Twin Paradox Resolution -- Inertial frames

[Jun Pan]

No, this is wrong precisely because of the simultaneity of relativity. The spaceship does not have the same rest frame and applying the time dilation formula without taking relativity of simultaneity into account, you will be missing a large part of the time for the Earth observer. The spaceship changes system and that changes which event on the Earth is simultaneous with the turnaround.

1, For Earth (x,y,z,t) chosen as 'rest' frame, spaceship (x',y',z',t') move away at speed v; for spaceship(x',y',z',t') chosen as 'rest' frame, Earth (x,y,z,t) moves away at speed -v. 2, When t=t'=0, let x=x'=0, or let two coordinate systems have the same start point. 3, According to the principle of relativity, physical law should be the same in each system except the v change sign and they have simultaneous time at the beginning.
So I do not think there is problem of simultaneity. If the twin on the Earth thinks the twin in the spaceship is younger, the twin in the spaceship also thinks the twin on the Earth is younger. Otherwise, the principle of relativity would be violated. I think there maybe problem in your geometry view. From (x,y,z,t), the turning point of spaceship is the halfway point (xh) in (x,y,z,t). From (x',y',z',t'), the corresponding point of xh in (x',y',z',t') has larger value than xh, written as x'(xh) (x'(xh)>xh). However x'(xh) is the coordinate in (x',y',z',t') and it seems to me in your geometry view, x'(xh) is treated as it is in (x,y,z,t). I think this is the cause of 'simultaneity problem' you mention, which I think is not right.

 

[Orodruin]

When t=t'=0, let x=x'=

You cannot generally define t=t'=0. That t=t' holds only at one event along t=0, the event x=0.

According to the principle of relativity, physical law should be the same in each system except the v change sign and they have simultaneous time at the beginning.

Wrong again. You are misinterpreting the relativity principle. Events being simultaneous is not a physical law - it is in direct contradiction to the law of the speed of light being invariant.

 

[Dale]

The two coordinate systems are symmetrical.

No. They are not. One system is inertial and the other is not. Both observers can identify and agree which is inertial. Thus they are not symmetrical.

If to chose the spaceship as the rest inertial frame

This is impossible. The spaceship's rest frame is not inertial

 

[Jun Pan]

No. They are not. One system is inertial and the other is not. Both observers can identify and agree which is inertial. Thus they are not symmetrical.

During the period of acceleration, the observer in system 1 thinks the system 1 is inertial while the system 2 is not inertial system, and the observer in system 2 thinks system 2 is inertial while the system 1 is not inertial. This situation is symmetrical for two systems.

 

[Orodruin]

During the period of acceleration, the observer in system 1 thinks the system 1 is inertial while the system 2 is not inertial system, and the observer in system 2 thinks system 2 is inertial while the system 1 is not inertial. This situation is symmetrical for two systems.

No. Again you are simply wrong. Proper acceleration is something measurable - it is what an accelerometer measures. The observer going away can objectively deduce that it is accelerating by looking at an accelerometer. Anyway, the problem is not about the acceleration, it is about the geometry of space-time.

If you agree that the green curve is longer in this image

then you must also agree that the observer going away will be younger at the reunion. The mathematics is exactly equivalent.

 

[Jun Pan]

The observer going away can objectively deduce that it is accelerating by looking at an accelerometer.

When the observer going away looks at an accelerometer to deduce that the spaceship is accelerating means this observer does not choose the spaceship as the rest frame, but instead chose other frame such as the frame the Earth rests as the rest frame.

If you agree that the green curve is longer in this image

Yes, the green line is longer. However in different coordinate system, different observer chooses the green line as the world line of his or her move.

 

[Jun Pan]

You cannot generally define t=t'=0. That t=t' holds only at one event along t=0, the event x=0.

Yes. When the spaceship rests on the earth, in the Earth coordinate (x,y,z,t), the simultaneity is defined without any doubt in the Earth coordinate system (x,y,z,t). So the moment the spaceship flies can be chosen as t=0 in (x,y,z,t), t'=0 in (x'.y',z',t'),and choose the spatial point x=0 in (x,y,z,t), x'=0 in (x',y',z',t').

Events being simultaneous is not a physical law - it is in direct contradiction to the law of the speed of light being invariant.

It is supposed to me simultaneity is described by light ray propagation which links to the time space positions and so links to the movement of objects.

 

[Dale]

During the period of acceleration, the observer in system 1 thinks the system 1 is inertial while the system 2 is not inertial system, and the observer in system 2 thinks system 2 is inertial while the system 1 is not inertial. This situation is symmetrical for two systems.

This is simply untrue. An inertial system is one which obeys the law of inertia. That is that a force-free object (0 accelerometer reading) travels in a straight line at a constant speed.

Both observers agree that system 1 is inertial and that system 2 is not.

 

[vanhees71]

The proper time of each of the twin is a scalar and thus there's never ever a paradox about the age of the twins within special or general relativity. It's just a simple integral along the trajectory of each twin,
$$\tau = \int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu}},$$
which is a generally covariant expression, i.e., a scalar. That's all there is to say about the "twin paradox".

Usually all the geometric diagrams are more confusing than helping, because first you have to learn how to read a Minkowski diagram (in the special relativistic case), i.e., to say good bye to your good old Euclidean view on the plane where it's drawn!

 

[Orodruin]

Yes. When the spaceship rests on the earth, in the Earth coordinate (x,y,z,t), the simultaneity is defined without any doubt in the Earth coordinate system (x,y,z,t). So the moment the spaceship flies can be chosen as t=0 in (x,y,z,t), t'=0 in (x'.y',z',t'),and choose the spatial point x=0 in (x,y,z,t), x'=0 in (x',y',z',t').

It is supposed to me simultaneity is described by light ray propagation which links to the time space positions and so links to the movement of objects.

You have had it explained to you by several experts and still do not understand where you are going wrong. Your last sentence is very incoherent and does not add any kind of information on how you are actually thinking. You seem to be stuck in a Newtonian view with absolute simultaneity or to hold the misconception that simultaneous events are events whose light arrives at some arbitrary point at the same time - this is not what we mean when we talk about simultaneity.

I suggest you start from the beginning and study the basic concepts of special relativity because, unfortunately, it is clear to me that you have missed some very crucial and basic points.

 

[Orodruin]

So the moment the spaceship flies can be chosen as t=0 in (x,y,z,t), t'=0 in (x'.y',z',t'),and choose the spatial point x=0 in (x,y,z,t), x'=0 in (x',y',z',t').

The crucial point here is that events for which t=0 but x ≠ 0, you will no longer have t' = 0. This is the essence of simultaneity of relativity.

 

[PeterDonis]

When the observer going away looks at an accelerometer to deduce that the spaceship is accelerating means this observer does not choose the spaceship as the rest frame

This is not correct. Accelerometer readings are direct observables; all observers agree on what a given accelerometer reads at a given event. There is no need to choose a frame in order to look at an accelerometer and see what it reads.

The fact that the spaceship's accelerometer reads nonzero for at least some portion of its trip, whereas the stay-at-home twin's accelerometer always reads zero, is a physical asymmetry between them, regardless of anyone's choice of frame.