Proper Time : Constant Velocity Clock vs Constant Acceleration Clock

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Homework Statement


3 clocks at to
A clock on Earth
B clock above Earth 4.66 * 1014 meters
C clock above Earth 3.30 * 1013 meters
B accelerates to .6 c and arrives at Earth when Earth clock reads 2.6 *106 sec
= 30 days With velocity 1.8 *10^8 m/sec with gamma = .8 B clock reads 2.07 * 10^6 sec = 24 days
C clock travels to Earth with constant acceleration in respect to Earth frame of 9.81 m/sec2
and arrives at the same time as B . C clock velocity 2.5 * 10^7 m/sec
What is the proper time on Clock C ?

Homework Equations


Integral to to 2.6 * 106 sec. sqrt [ 1-v(t)2/c2] dt
So Int to to t1 sqrt [ 1-1.01*10-15t2] dt

The Attempt at a Solution

I put Sqrt{1-ax2] in The Integrator
and plugged in values and got 15 days proper time on C clock ?
 
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morrobay said:

Homework Statement


3 clocks at to
A clock on Earth
B clock above Earth 4.66 * 1014 meters
C clock above Earth 3.30 * 1013 meters
B accelerates to .6 c and reaches Earth when Earth clock reads 2.6 *106 sec
= 30 days . with gamma = .8 B clock reads 2.07 * 10^6 sec = 24 days
C clock travels to Earth with constant acceleration in respect to Earth frame of 9.81 m/sec2
and arrives at the same time as B
What is the proper time on Clock C ?

Homework Equations


Integral to to 2.6 * 106 sec. sqrt [ 1-v(t)2/c2] dt
So Int to to t1 sqrt [ 1-1.01*10-15t2] dt

The Attempt at a Solution

I put Sqrt{1-ax2] in The Integrator
and plugged in values and got 15 days proper time on C clock ?

The evaluation of the above integral =
1/2x sqrt [ 1-ax2] + sin-1 ( sqrt a) x/2 sqrt a

Note: I am posting this problem because there are endless discussions on proper time
but not many numerical answers
 
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morrobay said:
The evaluation of the above integral =
1/2x sqrt [ 1-ax2] + sin-1 ( sqrt a) x/2 sqrt a

For the above problem the values for proper time :
x = 2.6 * 106 sec
a= 1.01 * 10 -15
1/2 x = 1.3 *106
x2 = 6.76 * 1012 sec2
a1/2= 3.17 *10-7
2(a)1/2 = 6.35 *10-7
With these values in the above evaluation:
1/2x [1-ax2]1/2 = 1.29 *106 sec
sin-1 (a)1/2 x / 2 a1/2
= sin-1 .824 = 55
so 55/6.35*10-7 = 8.66*107 added to term on left
1.29*106 sec = 87.8*10^6 sec and is not correct since it is more than A clock ?
 
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morrobay said:

The Attempt at a Solution

I put Sqrt{1-ax2] in The Integrator
and plugged in values and got 15 days proper time on C clock ?
All your work is good. But it seems like you are getting bad values out of your numerical integration routine. Perhaps it is a numerical precision problem.

I plug the same integral into Mathematica and get 2.588E6 s = 29.95 days.
 
Would you expect that the acellerating clock C in the original problem would have
essentially the same proper time as clock A ( 30 days proper time ) ?
Also clock B had 24 days proper time with .6c

note correction : a = 1.07 * 10 ^-15 from v(t^2)/c^2 = (9.81)^2 m/s^2 / 9*10^16
but does not change values too much.
Yes I am having a few numerical precision problems.
 
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morrobay said:
Would you expect that the acellerating clock C in the original problem would have
essentially the same proper time as clock A ( 30 days proper time ) ?
Yes. 3.3E13 m / 2.6E6 s is only an average speed of .04 c which corresponds to an average time dilation factor less than 1.001, so I would expect the clock to not be significantly time dilated overall.