
#1
Mar2111, 10:56 AM

P: 125

Say if two synchronised clocks are at either end of a spacecraft. when the spacecraft accelerates, the clocks accelerate at different speeds due to length contraction and so are not synchronised post acceleration, according to a non accelerating observer who considered the clocks synchronised before their acceleration.
How do I calculate the variation between the clocks according to this observer after the period of acceleration ceases? 



#2
Mar2111, 05:18 PM

Sci Advisor
Thanks
P: 3,853

Well OK, but this is gonna be a real mess. The world line of a uniformly accelerating particle is a hyperbola. Its equation is given parametrically by
x(τ) = (c^{2}/a) cosh(aτ/c), t(τ) = (c/a) sinh(aτ/c) where a is the acceleration and τ is the proper time. To help convince you of that, notice that x^{2}  (ct)^{2} = (c^{2}/a)^{2} = const, showing that it is indeed a hyperbola. Also for small τ, x ~ (c^{2}/a)(1 + (aτ/c)^{2}/2 + ...) = c^{2}/a + aτ^{2}/2 + ..., showing that a is the acceleration. What you want to know is how much proper time has elapsed at any time t. So you need to turn the t(τ) equation inside out and solve for τ, getting: τ = (c/a) sinh^{1}(at/c) This value will be different for the clocks at the nose and tail of the spacecraft, since they undergo different accelerations. Acceleration at the tail is bigger. How much bigger depends on exactly how you fire your rockets. A natural way to do it is to keep the proper distance L between nose and tail constant, so the occupants do not get stretched! Then nose and tail follow hyperbolas with the same origin. The path of the tail has radius c^{2}/a, while the path of the nose has radius c^{2}/a' = c^{2}/a + L, and from this you can determine the relationship between a and a'. 


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