Twin Paradox (thorough explanation needed)

In summary, the twin paradox is not actually a paradox because it involves a frame of reference change due to the acceleration of the spaceship. This can be further illustrated through a thought experiment involving a spaceman traveling to a star and staying there, while trying to determine the current year on Earth. It is important to note that the separation between the Earth and spaceship at the time of acceleration is what matters in this scenario.
  • #176
DaleSpam said:
Hi LightNg, welcome to PF,

It does break the symmetry since an accelerometer aboard one ship will detect the acceleration and an accelerometer aboard the other ship will not. That is asymmetric. This highlights the difference between coordinate acceleration (which is symmetric) and proper acceleration (which is asymmetric).
Yeah, and more generally the time dilation equation relating clock rate to speed only works in inertial frames, and it can be taken as part of the definition of an inertial frame that any object with a constant position coordinate (one which is "at rest" in that frame) should show zero reading on an accelerometer (not experiencing any G-forces).
 
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  • #177
Thanks for the replies! :)

DaleSpam said:
Hi LightNg, welcome to PF,

It does break the symmetry since an accelerometer aboard one ship will detect the acceleration and an accelerometer aboard the other ship will not. That is asymmetric. This highlights the difference between coordinate acceleration (which is symmetric) and proper acceleration (which is asymmetric).

Yeah, and more generally the time dilation equation relating clock rate to speed only works in inertial frames, and it can be taken as part of the definition of an inertial frame that any object with a constant position coordinate (one which is "at rest" in that frame) should show zero reading on an accelerometer (not experiencing any G-forces).

Oh. So the present of G-force should be taken into consideration.

So how about this one hypothetical example:

The twins are now in somewhere in space(vacum) separated by 1 meter with each other. There is not any planet near by. They are not in any spaceship. They only wear spacesuits in space(vacum). They experience no g-force now. They both have an accelerometer than show reading of no g-force. They both also have a clock with each of them. And they have synchronized their clocks with each other.

Now suppose twin A is on the left side, twin B on the right side.

And now suddenly there is a strong gravitational field appear to the left of 10km of twin A, and the gravitational field pulls twin A to accelerate to the left in coordinate acceleration to the speed of 0.3C in says, 10 seconds.

And also there is another gravitational field with the same strength appear to the right of 10km of twin B, and also pulls twin B to accelerate to the right in coordinate acceleration to the speed of 0.3C in 10 seconds.

The 2 gravitational forces cancel each other out in the middle points of the 2 twins. So the left gravitational force only pulls the twin A but not twin B, and the right gravitational force only pulls the twin B but not twin A.

And after the twin A and twin B accelerates to 0.3C in opposite directions, the 2 gravitational forces disappear. So now twin A and twin B travel in constant speed of 0.3C in opposite directions.

So after twin A and twin B travels for 1 light year in opposite directions, their distance is 2 light years.

So suppose now there is very strong gravitational field appears in the middle point of the twins. And it pulls twin A and twin B to deaccelerate, to halt, and accelerate to 0.3C back toward the middle graviational field. (all in coordinate acceleration)

After twin A and twin B accelerate to 0.3C back toward the middle point, the middle gravitational field also disappear. So now twin A and twin B travels in constant speed of 0.3C towards each other. Eventually they meet again at the middle point.

So now. In this example there is no g-force. Twin A and twin B appears to be symmetry. Their accelerometers show g-force at all. Now which one of them have the clock run slower?



And there is another example in my original post that is not answered yet.
And another simpler question. Consider there is 2 object A and B. And they exists right from the beginning of the universe (forget about Big Bang). And right from the beginning of the universe, from the point of view of object A, object B is, and always, traveling at the contant speed of 0.5C, never accelerate or deaccelerate. So which object experience time slower? A or B?

In this example object A and object B also appear to be symmetry. There is no acceleration or deacceleration at all.

I would like to see answer to this example as well.
 
  • #178
If there's a gravitational field, then in relativity you need the theory of general relativity which explains gravity in terms of curved spacetime, not special relativity which assumes non-curved spacetime. And no coordinate system covering a substantial region of curved spacetime qualifies as "inertial" so the SR time dilation equation won't apply (though in a very small region a freefalling observer can have a "locally inertial frame" according to the http://www.einstein-online.info/spotlights/equivalence_principle). As for your second question:
LightNg said:
And another simpler question. Consider there is 2 object A and B. And they exists right from the beginning of the universe (forget about Big Bang). And right from the beginning of the universe, from the point of view of object A, object B is, and always, traveling at the contant speed of 0.5C, never accelerate or deaccelerate. So which object experience time slower? A or B?
In this case there is no objective frame-independent answer to who experiences time slower, in the rest frame of object A it is B's clock that's running slower, while in the rest frame of object B it is A's clock that is running slower. This is related to the relativity of simultaneity, which says different frames disagree about which events separated in space are "simultaneous". Suppose the two observers pass by each other at some point, and both their clocks read T=100 years at that moment. Then in the frame of object A, 50 years earlier when his clock read T=50, that event was simultaneous with the event of object B's clock reading 100 - 0.866*50 = 56.7 years, so according to object A, object B only aged 0.866 times as much as he did between the time his clock said T=50 and the time they met when his clock read T=100. But in object B's frame, the event of his clock reading 50 was simultaneous with A's clock reading 56.7, and the event of his clock reading 56.7 was simultaneous with A's clock reading 62.5, so in B's frame it is A who is only aging 0.866 times as much as B.
 
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  • #179
Ok now i see the picture. In the rest frame of object A, when A's clock read T = 200 years (100 years in the future), B only experiences 86.6 years in the future.
And vice versa.

Thanks a lot.

By the way, relativity of simultaneity only apply in non curved spacetime?
 
  • #180
LightNg said:
By the way, relativity of simultaneity only apply in non curved spacetime?
In curved spacetime you no longer have global inertial frames as I said, but there are an infinite variety of non-inertial coordinate systems you can choose where the laws of GR still work (see the comments here on "diffeomorphism invariance"), so simultaneity is still relative to your choice of coordinate system.
 
  • #181
JesseM said:
[...]

You are a fairly good approximation of a "perpetually inertial observer". So, according to your previous comments, you shouldn't have a strong preference for which inertial frame you personally (usually) choose for your own "point-of-view" (POV).

You could choose to use the ordinary wristwatches and measuring tapes that you can buy at numerous retail stores here on Earth (which provide the time and spatial coordinates which Einstein chose to use for any given one of his inertial frames, in his 1905 paper, OR you could choose to use the coordinates of some particular inertial frame that is moving at some given constant velocity very near the speed of light, relative to the (approximately) inertial frame of the Earth.

If you make the latter choice, you will need to acquire and wear a wristwatch that doesn't tick at the same rate that my Timex ticks at. And you will need a measuring tape that is different from those that I've bought at Lowe's hardware store. In fact, you will need to buy one tape for measuring distances along the direction of relative motion, and another tape for measuring distances perpendicular to that direction (and even additional tapes, if you want to measure distances along various other angles to the direction of relative motion).

Presumably, from your previous comments, you really don't have much preference for which of the above two alternative POV's you decide to choose ... they are both equally good.

So, which of those two types of wristwatch have you been using, during your life, so far? Which type of measuring tape?

Mike Fontenot
 
  • #182
Mike_Fontenot said:
[...]
Mike's even worse off than I thought he was.
 
  • #183
Mike_Fontenot said:
You are a fairly good approximation of a "perpetually inertial observer". So, according to your previous comments, you shouldn't have a strong preference for which inertial frame you personally (usually) choose for your own "point-of-view" (POV).

You could choose to use the ordinary wristwatches and measuring tapes that you can buy at numerous retail stores here on Earth (which provide the time and spatial coordinates which Einstein chose to use for any given one of his inertial frames, in his 1905 paper, OR you could choose to use the coordinates of some particular inertial frame that is moving at some given constant velocity very near the speed of light, relative to the (approximately) inertial frame of the Earth.

If you make the latter choice, you will need to acquire and wear a wristwatch that doesn't tick at the same rate that my Timex ticks at. And you will need a measuring tape that is different from those that I've bought at Lowe's hardware store. In fact, you will need to buy one tape for measuring distances along the direction of relative motion, and another tape for measuring distances perpendicular to that direction (and even additional tapes, if you want to measure distances along various other angles to the direction of relative motion).
Or I could just use the ruler and clock at rest relative to me, and do some simple calculations to figure how any given ruler/clock reading translates into the coordinates of a frame moving at high velocity relative to me. If you want to restrict me to using a coordinate system where I can't do any calculations of this sort, where the coordinates of a given event must be identical to the reading on some physical ruler and clock that were right next to the event when it happened, that would be a new "rule" you never mentioned before (which is itself just an arbitrary aesthetic choice), and applying the same criterion to your CADO system would show it would actually be quite complicated to construct a set of ruler/clocks whose readings matched that system, arguably more so than some other non-inertial systems where (as with CADO) you're at rest at the origin.
 
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  • #184
JesseM said:
If there's a gravitational field, then in relativity you need the theory of general relativity which explains gravity in terms of curved spacetime, not special relativity which assumes non-curved spacetime. And no coordinate system covering a substantial region of curved spacetime qualifies as "inertial" so the SR time dilation equation won't apply (though in a very small region a freefalling observer can have a "locally inertial frame" according to the equivalence principle). As for your second question:

Doesn't mean to interrupt the conversation, but I have another question.

Suppose the hypothetical example of the twin A and twin B given by me, doesn't involve gravitational field, but instead the gravitational field is replaced by electromagnetic force.

So say the twin A and twin B are actually just 2 magnets (that each has an accelerometer and a clock attached to it), they are not human.
So instead of being pulled by gravity, the magnets(A and B) are pulled by other strong big magnets (that appear and disappear), so the magnets (A and B)depart from each other, then reunite in the middle point again because of magnetic force.

So now, this is non curved space time, so time dilation equation applies. And still no g-force in effect. It appears that A and B is symmetry here. So now which clock run slower? A or B?

Or will both clock run slower? That is, relativity of simultaneity is in effect?
But if relativity of simultaneity is in effect, what the clocks of A and B show when they reunite? Note that the clocks have been synchronized in the beginning of the example, when A and B are 1 meter apart.
 
  • #185
LightNg said:
Doesn't mean to interrupt the conversation, but I have another question.

Suppose the hypothetical example of the twin A and twin B given by me, doesn't involve gravitational field, but instead the gravitational field is replaced by electromagnetic force.

So say the twin A and twin B are actually just 2 magnets (that each has an accelerometer and a clock attached to it), they are not human.
So instead of being pulled by gravity, the magnets(A and B) are pulled by other strong big magnets (that appear and disappear), so the magnets (A and B)depart from each other, then reunite in the middle point again because of magnetic force.

So now, this is non curved space time, so time dilation equation applies. And still no g-force in effect.
In this case, the accelerometer would measure a nonzero g-force, the only way to not feel g-forces is if there aren't any non-gravitational forces acting on you (either inertial motion in flat spacetime, or freefall in a gravitational field)
 
  • #186
LightNg said:
So now. In this example there is no g-force. Twin A and twin B appears to be symmetry. Their accelerometers show g-force at all. Now which one of them have the clock run slower?
This example is completely symmetric, so their clocks will be equal when they meet again. However, as JesseM mentioned, actually calculating this would require GR and the simplified equations of SR would not apply.

LightNg said:
So now, this is non curved space time, so time dilation equation applies. And still no g-force in effect. It appears that A and B is symmetry here. So now which clock run slower? A or B?
The accelerometers would read non-zero forces, but there is still symmetry so the clocks will be equal when they meet again.
 
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  • #187
So, for mutual time dilation to occur without contradiction, I take that it is not possible for 2 objects (object A and object B in my example) to meet again after they depart, at least in the current framework of SR?

What if suddenly a magical portal appear in front of object B, and object B crosses the portal and meet with object A again?
Then what will their clocks read? Hahahaa
 
  • #188
LightNg said:
So, for mutual time dilation to occur without contradiction, I take that it is not possible for 2 objects (object A and object B in my example) to meet again after they depart, at least in the current framework of SR?
Correct, this is a simple geometrical fact. In a flat space two straight lines which are initially diverging can never intersect.
 
  • #189
DaleSpam said:
Correct, this is a simple geometrical fact. In a flat space two straight lines which are initially diverging can never intersect.

I know bout this geometry, But is there any possibility, or is there any observed phenomena in the universe, that such line be not continuous in space?
Specifically, that the observed movement of an object is not continuous in space?

Ok this is not quite relevant to the topic of this thread, so someone may want to move this post to other thread/forum.
 
  • #190
LightNg said:
So, for mutual time dilation to occur without contradiction, I take that it is not possible for 2 objects (object A and object B in my example) to meet again after they depart, at least in the current framework of SR?

What if suddenly a magical portal appear in front of object B, and object B crosses the portal and meet with object A again?
Then what will their clocks read? Hahahaa
Well, in this case the answer would depend on where in time the portal deposited object B! Because of the relativity of simultaneity, you can't just say that the portal transports B to A's "current" location, because there is no objective-frame independent fact about which event on A's worldline (which reading on its clock) is simultaneous with the event of B first reaching the position of the portal.
 
  • #191
LightNg said:
I know bout this geometry, But is there any possibility, or is there any observed phenomena in the universe, that such line be not continuous in space?
Specifically, that the observed movement of an object is not continuous in space?
Not yet. Even for quantum mechanical systems the wavefunction is continuous in space and time.
 

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