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neopolitan
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For the twin paradox to be considered a true paradox the framing of the scenario must be stringent, that is to say we cannot permit assumptions to be ignored. Therefore I must start with a short description of the twin paradox followed by identification of the inherent assumptions.
From Einstein Light (http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm):
(The author goes on to explain that asymmetry resolves the paradox, this is an not entirely satisfactory explanation.)
There are a few assumptions, which are perhaps only obvious when one takes time to search for them.
"(S)ome distant location" appears sufficiently vague as to avoid creating problems but an inherent assumption is that this location shares the same frame as Joe.
By placing Joe on Earth we hide the other assumption, which is that we also share the same frame as Joe.
The at-a-distance observations of each other's clock potentially hides an assumption of instantaneous transmission of information. I doubt the author intended that, but it must be remembered that information is not transmitted instantaneously.
"(Jane) decelerates and returns" is distracting. As the author correctly points out this is a point of asymmetry. But a similar scenario (to be shown shortly) while show that it doesn't matter which decelerates and changes direction - Jane or the entire universe. It is generally assumed that the period during Jane changes direction is insignificant enough to ignore.
Finally, "Jane travels in a straight line at a relativistic speed v" begs the question "relativistic speed v relative to what?" The assumed answer is "relative to both Joe and the distant location" (and we the readers). This is a direct consequence of the assumption that Joe and the distant location share the same frame (and that we also share that frame).
Let me provide an analogous scenario.
Joe floats in a space suit with two clocks.
Jane sits at one end of an extremely long structure with another two clocks. At the other end of the structure is a beacon. According to Jane, the structure has a length of L. Joe knows this.
Jane and Joe pass each other twice, at relativistic velocities of v and -v. Joe and the beacon pass each other twice, also at relativistic velocities of v and -v (Jane and the beacon are fixed to the same structure and hence share the same frame).
Four events are noteworthy:
1. Joe and Jane are collocated as they pass for the first time. Their clocks begin measuring time elapsed.
2. Joe and the beacon are collocated as they pass for the first time. Joe's clocks are paused and the beacon sends a message to Jane's clocks to pause.
3. Joe and the beacon are collocated as they pass for the second time. Joe's clocks restart measuring time elapsed and the beacon sends a message to Jane's clocks to resume measuring time elapsed.
4. Joe and Jane are collocated as they pass for the second time. Their clocks stop measuring time elapsed and Joe and Jane exchange one of their clocks. Neither consults the other as they each attempt to work out what the other's clock will read.
Observe that I do not say who reverses direction. For the purposes of the mind experiment, we can say that both Joe and Jane were anaethetised while one of them reversed direction, but neither knows which of them has now changed direction relative to the third observer (the reader).
There is an asymmetry in this scenario, but Jane and Joe cannot determine on whose part that asymmetry lies.
Jane's calculations:
Joe is in motion relative to Jane. Jane calculates that Joe must take a period of 2L/v to travel between events 1 and 2 and events 3 and 4. She further calculates that because Joe is in motion, his clocks will run slow and will show a time elapsed of γ.2L/v where γ = sqrt (1-v^2/c^2). She can check her clock and see that the first period elapsed (between event 1 and event 2) was L/v + L/c and the second period elapsed (between event 3 and event 4) was L/v-L/c for a total of 2L/v.
Joe's calculations:
Jane is in motion relative to Joe. Joe therefore calculates that Jane's structure is foreshorted by a factor of γ. Therefore the time elapsed while the entirety of the structure passes twice will be γ.2L/v. Sure enough, he checks his clock and sees that this is the case.
Working out what Jane's clock will read is a little more complex. Joe knows that not only is Jane's structure foreshortened, but her clocks will also run slow by a factor of γ.
The first period elapsed can therefore be calculated as follows (noting that Jane's relative motion is in the same direction as the message from the beacon to Jane's clocks):
t1=γ.(γ.L/v + γ.L/(c-v))=γ^2.(L/v + L/(c-v))
=γ^2.(L/v.(c^2-v^2)/(c^2-v^2) + L(c+v)/(c^2-v^2))
=γ^2.(c^2.L/v - Lv + Lc + Lv)/(c^2-v^2)
=γ^2.(c^2.L/v + Lc)/(c^2-v^2)
but since γ^2 = 1 - v^2/c^2 = (c^2 - V^2)/c^2,
t1=(c^2 - V^2)/c^2 . (c^2.L/v + Lc)/(c^2-v^2)
=(c^2.L/v + Lc)/c^2
=L/v + L/c
The same process can be used to calculate that the second period elapsed is L/v-L/c. The total time elapsed on Jane's clock, as calculated by Joe, will be 2L/v - precisely the same as calculated by Jane.
There is no disagreement and there is no paradox, merely a poorly frame scenario.
cheers,
neopolitan
(If you want to read more, try http://www.geocities.com/neopolitonian/lightclock.doc and http://www.geocities.com/neopolitonian/sr.doc)
From Einstein Light (http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm):
Jane and Joe are twins. Jane travels in a straight line at a relativistic speed v to some distant location. She then decelerates and returns. Her twin brother Joe stays at home on Earth. The situation is shown in the diagram, which is not to scale (see link above).
Joe observes that Jane's on-board clocks (including her biological one), which run at Jane's proper time, run slowly on both outbound and return leg. He therefore concludes that she will be younger than he will be when she returns. On the outward leg, Jane observes Joe's clock to run slowly, and she observes that it ticks slowly on the return run. So will Jane conclude that Joe will have aged less? And if she does, who is correct? According to the proponents of the paradox, there is a symmetry between the two observers, so, just plugging in the equations of relativity, each will predict that the other is younger. This cannot be simultaneously true for both so, if the argument is correct, relativity is wrong.
(The author goes on to explain that asymmetry resolves the paradox, this is an not entirely satisfactory explanation.)
There are a few assumptions, which are perhaps only obvious when one takes time to search for them.
"(S)ome distant location" appears sufficiently vague as to avoid creating problems but an inherent assumption is that this location shares the same frame as Joe.
By placing Joe on Earth we hide the other assumption, which is that we also share the same frame as Joe.
The at-a-distance observations of each other's clock potentially hides an assumption of instantaneous transmission of information. I doubt the author intended that, but it must be remembered that information is not transmitted instantaneously.
"(Jane) decelerates and returns" is distracting. As the author correctly points out this is a point of asymmetry. But a similar scenario (to be shown shortly) while show that it doesn't matter which decelerates and changes direction - Jane or the entire universe. It is generally assumed that the period during Jane changes direction is insignificant enough to ignore.
Finally, "Jane travels in a straight line at a relativistic speed v" begs the question "relativistic speed v relative to what?" The assumed answer is "relative to both Joe and the distant location" (and we the readers). This is a direct consequence of the assumption that Joe and the distant location share the same frame (and that we also share that frame).
Let me provide an analogous scenario.
Joe floats in a space suit with two clocks.
Jane sits at one end of an extremely long structure with another two clocks. At the other end of the structure is a beacon. According to Jane, the structure has a length of L. Joe knows this.
Jane and Joe pass each other twice, at relativistic velocities of v and -v. Joe and the beacon pass each other twice, also at relativistic velocities of v and -v (Jane and the beacon are fixed to the same structure and hence share the same frame).
Four events are noteworthy:
1. Joe and Jane are collocated as they pass for the first time. Their clocks begin measuring time elapsed.
2. Joe and the beacon are collocated as they pass for the first time. Joe's clocks are paused and the beacon sends a message to Jane's clocks to pause.
3. Joe and the beacon are collocated as they pass for the second time. Joe's clocks restart measuring time elapsed and the beacon sends a message to Jane's clocks to resume measuring time elapsed.
4. Joe and Jane are collocated as they pass for the second time. Their clocks stop measuring time elapsed and Joe and Jane exchange one of their clocks. Neither consults the other as they each attempt to work out what the other's clock will read.
Observe that I do not say who reverses direction. For the purposes of the mind experiment, we can say that both Joe and Jane were anaethetised while one of them reversed direction, but neither knows which of them has now changed direction relative to the third observer (the reader).
There is an asymmetry in this scenario, but Jane and Joe cannot determine on whose part that asymmetry lies.
Jane's calculations:
Joe is in motion relative to Jane. Jane calculates that Joe must take a period of 2L/v to travel between events 1 and 2 and events 3 and 4. She further calculates that because Joe is in motion, his clocks will run slow and will show a time elapsed of γ.2L/v where γ = sqrt (1-v^2/c^2). She can check her clock and see that the first period elapsed (between event 1 and event 2) was L/v + L/c and the second period elapsed (between event 3 and event 4) was L/v-L/c for a total of 2L/v.
Joe's calculations:
Jane is in motion relative to Joe. Joe therefore calculates that Jane's structure is foreshorted by a factor of γ. Therefore the time elapsed while the entirety of the structure passes twice will be γ.2L/v. Sure enough, he checks his clock and sees that this is the case.
Working out what Jane's clock will read is a little more complex. Joe knows that not only is Jane's structure foreshortened, but her clocks will also run slow by a factor of γ.
The first period elapsed can therefore be calculated as follows (noting that Jane's relative motion is in the same direction as the message from the beacon to Jane's clocks):
t1=γ.(γ.L/v + γ.L/(c-v))=γ^2.(L/v + L/(c-v))
=γ^2.(L/v.(c^2-v^2)/(c^2-v^2) + L(c+v)/(c^2-v^2))
=γ^2.(c^2.L/v - Lv + Lc + Lv)/(c^2-v^2)
=γ^2.(c^2.L/v + Lc)/(c^2-v^2)
but since γ^2 = 1 - v^2/c^2 = (c^2 - V^2)/c^2,
t1=(c^2 - V^2)/c^2 . (c^2.L/v + Lc)/(c^2-v^2)
=(c^2.L/v + Lc)/c^2
=L/v + L/c
The same process can be used to calculate that the second period elapsed is L/v-L/c. The total time elapsed on Jane's clock, as calculated by Joe, will be 2L/v - precisely the same as calculated by Jane.
There is no disagreement and there is no paradox, merely a poorly frame scenario.
cheers,
neopolitan
(If you want to read more, try http://www.geocities.com/neopolitonian/lightclock.doc and http://www.geocities.com/neopolitonian/sr.doc)
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