by Corry Shores

*Duration and Simultaneity*

*c*-

*v*)

*c + v*)

*r*, the relative speed of the light ray.

*v*. Immediately upon their parting ways, S' contracts in the direction of its motion. And all things that are not perpendicular to this direction are likewise contracted. So if S is a sphere, S' becomes and ellipsoid. "This contraction explains why the Michelson-Morley experiment gives the same results as if light had a constant speed equal to

*c*in all directions" (5b).

*l*no matter which way it goes, and so the total journey is 2

*l*. And, this terrestrial observer always finds the same speed for light,

*c*. Again recall the time formula.

*l*/

*c*

*c.*So according to the ether observer's clock, the time it takes the beam to go from O to B to O:

*l*/

*c*, which is the same for both the beams along OB and OA. Hence time moves slower on the earth than it does in the motionless ether. The second-hand on the earth-clock takes longer to pass than does the second-hand on the ether clock.

*v*. We will continue this presumption that S' contracts in the direction of its motion, all while its time expands. Let's imagine that there is a person in motionless system S. He is able to see the second-hand of a clock in system S'. He is able to see time slow down, "like an elastic band being stretched, like an arrow seen under a magnifying glass" (6ab). And yet, the mechanisms in the S'-system clock have not changed their function. "It is not because clocks go more slowly that time has lengthened; it is because time has lengthened that clocks, remaining as they are, are found to run more slowly" (6b). Thus, "as a result of motion, a longer, drawn-out, expanded time comes to occupy the spatial interval between two positions of the clock hand" (6bc).

*t'.*Bergson calculates that he will have to add the following value to it, in order to synchronize his clock-value with O's clock value, at that moment the light beam reaches mirror A. This adjustment will allow the two clocks to synchronize from the perspective of the ether observer.

*c + v*to

*c - v*. He sees that the ground observers who find an equal proportion make an error in their speed assessments for the departing and returning journeys. However, the errors compensate for each other, which is why they never notice the difference.

*o*in systems S' coincides with clock C

*o*in system S. Both clocks are at zero when the systems diverge. System S' moves in a straight line. Let's say that there are two clocks in system S' that are separated just as they are in the Michelson-Morley experiment. We call the second clock C'

*1*. Let's say that the people running the clocks synchronize without taking the ether-observer's perspective. So they divide the total trip in half. But the person in stationary system S knows that when the clock-operators do this, really clock C'

*1*lags behind. So when clock C'

*1*reads time

*t*', there really was a time elapse (from the stationary perspective) of

*1*. He needs to adjust that time again, to account for the fact that time passed at a different rate for clock C'

*1.*

*l*[this seems to be the same

*l*as in our previous experiment]. They are clocks C'

*0*, C'

*1*, C'

*2*, and so on. Let's image back when S and S' still coincided. They were both motionless in the ether. So when the people in system S' try to synchronize their clocks in that Michelson-Morley way, they will not need to adjust their readings. In fact, the light beam does indeed take the same amount of time to go back and forth from the mirror. Now S and S' separate. But the people in S' are not aware they are in motion. They think each of their synchronized clocks display real simultaneities when their clock-hands all point to the same number. The ether-observer learned that when light went as far as

*l,*it needed to be turned back

*l*away. So it will need to be turned back by

*v*. We then had to make the following modifications in S' so that its temporal and spatial measurements reflected the stable reality of system S.

*l*we mean their distance apart computed in the direction of motion of their system, that is, the distance between the two planes, perpendicular to this direction, which pass through each of them respectively" (11a).

*1*, C'

*2*, C'

*3*point to the same time when there are three different moments? But at the different moments at which they point to the same time in my system, events occur at points C'

*1*, C'

*2*, C'

*3*of my system which were legitimately designated contemporaneous in system

*S*; I shall then still agree to call them contemporaneous in order not to have to take a new view of the relations of these events first among themselves, and then with all the others [11-12]. I shall thereby preserve all their sequences, relations and explanations. In naming as succession what I called simultaneity, I would have an incoherent world or one built on a plan utterly different from yours. In this way, all things and all relations among things will retain their size, remain within the same frames, come under the same laws. I can therefore act as if none of my lengths had shrunk, as if my time had not expanded, as if my clocks agreed. So much, at least, for ponderable matter, for what I carry along with me in the motion of my system; drastic changes have occurred in the temporal and spatial relations of its parts, but I am not, nor need to be, aware of them" (12b).

*x*,

*y*,

*z*from three given planes at right angles, which we shall declare motionless, and which will intersect on axes

*OX, OY, OZ "*(13ab).

*x', y', z'*from the three planes intersecting on those lines that every point in my system will be defined. Since it is from your motionless point of view that the global representation of the All has to be framed, I must find a way to relate my observations to your axes OX, OY, OZ, or, in other words, to set up equations by means of which I shall once and for all be able, knowing

*x', y', z',*to calculate

*x, y, z*. But this will be easy, thanks to the information you have just given me. First, to simplify matters, I shall assume that my axes

*O'X', O'Y', O'Z'*coincided with yours before the dissociation of the two worlds S and S' (which for the clarity of the present demonstration it will this time be better to make completely different from one another), and I shall also assume that

*OX*and, consequently,

*O'X'*denote the actual direction of motion of S'. This being so, it is clear that planes

*Z'O'X'*and

*X'O'Y'*simply glide over planes

*ZOX*and

*XOY*respectively, that they ceaselessly coincide with them and that consequently

*y*and

*y'*are equal, as are

*z*and

*z'*" (13c).

*x*. If, from the moment O' leaves O, I compute a time

*t'*on the clock at point

*x', y', z',*I naturally think of the distance from this point to plane ZOY as equal to

*x' + vt'*. But in view of the contraction to which you call my attention, this length x' + vt' would not coincide with your

*x*but with

*x*is

*t'*, which has elapsed for me and which my clock at point

*x', y', z'*shows me, is different from yours. When this clock gave me the

*t'*reading, the time

*t*shown by yours was, as you stated,

*t*which I shall show you. For time as for space, I shall have gone over from my point of view to yours" (13-14).

*We call the coordinate values for M:*

*x, y, z*. And for M':

*x', y', z'.*We see from the picture that M' retains it position relative to the

*y*and

*z*axes. It has only advanced along the

*x*axis. Again, recall the distance formula:

*t'*. Hence we have the following equalities:

*x = x' + vt'*

*y = y'*

*z = z'*

*t = t'*

*1*. Because it also coincides with system S, it will diverge in a direction away from the clock C

*o*in system S. So the line C'

*oC'1*indicates the direction of the motion that S' travels.

*x*with this formula:

*t*with

*v + v'.*So imagine that S is a river bank. Boat S' sails down it at speed

*v*, relative to the riverbank. There is a passenger on the boat who also walks at speed

*v*across the boat's deck. The speed from the shore is not

*v + v'*, but rather a speed less than that.

*v*and he shines a light. Will the light beam then go

*c + v*? No, it will just go

*c*. We will substitute

*c*for

*v'*in the equation from above.

*x', y', z', t',*and

*v'*knowing just

*x, y, z, t, v''*, by using the following formulas:

*Duration and Simultaneity*. Ed. Robin Durie. Transl. Mark Lewis and Robin Durie. Manchester: Clinamen Press, 1999.

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