<author/>
</titleStmt>
<extent/>
<publicationStmt>
<distribution>Tufts Technology Services</distribution>
<address>
<addrLine>Tufts University</addrLine>
<addrLine>16 Dearborn Road</addrLine>
<addrLine>Somerville, MA, 02155</addrLine>
</address>
<idno/>
<availability status="free">
<p>This publication is freely available for scholarly or educational use.</p>
</availability>
</publicationStmt>
<sourceDesc>
<recordingStmt>
<recording type="audio" dur="695940ms">
<date/>
<equipment>
<p/>
</equipment>
<respStmt>
<resp/>
<name/>
</respStmt>
</recording>
</recordingStmt>
</sourceDesc>
</fileDesc>
<encodingDesc>
<editorialDecl>
<stdVals>Standard date values are given in ISO form: yyyy-mm-dd.<p/></stdVals>
</editorialDecl>
<classDecl>
<taxonomy id="LCSH">
<bible>
<title>Library of Congress
English.
__
____So Newton knows how to do uniform circular motion. And I'm going to do it the way Hooke did it. I mean, Hooke. The way Huygens did it. But Newton is perfectly able to do that as well and does a version of this in de motu. So, we have a centripetal force retaining a body in uniform circular motion, and we know it varies as the distance BD, over the times squared.__
__
____Why? Because BD is the displacement from the tangent to the circle, and the displacement from a force occurs over times squared, at least at the very beginning, of such a thing. But by Euclid, 336, BD is the mean proportion between, excuse me. AB is the mean proportion between BD and DE.__
__
____So we can replace BD, by AB squared over BE. AB squared, as you take the limit as B approaches A, becomes the same as AD, which is a measure of distance along the arc. BE as a measure becomes the diameter, or radius, since factors are constantly, can drop out.__
__
____So we end up with AD over Delta T squared, which is a distance over time squared, which is the velocity squared over R, or R over P squared. That's the solution. And the key to the solution is this displacement from where the body would be had it stayed on the tangent, to where it is.__
__
____And that displacement takes place because of a force over a time squared. Fair enough? And now the issue, how to generalize from uniform circular motion to arbitrary curvilinear motions, like Kepler's ellipse, and I've already showed you the way they were trying to do it, just directly, circle after circle; has too many degrees of freedom.__
__
____All right. So the natural way to do it, once Hooke gives his proposal, is to draw the same diagram. But now, not with a circle, with an arbitrary curve. And have a little segment, QR, and have a force directed toward the center, S. And have that force in times square, give you the displacement of QR as the point R approaches P.__
__
____This has set it up exactly the same way as before. But now you have a problem. We're gonna be working here geometrically. If we're gonna work geometrically we need a geometrical way of representing time. Okay. What--in a circle, uniform circle, we have three candidates. The two candidates that were always used.__
__
____Equal angles and equal time, so the angle can be used. Or equal arcs and equal times. There's always a third but nobody talked about it. In equal times. So, having to find a way to represent time, Newton turns to the so-called, what I call, the first key to it all, Theorem One is proposition one in the Principia.__
__
____In everything Newton does it's always the first theorem because it provides a representation of time. If all departures of a body from uniform motion in a straight line are directed toward a single point in space, S, the external force on the body is centripetal, in other words. Then the body sweeps out equal areas in equal times with respect to S, which means area can represent time.__
__
____Area swept out can represent time. The proof of it. Suppose a body is moving from A to B and, is struck by a force at B pointed toward S. Deflecting at the capital C. First consider what happens if the force doesn't act in B, it continues to little c.__
__
____Look at the two triangles SBc and SAB. It may not be obvious to you, so I'll give you the argument. They are equal in area because each of them is half the bigger triangle. The triangle ScA. What's the grounds for that? Extend this, drop from little c to perpendicular.__
__
____You can drop it here, and you will see immediately thatvbecause these two are equal, the height from here down to here has to be half the height from the little c down to here. Which means the area of the bottom triangle is half the other area. Go ahead.__
__
____You could do it perhaps more perspicuously by dropping a perpendicular from S, and it's the common height.__
__Yeah, I can do it more than one way, I can, I understand. Now, that gets us the equality without the force. Now we have to show SBc and SBC are equal in area, and now Marius's point becomes exactly the point I want to use.__
__
____Extend this, drop perpendiculars from this point and this point, because this line is parallel to this one, those two perpendiculars have to have the same height. So those two triangles have the same area. Therefore the triangle sb capital C has the same area as the triangle sb capital A.__
__
____Okay, now Newton says, and the same argument repeats for each subsequent triangle. Now let the number of the triangles increase, the frequency with which the impulse hits increase. Remember how he did the circle originally, bouncing walls? He's doing a similar thing here. Until you approach a smooth curve, and you will still be sweeping out equal areas in equal times.__
__
____Because you sweep out equal areas in equal times, all if it's impulse forces instead of continuous forces. The proof is a little bit controversial. Different people have said different things. I put on supplementary material an article that I refereed, as you'll see when you look at it, by Bruce Porcio, a mathematician who has largely straightened out this proof.__
__
____Newton could not have proved that you get what you need to in the limit, but you can now. It takes more math. Nobody at the time questioned this proof. This proof was considered just fine. There's an interesting question. It's so straightforward, why did nobody else discover it? And I can give you a series of possible reasons and just stop with that for now.__
__
____One is, people weren't thinking of impulse approximations to smooth curves. Newton had thought of that from when he was very young. So that's part of it. Second, nobody knew the solution for circular motion except Huygens, the actual proof for circular motion except Newton and Huygens. So they would not have even started thinking about how to generalize it, the way the two of them might have.__
__
____So that's a further thing. Until that proof comes out, other people weren't thinking that way. Third, at this time nobody, in fact, was using the area of rule to compute planets. Horrocks was long dead. While Mercator pointed out that it gives you good results. Mercator himself had replaced it with an alternative.__
__
____And of course Buyo had as well. So maybe nobody was quite looking at this in the way that Hooke, and or Halley, managed to get Newton to. Regardless, it's a remarkably easy proof at the time versus virtually everything else in De motu. It's probably the easiest proof in all of De motu and the question, why didn't others discover it?__
__
____I think the obvious answer is somehow or another, they didnâ€™t ask themselves the right question. Once he had this, then he could go back to this problem and he could change it. Because now he knows its area, whats the area of SQP? Well, as Q approaches P, it becomes a triangle of height QT and base SP, and since what we want is time square, we want area square, so therefore, the force is going to be proportional to the limit of QR, over QT squared times SP squared, as Q or R approaches P.__
__
____And two or three comments. This is the fundamental enabling theorem for all the rest of what Newton does in the first edition of the Principia. In the second edition, he gives two ways of doing it. One, working from curvature, the way I did when I talked about oscillating circles, and this way.__
__
____So they become both ways, they play off against one another, their mathematically equivalent. But in the first addition, this is it. That's first comment. Second comment, no, this is totally general this has nothing to do with inverse squared. Give me any curvilinear path in which you're sweeping out equal areas about A, I have to say it, it can be any curvilinear path in which all of the deviation from uniform motion in a straight line is always directed toward a single point in space.__
__
____Give me that. I automatically get the area rule. I automatically get this as the criterion for how the force has to vary along the trajectory, at every point. Let me go one step further. The formula at the bottom is due to Johann Bernoulli, approximately 1710. What he does, he says is much better to do this whole thing symbolically rather than geometrically, and he works out the proper formula for doing this.__
__
____This is it. Namely, R here is the variable it's where, in polar coordinates, Theta is the independent variable, R is the distance from S, and that gives it to you. It's the modern way of doing it, and you'll find it in most textbooks that way today.__