We are in a given isor and observe, instead of just particles moving, *clocks* moving. These clocks move in arbitrary ways, but at each instant we can define an isor that moves as the clock. What we want is to compare the reading if this clock during this instant with the reading of our clock for the same interval.
For us, the clock moves dl = (dx² + dy² + dz²)¹⸍² in an interval of time dt or cdt if we prefer to keep things in spatial units.
For the clock itself, it does not perceive self-motion, so dl' = 0. However, the intervals must be the same, so dl² - c²dt² = c²dt'².
Considering that dl²/(cdt)² = (v/c)² we get that
dt' = ds/c = dt·(1-v²/c²)¹⸍² .
We can also integrate this, always having an isor defined for each instant (for the clock), and then
Δt' = ∫ds/c =∫dt·(1-v²/c²)¹⸍² .
The integral is taken along the world line of the clock.
We call Δt' the *proper time* of the clock or the proper time of the object moving along it. Proper time reading is always less than *coordinate time* (our Δt). So moving clocks tick slower than clocks staying with us.
But now you can say that for the moving clock it is YOU who moves, so YOU should be fall behind. This is not true.
Suppose the moving clock passes through our clock at t=0, and they can sync themselves there. Now, in order to establishing a comparison between the two clocks, we must compare them again. This can happen if one of them turns back (accelerating) so that it can meet the other. It will be the one accelerating the one falling behind back in time. So the situation is not symmetric. Or, alternatively, if the moving clock does not accelerate, it will never encounter our clock again, but we could have placed another static clock out there to make the second comparison. However, this distant clock is not our clock. So we require two clocks while the moving object needs only one clock: not a symmetric situation either.
The clock that lags is the one being compared to several static clocks or the one accelerating (so that its frame will not be inertial). And remember, *laws of nature are the same ONLY for isor*. So the arguments we apply for isor are not valid for the sor of the moving clock (it if accelerates).
Notice as well that Δt' will always be smaller or equal to Δt. It is equal if the moving clock is not moving. If the clock is moving with non-uniform motion in a closed (spatial) path, meaning for example that it departs from x=0 and returns to x=0, then the world line of the clock is a curve passing twice through x=0. If we compare this with a clock staying at x=0 we arrive to the conclusion that in order to maximise its proper time, a clock must stay at rest, or equivalently, with uniform motion.
In Euclidean geometry, the smallest distance between two points is a straight line connecting them. In this geometry, however, we say that the greatest proper time between two events is a straight line connecting them.