Proper Time : Constant Velocity Clock vs Constant Acceleration Clock

AI Thread Summary
The discussion centers on calculating the proper time for Clock C, which accelerates towards Earth at 9.81 m/s², arriving simultaneously with Clock B, which travels at 0.6c. Clock B records 24 days while the Earth clock shows 30 days. The integral used for the calculation yields a proper time of 15 days for Clock C, but numerical precision issues are noted. A recalculation using Mathematica suggests that Clock C's proper time should be closer to 29.95 days, indicating it would not be significantly time-dilated compared to Clock A. The overall expectation is that Clock C's proper time aligns closely with Clock A's due to its average speed being only 0.04c.
morrobay
Gold Member
Messages
1,116
Reaction score
1,693

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 ?
 
Last edited:
Physics news on Phys.org
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
 
Last edited:
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 ?
 
Last edited:
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.
 
Last edited:
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top