Henri Bergson: Half-relativity – Lorentz Equations (11)

”There is, however, one case where I will have to take your indications into account and modify my measurements. It is when it is a question of constructing an integral mathematical representation of the universe, I mean of everything that happens in all the worlds that move relative to you with all speeds. To establish this representation which would give us, once complete and perfect, the relationship of everything to everything, it will be necessary to define each point of the universe by its distances x, y, z to three determined rectangular planes, which will be declared immobile, and which will intersect according to axes OX, OY, OZ. On the other hand, the axes OX, OY, OZ which will be chosen in preference to all the others, the only axes which are truly and not conventionally immobile, are those which will be given to yourself in your fixed system.

Now, in the moving system in which I find myself, I relate my observations to axes O’X’, O’Y’, O’Z’ that this system carries with it, and it is by its distances x’, y’, z’ to the three planes intersecting along these lines that every point of my system is defined in my eyes. Since it is from your point of view, immobile, that the global representation of the Whole must be constructed, I must find a way to relate my observations to your axes OX, OY, OZ, or, in other words, that I establish once and for all formulas by means of which I can, knowing x’, y’ and z, calculate x, y and z.

But this will be easy for me, thanks to the indications that you have just provided me. First, to simplify things, I will suppose that my axes O’X’, O’Y’, O’Z’ coincided with yours before the dissociation of the two worlds S and S’ (which it will be better, for the clarity of the present demonstration, to make this time completely different from each other), and I will also suppose that OX, and consequently O’X’, mark the very direction of the movement of S’. In these reconciliations, it is clear that the planes Z’O’X’, X’O’Y’, only slide respectively on the planes ZOX, XOY, that they constantly coincide with them, and that consequently y and y’ are equal, z and z’ also. It then remains to calculate x. If, since the moment when 0′ left 0, I have counted on the clock which is at the point x’,y’,z’ a time t’, I naturally represent to myself the distance from the point x’y’z’ to the plane ZOY as equal to x’ + vt’.

But, given the contraction that you point out to me, this length x’ + vt’ would not coincide with your x; it would coincide with x√(1 – v²/c²). And consequently what you call x is (1/√(1 – v²/c²))(x’ + vt’) . There it is the problem solved. I will not forget, moreover, that the time t’, which has elapsed for me and which my clock placed at the point x’,y’,z’ indicates to me, is different from yours. When this clock gave me the indication t’, the time t counted by yours is, as you said, (1/√(1 – v²/c²))(t’ + vx’/c²). Such is the time t which I will mark for you. For time as for space, I will have passed from my point of view to yours.”

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