If we have two objects (clocks) in (empty) space and they separate (the distance between them increases) at relativistic speeds then get back together (just for easier comparison of their times), what determines which clock would advance more (tick faster) and which less?
In other words: Let's name one object A, the other object B. In case one, object A stands still and object B moves away (then returns). In second case object B is still and B moves.
Does the above make sense?
If there is nothing else around, can we claim, that one object is standing still and the other moves?
Update: This is basically the twin paradox. Does it make any change if the moving object does not turn around? Just stops at some distance? Or never stops?
Best Answer
Beginners to special relativity generally learn the equation for time dilation:
$$ t' = \frac{t}{\gamma} \tag{1} $$
where $\gamma$ is the Lorentz factor:
$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$
For any $v > 0$ the Lorentz factor is greater than one and hence $t' < t$, which is what we mean by time dilation.
And this is all very well, but equation (1) is a special case that applies only for uniform unaccelerated motion, and it's absolutely no use for understanding the twin paradox. Indeed, the twin paradox only arises because people mistakenly apply equation (1) when it isn't appropriate. To understand the twin paradox properly we have to go a bit deeper into how special relativity works.
Special relativity is a geometrical theory of spacetime, just like general relativity in fact though considerably simpler. The basic equation for special relativity is the Minkowski metric:
$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 \tag{2} $$
Suppose I set up a coordinate system $(t, x, y, z)$ with myself stationary at the origin and the time $t$ measured with my clock and the distances $x$, $y$ and $z$ measured with my rulers. What the Minkowski metric says is that if I observe an object to move a distance $(dx, dy, dz)$ in a time $dt$ then the proper time $d\tau$ can be calculated using equation (2).
Why would we do this? Well firstly the proper time is an invariant i.e. every observer in any coordinate frame moving, rotating, accelerating or whatever will calculate the same value for the proper time $d\tau$. Secondly the proper time is just the time shown on a clock carried by our moving object.
The invariance of the proper time is an assumption and actually it's the key assumption in special relativity. You just have to accept that's how SR works. However we can easily prove the second point i.e. that the proper time is the time shown on the moving object's clock. In the moving object's rest frame it is stationary so $dx = dy = dz = 0$ and therefore equation (2) simplifies to:
$$ c^2d\tau^2 = c^2dt^2 $$
or just:
$$ d\tau = dt $$
So in the object's rest frame the proper time $d\tau$ is equal to the time measured by the object's clock $dt$. But we've just said that $d\tau$ is an invariant and is the same for all observers. That means for all observers the proper time is the time shown on the moving objects clock.
The point of all this is that we now have a way to calculate time dilation because we can use equation (2) to calculate the proper time. Let's do this for the simple case of uniform unaccelerated motion and check we get equation (1). We'll assume the object passes us at time zero with velocity $v$, and for convenience we'll assume it's moving along the $x$ axis so $dy = dz = 0$. The equation (2) becomes:
$$ c^2d\tau^2 = c^2dt^2 - dx^2 $$
But if the object is moving at velocity $v$ this means that $dx/dt = v$ because that's what we mean by velocity. So $dx = vdt$, and substituting this into the equation about we get:
$$ c^2d\tau^2 = c^2dt^2 - v^2dt^2 $$
and a quick rearrangement gives:
$$ d\tau^2 = dt^2(1 - v^2/c^2) $$
or:
$$ d\tau = \frac{dt}{\frac{1}{\sqrt{1 - v^2/c^2}}} $$
which is just equation (1).
This is a good point to stop and summarise where we've got to. We've stated that the Minkowski metric, equation (2) above, is the core equation in special relativity, and that it can be used to calculate the proper time $d\tau$, which is just the time measured by the moving object's clock. And this is what you asked, i.e. how do we calculate how much time has elapsed for the moving object. So in principle at least we've answered your question.
But can we do the calculation for an accelerating observer like the twin? And the answer is that yes we can. The maths will get harder by the same basic principle applies. We'll do it for a simplified case. Suppose at time zero our object starts accelerating at a constant (in our frame) acceleration $a$. Then in our frame the distance is given by the equation we all learned in school:
$$ x = \tfrac{1}{2}at^2 $$
and differentiating gives:
$$ dx = atdt $$
and as before we substitute this into the Minkowski metric to get:
$$ c^2d\tau^2 = c^2dt^2 - a^2t^2dt^2 $$
or:
$$ d\tau = dt\sqrt{1 - \frac{a^2}{c^2}t^2} $$
Integrating this turns out to be rather sticky, so we'll assume that $at \ll c$ in which case we can approximate the square root as;
$$ d\tau \approx dt\left(1 - \frac{a^2}{2c^2}t^2\right) $$
And we can integrate this to get the elapsed time for the accelerating object:
$$ \tau \approx t\left(1 - \frac{a^2}{6c^2}t^2\right) $$
And we find that $\tau \lt t$ so the time for the accelerating object is dilated - the accelerating twin ages less than us.