Fig. 19. Salviati’s Pendulum in Galileo.
Galileo, Dialogues Concerning Two New Sciences
Part 3 (Third Day: Naturally Accelerated Motion)
MCKEON: In our last discussion we made a transition from the analysis of uniform motion to motion uniformly, or naturally, accelerated. In the discussion we examined the fact that we proceeded in the same fashion and, in turn, based our new definition, the definition of naturally accelerated motion, on the analysis of uniform motion. We then proceeded to a consideration of two separate difficulties, both again arising from our basic assumptions. First was the question of whether you could ever get from motion to rest or from rest to motion, and this turned on our assumption about continuity. The second turned on the question of whether uniformly or naturally accelerated motion is proportionate to space as well as to time; and the answer to that, as far as we got, followed straight from a consideration of the proportion we set up in our equation. We, therefore, reached the point at which, Salviati having given his answer and a supporting example, Sagredo complains that the answer is a little bit too easy. Salviati properly observes that in the refutation of a fallacy, you anger those who oppose the possible answer. Then we go on to another answer, one which Salviati says will increase the probability of our definition’s single assumption by an experiment. This last quotation is from page 169 and I’m now going on to pages 170–71.
My first question has to do with the nature of this experiment, of the way in which we set up all our proofs. Is Mr. Brannan here?
BRANNAN: Yes.
MCKEON: If we were going more slowly, I would ask you about the way in which Sagredo gets his figure at the top of page 170. It’s quite important.
BRANNAN: I’m pulling my thoughts together. . . . Well, he shows that the . . .
MCKEON: No. Let me warn you again: don’t tell me what he shows. He has already demonstrated something which has been accepted as probable; he is now going to perform an experiment which will increase its probability. Tell me what kind of an experiment it is and how it will help us out.
BRANNAN: The experiment is a pendulum which shows the same thing that the experiment with the inclined plane showed.
MCKEON: But he hasn’t come to the inclined plane yet, so it couldn’t possibly show the same thing. He comes to the inclined plane later.1 What is it that he has to prove?
BRANNAN: Oh, the way it’s done is the previous proof. He says it’s merely the logic of falling down a . . .
MCKEON: Have you read this, Mr. Brannan? Are you trying to . . .
BRANNAN: No, I’ve read it. I just don’t remember the . . .
MCKEON: We’re not dealing with things falling down. Mr. Kahners?
KAHNERS: Well, first, what he ends up with is that he hasn’t really proved he had different types of motion. He showed . . .
MCKEON: What has he proved thus far?
KAHNERS: Well, he hasn’t. So far he’s talked about the . . .
MCKEON: He has proved something. What is it he’s proved?
STUDENT: I think he’s established the notion of inertia. His definition corresponds to reality.
MCKEON: Mr. Stern?
STERN: Yes. Well, it seems to me what he’s trying to establish and affirm is that, as he says, it’s not a function of space but rather of time that will determine the acceleration. I think the way he’s doing it is by not really talking about time but about how the angle of inclination here has no real effect on the momentum; rather, it’s a function of the height . . .
MCKEON: No, I think you’re going too far, Mr. Stern, because to give an argument, you need to give the steps. Let me indicate where Mr. Brannan got into trouble. What we have proved thus far is that the proportion of acceleration is not relative to space but relative to time. Obviously, the two interlocutors are a little disturbed, and it comes out pretty flatly. We then make a transition in which we repeat the definition—I’m giving the beginning of an answer—and we say that this definition, being established, only makes a single assumption. It’s this assumption that the inclined plane will explain. That is, he says, “The speeds acquired by one and the same body moving down planes of different inclinations are equal when the heights of these planes are equal” [169]. I would expect, therefore, that the first step which we would have to take in order to indicate what it is we need to prove would be to show what relation this assumption has to what it is that we’ve just proved. That’s why I said Sagredo brings in his figure. That leads, secondly, to Salviati’s reply that this geometric demonstration is “plausible; but I hope by experiment to increase the probability . . .” [170]. Since we’re talking about proof, what we’re interested in is how our original proof, namely, regarding what the acceleration is relative to, is related to the Sagredo figure, which is an inclined plane, and Salviati’s figure, which is of a pendulum. Now, I don’t want a subtle examination of them. They’re nice proofs if you go into them in some detail. What I would like would be a characterizing proposition about what’s going on here. But these are the two steps that I had in mind when I asked my initial question.
STERN: Well, I understand. I would accept that that’s exactly what it is. That’s what I understand it to be. What he’s talking about here is, given the height, you have the same momentum, even though it did go over the inclined plane. Therefore, the inclinations may differ, but the acceleration is a function of . . .
MCKEON: Well, let me stop you again because, you see, what I’m interested in is how you prove anything. Sagredo begins by saying, “Your assumption appears to me so reasonable that it ought to be conceded without question,”—that is, we’re again down to a place where we begin with definitions which people grant—“provided of course there are no chance or outside resistances . . .” [170]. Well, that’s the normal hypothesis: abstract from anything that might interfere, from the surface of the ball or the surface of the inclined plane or the atmosphere or anything else. What is it, then, that he is now saying? And, since in the uniform motion example we have said that Salviati did the same kind of thing, why at this point doesn’t he say, “Fine, I’m glad to see you’re doing it the way I do"?
STERN: Doesn’t Salviati conclude that this is acceptable?
MCKEON: Exactly.
STERN: Well, I think that perhaps the reason is that it does depend upon these variables being removed.
MCKEON: No, no. This is always the case. It is not the case that all objects fall with equal acceleration, for instance, if you don’t have an awfully controlled situation and drop a light object and a heavy object. [McKeon drops a book on the floor.]
STERN: Well, in terms of what you’re talking about, this plane on which things are moved is, therefore, involved . . .
MCKEON: Let me throw this open to the class. [McKeon again drops a book on the floor.] Does anyone see what I’m driving at? . . . In other words, I’m asking, What does the pendulum add to this demonstration? Salviati would have objected to it if it was false; he finds it plausible. But he wants to increase the plausibility. And my initial question is, What is it that he adds when he agrees to the plausibility?
STUDENT: More evidence?
MCKEON: Well, let’s ask this. Does he do this? This is the point at which he says he did an experiment with a pendulum. Did he?
STUDENT: No.
MCKEON: What did he do?
STUDENT: He theoretically constructed an experiment.
MCKEON: What does he actually set up?
STUDENT: He imagined.
MCKEON: Imagined what?
STUDENT: This page.
MCKEON: He imagined a piece of paper representing a wall, imagined a pin in the wall, imagined a string on the pin (see fig. 19).2 I mean, there’s nothing wrong with this, but the important thing is that this is not an experiment in the sense that he did anything. This is something which everyone will see as soon as they begin imagining the pendulum on the page.
STUDENT: Could the difference be that he’s using this experiment to provide a causal explanation.
MCKEON: No, the causes we will worry about later.3 Yes?
STUDENT: Well, doesn’t this set up a postulate from which the answer would be derived and he . . .
MCKEON: No, the only postulate we’re going to have is the postulate that we’ve stated. That’s our postulate. This is a conclusion. A conclusion is different from a postulate since it follows from a postulate, whereas a postulate is set down.
Do you all at least feel that there’s something that you ought to see here, or is it a mystery what I’m trying to define? [L!] Let me answer it because maybe then you will see. In Sagredo’s example, all that you do is to give your planes and the perpendicular and to conclude that at points A, D, and B there will be equal momenta, equal impeta (see fig. 20).4 The example that we then go on to, the experiment, translates that possession of equal momenta into something which would be readily apparent; namely, if it has equal momentum, it will rise to equal heights. Isn’t this the difference? In other words, all that Sagredo would give you would be an inference which is perfectly correct—it follows—but whether or not what follows is, in fact, the case is not apparent. The experiment that Salviati sets forth gives you a criterion by which to find out what it would mean if that is the case; and even though he didn’t make a pendulum, the imagined pendulum answers the question. Does this seem to you, now, like a reasonable question and a reasonable answer? Yes?
DEAN: I still don’t understand the difference in kind from that answer and saying, Well, we’ll look at A, B, and D and find out if the bodies are there at the same time.
MCKEON: Sagredo doesn’t seem different than Salviati?
DEAN: Well, Sagredo’s demonstrated what equal momentum would amount to in terms of different positions, and, as we said, at a given point the ball will rise at a given momentum. Salviati’s illustration shows that it will have an arc equal to the arc from another position. Were it not equal, there would be a reason for it because it wouldn’t have a sufficient momentum at a given point and so it wouldn’t end the arc the same.
MCKEON: Yes?
STUDENT: Doesn’t he say, Look and see what happens to the arcs when the pin is moved, and look and see where the balls are when they stop?
MCKEON: Yes?
STUDENT: I don’t see the difference in kind between the two experiments either. In other words, the second experiment seems to have been constructed to further explicate the first.
STUDENT: They seem similar to me, too.
MCKEON: They are similar in that you can see how you go from rest to motion and from motion to rest. This is there, but does this answer the question that Mr. Dean found so disturbing?
STUDENT: Isn’t there a difference in the curved line of the pendulum and the straight line of the inclined plane?
MCKEON: The curved plane will come in, but that doesn’t matter here. That is, the curvedness of the line taken by the bob of the pendulum makes no difference. . . . Well, let’s leave it here. This is still an answer to the question, even though we now have to come away from it.
STUDENT: I’m still confused by the difference.
MCKEON: You don’t see any difference? . . .
STUDENT: Well, the only thing I can think of is that when you were demonstrating the definition . . .
MCKEON: No, no. I think you’ve got to begin by saying, This is not a demonstration of the previous proof; this is another question. Notice, he begins by saying, “This definition established, the Author makes a single assumption . . .” [169]. In other words, we’re now going on further; we are not by this experiment establishing the proof concerning the distance being proportionate to time . . .
STUDENT: Yes, I know that, but we didn’t talk about how to measure the forces involved, even with the different questions that he brings up.5
MCKEON: We are talking about speed, as we were before. The final speed of one body moving down a plane at different inclinations will be equal to the heights of the plane. We have three variables: speed, distance—the height is a distance—and time. Now, in terms of our definition, namely, that the distance is proportionate to time, which has been proved, we are arguing that the final speed acquired will be the same for any given height, no matter what the paths taken. This is an assumption we are making, and we are now going to prove it, as we have proved the previous one, by analyzing our variables again. We are not deducing it from the definition; we are, in terms of the definition, making this assumption to prove the diagram. But the important thing is that we’re introducing no new variables; we’re merely asking another question.
If this is the case, what we have been talking about thus far has been the descent of a freely falling body; and we’ve been arguing about whether the successive stages are accelerated relative to time or to distance. We’ve said time. We are now asking, with respect to this descent, What about if it rolls down a plane instead of falling? And we’re saying, The velocity at the three points A, D, and B in Sagredo’s figure will be the same if the distance CB is the same. And this is what we want to prove. And what Salviati does is to do it directly.
Let’s first make clear what it is that Salviati’s experiment adds that the first did not have. . . . I thought when I asked that question I was giving the answer away. It had velocity or speed, distance, and time. All our assumption says is that the velocities at A, D, and B will be the same. Is there any other word that comes into both proofs?
STUDENT: Momentum.
MCKEON: Momentum—notice, momentum isn’t on our list—impeto. The demonstration, according to Sagredo, is one that reason will see. All that reason would say would be that, analyzed in terms of this, it is reasonable that the same body, if there’s no external interference, will reach the three points, namely, A, D, and B, with the same momentum. That’s plausible enough: it would have the same momentum simply because there isn’t any source of addition or subtraction. But we need a way of knowing what we mean by momentum. If we define momentum acquired by descent to be that by means of which an ascent to the same height is possible, we have a way of knowing whether the momenta are the same or not. Now, instead of a purely negative proof, namely, one stating that there isn’t any way there could have been an increase or a decrease and, therefore, it’s reasonable that they would be the same no matter how you rolled or dropped the ball, we will turn around and ask, Is there anything we can imagine which would involve the possibility of a momentum being used to get back to the start? Notice, if we’re dealing with balls, the problems of elasticity are so great that we’d never get anywhere; therefore, the use of the pendulum is a good way of avoiding these problems. The pendulum is so arranged that the pins stuck in the wall will stop the swing of the pendulum before it’s completed at different points and thereby allow you to compare what the swing would be whether uninterrupted or interrupted. Consequently, you get your three ascents which would roughly duplicate the problem of the ball falling or the ball rolling. All right?
STUDENT: Is this a question of where the lack of definition would matter in the first case, or at least the lack of it being measured properly?
MCKEON: I don’t know what you mean by a definition of measurement; that is an additional variable. This is a variable which is of the order of a force; and when I earlier stopped the question of the cause being brought in, it was because cause is explicitly brought in later and a force is of the order of a cause. All we’ve been dealing with thus far have been bodies in motion; therefore, we had velocity, we had time, we had space. Now we’re asking something about what happens when changed conditions are imposed on a body in motion while it is still in motion and we want to identify that. It’s the cause of the next motion. And, therefore, there would be no way of saying that the force is the same or different unless you could compare what the force did.
STUDENT: That is, this is a variable we’re dealing with?
MCKEON: Yes.
STUDENT: Would it be possible to get the proper tools to do it with the first case?
MCKEON: Notice what you would have to have. The proper tools would have to be instruments that would be hit by the balls and would register force. I think we could probably do that now, but even that would be fairly complicated. It would be inconceivable at this earlier point in time. And it would not be as good since all you would be doing, then, would be giving an instrument reading; whereas here you are having an experimental indication of the force using itself up. The experiment has to be practical even if you did have a force-measuring instrument.
STUDENT: Aren’t you using as a variable a minimum force to give you a way of talking about acquiring equal speeds? It seems to me you’re explaining the first case, but the assumption here is in terms of speed.
MCKEON: No, the assumption is in terms of cause of speed. In general, you need to take into account the plumb’s momentum, that is, the mass as well as the time and space.
STUDENT: Yes, but doesn’t that create—I’m asking the same question now which I asked before—doesn’t that create your conflict, that is, involving mass with velocity to give you a way of looking to see if it’s part of the speeds? I mean . . .
MCKEON: I think . . .
STUDENT: It seems to me that the question is, Do they acquire equal speeds?
MCKEON: In general, whenever you have anything, you have a cause; but the examination of the phenomenon and examination of its source are totally different things. If I were to say something that embarrassed you, this might be a cause of your blushing, but the phenomena of your blushing and my saying something embarrassing to you are quite different. It’s entirely possible that you blushed for other reasons. Consequently, the question would need a further examination because my variables are not precisely the same. We’ve chosen a better one here, that is, a continuation of motion, the impetus. Let me put it another way. The whole point of the analysis is to get the concept of impetus. If impetus were the same as velocity, there’d be no reason for Galileo working on this.
STUDENT: But why does it come out of an assumption which doesn’t seem to be, on the face of it, needed? I mean, isn’t that a backwards way of proving . . .
MCKEON: No. On the contrary, you can’t possibly explain what this assumption is, namely, that the bodies rolling down a plane and falling would have the same velocity, except for the concept of impetus. We have added a variable.
All right, having painfully and slowly differentiated these two proofs, let’s now ask, How’s this demonstration accepted? What is the reaction, in other words, that we get from Sagredo in the further conversation? Mr. Rogers? . . . Mr. Flanders? . . . Miss Frankl?
FRANKL: The experiment is proof enough that the statement can be taken as a postulate and . . .
MCKEON: Just a moment. What do you mean “the statement” says? The original statement you say is a postulate?
FRANKL: It’s a postulate that the Author lays down.
MCKEON: Let’s take a look at what Sagredo says; it’s on page 172. What is it that he says?
FRANKL: He argues that . . .
MCKEON: “The argument seems to me so conclusive and the experiment so well adapted to establish the hypothesis that we may . . . consider it as demonstrated.” Tell me about the argument, the experiment, and what is demonstrated.
FRANKL: You’re asking me?
MCKEON: Remember, what Salviati said about Sagredo’s argument is that it was plausible and he wants “to increase the probability to an extent which shall be little short of a rigid demonstration” [170]. Therefore, Sagredo’s comment is in recognition of this. Notice, it will still be a little short of a rigid demonstration. Sagredo says, “The argument . . . [is] so conclusive and the experiment so well adapted” that he will be willing to “consider it as demonstrated," although it is still short of rigid demonstration.
FRANKL: The argument is the thought that equal height gives equal momentum to a body. And the experiment was the pendulum, which shows the momentum would be equal because the dropping of the bob would equal the height it returned to. The momentum could be used in the . . .
MCKEON: I think you may be falling into the classic error that led me to ask the question, Why? That’s what he said, but I’m not asking what’s said in the book. I’m asking, rather, What is the relation between an experiment and an argument? What is the conclusiveness of an argument, this one being almost conclusive? What are they driving at? What would a conclusive demonstration be?
FRANKL: If all the experiments would have an equal momenta?
MCKEON: Yes?
STUDENT: Wouldn’t it be one that you could do completely in geometry?
MCKEON: Eventually, as we go along, you’ll notice that he will be speaking of a system of results that come out of the inferences. A conclusive demonstration, if you had systematized all of this, would prove it geometrically from the assumptions you had made. Well, now let me go on. What is an experiment as opposed to the demonstration, Miss Marovski?
MAROVSKI: I thought there was a greater degree of freedom with a curved plane than . . .
MCKEON: No, leave this out. This I don’t think is relevant.
MAROVSKI: Well, you have a more conclusive demonstration when you proved it in terms of Sagredo’s proof, with the rolling ball, because the angle to do that would have been in terms of . . .
MCKEON: No, you couldn’t have proved it that way because he had not defined momentum; there was no way in Sagredo’s proof of doing it. Even if he had been able to construct an experiment that didn’t involve curved lines, he would only have approximated the demonstration. Why is this?
FRANKL: A demonstration reaches the truth directly, whereas an experiment has to be interpreted . . .
MCKEON: No, no. I think this does it too psychologically. Maybe I’m asking too subtle a question in these terms, but I think that it is apparent. A demonstration would be the systematic consequences that follow from the hypotheses or the postulates set down. An experiment is the concrete indication of circumstances in which this would occur. It’s the difference between a proof which is universal—a geometric demonstration is always universal—and the construction of something which would be an example of that and is, therefore, a good approach. You notice, it doesn’t make any difference whether it’s an imagined experiment or a worked-out experiment: if the imagined experiment runs into difficulties, then you actually put a pendulum up. But the difference between the experiment with the pendulum and the other is that Sagredo is moving in terms of a demonstration; and one difficulty is that all he can do is to say, It’s reasonable, since the velocities can be demonstrated to be the same, that the impetus is. The second gives you a pendulum of particular size and an indication of heights which accompany it; consequently, you are proceeding inductively from instances—it’s a concreteness—as opposed to the generality of the first proof.
Well, let’s go on. As I say, maybe I shouldn’t have sprung that one on you at this time. This is the point, as a matter of fact, where the curved planes come in. Miss Marovski, you’ve been worried about it. Suppose you tell us about how this comes in. Why does Salviati bring it up?
MAROVSKI: Well, he says he does agree with him, though he doesn’t think that accelerated motion which occurs on a plane surface creates a momentum equal to that on a curved.
MCKEON: My edition doesn’t say that he agrees with Sagredo.
MAROVSKI: Well, I don’t think that he still thinks . . .
MCKEON: Well, all that he says is, We’ve proved what we wanted to; consequently, let’s not get subtle about this because what we’re going to do is to apply the principle that we are establishing—and we’ve established it with our curves—to “plane surfaces, and not upon curved, along which acceleration varies in a manner greatly different from that which we have assumed for planes” [172]. What I would interpret he is saying here is that curved or straight, the momentum is exactly the same; but having used our curve, which is the best way of establishing the equality of the momentum, let’s not stick to this because we’re now going to go on to questions of acceleration, and acceleration on a curve is different than the acceleration on the plane.
MAROVSKI: If we were to disagree with the postulate, the actual truth would be proven by the . . .
MCKEON: NO. We have demonstrated our postulate; we will not disagree with our postulate. That’s why he brought it in.
MAROVSKI: I don’t understand what you’re saying, then.
MCKEON: Well, my question has to do with what it is that Salviati thinks we are now turning to and what he has used with respect to his assumptions thus far.
MAROVSKI: Obviously, he’s going to talk about acceleration on a plane.
MCKEON: Well, now, what is the similarity and difference? Let me read you the sentence that would be relevant to the whole problem, the beginning of Salviati’s second paragraph. “So that, although the above experiment shows us that the descent of the moving body through the arc CB confers upon it momentum just sufficient to carry it to the same height through any of the arcs BD, BG, BI, we are not able, by similar means, to show that the event would be identical in the case of a perfectly round ball descending along planes whose inclinations are respectively the same as the chords of these arcs” [172; and see fig. 19]. What is it that we have shown, and what is it that we need to show? “It seems likely, on the other hand, that, since these planes form angles at the point B, they will present an obstacle to the ball which has descended along the chord CB, and starts to rise along the chord BD, BG, BI” [172].
MAROVSKI: Well, I thought it was obvious that you have to get an experiment using the curved line of the pendulum to prove the assumption before it could apply.
STUDENT: Well, it suggests that in terms of the momentum, some is going to be lost in hitting the opposite plane; that is, you can’t purposely put in something to interfere with the ball’s motion.
MCKEON: Yes?
STUDENT: Isn’t the pendulum a proof of what we need?
MCKEON: We have proved by our pendulum experiment that the momentum is the same. We put a pin at F, a pin which will hit the chord and give you the arc BI. This arc and the chord of the arc would be exactly the same. The problem comes if you were rolling the ball down one of the chords of the arcs and up a cord on the other side. It would be unlike the pendulum because the rolling ball, whatever you do, when it hit the other plane at B would not simply go up because an obstacle is presented which is almost perpendicular. The ball wouldn’t roll up, whereas the pendulum would swing up. Consequently, it’s not a question of the curved or the straight line. Let me read it to you again: “It seems likely, on the other hand, that, since these planes form angles at the point B, they will present an obstacle to the ball which has descended along the chord CB, and starts to rise along the chord BD, BG, BI.” Notice what we are doing: we’re going to run the same ball down CB, and it’s going to have to go up these various planes, BI, BG, BD. How can we construct the planes so that the resistance that will meet the ball on the first, BI, which is rather steep, will be the same as the one that will strike the ball on the last, BD, which is not so steep? And this is all that’s involved. The momentum part is directly involved; therefore, he has proven this by demonstration and he’s not leaving it as a postulate. But we are now going on to a question which is different, in part because we want to deal with straight-line motions and not with curved, and in part because the differences of the acceleration of curved motions will be somewhat more complex than the differences of the acceleration of a body moving on a flat plane.
All right, if this is our transition, we now start a series of demonstrations. Is Mr. Davis here?
DAVIS: Yes.
MCKEON: Since we have only a few minutes left, suppose you take the first two propositions together. This will save you from the temptation to tell me what they contain and let you tell me what he’s trying to show. What, in other words, is the step that he goes through from I to II? . . . We have six propositions again, and if we can get the line of these six going as we got the line of the six on uniform motion going, we can be clear about what’s going on here. . . .
Let me give you the answer to the first as a kind of model; then you tell me the second one. In the first, he is making the transition from the uniform to the accelerated. And since we are dealing with uniformly accelerated motion, he is demonstrating here that the time—and notice, it’s the time, not the space—in which any space is traversed by a body in uniformly accelerated motion is equal to the time in which the same space would be traversed by uniform motion at a velocity half of the greatest. And this we could get from our mere equations, couldn’t we? In other words, if you drop a body with uniformly accelerated velocity for a given space, and if you then take one half of the final velocity of that acceleration and drop the same body with that uniform velocity through the same space, they take the same time.
All right, if this is our proposition of transition, what are we going to do in proposition II?
DAVIS: Well, I suppose the relation of the space would be defined by giving the result of the times.
MCKEON: All right. We’ll take uniformly accelerated motion and pick off any piece of it—this is still that any piece business. For any piece we can set up a relation between the space and the time. How will the spaces within that vary with respect to the time?
DAVIS: Well, we say it’s the proportion of the squares of the time intervals.
MCKEON: How do we know that?
DAVIS: How do we demonstrate it?
MCKEON: Not how do we demonstrate it. The two people listening to him expound would grant him this. Can you tell me how we would know it in terms of what we have done before?
DAVIS: Well, yes. Because in the case of uniform motion, there were cases in which the two variables we had before are now identical to the squares; both time, instead of time and velocity . . .
MCKEON: Don’t be so vague about it. I mean, in our previous proposition we related uniformly accelerated motion beginning at rest down to a given point with the uniform motion which had half the final accelerated velocity. We’re now going to compare the distances and they will vary according to the square of the times. We know our transition is from uniform motion—what’s uniform motion tell us distance is?
STUDENT: Velocity times time.
MCKEON: O.K. That is, this is a ratio for the purpose of the equation, d = v X t. What’s velocity?
STUDENT: Proportional to the time.
MCKEON: Well, velocity is acceleration times time; consequently, if we want a similar ratio, we can substitute another t, namely d = t X t. Consequently, you would always have the proportion from uniform velocity of distance as the square of the time. This is one of the reasons why all of our equations in elementary dynamics will come out as reversible because the time is squared, and when you take the square root of the squared time, you get plus or minus, that is, you can go in different directions. But when you get to problems of momentum, you can’t do this. This is one of the reasons why the Second Law of Thermodynamics is not reversible: it’s because you’ve got only a t in it.
STUDENT: Could you say again what you said about the substitution of the second t?
MCKEON: Well, the d equals vt. If, then, you are trying to determine with respect to distance equals vt what the variation for any value of d would be with respect to t, you’ve got a hidden t in the v. Consequently, it would come out in a ratio that states that d varies according to the square of the time. And the long demonstration here in proposition II is, in fact, tied onto the fourth proposition of uniform motion in this fashion.
Well, our time is up, but let me call your attention to something. We’ll begin with corollary I next time because I want you to notice what happens there. What we’ve just discussed means—and this is one of the statements that he made quite early—that the successive spaces traversed in uniformly accelerated motion will bear a ratio of odd integers, 1, 3, 5, 7, 9, and so on; while the times will be units, that is, 1, 1, 1. Consequently, if you take the total distances instead of the increments that I’ve set up, you have the sequence for the times, 1, 2, 3, 4, 5, that is, the natural number sequence, and for the spaces, 1, 4, 9, 16, 25, that is adding each increment. I suggested that this sequence begins to look something like the Platonic one. Do you remember what the Platonic sequence was?
FRANKL: 1, 4, 8, 16.
MCKEON: I know, but what were they?
FRANKL: A harmonic progression?
MCKEON: He was proceeding by taking 2 and multiplying it by 2 all the way, taking 3 and multiplying it by 3 all the way. Thereafter, he began to look for geometric and arithmetic proportions and you can eventually come out with proportions. In other words, he was setting up proportions according to a regular scheme. Here we are taking the natural numbers and the odd integers and making a simple mélange. I don’t think that there’s any subtle point to be made about it. The result of Galileo’s gave him the dynamics which has had a long history in the setting up of modern physics. I think that it would be possible, if one looked at the sequence that Plato dealt with, to see that there are similar periodical interrelations that have not been always brought into contact with the observational phenomena that they illustrate. In any case, we’ll spend at least one more discussion still on Galileo. We ought to get to Newton pretty soon, but we’ll begin with the corollary.
STUDENT: Well, how much should we read? The whole Third Day?
MCKEON: We will discuss only as far as we can go. I doubt whether we will conclude the Third Day, but we will probably get to page 200, certainly up to page 190. From that point on, the philosophic aspects are not that clear.
