B What is the effect of the twin paradoxon?

[PeterDonis]

As the twins move past each other, they set the clocks directly in front of them - and thus their entire row of clocks - to "zero".

This is impossible as you describe it. You cannot magically set an entire row of clocks instantaneously. Each twin can set the clock directly in front of them that is moving with them; but they can only propagate that setting to other clocks moving with them at the speed of light.

 

[Nugatory]

To compare the age (time elapsed since meeting) of the twins (by the "observer" described above), it is not necessary for them to be at the same place. The twins do not have to pull each other's beards. It is sufficient for them to rest in the same inertial frame, even if they are far apart in space.

This is simply wrong. Even when they are at rest in the same inertial frame (a frame-independent way of expressing this constraint would be to say that their relative velocity is zero) if they are not colocated then different frames will calculate and observe A to be older than B, A to be younger than B, or both to be the same age.

 

[Dale]

An "observer" is an ideal, massless, imaginary measuring station at the center of a chosen frame of reference.

No, again, this is exactly the issue I am addressing. An observer is anything that can make an observation, whether that is a person or a measuring device or a particle or whatever. Describing them as “ideal” is fine (“massless” is not).

But the frame of reference is separate from the observer. For an inertial observer there exists an inertial reference frame, called the observer’s frame, in which it is at rest and at the origin. But the reference frame is not part of the observer any more than your shoes are part of you.

An observer may use any reference frame, even more easily than you may change your shoes. That is the principle of relativity.

In this analysis I am deliberately dissociating the observers and the reference frames. I am deliberately using reference frames where the observers are not at rest. This is the principle of relativity.

Your diagrams do not show what I have described.

They describe exactly this:

I have simplified the example: it is enough for one twin to switch to the other's frame of reference once to make him age less

As shown in the diagrams this specific claim of yours is false.

This is what I said

You said a lot. That is not what I was reacting to. I was reacting to the specific statement I quoted. That is why the quote feature exists.

To compare the age (time elapsed since meeting) of the twins (by the "observer" described above), it is not necessary for them to be at the same place. The twins do not have to pull each other's beards. It is sufficient for them to rest in the same inertial frame, even if they are far apart in space

Not if you want the comparison to be frame invariant, as I showed.

to determine which time the clock of the clock line of B

The issue that you are still neglecting is that neither B nor A must use a line of clocks at rest with respect to them.

you are free to choose an inertial frame where either twin is older or where they are the same age

No

So you reject the principle of relativity. That is definitely “personal theory” territory.

 

[Peter Strohmayer]

He showed you explicitly how. I strongly advise you to be careful, as you are getting close to a warning at this point.

So you reject the principle of relativity. That is definitely “personal theory” territory.

Did the "age of twin A" mean the proper time of twin A elapsed since the meeting?
Was the proper time of a twin (his age) invariant?

 

[Peter Strohmayer]

This is impossible as you describe it.

Is it possible that each twin synchronizes its line of clocks according to Einstein's method? Is it possible that each of the twins considers the time, which their clocks indicate, when they pass each other later, as "zero point" of their time counts?

 

[Dale]

Did the "age of twin A" mean the proper time of twin A elapsed since the meeting?
Was the proper time of a twin (his age) invariant?

The proper time along any timelike worldline between two fixed events is indeed invariant.

The problem is that by having them meet only once you are only fixing one event. The other events are not fixed but depend on the frame-variant simultaneity convention.

In other words although $d\tau$ is unequivocally invariant, $\tau =\int_a^b d\tau$ depends on both of the end points $a$ and $b$. If both are invariant then $\tau$ is also invariant. But if either depend on the reference frame then $\tau$ also does.

They must meet twice in order to fix the two events $a$ and $b$.

Is it possible that each twin synchronizes its line of clocks according to Einstein's method?

Yes, and it is also possible that each twin uses an Einstein synchronized line of clocks that they are moving relative to. That is part of the principle of relativity that you seem to be missing.

 

[vanhees71]

An "observer" is an ideal, massless, imaginary measuring station at the center of a chosen frame of reference.
It leads to misunderstandings, to call a mass point an "observer", which cannot observe anything, but only produce a world line.

An "observer" is an ideal MASSIVE point, moving along a TIMELIKE worldline. You can define a local (in general non-inertial) non-rotating frame of reference by Fermi-Walker transport of a tetrad along this worldline. Note that "non-roting" is meant wrt. the observer. In general relative to a fixed inertial frame of reference this reference frame will be rotating. That's the famous Wigner rotation giving rise to the Thomas precession of spins.

 

[Peter Strohmayer]

you are only fixing one event. The other events are not fixed but depend on the frame-variant simultaneity convention

Again, I strongly advise you to be careful. You are getting close to a warning.

Isn't what happens above (#46 and #48)(ct0=0, x0=0; ct'=0, x'0=0) an event a?

Isn't what happens above (ct1=5, x1=3; ct'1=4, x'1=0) an event b?

Doesn't it follow from event B (because of ct'1=4) that twin A (green) has aged 4 years from event a to event b (proper time)?

Doesn't it follow from event B (because of ct1=5) that twin B (red) has aged 5 years from event a to event b (proper time)?

Is this true, even though twin B (red) (as a mass point at x=0) is not personally present at event b (although the world line of twin B does not intersect the world line of twin A)?

If yes, does this have something to do with the fact that the clock of twin B and the clock of his line of clocks which meets twin A at event b when he jumps (and equalize the relative speed of the meeting clock by recoil acceleration) run synchronously?

Wasn't the age difference caused by the one-time recoil acceleration?

Do the twins age at the same rate after event b?

Would this mean that without acceleration the age difference does not change after event b?

Is the elapsed proper time of A (green) of 4 years and the elapsed proper time of B (red) of 5 years between the events a and b the same from the point of view of all reference frames?

 

[Nugatory]

Isn't what happens above (#46 and #48)(ct0=0, x0=0; ct'=0, x'0=0) an event a?

Yes.

Isn't what happens above (ct1=5, x1=3; ct'1=4, x'1=0) an event b?

Yes.

Doesn't it follow from event B (because of ct'1=4) that twin A (green) has aged 4 years from event a to event b (proper time)?

Yes, because events A and B are both on twin A's worldline (and assuming that twin A's path between the two events is an inertial straight line).

Doesn't it follow from event B (because of ct1=5) that twin B (red) has aged 5 years from event a to event b (proper time)?

No. Event B is not on twin B's worldline so twin B's aging between events A and B is not defined (The spacetime interval between events A and B is of course defined, but it has nothing to do with twin B's age or proper time). If we choose an event C somewhere on twin B's worldline then the proper time between events A and C would be how much twin B aged between those two events.

Is this true, even though twin B (red) (as a mass point at x=0) is not personally present at event b (although the world line of twin B does not intersect the world line of twin A)?

No, because of the "no" answer above.

If yes, .....

We don't need to go here because the "If yes" condition is not satisfied.

Do the twins age at the same rate after event b?

"Same rate" is not a meaningful concept here. Post #11 of this thread explains why.

Would this mean that without acceleration the age difference does not change after event b?

It's not defined, and again see post #11.

Is the elapsed proper time of A (green) of 4 years and the elapsed proper time of B (red) of 5 years between the events a and b the same from the point of view of all reference frames?

All frames agree about the elapsed proper time between events A and B, and that this proper time is the amount by which twin A aged (because both events are on A's worldline and A followed a straight path between them). All frames also agree that this proper time has nothing to do with twin B's age.

 

[Peter Strohmayer]

Event b is not on twin B's worldline so twin B's aging between events a and b is not defined

Is it not permissible, then, to conclude that two children born at the same time in different places will always be the same age at any given time, from the point of view of the frame of reference in which they rest, because their world lines do not cross?

 

[Ibix]

Is it not permissible, then, to conclude that two children born at the same time in different places will always be the same age at any given time, from the point of view of the frame of reference in which they rest, because their world lines do not cross?

Using that frame only, yes. Using other frames, no. This is the point of the twin paradox: the twins meet, separate, and meet again, and everyone agrees on their accumulated age difference between the meetings. In your example, almost everyone disagrees.

 

[Peter Strohmayer]

Using that frame only, yes. Using other frames, no.

Is not the age of the children (their proper time) an invariant?

 

[Ibix]

Is not the age of the children (their proper time) an invariant?

There age when? It's the "when" that is not invariant unless they meet.

 

[Sagittarius A-Star]

Is not the age of the children (their proper time) an invariant?

two children born at the same time in different places

And because they were born at different places, it is not invariant, if they were born at the "same time".

 

[Nugatory]

Is it not permissible, then, to conclude that two children born at the same time in different places will always be the same age at any given time, from the point of view of the frame of reference in which they rest, because their world lines do not cross?

It's neither more nor less permissible than concluding that one of them will always be older than the other when we calculate "ages" using coordinates in which they are both moving at a constant non-zero velocity - it all depends on our essentially arbitrary choice of coordinates.

When you inserted that additional "from the point of view" qualifier you redefined "same age" so that it is frame-dependent and no longer has any physical significance, and furthermore chose a redefinition that makes it true by definition: define "same age" to mean "has the same time coordinate" and then choose time coordinates that advance at the rate as proper time and of course they will age at the same rate.

 

[robphy]

Is not the age of the children (their proper time) an invariant?

The proper-time age of a child is the elapsed time on its wristwatch between two events A_1 and A_2 on that child's worldline (akin to an arc length along a specific curve between two points). For two specified events along that worldline, that elapsed proper-time is invariant.

To compare the ages of two children,
you have to specify events A_1 and A_2 on child-A's worldline
and events B_1 and B_2 on child-B's worldline.

For distinct spacelike-separated events A_1 and B_1 ,
not all observers will regard A_1 and B_1 as simultaneous (i.e. as "being at the same time")
.... and similarly for A_2 and B_2.

 

[PeroK]

When I first studied SR, I assumed that acceleration must be the ultimate cause of the differential ageing in twin paradox. Then I learned about Minkowski spacetime. Then someone on this forum posted the idea of replacing the turnaround with the communication of a clock reading to an inbound spacecraft. And it was clear that acceleration was only a physical constraint and a geometric red herring.

This process of continually updating one's knowledge of a subject is called learning. It's the opposite of religiously adhering to an established view in the face of evidence to the contrary.

 

[Dale]

born at the same time in different places

This is a frame variant concept.

from the point of view of the frame of reference in which they rest

In that frame it is true. Not in other frames. They do not need to use the frame in which they are at rest.

Do you accept the principle of relativity?

 

[Dale]

And it was clear that acceleration was only a physical constraint and a geometric red herring.

Yes. The acceleration breaks the symmetry, but it is not the only way to break the symmetry. Acceleration is neither necessary nor sufficient for different aging, so it cannot be the cause.

 

[Dale]

Is not the age of the children (their proper time) an invariant?

As I mentioned previously $d\tau$ is an invariant. $\tau=\int_a^b d\tau$ is invariant only if $a$ and $b$ are.

 

[Peter Strohmayer]

And because they were born at different places, it is not invariant, if they were born at the "same time".

When you inserted that additional "from the point of view" qualifier you redefined "same age" so that it is frame-dependent and no longer has any physical significance,

For distinct spacelike-separated events A1 and B1 ,
not all observers will regard A1 and B1 as simultaneous

This process of continually updating one's knowledge of a subject is called learning. It's the opposite of religiously adhering to an established view in the face of evidence to the contrary.

This is a frame variant concept.

τ=∫abdτ is invariant only if a and b are.

From the point of view of reference system B, event a is simultaneous with event a' and event b is simultaneous with event b'.

Does an invariant proper time elapse for a mass point B moving - from the point of view of reference system B - on a straight world line on the time axis from event a to event b?

Does no invariant proper time elapse for a mass point B' moving - from the point of view of reference system B - on a straight world line parallel to the time axis from event a' to event b'?

Does this have to do with the fact that the events a and a' and the events b and b' occur simultaneously from the point of view of the reference system B, but not from the point of view of other reference systems?

 

[Dale]

Asked and answered. This thread is closed.