Elapsed time on accelerating clock

In summary, Clock C is on a rocket accelerating at 9.81 m/s/s in the inertial frame of Clock A. After 105.6 days in Clock A's frame, the proper time shown on Clock C would be 9.124 * 10^6 s, as calculated by solving 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.
  • #1
morrobay
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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
Clock B reads 84.5 days
Just before Clock C rocket with v= 8.94 (10^7) m/s strikes Earth what time does it show?
 
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  • #2
morrobay said:
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
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
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
morrobay said:
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.
If you calculate the velocity as a function of time v(t) of C in the frame of A, then between two coordinate times [tex]t_0[/tex] and [tex]t_1[/tex] in A, the elapsed time on C would be [tex]\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt[/tex]. 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 [tex]\int_0^{9.124 * 10^6} \sqrt{1 - 96.24*t^2 / 8.98755 * 10^{16}} \, dt[/tex] = [tex]\int_0^{9.124 * 10^6} \sqrt{1 - 1.071 * 10^
{-15} *t^2} \, dt[/tex]. You can enter Sqrt[1 - a x^2] into the integrator to see how to evaluate this integral.
 
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  • #5
JesseM said:
If you calculate the velocity as a function of time v(t) of C in the frame of A, then between two coordinate times [tex]t_0[/tex] and [tex]t_1[/tex] in A, the elapsed time on C would be [tex]\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt[/tex]. 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 [tex]\int_0^{9.124 * 10^6} \sqrt{1 - 96.24*t^2 / 8.98755 * 10^{16}} \, dt[/tex] = [tex]\int_0^{9.124 * 10^6} \sqrt{1 - 1.071 * 10^
{-15} *t^2} \, dt[/tex]. You can enter Sqrt[1 - a x^2] into the integrator to see how to evaluate this integral.

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
morrobay said:
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 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 . 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.
 
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Related to Elapsed time on accelerating clock

1. How is elapsed time measured on an accelerating clock?

Elapsed time on an accelerating clock is measured using a special type of clock called an "accelerometer clock". This clock uses an accelerometer, which is a device that measures acceleration, to track the movement and acceleration of the clock. The elapsed time is then calculated based on the acceleration data and the clock's initial position and velocity.

2. Does time pass differently on an accelerating clock compared to a stationary clock?

Yes, time does pass differently on an accelerating clock compared to a stationary clock. This is due to the principles of special relativity, which state that time is relative to the observer's frame of reference. In other words, an observer on an accelerating clock will experience time passing slower compared to an observer on a stationary clock.

3. Can the elapsed time on an accelerating clock be negative?

No, the elapsed time on an accelerating clock cannot be negative. Since time is a scalar quantity, it can only have positive values. However, the rate at which time passes on an accelerating clock can be negative, meaning it is slowing down or going backwards compared to a stationary clock.

4. How does the speed of an accelerating clock affect the elapsed time?

The speed of an accelerating clock affects the elapsed time in a complex manner. As the speed of the clock increases, the rate at which time passes on the clock will decrease. However, as the acceleration of the clock increases, the rate of time will also decrease, but at a faster rate. Therefore, the elapsed time on an accelerating clock is affected by both speed and acceleration.

5. Is there a limit to how much time can be elapsed on an accelerating clock?

No, there is no limit to how much time can be elapsed on an accelerating clock. As long as the clock is accelerating and the observer is moving along with it, the clock will continue to measure elapsed time. However, due to the principles of special relativity, the elapsed time on an accelerating clock can never exceed the elapsed time on a stationary clock for the same event.

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