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Science Forum Index » Physics - Relativity Forum » Time dilation scenarios
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| Guest |
Posted: Thu May 01, 2008 4:35 pm |
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Ok. Let's try another one. I'll use some of Tom Roberts language for
writing this scenario.
Initially, spaceships A and B are at rest in inertial frame S1 but
separated by a significant distance. Their on-board clocks are
synchronized with each other. At a prearranged time, tA=0, indicated
on clock A (and subesequently on clock B), spaceship A accelerates
with a constant acceleration, a, toward B for a time t1, then coasts
for a time t2. It then reverses the acceleration for another period
equalling 2 x t1, coasts for t2 and accelerates for the original t1
again. This brings it back into rest wrt the inertial frame S1, but
now the spaceship is located at the midpoint between the two starting
positions. At the same prearranged time, B follows the mirror image of
A's movement profile, moving towards A, and finishing up right next to
A. Has a difference been created between their proper times?
I am suspecting that, due to the symmetry of this problem (i.e. you
can reverse the labels A and B without changing the physics of the
scenario), there can be no difference between tA and tB when they
meet. Is this correct? |
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| Dono |
Posted: Thu May 01, 2008 4:57 pm |
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Guest
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On May 1, 7:35 pm, matscience...@yahoo.com wrote:
Quote: Ok. Let's try another one. I'll use some of Tom Roberts language for
writing this scenario.
Initially, spaceships A and B are at rest in inertial frame S1 but
separated by a significant distance. Their on-board clocks are
synchronized with each other. At a prearranged time, tA=0, indicated
on clock A (and subesequently on clock B), spaceship A accelerates
with a constant acceleration, a, toward B for a time t1, then coasts
for a time t2. It then reverses the acceleration for another period
equalling 2 x t1, coasts for t2 and accelerates for the original t1
again. This brings it back into rest wrt the inertial frame S1, but
now the spaceship is located at the midpoint between the two starting
positions. At the same prearranged time, B follows the mirror image of
A's movement profile, moving towards A, and finishing up right next to
A. Has a difference been created between their proper times?
I am suspecting that, due to the symmetry of this problem (i.e. you
can reverse the labels A and B without changing the physics of the
scenario), there can be no difference between tA and tB when they
meet. Is this correct?
Enough, bozo.
This website gives you all the tools to calculate all the imaginable
scenarios:
http://en.wikipedia.org/wiki/Twin_paradox#Accelerated_rocket_calculation
Now, go away. |
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| Dono |
Posted: Thu May 01, 2008 5:05 pm |
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Guest
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On May 1, 7:35 pm, matscience...@yahoo.com wrote:
Quote: Has a difference been created between their proper times?
none |
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