# Elapsed time on accelerating clock

1. Feb 29, 2008

### morrobay

Consider three clocks A,B and C all set at t=0 at t=0
Clock A is stationary on earth
Clock B is on a rocket at a distance of 16.43 (10^14) meters away from earth and accelerates instantaneously to .6c and arrives at earth. Clock A time = 105.6 days.
Clock C is on a rocket at a distance of 4.08 (10^14) meters from earth and at t=0 starts accelerating in the direction of earth with acceleration of 9.81m/s/s
From v= sq rt (2) (acceleration) (distance). And distance = 1/2 (acceleration) (tt). And ds/dt = (acceleration) (t) It takes the Clock C rocket 105.6 days Clock A time to reach earth that is an elapsed time on Clock A of t= 105.6 days when rocket with Clock C arrives.
Clock B which traveled , distance /rate , also arrives at Clock A time of 105.6 days. The elapsed time on Clock B is, (1/gamma )(Clock A) = .8 (105.6) = 84.5 days
So Clock A reads 105.6 days when all three clocks are back at x=0
Just before Clock C rocket with v= 8.94 (10^7) m/s strikes earth what time does it show?

2. Feb 29, 2008

### DrGreg

Before anyone answers, we need to clarify how the acceleration is being measured. From the equations you have given, it seems you are measuring it relative to A.

However, you might mean the acceleration measured by C (the "proper acceleration", which is the "G-force" that would be felt by a person attached to the clock). If that's what you meant, we would need to change the starting distances to arrange for all three clocks to meet at the same time and place.

So which of these two accelerations do you mean?

3. Feb 29, 2008

### morrobay

Yes, the acceleration is beieng measured relative to A.
My question is what is the proper time on Clock C when it arrives at earth at the same time as Clock B (as above ) when Clock A shows elaped time of 105 days.

4. Feb 29, 2008

### JesseM

If you calculate the velocity as a function of time v(t) of C in the frame of A, then between two coordinate times $$t_0$$ and $$t_1$$ in A, the elapsed time on C would be $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt$$. If the elapsed time in A's frame is 105.6 days = 9.124 * 10^6 s, and v(t) in A is 9.81*t m/s (so v(t)^2 = 96.24*t^2 m^2/s^2), then to calculate the elapsed time you'd evaluate the integral $$\int_0^{9.124 * 10^6} \sqrt{1 - 96.24*t^2 / 8.98755 * 10^{16}} \, dt$$ = $$\int_0^{9.124 * 10^6} \sqrt{1 - 1.071 * 10^ {-15} *t^2} \, dt$$. You can enter Sqrt[1 - a x^2] into the integrator to see how to evaluate this integral.

Last edited: Feb 29, 2008
5. Feb 29, 2008

### morrobay

Clock C on the rocket is accelerating at 9.81 m/s/s from t =0 on all clocks A.B C
until it arrives at earth. See original post. It is accelerating for the total time relative to clock A , That is from t=0 to t= 105.6 From the equations listed in my original post it is shown that the time and distance with constant acceleration by rocket with Clock C , that it arrives at earth when Clock A shows t= 105.6 days. My question is : what is the
proper time reading on Clock C on arrival.

6. Mar 1, 2008

### JesseM

It can't be accelerating at that rate according to all observers. A constant rate of "proper acceleration" (constant G-force as experienced by an observer moving along with the accelerating clock) leads to a non-constant rate of coordinate acceleration in the inertial frame of an observer, and vice versa. See http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]. My calculations above are assuming that you mean the clock is accelerating at 9.81 m/s^2 in the inertial frame of A, starting at a speed of 0 in this frame; if this is indeed what you meant, you can find the answer to your question about the time shown on C after 105.6 days in A's frame by solving the integral I posted.

Last edited by a moderator: May 3, 2017