Twins' paradox and acceleration

[neopolitan]

"twins' paradox" and acceleration

Sorry to create another post with the twins' paradox as a central issue but I would like to explore one issue involved in it.

Specifically, I have been told elsewhere that I am misunderstanding a claim about the twin paradox.

As I understand it, the twins' paradox is not a paradox at all. There is a supposed symmetry which either does not exist or which is broken. The symmetry can be broken by acceleration which changes the rest frame for one observer, usually three times:

travelling twin (acceleration 1) : rest -> v (relative to stationary twin)
travelling twin (acceleration 2) : v -> -v (relative to stationary twin)
travelling twin (acceleration 3) : -v -> rest (relative to stationary twin)

My point of view is that it is this break in symmetry which leads to the apparent paradox (not real paradox). A break in symmetry due to another cause would lead to the same result.

But, as I understand it, some claim that it is the actual acceleration itself leads to less proper time experienced. In this claim (as I have perceived it), whichever observer undergoes acceleration experiences less time elapsed, irrespective of distances traveled or timing of accelerations. In other words, the acceleration modifies the world line of the affected observer.

So, my question is:

Which is true?

1. the symmetry break causes one observer to experience less time elapsed than the other, and which experiences less time elapsed is dependant on the scenario as a whole

-OR-

2. the acceleration causes the accelerated observer to experience less time elapsed than the other, and the symmetry break is just another symptom of the acceleration

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

I do hope the question is sufficiently clear. I would appreciate getting a few people to respond. While physics is not democratic, it is however useful to see what the majority view is.

cheers,

neopolitan

 

[granpa]

if we make the simplifying assumption that one really is at rest and the other really is moving then relativity shows that the stationary one sees the moving one as length contracted and time dilated because he really is.

the moving one sees the nonmoving one as length contracted and time dilated because the moving one experiences a loss of simultaneity. this illusion is lost when he stops.

more complex versions can be created where both are moving but the results are always the same. the one that accelerates is the one that ages less.

 

[matheinste]

Quote

----whichever observer undergoes acceleration experiences less time elapsed, irrespective of distances traveled or timing of accelerations. In other words, the acceleration modifies the world line of the affected observer.-----

Yes but the acceleration itself does not affect the time dilation but the fact that it causes a different spacetine path to be taken does.

Thats how i see it.

Matheinste.

 

[robphy]

So, my question is:

Which is true?

1. the symmetry break causes one observer to experience less time elapsed than the other, and which experiences less time elapsed is dependant on the scenario as a whole

-OR-

2. the acceleration causes the accelerated observer to experience less time elapsed than the other, and the symmetry break is just another symptom of the acceleration

Here is the spacetime-geometric interpretation of the situation... which hopefully amplifies the key points.

The elapsed-proper-time of an observer (as read off from his wristwatch) is the arc-length of his worldline in spacetime.

Fix once and for all event A (the separation event) and event B (the reunion event),
as if one marked two points on a piece of paper.

There are many ways to experience both A and B.
That is, there are many worldlines that pass through both A and B.

In Minkowski spacetime,...

- there is one inertial observer that can experience both A and B, and that observer records the longest elapsed proper-time (as read off from his wristwatch). That is, there is one geodesic (one straight worldline) that meets both A and B, and this worldline has the longest arc-length (using the Minkowski metric)

- every other observer experiencing both A and B will not be an inertial observer, and such an observer will record a shorter elapsed proper-time (as read off from his own wristwatch). That is, every other worldline meeting both A and B will not be a geodesic and will not be a straight worldline... and thus must have at least a portion that is curved (a smooth portion representing a finite acceleration, or a kink representing an impulsive acceleration).
--Note that a kink itself will not contribute any additional arc-length to the worldline.
(That is, the impulsive acceleration at an event (say, a turnaround event) will not change (i.e., "will not cause a jump in") the reading of that observer's wristwatch experiencing that event. [Slightly off topic... what will change are the sets of distant events that this observer will regard as simultaneous. Possibly interesting... but secondary... and often a distraction from the key point.])
--[He will be a non-inertial observer and will be able to detect this because an ice cube placed over a spot on his frictionless table will be displaced at some time during the trip from A to B.]

The supposed paradox is due to ignoring the distinction between the inertial and noninertial observer... and falsely thinking that it is sufficient to consider "being at rest (in your own reference frame)". Said another way... "being at rest (in your own reference frame)" does not make you an inertial observer... especially when an ice cube on your table suddenly gets displaced.The symmetry break (between inertial and noninertial) is the "presence of an acceleration (worldline curvature) somewhere during the trip" for the noninertial observer. Neither are causes of the shorter-elapsed-proper-time from A to B... they are correlated with the shorter-elapsed-time because they indicate that a noninertial (i.e. nongeodesic) worldline was used to experience both A and B.

 

[neopolitan]

robphy,

It is not quite how I would have expressed it, but I agree entirely.

Just to make sure that I am agreeing with the right thing, is this right? ...

I observe two other observers, Fred with a velocity of v (relative to me), George with a velocity of 2v (relative to me).

At some stage George stops (relative to me) and Fred catches up.

In this case, Fred is inertial the whole time and, when they meet up, Fred will have a clock which reads more elapsed time than George's.

Alternatively, Fred speeds up to 3v and catches up to George.

In this case, George is inertial the whole time and when they meet up, George's clock will read more than Fred's.

cheers,

neopolitan

 

[Dale]

The nice thing about the spacetime geometric approach is that you can perform the analysis in any reference frame, using any coordinate system, with any number of "twins", for any arbitrary scheme of acceleration and deceleration, using SR or GR, and always get the right answers.

I always prefer simple general answers to complicated case-specific answers. The "acceleration based" explanations are very dissatisfying once you have been exposed to the geometric explanation.

 

[MeJennifer]

The main reason people have so much trouble with things like the twin scenario is because they have a hard time letting go of absolute time. Also the notion of clocks slowing down actually perpetuates the idea of absolute time. The fact is that there is no such thing as absolute time, all observers have their own measure of time and all their clocks run at the same rate.

The twin case is a situation where two people want to go from A to B using different paths in spacetime. The straightest possible path takes the most time while all other paths take less time. A person who accelerates does not travel on the straightest possible path and thus takes less time for his trip from A to B.

 

[neopolitan]

The nice thing about the spacetime geometric approach is that you can perform the analysis in any reference frame, using any coordinate system, with any number of "twins", for any arbitrary scheme of acceleration and deceleration, using SR or GR, and always get the right answers.

I always prefer simple general answers to complicated case-specific answers. The "acceleration based" explanations are very dissatisfying once you have been exposed to the geometric explanation.

Again I agree.

I have revisited the scenario where you have two buoys communicating with each other and moving past a "stationary" observer. The idea was to simulate the supposed "rest frame" view point of the traveling twin (ie the idea that the entire universe moved in one direction until the traveling twin arrived at the destination, then the entire universe changed direction so that the traveling twin would end up back at the point of departure).

In their rest frame, two buoys (Buoy 1 and Buoy 2) are a distance of L apart, with Buoy 1 intially collocated with a third buoy (Buoy 3). But when both have a velocity of v, relative to Buoy 3, then according to Buoy 3, they have a separation of L' = L. sqrt (1 - v^2/c^2).

...----- v ---->
2 __________________ 1
_........ 3

Let's assume all start off at rest. Buoy 1 sends a signal to Buoy 2 and after a period of L/c, both as good as instantaneously achieve a relative velocity of v (relative to Buoy 3).

According to Buoy 3, it takes a period of L/v before buoy 2 arrives.

According to Buoy 2, it takes a period of L'/v before buoy 3 arrives.

Buoy 2 immediately changes direction (as good as instantaneously changing relative velocity to -v). From now on, Buoy 2 can sail off into the sunset.

After a period of L/v after the initial massive acceleration, Buoy 1 assumes that Buoy 2 has passed Buoy 3 - note that this is a period of L/v on Buoy 1's clock. Buoy 1 then accelerates so as to travel back to Buoy 3 with a relative velocity of -v.

According to Buoy 1, the total period spent traveling will be 2L/v.

According to Buoy 3 though, the total period spent waiting for Buoy 1 to come back would be greater. Careful working through of the maths should reveal that the time elapsed for Buoy 3 before Buoy 1 comes back is (L/sqrt (1 - v^2/c^2))/v. The cause of the longer elapsed time is in part due to the fact that Bouy 1 actually traveled further, waiting longer before turning back (Buoy 1 and Buoy 2 lost their fixed distance when Buoy 2 turned around after a period of L'/v).

Additionally, Buoy 3 was inertial the whole time. The geometric explanation would probably reveal this far more simply.

I fully accept that anytime that I have argued anything else than this, I was wrong.

cheers,

neopolitan

 

[Dale]

I wouldn't say that the "acceleration based" explanations are wrong. They are just unnecessarily complicated and not very general.