A spacecraft flies away from earth with a speed of 4.8 million meter per second relative to the earth, and then returns at the same speed. The spacecraft carries a clock that has been synchronised with a clock on earth, When the spacecraft returns to its starting point 1 year later as measured by the clock on earth, what is the difference in elapsed times in the two clocks?

Ok, so the problem for me here is determining the proper time, I figured that since the spacecraft is accelerating the correct proper time $t$ should be the time on earth ( 1 year ) and $t'$ should be the time elapsed on the rocket. But when I use the formula for time dilation this way I get the wrong answer however if I assume that the proper time is the one on the spacecraft I get the right answer. My question is why is the proper time the one on the spaceship

Both proper times are correct. But you should decide which one to use depending on what you want to compute. Proper time is the time which the observer's clock measures. Thus, Earth proper time is what Earth-based clock measures, and spacecraft's proper time is spacecraft clock measurement.

For Earth-based observer you're already given the proper time: it's what the clock on Earth measures. For spacecraft you have another proper time. This proper time is sum of proper times from flying away and going back.

To understand this better, let's make a drawing:

Here the green line is world line of spacecraft, red one is of Earth, and the dashed lines represent the light cone. It's obvious how to find the proper time of the Earth, since it's stationary. Just measure the length of its world line. How do we now compute proper time of spacecraft? We just compute the "length" of its world line. BUT: as space-time is not Euclidean, namely we have $d\tau^2=c^2dt^2-dx^2$, we should take intervals instead of lengths. It becomes important for moving objects. So, we split the world line of spacecraft into two pieces, where the motion is unaccelerated, and get:

$$c\Delta\tau_1=\sqrt{c\Delta t_1^2-\Delta x_1^2},$$ $$c\Delta\tau_2=\sqrt{c\Delta t_2^2-\Delta x_2^2},$$

$$\Delta\tau_\Sigma=\Delta\tau_1+\Delta\tau_2,$$

$\Delta\tau_\Sigma$ is the proper time of spacecraft. So, spacecraft clock will read $\Delta\tau_\Sigma$. You now just have to find the difference between Earth proper time $t$ and spacecraft proper time to get solution of the problem.

From this you can also derive time dilation formula if you wish.

The proper time of an observer is the time shown on that observer's clock. This is because the observer's frame looks locally like flat space so:

$$d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$$

and (obviously) in the observer's frame the observer is at rest, so $dx$, $dy$ and $dz$ are all zero and therefore $d\tau = dt$.

• i know but if i can pick the clock on earth to be the one that i will take the proper time of, why should i pick the one on the space craft ? – AD HD Aug 10 '14 at 14:57
• @ADHD: The two proper times aren't equal. As soon as you introduce acceleration (or spacetime curvature) the proper time is path dependant. It's only for unaccelerated motion that the proper time between any two spacetime points is path independant. – John Rennie Aug 10 '14 at 15:04
• To be precise, the proper time between any two spacetime points is path-dependent regardless of acceleration or curvature, but in flat spacetime inertial motion provides a specific path between (timelike-separated) points that we conventionally select to be the standard. – Stan Liou Aug 10 '14 at 15:12
• This difference in the proper times is of course the origin of the twin paradox - not that it's a paradox to anyone who knows how to integrate a line element. – John Rennie Aug 11 '14 at 9:43