Ardeshir Mehta, N.D.
Normally the "Twin Paradox" is discussed in terms of acceleration: one of the twins has to turn around, and therefore be accelerated -- either positively and/or negatively -- in order to return to the other twin, so that the ages of the two can be compared in the same reference frame.
But this is not really necessary. If we use stop watches, we can compare a delimited period in the life of each twin.
Indeed we need not have twins at all: all we need is two identical stop watches. So let's discuss a non-accelerated "Twin Paradox" using stop watches, as follows:
Let two identical and very large spaceships, A and B, pass by one another at high speed in a straight line. Let each spaceship carry an identically-constructed stop watch. Let the readings of the stop watches initially be 0.00, and let the stop watches not yet be ticking.
As the two spaceships pass one another, let a fine antenna on one spaceship briefly touch another fine antenna on the other, and let the contact between the two antennae cause a switch to be thrown in each spaceship, setting the stop watches on board the spaceships ticking simultaneously. Since the spaceships are very large and heavy, the momentary and light contact between the antennae causes only a negligible, and perhaps even non-measurable, slowing down of the spaceships.
Let there be a very long and light string floating around in space just at the location where the spaceships are expected to pass by each other. Let the speed of the string relative to each spaceship be exactly half the relative speed between the spaceships themselves, so that each spaceship is moving relative to the string at an equal speed but in opposite directions. (The string's location and relative speed can of course easily be calculated and adjusted in advance, since the spaceships, being large, can be observed from a great distance.) Let there be a ring tied to each end of the string. Let the rings be positioned so that they are close together but not overlapping. Let there be a hook on each of the spaceships, and let each hook engage one of the rings as the spaceships pass by one another. In this manner, as the spaceships move apart, the loose string is drawn tight due to the movement of the spaceships away from one another.
At the exact mid-point of the string, let the string be weakened, so that when the spaceships draw the string tight, the string eventually snaps just at the weakened point in the very middle of the string.
Let the hooks on each spaceship be connected to triggers on the spaceships, so that when the string snaps, the resulting jerk sends a signal to the stop watch carried aboard each of the spaceships, causing the stop watches to stop ticking.
Of course there may well be a perceptible and measurable time difference between the moment the string snaps, on the one hand, and on the other, the moment the hook on each spaceship triggers the stop watch carried on board to stop: because the tension of the string must take a finite amount of time to travel the distance along the string to the snapping point from the ring (and the hook) at the end of the string. But since the snapping point is smack bang dead-centre on the string, this time difference, if there is any, must be exactly the same for each of the spaceships.
So what we have is two stop watches, each of which measures the same amount of time, regardless of reference frame: viz., the time between (a) the moment the antennae of the spaceships touch each other, causing the stop watches to start ticking, and (b) the moment the snapping of the string jerks the hook on each of the spaceships, causing the stop watches to stop ticking.
Note that the touching of the antennae, being just one single event taking place very close to the spaceships, has to be simultaneous for each of the spaceships. The snapping of the string is also, of course, just one event, and although this event takes place at a considerable distance from the spaceships, the signal conveying the tension of the string from the hook to the snap must take the same amount of time for each spaceship, because the snap takes place right at the mid-point of the string, and because the mid-point of the string is moving at exactly the same speed relative to each space-ship.
Now bring the spaceships home and compare the readings of the stop watches, which will now no longer be ticking. What should the readings be?
Note that according to the Principle of Relativity, there is no way whatsoever to tell which of the spaceships was moving and which was stationary, or even which of them was moving slower and which faster. All we can possibly know is the relative speed between the two spaceships.
If we take it that it was spaceship A that was moving and spaceship B that was stationary, then if we use the Lorentz transformation formulae to calculate the times registered by the clock, the time tA registered by the stop watch aboard spaceship A will be shorter than the time tB registered by the stop watch aboard spaceship B.
But if we assume that it was spaceship B that was moving and spaceship A that was stationary, then the reverse will be the case!
In other words, the readings of the stop watches should show each less than the other ... for the same amount of time!