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Bell spaceship paradox |
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| Jun20-12, 07:31 AM | #1 |
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Bell spaceship paradox
Hello,
Suppose, A and B spaceships are moving with constant acceleration. There is a string tied to spaceships center to center. A is on left of B and B is on right of A. C is outside observer. As speed increase, A sees that B going further ahead, and B sees A going further behind. So, from ships PoV, string would be broken. But, from C's PoV, distance between A and B remains constant. So, I cannot understand, how string would be broken from C's PoV? Thanks. |
| Jun20-12, 07:42 AM | #2 |
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From C's PoV the string gets length-contracted. So the (constant) distance between A and B is greater than the (contracting) length of the string. Therefore the string snaps.
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| Jun20-12, 07:44 AM | #3 |
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This--Bell's Spaceship paradox--has been discussed many, many times here. Do a search to find the relevant threads. |
| Jun20-12, 08:17 AM | #4 |
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Bell spaceship paradox |
| Jun20-12, 08:32 AM | #5 |
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DaleSpam, Doc Al, GAsahi
The A and B starts simultaneously and continue with simultaneous constant acceleration in C's frame. But, not in spaceship's frame. The simultaneous changes of speed in C's frame remains spaceships at constant distance in C's frame. The distance remains equals for all journey even the distance is equal to the distance before journey. From spaceships' frame, distance between them increased than distance before journey. Because, B starts and changes speed before A in spaceships' frame. Length contraction cannot applied to only string. It can also apply to the space. If space between ships not contracted than how only string can? |
| Jun20-12, 08:41 AM | #6 |
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| Jun20-12, 08:58 AM | #7 |
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| Jun20-12, 10:51 AM | #8 |
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You can replace the string with a chain, and consider the individual chain links as rigid elements, which are all contracting as the chain accelerates:
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| Jun20-12, 11:21 AM | #9 |
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| Jun20-12, 12:04 PM | #10 |
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| Jun21-12, 06:21 AM | #11 |
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a) Moving object is shorter than at rest (the usual interpretation) b) Space is elongated for a moving object Interpretation b) sounds a bit weird. But when you go into a rotating frame, rulers at rest measure a circumference greater than 2πr. You cannot claim that the rulers are contracted, because they are at rest. You must conclude that the space along the circumference is elongated. |
| Jun21-12, 06:54 PM | #12 |
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| Jun21-12, 06:56 PM | #13 |
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| Jun22-12, 12:00 AM | #15 |
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| Jun22-12, 02:52 AM | #16 |
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Ok.... Thanks guys.. for detailed explanation.
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| Jun25-12, 01:54 PM | #17 |
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Recognitions:
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I usually recommend <http://arxiv.org/abs/0906.1919> for this.
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