earth. They may, for example, differ by a few minutes if the acceleration phases lasted only a few minutes.

Given that the second watch was switched off during the critical acceleration phases, how do you explain this? Using your pseudo-gravitational explanation?
 
  • #59
PeroK said:
Given that the second watch was switched off during the critical acceleration phases, how do you explain this? Using your pseudo-gravitational explanation?
You find the answer already in my posting #57. Let me summarize it: The pseudo-gravitational time dilation relates to the stationary twin's watch tick-rate (in the rest frame of the traveling twin) and not to the tick-rate of the travelling twin's watch.

While the inertial phases, the stationary twin's watch ticks slow (1/"Gamma"), and while the short turn-around phase, the stationary twin's watch ticks very fast (over-compensating the slow tick-rates of the inertial phases), all in the rest frame of the traveling twin.
 
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  • #61
PeroK said:
Given that the second watch was switched off during the critical acceleration phases, how do you explain this? Using your pseudo-gravitational explanation?
The point is to use pseudo-gravity to regard the stay-at-home twin as above the traveller during the turnaround. Thus the stay-at-home's clock runs fast and elapses more time than the traveller.

I don't think the explanation is particularly intuitive because you need to be quite careful by what you mean by "during" the turnaround. That goes double for instantaneous turnaround - what's the pseudo-gravitational potential associated with an infinite acceleration? But if you are willing to do the mucking around it ought to work.
 
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  • #62
Ether theory is physically distinct from special relativity. The theories agree only up to first order in ##\beta=v/c## (roughly speaking). As you point out yourself ether theory has been ruled out at for at least one century ago with many experiments (Michelson-Morley, Trouton-Noble, Doppler effect on light,...). A nice review can be found here:

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html
 
  • #63
This whole "pseudo-gravity" explanation seems supremely unhelpful. Telling someone who is struggling with special relativity that understanding it requires understanding general relativity, kinda-sorta, does not sound like the easiest path. The fact that thousands of people get it without GR or GR-like or pseudo-GR establishes this.

(It might possibly be usable in reverse - starting from SR to get to GR as motivation - but it's certainly not often adopted this way)

"Acceleration" cannot be the answer to "why is the aging different" because it is possible to set this up with multiple travelers looking at clocks through windows with nobody accelerating at all. The answer is that these are different paths through spacetime, and they are different just like different paths between A and B in space (say LA to San Francisco via Baltimore) have different lengths.

"Acceleration" is only an answer to the question "I don't want to do this with windows and clocks - I want to compare two clocks, one staying home, and one that goes on the trip. How do I tell them apart?"
 
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  • #64
Sagittarius A-Star said:
That's all fine, if you teach relativity to physics students. But the OP asks for a laymen explanation of the typical "twin paradox" scenario and states wrongly, that a scenario without real gravity would be symmetrically.

I think, for laymen an explanation with 4-vectors and 4D-spacetime may be too difficult to understand. An explanation comparing the empirical viewpoint of both twins maybe didactically better for an explanation to laymen.
The diagrams actually are intended for the lay person (after some introduction),
but the words I used for my reply weren't necessarily addressed to the layperson.
I see now I used "pseudo-riemannian" inappropriately to make my point
since Galilean is also pseudo-riemannian.

My main point is that
the way we use the position vs time graph in PHY 101 is only an approximation to what is really going on.
In fact, it is not well-appreciated that the position-vs-time graph is already a non-euclidean geometry.

Here's are the real measurements on a position-vs-time graph.
In this graph, the red lines are perpendicular to each other (in all three geometries mentioned earlier) and the blue segment is the hypotenuse. You can introduce units to redefine variables to make the sides have the same units... but that doesn't change the underlying geometrical relationships.

In PHY 101, we treat the blue segment as having length 1 (already in violation of Euclidean geometry)
but it's really length (approximately) 1-(5\times 10^{-17}). (Someday, we'll have a wristwatch that will measure this.)
The event that is "Length 1 along the blue segment" occurs to the right of the red vertical line.
This is time-dilation.

1595162404204.png
 
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  • #65
I might be repeating something that was aleardy said, but here is a scenario that might help with the issue of the acceleration. The two twins depart together in the same direction. One of them turns back first, the second later. They both return to the original place, the one that turns first will wait there for the second one. This way they accelerate in the begining, during the turn, and at the end in exactly the same way. So the accelerations were the same for both. Just one turns later. When they meet they compare age and it turns out that they are not the same age. So acceleration cannot be the reason for the different aging.
 
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  • #66
martinbn said:
I might be repeating something that was aleardy said, but here is a scenario that might help with the issue of the acceleration. The two twins depart together in the same direction. One of them turns back first, the second later. They both return to the original place, the one that turns first will wait there for the second one. This way they accelerate in the begining, during the turn, and at the end in exactly the same way. So the accelerations were the same for both. Just one turns later. When they meet they compare age and it turns out that they are not the same age. So acceleration cannot be the reason for the different aging.
The observed effect on another observer depends on the distance between the observers. The greater the relative distance, the greater the effect on the clocks. This is why the accelerations at the start and end of the twin's travel has very small effect, but the accelerations when he turns around has a large effect. Otherwise, the acceleration effects would largely cancel out. So your example does not remove the proposed effect of acceleration.
 
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  • #67
martinbn said:
I might be repeating something that was aleardy said, but here is a scenario that might help with the issue of the acceleration. The two twins depart together in the same direction. One of them turns back first, the second later. They both return to the original place, the one that turns first will wait there for the second one. This way they accelerate in the begining, during the turn, and at the end in exactly the same way. So the accelerations were the same for both. Just one turns later. When they meet they compare age and it turns out that they are not the same age. So acceleration cannot be the reason for the different aging.
It's also interesting that this question has been investigated experimentally:

http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Clock_Hypothesis

The upshot is that indeed even at very high proper accelerations against an inertial frame "proper clocks" like lifetimes of unstable particles (muons) show the "proper time", defined by ##\mathrm{d} \tau=\sqrt{1-\beta^2} \mathrm{d} t##, i.e., independent of acceleration.
 
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  • #68
martinbn said:
So acceleration cannot be the reason for the different aging.
Acceleration does not directly effect the different aging. But it has an indirect effect on it, because for example in the inertial rest frame of the stationary twin, the integral over time of the acceleration of the traveling twin equals the change of his/her velocity.

As @FactChecker stated correctly in posting #66, in the standard "twin paradox" scenario, this indirect effect is greater, if the distance of the turnaround from Earth is greater:

1) Reason, described in 4D-spacetime: If this spatial distance is greater, then the proper time of the traveling twin is smaller, because of the minus-sign in the formula for invariant spacetime-distance.

2) Reason, described in the rest frame of the stationary twin: If this spatial distance is greater, then the traveling twin travels longer with an almost constant slow tick-rate of his/her watch (1/"Gamma").

3) Reason, described in the rest frame of the traveling twin: If this spatial distance is greater, then the pseudo-gravitational potential of the remote "stationary" twin is greater, and therefore also the tick-rate of his/her watch is greater - while the short "turnaround"-time.
 
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  • #69
Sagittarius A-Star said:
Acceleration does not directly effect the different aging. But it has an indirect effect on it, because for example in the inertial rest frame of the stationary twin, the integral over time of the acceleration of the traveling twin equals the change of his/her velocity.

As @FactChecker stated correctly in posting #66, in the standard "twin paradox" scenario, this indirect effect is greater, if the distance of the turnaround from Earth is greater:

1) Reason, described in 4D-spacetime: If this spatial distance if greater, then the proper time of the traveling twin is smaller, because of the minus-sign in the formula for invariant spacetime-distance.

2) Reason, described in the rest frame of the "stationary" twin: If this spatial distance if greater, then the traveling twin travels longer with an almost constant slow tick-rate of his/her watch (1/"Gamma").

3) Reason, described in the rest frame of the traveling twin: If this spatial distance if greater, then the pseudo-gravitational potential of the remote "stationary" twin is greater, and therefore also the tick-rate of his/her watch is greater - while the short "turnaround"-time.

That's a bit like being given a speeding ticket that, instead of saying you were driving too fast, provides an acceleration profile that implies you were drving too fast.

By that argument, acceleration is an indirect cause of speeding. That may be a true statement, but it would make for a complicated law as to if and when you got a speeding ticket.
 
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  • #70
PeroK said:
By that argument, acceleration is an indirect cause of speeding. That may be a true statement, but it would make for a complicated law as to if and when you got a speeding ticket.
I have to admit that this is a good point. But I think that to say that turning around causes the effect but that acceleration does not is quibbling about semantics. Turning around is acceleration.
 
  • #71
Sagittarius A-Star said:
Acceleration does not directly effect the different aging. But it has an indirect effect on it,
I agree. To me, the acceleration does not cause the time dilation, but it does resolve the paradox. The paradox is not about calculating the amount of time dilation, it is about the symmetry.

The confused student has learned that “motion is relative” and therefore thinks that the effects should be the same for each twin because of the symmetry. The acceleration breaks that symmetry and thereby resolves the paradox. Any proposal to avoid the acceleration always introduces some other asymmetry.
 
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  • #72
FactChecker said:
I have to admit that this is a good point. But I think that to say that turning around causes the effect but that acceleration does not is quibbling about semantics. Turning around is acceleration.
There is more to it than this. Using only the relative positions of the twins, there is no way to mathematically determine which twin is moving and which is stationary. Some reference to an external object or force is necessary to break the mathematical symmetry of the two twins. The most obvious way is to say that the traveling twin feels acceleration when he turns around.
 
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  • #73
PeroK said:
That may be a true statement, but it would make for a complicated law as to if and when you got a speeding ticket.
I would ague with relativity, that I should get no ticket. In my rest frame, I had speed Zero. Only the speed-measurement equipment moved too fast backwards towards me. :cool:
 
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  • #74
FactChecker said:
Using only the relative positions of the twins, there is no way to mathematically determine which twin is moving and which is stationary. Some reference to an external object or force is necessary to break the mathematical symmetry of the two twins. The most obvious way is to say that the traveling twin feels acceleration when he turns around.

As has been emphasised many times, you get the same results without acceleration, by simply measuring the spacetime distance along two joined inertial paths. And that is simply spacetime geometry.

The analogy with the triangle inequality applies. Suppose you wanted to confirm that the spatial distance along two edges of a triangle is larger than the distance along the remaining edge. In order to follow the non-straight path, you must accelerate to change direction at the intermediate vertex.

You would then indirectly attribute the triangle inequality to acceleration. The amount of acceleration would define the change in angle and, indirectly, tell you the length of the overall path.

This overlooks the simpler explanation that it was nothing to do with the acceleration at the vertex: it was a simple case of ##|AB| + |BC| > |AC|##.
 
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  • #75
PeroK said:
As has been emphasised many times, you get the same results without acceleration, by simply measuring the spacetime distance along two joined inertial paths. And that is simply spacetime geometry.
Using only the relative positions, there is no way to mathematically determine which twin stays on one inertial path and which changes to a different inertial path. Something else is needed to distinguish between the twins. Saying that one twin turns around and switches to a different inertial path is just another way to say that he accelerated.
 
  • #76
PeroK said:
As has been emphasised many times, you get the same results without acceleration, by simply measuring the spacetime distance along two joined inertial paths. And that is simply spacetime geometry.
Can you give me an example where an observer did not follow an inertial path and yet did not accelerate?
 
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  • #77
FactChecker said:
Using only the relative positions, there is no way to mathematically determine which twin stays on one inertial path and which changes to a different inertial path. Something else is needed to distinguish between the twins. Saying that one twin turns around and switches to a different inertial path is just another way to say that he accelerated.
If you define your geometry to be hyperbolic, flat spacetime then you can show mathematically that for timelike paths in hyperbolic geometry you have ##|AC| < |AB| + |BC|##.

That's the mathematics. There are no complications in using hyperbolic geometry.

To do a physical experiment you need some way to establish an inertial reference frame. Any one will do. You could pick any frame in which the Earth and distant planet are moving inertially. The Earth frame is the simplest, but it works out equally in all inertial reference frames. That's where physics enters. Mapping the mathematical hyperbolic geometry to a physical scenario.

It's true that a physical object in flat spacetime cannot change frames without accelerating, but that constraint can be removed if you simply measure time along spacetime paths (swapping physical clocks at the turnaround).
 
  • #78
FactChecker said:
Can you give me an example where an observer did not follow an inertial path and yet did not accelerate?
PeterDonis said:
(And in fact, there are versions of the scenario where even the turnaround does not require acceleration: for example, the traveling twin could pass close enough to some large, distant planet or star to "slingshot" around it and be heading back towards Earth, and stay in free fall the whole time.)
 
  • #79
FactChecker said:
Can you give me an example where an observer did not follow an inertial path and yet did not accelerate?
If you want to remove acceleration from the experiment, you need to swap physical clocks at the turnaround. Or, you simply study the problem mathematically using hyperbolic geometry.
 
  • #80
PeroK said:
As has been emphasised many times, you get the same results without acceleration, by simply measuring the spacetime distance along two joined inertial paths. And that is simply spacetime geometry.

The analogy with the triangle inequality applies. Suppose you wanted to confirm that the spatial distance along two edges of a triangle is larger than the distance along the remaining edge. In order to follow the non-straight path, you must accelerate to change direction at the intermediate vertex.

You would then indirectly attribute the triangle inequality to acceleration. The amount of acceleration would define the change in angle and, indirectly, tell you the length of the overall path.

This overlooks the simpler explanation that it was nothing to do with the acceleration at the vertex: it was a simple case of ##|AB| + |BC| > |AC|##.
But it is indeed true that in order to have different curves connecting the same two spacetime points at least one of the worldlines cannot be a straight line in Minkowski space, i.e., at least one must be accelerated against the inertial reference frames.

But it's also generally true: If both observers are accelerated wrt. the inertial reference frames, along worldlines with the same initial and final points, in general measure different proper times it takes for their travel.

The best analogon to Euclidean geometry is indeed that there of course also the lengths of curves connecting the same two points may be of different length.

The largest proper time you get always for the straight line, as one sees easily by using the variational principle. In a sense you have also a triangle rule in Minkowski space, but the straight line here is the longest not the shortest connection (for time-like curves only of course).
 
  • #81
PeterDonis said:
And in fact, there are versions of the scenario where even the turnaround does not require acceleration: for example, the traveling twin could pass close enough to some large, distant planet or star to "slingshot" around it and be heading back towards Earth, and stay in free fall the whole time.) So it is not true to say that the acceleration of the traveling twin is "far greater" than that of the twin who remains on earth.
It seems to me that slingshotting around a distant star brings an entirely different geometry into the situation. I don't think that SR applies in that situation.
 
  • #82
I also think that's GR. But also in GR the maximum proper time among all time-like curves connecting to points is reached for a geodesic connecting the two points.
 
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  • #83
vanhees71 said:
But it is indeed true that in order to have different curves connecting the same two spacetime points at least one of the worldlines cannot be a straight line in Minkowski space, i.e., at least one must be accelerated against the inertial reference frames.

But it's also generally true: If both observers are accelerated wrt. the inertial reference frames, along worldlines with the same initial and final points, in general measure different proper times it takes for their travel.

The best analogon to Euclidean geometry is indeed that there of course also the lengths of curves connecting the same two points may be of different length.

The largest proper time you get always for the straight line, as one sees easily by using the variational principle. In a sense you have also a triangle rule in Minkowski space, but the straight line here is the longest not the shortest connection (for time-like curves only of course).

What I'm saying is this:

First, in Euclidean geometry (with the usual definition of distance) we have the triangle inequality. Take three points in the plane: ##A = (0,0), \ B = (1, 1), \ C = (0, 2)##. We have:
$$|AB| = \sqrt 2, \ \ |BC| = \sqrt 2, \ \ |AC| = 2$$
And we see that, indeed:
$$|AC| < |AB| + |BC|$$
This is generally called the triangle inequality.
Now, in hyperbolic geometry with ##ds^2 = dt^2 - dx^2## we have three points:
$$A = (0, 0), \ \ B = (t, vt), \ \ C = (2t, 0), \ \ (0 < v < 1)$$
Then we have:
$$|AB| = t\sqrt{1 - v^2} < t, \ \ |BC| = t\sqrt{1 - v^2} < t, \ \ |AC| = 2t$$
And, we see that:
$$|AC| > |AB| + |BC|$$
This is generally called the twin paradox.

Now, you could demonstrate the triangle inequality (to show that we have approximately Euclidean geometry on the surface of the Earth) in a real experiment by pacing out the sides of a triangle. And you could, if you wanted, attribute the difference in lengths measured to the acceleration when you changed direction at the intermediate vertex. Rather than accepting the underlying Euclidean geometry of space.

Likewise, you could demonstrate the hyperbolic geometry of spacetime by doing a real twin paradox experiment. And, if you wanted, you could again attribute the difference in times measured to the acceleration when you changed direction. Again, rather than accepting the underlying hyperbolic geometry of spacetime.
 
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  • #84
vanhees71 said:
I also think that's GR. But also in GR the maximum proper time among all time-like curves connecting to points is reached for a geodesic connecting the two points.
And we know that the twin who slingshots around a large mass would age faster, so it is not immediately clear to me that the twins would not have identical ages when they get back together.

Vanadium 50 said:
None of that is correct. You can set this up entirely with people looking at clocks through windows without anyone accelerating.
I think that those scenarios only add a complication to the paradox, they do not resolve it. In an identically symmetric way, the "traveling" twin can imagine the Earth twin switches his clock to another inertial frame that comes back to the "traveling" twin and makes the Earth twin younger. So ther is still a paradox. Once again, the simplest way to break the logical symmetry is to say that the traveling twin feels acceleration.
 
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  • #85
Dale said:
I agree. To me, the acceleration does not cause the time dilation, but it does resolve the paradox. The paradox is not about calculating the amount of time dilation, it is about the symmetry.

The confused student has learned that “motion is relative” and therefore thinks that the effects should be the same for each twin because of the symmetry. The acceleration breaks that symmetry and thereby resolves the paradox. Any proposal to avoid the acceleration always introduces some other asymmetry.
In my post I gave an example where the accelerations are the same for the two. The point was that it cannot be the acceleration that makes the difference.
 
  • #86
martinbn said:
In my post I gave an example where the accelerations are the same for the two. The point was that it cannot be the acceleration that makes the difference.
Your example is flawed. You ignore that the distance between the observers has a great effect on the acceleration. The greatest effect is when the observers are widely separated.
 
  • #87
martinbn said:
In my post I gave an example where the accelerations are the same for the two. The point was that it cannot be the acceleration that makes the difference.
Even then the acceleration profiles were different. Both twins agree which twin is the early accelerating and which is the late accelerating twin. The acceleration still eliminates the symmetry.

Again, (proper) acceleration does not cause the differential aging, but it does break the symmetry.
 
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  • #88
FactChecker said:
So ther is still a paradox.
One thing is true: there is only a paradox for those who do not understand SR. Mostly the general public. You would be struggling to find a physicist who thinks there is a paradox that needs an elaborate, pseudo-gravitational explanation.
 
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  • #89
FactChecker said:
It seems to me that slingshotting around a distant star brings an entirely different geometry into the situation. I don't think that SR applies in that situation.
The calculation is the same regardless of the geometry: We integrate the quantity ##g_{uv}dx^udx^v## along the path. The difference is that in the flat spacetime of SR (and if we choose our coordinates in the most natural way) the ##g_{uv}## are all either 1 or -1 so we don't bother to write them down and the integral becomes so trivial that we don't notice it.
 
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  • #90
FactChecker said:
And we know that the twin who slingshots around a large mass would age faster
Huh?

The twin who slingshots around a large mass is lower in a potential well. So he ages more slowly. And he is moving faster. So he ages more slowly.

Both of these are coordinate-dependent heuristics for time dilation, but the result should hold for differential aging: Shorter elapsed time for the slingshotting twin.
 
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  • #91
Dale said:
Even then the acceleration profiles were different. Both twins agree which twin is the early accelerating and which is the late accelerating twin. The acceleration still eliminates the symmetry.

Again, (proper) acceleration does not cause the differential aging, but it does break the symmetry.
Of course there is no symmetry, but the experienced acceleration is the same for the two. It is not the acceleration that makes it asymmetric.
 
  • #92
jbriggs444 said:
The twin who slingshots around a large mass is lower in a potential well. So he ages more slowly. And he is moving faster. So he ages more slowly.

Both of these are coordinate-dependent heuristics for time dilation, but the result should hold for differential aging: Shorter elapsed time for the slingshotting twin.
Sorry. I stand corrected.
 
  • #93
martinbn said:
Of course there is no symmetry, but the experienced acceleration is the same for the two. It is not the acceleration that makes it asymmetric.
The relative distances when the acceleration occurs is different. So it is the acceleration and the timing/relative distance of it that breaks the symmetry.
 
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  • #94
FactChecker said:
The relative distances when the acceleration occurs is different. So it is the acceleration and the timing/relative distance of it that breaks the symmetry.
That's a remarkably uncovariant explanation.
 
  • #95
FactChecker said:
The relative distances when the acceleration occurs is different. So it is the acceleration and the timing/relative distance of it that breaks the symmetry.
What do you mean the relative distences are different? Distances to what?
 
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  • #96
martinbn said:
Of course there is no symmetry, but the experienced acceleration is the same for the two. It is not the acceleration that makes it asymmetric.
I disagree that the acceleration is the same. Acceleration is a vector valued function of time and those functions are not the same. The acceleration is indeed asymmetric.

Again, my claim isn’t that the acceleration causes the aging, just that it breaks the symmetry. With no other information besides their accelerometer readings the twins agree who accelerated early and who accelerated late. So the acceleration itself does break the symmetry.
 
  • #97
martinbn said:
What do you mean the relative distences are different? Distances to what?
The relative distance between two points is the distance from one point to the other.
 
  • #98
PeroK said:
One thing is true: there is only a paradox for those who do not understand SR. Mostly the general public. You would be struggling to find a physicist who thinks there is a paradox that needs an elaborate, pseudo-gravitational explanation.
Einstein was a strong proponent of the pseudogravity explanation for most of his life. He never considered there was paradox, but his preferred explanation was pseudogravity. It could be made to work for every smooth (continuous second derivative, specifically) variant in SR, and generalizes to GR.

I do not like this approach, but I find it nonsensical to claim it is invalid.
 
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  • #99
FactChecker said:
The relative distance between two points is the distance from one point to the other.
A "point" in four dimensional space time is stretched into a chain of "events" known as a "world line".

Measuring the distance between two events is unambiguous. Measuring the distance between two world lines is ambiguous -- it depends on which pair of events you pick out.

The process of picking a particular world line through an event to be the "point" associated with that event is also a potential source of ambiguity.
 
  • #100
jbriggs444 said:
Huh?

The twin who slingshots around a large mass is lower in a potential well. So he ages more slowly. And he is moving faster. So he ages more slowly.

Both of these are coordinate-dependent heuristics for time dilation, but the result should hold for differential aging: Shorter elapsed time for the slingshotting twin.
Now, I'm confused. The geodesic equation is derived from the free-particle action principle with the Lagrangian
$$L=-m c \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}},$$
i.e., it's the curve between fixed initial and final points which minimizes
$$A=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda L=-mc \tau.$$
So ##\tau## is maximal for the geodesic (free-fall worldline) connecting two points, right?
 
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