0
$\begingroup$

Three clocks are started at exactly the same time on Earth. The first and second clocks are taken into the vacuum of space.

The first clock accelerates until it reaches 100,000m/s, then stays at this speed for about two years. It decelerates, turns around, and re-accelerates back to 100,000m/s to return to earth.

The second clock accelerates away from earth for one year at a constant 9.5 m/s/s, which would make its top speed 299,592,000 m/s - very close to the speed of light (have I done my maths right here?). It then decelerates for one year at the same rate, and does the whole thing in reverse, accelerating and decelerating back towards earth at 9.5m/s/s.

Both return journeys of the first and second clock take exactly the same amount of time - 4 earth years.

The third clock remains on earth.

I think the first clock would be the most ahead, the second clock would be in the middle, and the third clock - the one that 'stayed still' on earth - would experience the most time dilation (be running the slowest), due the fact it is in a gravitational field with an effective 9.81 m/s/s rate of acceleration. Is this right?

$\endgroup$
2
$\begingroup$

This is how you do the calculation.

The elapsed time on an observer's clock is called the proper time, $\tau$, and it is calculated by integrating the metric:

$$ c^2d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)c^2dt^2 - \frac{dr^2}{1-\frac{2GM}{c^2r}} - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 $$

In this case we'll assume all motion is radial so $d\theta = d\phi = 0$ and the equation simplifies to:

$$ d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)dt^2 - \frac{1}{c^2}\frac{dr^2}{1-\frac{2GM}{c^2r}} \tag{1} $$

The coordinates $t$ and $r$ are the Schwarzschild time and radial coordinates i.e. the time and radial coordinates for an observer far from the Earth. For the third clock that stays on Earth the calculation is easy since $dr = 0$ (i.e. it isn't moving) and equation (1) gives:

$$ d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)dt^2 $$

And this immediately integrates to:

$$ \tau_3 = \sqrt{1-\frac{2GM}{c^2r}}\,t \tag{2} $$

which is just the equation for time dilation in the gravitational field outside a spherically symmetric mass.

For the first clock we'll make two approximations:

  1. the time spent near the Earth is small compared to the total travel time

  2. we'll ignore the turnaround time at the far end of the trip

So in effect the third observer is moving at constant velocity $v$ at large $r$ where $2GM/c^2r$ is effectively zero. For the outbound journey the distance is simply given by $r = vt$ so $dr = vdt$, and substituting in equation (1) gives:

$$ d\tau^2 = dt^2 - \frac{v^2}{c^2}dr^2 $$

and this immediately integrates to:

$$ \tau_\text{out} = \sqrt{1 - \frac{v^2}{c^2}}\,\frac{t}{2} $$

where the time is $t/2$ because we're only considering half the journey. Since the journey is symmetric the outboud and return times are the same, so the total time for clock 1 is:

$$ \tau_1 = 2\tau_\text{out} = \sqrt{1 - \frac{v^2}{c^2}}\,t \tag{3} $$

which is just the expression for time dilation in special relativity.

The calculation for the second clock is really hard. We'll once again assume we can ignore the Earth's gravitational field since the rocket spends only a short time near the Earth. In that case the distance as a function of time is given by:

$$ r = \frac{c^2}{a} \left(\sqrt{1 + \frac{at}{c^2}} - 1\right) $$

where $a$ is the constant acceleration as measured on the rocket. Using this to substitute for $dr$ in equation (1) is somewhat involved so I'll cheat and just quote the result from Phil Gibbs' article on the relativistic rocket. For the first quarter of the journey we get:

$$ \tau_\text{qtr} = \frac{c}{a} \text{arcsinh}\left(\frac{a}{c}\frac{t}{4}\right) $$

Again assuming symmetry each quarter of the journey takes the same time, so the total time for the second clock is:

$$ \tau_2 = 4\frac{c}{a} \text{arcsinh}\left(\frac{a}{c}\frac{t}{4}\right) \tag{4} $$

I'll leave the last stage of the calculation to you. Equations (2), (3) and (4) give you the three times you need for the comparison.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.