A question on time dilation in relativity: Philosophy Forums

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A question on time dilation in relativity

A question on time dilation in relativity

jorndoe
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Posted 12/12/08 - 10:19 AM: Subject: A question on time dilation in relativity	#1
A question on time dilation in relativity. I realize this has been brought up before, the difference in my scenario are the details (intended to be close to physically realizable). A spaceship in orbit above Earth in geostationary orbit, is carrying a clock C_s. C_s is synchronized with another clock on Earth (on the surface), C_e. We use Earth as the reference (i.e. Earth is not "moving" in our inertial system). Well, C_s and C_e may just be stopwatches that are started when the spaceship departs. The spaceship now accelerates, say with a constant acceleration of a (e.g. in m/sÂ²). It continues to accelerate away from Earth, until it reaches a relative velocity of, say, 99% of lightspeed (i.e. 99% of 3Ã—10â¸ m/s). At this point it decelerates, at the same rate as before (a), until it comes to a standstill. And now the journey goes back, like before, but towards Earth instead. So the entire journey falls in the four phases described (accelerating away from Earth, decelerating, accelerating towards Earth, decelerating). Only the astronauts on the spaceship experiences acceleration (apart from the people on Earth experiencing Earth gravity). Because all phases have constant acceleration, it is fairly easy to calculate the relative speed at all times. Of course the question here is: What is the difference between C_s and C_e at the return of the spaceship? I suppose, inherit in my question is what kind of calculations would be involved (I think some integrals, but my physics is way to rusty). Intuitively, due to the reversal of the situation, first moving away, then back, I'd say C_s = C_e, but I'm not at all sure. In the Twin Paradox described at (thanks Kwalish Kid) http://math.ucr.edu/home/baez/phys...winParadox/twin_intro.html the scenario is different, and I think it may be sufficiently so to give a different (overall) result.
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Kwalish Kid
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Posted 12/12/08 - 12:44 PM:	#2
Your answer probably begins on the Equivalence Principle Analysis page: http://math.ucr.edu/home/baez/phys...R/TwinParadox/twin_gr.html IIRC, you are correct that you have to do some integrals over a collection of momentarily co-moving reference frames that cover the accelerating parts. I'm not sure how significant these acceleration sections are going to be in changing the overall results from the SR calculations. To answer this, one really needs to crack open a physics text unless one has all this stuff memorized, and I really don't have it all memorized.
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swstephe
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Posted 12/12/08 - 05:28 PM:	#3
The spaceship was already experiencing time dilation as soon as it got into geosynchronous orbit, (due to the difference in gravity -- a kind of acceleration). GPS satellites have to deal with the atomic clocks being out of sync or they would be useless, (http://en.wikipedia.org/wiki/Effects_of_relativity_on_GPS). You haven't given enough information to solve the problem. The biggest question is to know how long the acceleration took place. When I was about 12 years old, I remember spending weeks trying to figure out what speed a ship would have to accelerate to, in order to make a trip to the nearest star within a single lifetime, and maintain an artificial gravity through acceleration (1g or 9.8 m/s2). That was before I knew calculus.
I do seem to remember a process where you people ask me questions and I give you answers, and then I ask you questions and you give me answers, and that's the way we find out things. I think I read that in a manual somewhere. -- Haywood Floyd 2010

Death Monkey
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Posted 06/23/09 - 07:05 AM:	#4
Can't do the calculations without using GR. However, if we assume that the acceleration is low enough to neglect GR effects, we can do it using SR, in which case it is just an integral. Even then, the question is ambiguous. You said the acceleration was constant: According to whom? If the person in the spaceship says the acceleration is constant, then the person on Earth will not agree. On the contrary, he will observe that as the spaceship approaches the speed of light, the rate of acceleration decreases considerably. That said, without doing any calculations it is clear that much more time will pass on Earth than on the spaceship. DM
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Kwalish Kid
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Posted 06/23/09 - 09:05 AM:	#5
Death Monkey wrote: Can't do the calculations without using GR. However, if we assume that the acceleration is low enough to neglect GR effects, we can do it using SR, in which case it is just an integral. Even then, the question is ambiguous. You said the acceleration was constant: According to whom? If the person in the spaceship says the acceleration is constant, then the person on Earth will not agree. On the contrary, he will observe that as the spaceship approaches the speed of light, the rate of acceleration decreases considerably. That said, without doing any calculations it is clear that much more time will pass on Earth than on the spaceship. DM Indeed. There is a neat "paradox" based on this. If you tie a piece of string between two spaceships, one following the other and then you accelerate them at the same rate relative to some third, stationary observer, the string is guaranteed to break.
"I don't mind if Christians honor the moment by displaying, and singing about, reindeer (a hard species to find in the greater Jerusalem/Bethlehem area). Same for the pine trees that also don't grow in Palestine. I wish everybody joy of it." - Christopher Hitchens

swstephe
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Posted 06/23/09 - 09:32 PM:	#6
Kwalish Kid wrote: Indeed. There is a neat "paradox" based on this. If you tie a piece of string between two spaceships, one following the other and then you accelerate them at the same rate relative to some third, stationary observer, the string is guaranteed to break. You mean Bell's Spaceship Paradox? I thought the current consensus is that the string will not break. I'm kind of on the not-break side. If the distance between the ships remains constant, relative to an observer, why wouldn't a single spaceship stretch out in the middle, rather than flatten like a pancake? I think two ships would make a sandwich. Okay, now I'm hungry.
I do seem to remember a process where you people ask me questions and I give you answers, and then I ask you questions and you give me answers, and that's the way we find out things. I think I read that in a manual somewhere. -- Haywood Floyd 2010

Kwalish Kid
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Posted 06/24/09 - 08:28 AM:	#7
No, the string will break. In the reference frame of either ship, the ship that's ahead is accelerating sooner than the ship that's behind. (That's as long as they remain at a constant distance in the original rest frame.)
"I don't mind if Christians honor the moment by displaying, and singing about, reindeer (a hard species to find in the greater Jerusalem/Bethlehem area). Same for the pine trees that also don't grow in Palestine. I wish everybody joy of it." - Christopher Hitchens

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