Henri Bergson: Half-relativity – Michelson-Morley experiment (7)

How do you adjust two clocks located in different places to each other? By communication established between the two people responsible for the adjustment. However, there is no instantaneous communication; and, since all transmission takes time, we had to choose the one which takes place under invariable conditions. Only signals launched through the ether meet this requirement: any transmission by ponderable matter depends on the state of this matter and the thousand circumstances which modify it at each moment. It is therefore by optical signals, or more generally electro-magnetic, that the two operators had to communicate with each other. The man at point 0 sent the man at point A a ray of light intended to return to him immediately. And things happened as in the Michelson-Morley experiment, with this difference however that the mirrors were replaced by people. It had been agreed between the two operators at 0 and at A that the second would mark zero at the point where the hand of his clock would be located at the precise moment when the ray would reach him. From then on, the first only had to note on his clock the beginning and the end of the interval occupied by the double journey of the ray: it was in the middle of the interval that he located the zero of his clock, as long as he wanted the two zeros to mark “simultaneous” instants and the two clocks were now in agreement.

It would be perfect, moreover, if the signal path was the same on the way there and back, or, in other words, if the system to which the clocks 0 and A. are attached was immobile in the ether. Even in the moving system, it would still be perfect for the adjustment of two clocks 0 and B located on a line perpendicular to the direction of the path: we know in fact that, if the movement of the system brings 0 to 0′, the ray of light makes the same path from 0 to B’ as from B’ to 0′, the triangle OB’O’ being isosceles. But it is different for the transmission of the signal from 0 to A and vice versa. The observer who is at absolute rest in the ether sees clearly that the paths are unequal, since, in the first trip, the ray launched from the point 0 must run after the fleeing point A, while in the return trip the ray returned from point A finds the point 0 that comes to meet it. Or, if you prefer, he realizes that the distance OA, assumed identical in both cases, is crossed by light with a relative speed c – v in the first, c + v in the second, so that the travel times are in the ratio of c + v to c – v. By marking zero in the middle of the interval that the clock hand has traveled between the departure and return of the ray, we place it, in the eyes of our immobile observer, too close to the starting point. Let’s calculate the amount of the error. We said earlier that the interval covered by the needle on the dial during the double outward and return journey of the signal is 2l/c. If therefore, at the time of emission of the signal, we marked a provisional zero at the point where the needle was, it is at the point l/c of the dial that we will have placed the definitive zero M which corresponds, we say , to the definitive zero of the clock at A. But the immobile observer knows that the definitive zero of the clock at 0, to really correspond to the zero of the clock at A, to be simultaneous with it, should have been placed at a bridge which divided the interval 2l/c not into equal parts, but into proportional parts with c + v and c – v. Let us call x the first of these two parts. We will have

x/(2l/c – x) = (c + v)/(c – v)

and consequently

x = l/c – lv/c²

Which amounts to saying that, for the immobile observer, the point M where we marked the definitive zero is with lv/c² too close to the provisional zero, and that, if we want to leave it where it is, we should, to have real simultaneity between the final zeros of the two clocks, move back by lv/c² the final zero of the clock at A. In short, the clock at A is always late by one dial interval lv/c², on the ‘time which should mark. When the hand is at the point that we agree to call t’ (we reserve the designation t for the time of motionless clocks in the ether), the motionless observer says to himself that, if it really agreed with the clock at 0 , it would mark t’ + lv/c².