## 20 Nov 2009

### Distributed Durations. Ch.1 Half Relativity. Duration and Simultaneity. Henri Bergson

by Corry Shores
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[The following summarizes part of chapter 1 in Bergson'sDuration and Simultaneity. Paragraph headings are my own. My personal commentary is in brackets.]

Distributed Durations

Henri Bergson

Duration and Simultaneity

Ch. 1. Half Relativity

Bergson will discuss the Michelson-Morley experiment. He says that Einstein's special and general relativity theories are not based on this experiment. However, philosophers need to begin here in order to "grasp the true meaning of time in the theory of relativity." For, the experiment materializes the difficulty at hand by "stating the problem in concrete terms and also of spreading out the elements of its solution before our very eyes" (1).

We must turn our minds to earlier notions of time and space. There is a motionless ether, and as well there is absolute rest. We will base our first conception of time on the findings from the Michelson-Morley experiment. These findings will be "half-relativist, one-sided, and not yet Einstein's" (1d). For scientific purposes, physicists could very well ignore this experiment. But if physics would like to make philosophical claims about time, then we will need to attend to it.

Michelson set-up the experiment in 1881, and Morley repeated it in 1887.

There is a beam of light emanating from its origin at S, and it moves towards a thin glass plate, called O.

We see that plate O is positioned at a 45 degree angle to the direction of the light beam. The plate divides the light beam, sending one beam along its way, toward mirror A.

All the while, plate O has sent a perpendicular beam toward mirror B.

The distance between O and A has a certain measure. The distance between O and B has the same measure. When the light beams meet mirrors A and B, they reflect back to O. The beam coming from B goes through plate O towards lens M. And the beam from A bounces perpendicular from O also towards lens M. The lens looks towards O.

We will first suppose that there is no ether, even though physicists at this time believed there was. Recall that the distances OA and OB are equal. So, taking nothing else into account, they should both take the same amount of time to return to O. Thus the light waves would superpose perfectly and there would be no wave interference. No matter which way we rotate the device, the beams should still return in the same time at O. We could even rotate it 90 degrees.

We see that even though A and B switch places, there should still not be a difference in the amount of time it takes for the light to return to O.

Now we will take into account the earth's rotation. We will say that it is moving in this direction:

The earth is moving through the motionless either. Consider looking here at this animation. It shows swimmers moving like they do in the experiment, only their beginning-platform itself is moving through the water. The platform is moving to the left, all while the water is stationary. However, that has the effect of "pushing" things to the right, if you will. In our experiment, the earth is moving to the right, and not the left, but the effect is the same, only inverted. So consider a car with a sun roof. When we are in the car, we do not feel the resistance from the air, because it is moving with us. But if we stand-up through the roof-window, we feel the air pushing against us. For now we are supposing that the earth is moving through a static ether. Perhaps this might be a little like feeling wind resistance. There is another way to conceive this ether phenomenon. Light travels through the motionless ether. When we shine a light, we deposit it into this still medium. Light has a certain speed. Because the medium is always still, light will always go the same speed relative to the medium. But the earth in a sense will move under the light-beam. Consider first when the beam is traveling in the direction of the earth's motion. The earth will be moving forward under the light-beam. So the beam will need extra time to arrive at its destination, which is in motion relative to the ether that is propagating the light ray. But on the light-beam's return trip, the earth will be moving backward below the light beam. Hence the ray will need less time to arrive back at the glass plate.

So, we calculate its speed on the way to mirror A by subtracting the speed of the earth from the speed of light:

Light from O to A = (c - v)

But, for the trip back, we add together the earth's and light's speed.

Light from A to O = (c + v)

We can see in the animation below how very different the speeds are for the beam when it goes to A and when it returns back to O. [This animation is my own. As you see, it is not mathematically precise. I just hope it illustrates some basic aspects of the motion in its primitive way].

Now, recall that we calculate speed as distance over time.

To figure out time, we perform this algebraic operation on the formula:

We will now determine the time it takes for the light beam to go from O to A, then back from A to O. When we add them, we have:

Which we may operate upon to obtain:

Now we want to know the distance that the light beam traveled. Let's derive the formula

And let's apply our figures into the formula:

When something is in the numerator, that is the same as it being in the denominator's denominator. Hence we may convert this formula to:

We will now instead analyze the beam that diverges in the other direction toward mirror B. As we saw in the animation, its path is not equal to the length between O and B. Rather, it made a diagonal line.

[If we might excuse a shortcut, let's see a way to grasp how to find this speed. Consider that on the diagonal, the light is still moving at the speed of light. And on the horizontal, the device is moving at the speed of the earth. So to find the relative speed, let's think of it as we want to find the other leg of the triangle.

Here by the way is the image from the Michelson and Morley paper, "On the Relative Motion of the Earth and the Luminiferous Ether," available here.

We will use the Pythagorean theorem to find r, the relative speed of the light ray.

Recall the time formula.

We will now apply that to the light beam going from O to B, then back again from B to O.

We might also represent the motion with this diagram.

Hence we would say that the time it takes for the beam to go from O to B' is

And again, the total distance , O to B' to O' would be:

And recall the distance formula:

The speed is c, so the distance covered is:

He says this can be rendered this other way, perhaps somehow in a manner like the following:

So we would expect there to be a time delay. The beam from OA should return to the glass plate shortly after OB does. This would cause the light waves to be out of alignment, which should produce interference bands at lens M. But the experiment did not show them, nor did the experiments which faithfully repeated this one. Bergson writes: "Things happen as if the two double pasages were equal, as if the speed of light with respect to the earth were constant, in short, as if the earth were motionless in the ether" (4c).

Lorentz offers this explanation. Somehow line AO contracts. This allows the beam returning from mirror A to arrive fast enough to meet-up with the one from B. We are observers on earth moving at this speed. [So we are part of this contraction.] Hence we cannot see the contraction in the measurements from the earth. But if we could see the device from the perspective of the ether, we would be able to watch it contract.

Bergson has us consider two systems. System S is a motionless system in the ether. S' is a duplicate of this system. It begins united with the first system S. But then it breaks away in a straight line at speed 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).

Yet we also need to know something else. We always find the same speed for light here on earth. Our measurements do not seem to be affected by the complications involving the earth's speed with regard to the ether. Why is that?

Let's suppose the explanation will come from an observer in the motionless either (and not on the moving earth). He will point to the terrestrial observer. Both the terrestrial observer and the ether observer each have their own clocks beside them.

Now, the terrestrial observer only sees the light move from O to A or B, then it returns. The distance for the light is l no matter which way it goes, and so the total journey is 2l. And, this terrestrial observer always finds the same speed for light, c. Again recall the time formula.

So the time for the beam to go to and return from the mirror is

2l / c

But recall how we found the total (diagonal) distance of through the ether for the beam that travels from O to B and then back again.

Again,

And the speed of the light is c. So according to the ether observer's clock, the time it takes the beam to go from O to B to O:

But as we know, the terrestrial observers do not notice this. Their clocks show that it takes 2l / 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.

We again will consider two systems, S and S'. The system S is motionless in the ether. And system S' is a double of this system. They both begin unified. But immediately S' breaks off, flying away at speed 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).

So far, we have assumed something. The terrestrial observer follows the departure and the return of the light beam traveling from O to A then back to O. He only has a clock at point O. But what if he also had a clock at point A also? Perhaps he might only measure the speed of the light for the beam going from O to A, instead of O to A to O. No one yet has performed this experiment. But Bergson will still show we would find the same speed of light.

Bergson first addresses the issue of synchronizing two clocks separated at a distance. This requires that an individual at the one clock can communicate with the other one, to commonly set their clocks together. Yet, there is no instantaneous communication. Every transmission requires time. We need a sort of transmission that is under unchanging conditions. That way, we know in advance how to calculate for transmission delays.

When we transmit signals through matter with mass, the speed of the transmission will vary according to very complex and changing conditions. So, we need to transmit our signals through the ether, which does not have mass (it is not "ponderable matter"). We will use electromagnetic or light signals. The person at O sends a light beam to the mirror at A, just as in the Michelson-Morley experiment. When the beam reaches person A's mirror, he marks a zero at the point where the second-hand of his clock happens to be at that moment. The person at O keeps track of just two things: the time he sends the light beam, and the time it comes back. Because the distance is the same both ways, he knows that in the middle of these two time markings is the zero-point of the other person.

The problem with this arrangement is that the time may not have passed homogeneously throughout the light beam's journey. But clocks O and A are not both in systems lying in the motionless ether. Also, we are not dealing with the light moving perpendicular to the earth's motion, as with OB. In that case, the speed is the same both ways. But in OA, the speed is drastically different for each way. And if time changes with speed, then the person at O cannot suppose that A receives the light-beam half-way between the duration spanning between its dispatch and arrival. "The observer who is at absolute rest in the ether believes that the passages are unequal, since in the first journey the beam emitted from point O must chase after point A which is fleeing it, while on the return trip the beam sent back from point A finds point O coming to meet it" (7cd). So when O marks the halfway-moment, he is in error, it is "too close to the point of departure" (7d).

Bergson calculates the extent of his error. We are assuming that the light-beam takes a different length of time for each of its trips, there and back. The light goes slower on the way to A. Bergson finds that A needs to add a time-value to his zero. Let's say A receives the beam at a time-point we will call 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.

As we know, for those on earth, the light appears to take the same amount of time both directions. For earth observers, the halfway time-point of the light-beam's journey is in fact an equal division in half of the total journey duration, from O's measurement. This will be confirmed by A who has synchronized his clock with O.

However, the ether observer sees something different. The proportional value of duration AO and OA is the same as the ratio of 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.

So observers on earth will find that light moves the same speed both directions, regardless of whether they use one clock or two. However, the ether observer notices some mistakes. For him, both clocks are in motion. So already they are both moving too slow compared to his clock.

Now let's again begin with this assumption. System S and S' begin by coinciding. Then S' departs in another direction. When the split happens, clock C'o in systems S' coincides with clock Co 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

And the stationary observer already knows that each second that passes on system S is really

So the stationary observer cannot go by even the time measure that was adjusted on clock C'1. He needs to adjust that time again, to account for the fact that time passed at a different rate for clock C'1.

So the motionless observer in system S uses that formulation above to synchronize his own clock with the clocks in system S'. In the process, he is able to see that the people in system S failed to synchronize their clocks correctly.

Bergson now has us consider the linear path of system S. We place a series of clocks along that path at equal intervals of 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

So clock C'1 needs to be turned back by that amount. Yet clock C'2 is another l away. So it will need to be turned back by

In this way, "Simultaneity has changed from real to nominal. It has been incurvated into succession" (10cd).

Bergson now summarizes. We have been trying to understand how light could have the same speed for both the stationary and the moving observers. We spoke of moving system S' that diverged from motionless system S. S' moved at a speed of 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.

Modification 1. We learned that spatial lengths contracted in the direction of the system's motion. "The new length is proportional to the old in the ratio of

to unity" (10d).

Modification 2. We found that time in system S' stretches like a rubber-band. "The new second is proportional to the old in the ratio of unity to

" (10d).

Modification 3. The simultaneities in S become successions in S'. The exception are contemporaneous events that are perpendicular to the direction of the system's motion [their time and space distortions are equivalent]; they remain contemporaneous. Let's consider instead events in S' that are not perpendicular to the system's motion. And let's suppose that their respective duplicates in system S are contemporaneous. The events in S' however will not be contemporaneous. They "have separated in S' by

, seconds of system S', if by 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).

"In short, considered in space and time, system S' is a double of system S which, spatially, has contracted in the direction of its motion, and, temporally, expanded each of its seconds; and which, finally, has broken up into succession in time every simultaneity between two events whose distance apart has narrowed in space" (11b). Yet, the observer in S' does not notice these changes. Only the stationary observer can perceive them.

We will now use the names Peter and Paul to describe people in each system. Peter is in the motionless system. Paul is in the moving one. Peter says to Paul, "The moment you sparated from me, your system flattened out, your time swelled, your clocks disagreed. Here are the correction formulae which will enable you to get back to the truth." But Paul would merely reply, "I shall do nothing, because, if I used these formulae, everything in my system would, practically and scientifically, become incoherent. Lengths have shrunk, you say? But then the same is true for the metre that I lay alongside them; and, as the standard of these lengths in my system is their relation to the metre thus altered, that standard must remain what it was. Time, you say further, has expanded and you count more than one second while my clocks tick off just one? But, if we assume that S and S' are two copies of the planet earth, the S' second, like that of S, is by definition a certain fixed fraction of the planet's period of rotation; and say what you will about their not having the same duration, they both still last only one second. Have simultaneities become successions? Do all three docks [sic. probably, clocks] situated at points C'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).

Paul would also take note that the speed of light does not change even though his system is in motion. "Now, I must add that I regard these changes as fortunate. In fact, getting away from ponderable matter, what would not my predicament be with regard to light, and, more generally, electromagnetic events, had my space and time dimensions remained as they were! These events are not carried along in the motion of my system, not they. It makes no difference that light waves and electromagnetic disturbances originate in a moving system: the experiment proves that they do not adopt its motion. My moving system drops them off on the way, so to speak, into the motionless ether, which takes charge of them from then on. Even if the ether did not exist, it would be invented in order to symbolize the experimentally established fact of the independence of the speed of light from the motion of the source that emitted it. Now, in this ether, before these optical facts, in the midst of these electromagnetic events, you sit motionless. But I pass through them, and what you perceive from your fixed observatory happens to appear quite differently to me. The science of electromagnetism, which you have so laboriously built up, would have been mine to remake: I would have had to modify my once-established equations for each new speed in my system. What would I have done in a universe so constructed? At the price of what liquidation of all science would the soundness of its temporal and spatial relations have been bought! But thanks to the contraction of my lengths, the expansion of my time, the breakup of my simultaneities, my system becomes, with respect to electromagnetic phenomena, the exact imitation of a stationary system. No matter how fast it travels alongside a light wave, the latter will always maintain the same speed in relation to it, the system will be as if motionless with respect to the light wave. All is then for the best, and a good genie has arranged things this way."

Paul continues to say that he will need Peter's formulas in order to grasp the temporal relations between events occurring at different places in the cosmos. "There is, nevertheless, one case in which I have to take your information into account and modify my measurements. [12-13] This is the matter of framing a unified mathematical representation of the universe, that is, of everything happening in all the worlds moving with respect to you at every speed. In order to establish this representation which would give us, once complete and perfect, the relation of everything to everything else, we shall have to define each point in the universe by its distances 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).

"Moreover, axes OX, OY, OZ, which will be chosen in preference to all others as the only axes really and not conventionally motionless, will be given in your fixed system. But, in the moving system in which I happen to be, I shall relate my observations to axes O'X', O'Y', O'Z', which are borne along with this system; and, as I see it, it is by its distances 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).

We are then left to calculate 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

and consequently what you call x is

This solves the problem. I shall not forget, moreover, that the time 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,

Such is the time 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).

In fact, Paul could also have implemented Lorentz' "transformation equations" which do not even assume that system S is stationary.

Let's say in system S' there is a point M'.

Recall that system S' moves away from system S in the direction of O'X'.

[We will think for example that the x-coordinate for M' is 1, the y-coordinate is 3, and the z-coordinate is 1.] So we have been supposing that there is a single time and a space independent of time. 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:

We will say that the time that has passed is t'. Hence we have the following equalities:

x = x' + vt'
y = y'
z = z'

We are also assuming that "the same time unfolds in every system," hence

t = t'

But, recall we need to make adjustments for the contractions in lengths and the slowing of time. We only have local measurements for system S'. To rectify them to system S, we will need to modify them.

We will now consider another clock in system S'. It is clock C'1. Because it also coincides with system S, it will diverge in a direction away from the clock Co in system S. So the line C'oC'1 indicates the direction of the motion that S' travels.

Recall that we adjust x with this formula:

And recall also our formula for rectifying time. We will obtain the time t with

We will now imagine that point M' moves with uniform motion inside S', parallel to O'X' at speed v'. We measure it with x' / t'. That means we will divide

from

Bergson says that gives us:

This figure is less than 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.

The riverbank observer will get the same measurement as well. [Bergson provides the mathematics on pages 15-16.] Bergson will use this measurement to show that nothing can go faster than the speed of light. Let's suppose that the boat is traveling at 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.

So, Peter can then calculate x', y', z', t', and v' knowing just x, y, z, t, v'', by using the following formulas:

These equations are also known as the Lorentz transformation.

Bergson concludes by saying that we have worked through these formulae and thought experiments to "set the state for the analysis and demonstration that form the subject of the present work" (17a).

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

The original French version is available online at:

Image from the Michelson Morley article was obtained at:

Three-dimensional images based on an image from: