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If there is good definition of simultaneity, the earth coordinate system and the spaceship coordinate system are symmetrical and the so the time dilation effects are symmetrical for these two systems.
If there is good definition of simultaneity, the earth coordinate system and the spaceship coordinate system are symmetrical
the time dilation effects are symmetrical for these two systems.
Yes, the definition of simultaneity is coordinate-dependent. There are two coordinate systems considered here or the earth coordinate system (x,y,z,t) and the spaceship coordinate system (x',y',z',t'). Define when t=t'=0, x=x'=0. Spaceship is always at rest in the spaceship coordinate system. The two coordinate systems are symmetrical. So, (x',y',z',t') moves at v relative to (x,y,z,t), while (x,y,z,t) moves at -v relative to (x',y',z',t'), and the time dilation effects are symmetrical for the two systems.The definition of simultaneity is coordinate-dependent; it's not an absolute. And in the standard twin paradox, there is no single inertial frame in which the spaceship is always at rest, so the "earth coordinate system" and the "spaceship coordinate system" are certainly not symmetrical.
It is supposed to me when the two twins meet, the twin backs from the travel will not be older than the twin on the earth.The relative aging of the two twins when they meet up again, in the standard twin paradox, is not symmetrical; the traveling twin ages less. And this fact is independent of any choice of coordinates. So again I'm not sure what you mean here.
The twin going away will be younger! (Upon return.)It is supposed to me when the two twins meet, the twin backs from the travel will not be older than the twin on the earth.
I am sorry there is a mistake of my last post. It is supposed to me the twin going away will not be younger than the twin on the earth.The twin going away will be younger! (Upon return.)
Then you are simply wrong. Please read my insight A Geometrical View of Time Dilation and the Twin Paradox.I am sorry there is a mistake of my last post. It is supposed to me the twin going away will not be younger than the twin on the earth.
I have read your insight. If to chose the spaceship as the rest inertial frame, from the view of the twin in the spaceship, the twin on the earth is younger. When the two twins meet, neither one is younger.Then you are simply wrong. Please read my insight A Geometrical View of Time Dilation and the Twin Paradox.
No, you are wrong. You have not understood the geometrical perspective at all. Reread the part where I explain (in the standard geometry case) why the "relativity of the x-coordinate" solves the problem by showing that the point on the straight line with the same x' coordinate as the turning point is not the half-way point. The twin paradox works exactly the same way.I have read your insight. If to chose the spaceship as the rest inertial frame, from the view of the twin in the spaceship, the twin on the earth is younger. When the two twins meet, neither one is younger.
True, the turning point is the half way point observed from the earth (x,y,z,t). But it seems to me that the turning point observed from the spaceship (x',y',z',t') can not be regarded as not the half way point. If according to the (x,y,z,t), the turning point observed from (x',y',z',t') is not the half way point defined in (x,y,z,t), but should be the half way point defined in (x',y',z',t').No, you are wrong. You have not understood the geometrical perspective at all. Reread the part where I explain (in the standard geometry case) why the "relativity of the x-coordinate" solves the problem by showing that the point on the straight line with the same x' coordinate as the turning point is not the half-way point. The twin paradox works exactly the same way.
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 space ship changes system and that changes which event on the Earth is simultaneous with the turnaround.So at the turning point, the observer in the spaceship should observe the earth moves at the halfway point defined in the (x',y',z',t').
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.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 space ship changes system and that changes which event on the Earth is simultaneous with the turnaround.
You cannot generally define t=t'=0. That t=t' holds only at one event along t=0, the event x=0.When t=t'=0, let x=x'=
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.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.
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.The two coordinate systems are symmetrical.
This is impossible. The spaceship's rest frame is not inertialIf to chose the spaceship as the rest inertial frame
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. 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.
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.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.
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.The observer going away can objectively deduce that it is accelerating by looking at an accelerometer.
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.If you agree that the green curve is longer in this image
Yes. When the space ship 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').You cannot generally define t=t'=0. That t=t' holds only at one event along t=0, the event x=0.
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.Events being simultaneous is not a physical law - it is in direct contradiction to the law of the speed of light being invariant.
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.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.
Yes. When the space ship 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.
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.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').
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