Philosophy 167: Class 13 - Part 10 - De Motu Corporum in Gyrum: Problem 3.

Smith, George E. (George Edwin), 1938-

2014-12-02

Description
  • Synopsis: Further mathematical analysis of De Motu.

    Opening line: "This is for the Keplerian ellipse. So it's the same ellipse now, but we are sweeping out equal areas and equal times about a focus not about a center."

    Duration: 6:39 minutes.

    Segment: Class 13, Part 10.
This object is in collection Subject Genre Permanent URL
ID:
r494vx283
Component ID:
tufts:gc.phil167.154
To Cite:
TARC Citation Guide    EndNote
Usage:
Detailed Rights
view transcript only

This is for the Keplerian ellipse. So it's the same ellipse now, but we are sweeping out equal areas and equal times about a focus not about a center. And again from Appalonius, the first principle is the same. QX have to look at what these are. QX is to QT as PE is to PF.
Those are similar triangles. That's what that's from. Again all circumscribed parallelograms are equal in area. And the latus rectum, which is the line drawn through the focus perpendicular to the major axis. The lattice rectum is given by 2BC squared over AC. The latus rectum basically replaces the diameter.
When an ellipse becomes zero eccentricity, the latus rectum is the diameter of the resulting circle. So, in effect, the latus rectum collapses into the diameter, rather than the other way around. The latus rectum is defined for all conic sections, not just of ellipses. For parabolas, etc. Okay. Outline of the proof.
The first step. Newton has discovered a new feature of ellipses. And as Whiteside remarks, he seems pretty proud of having discovered this. EP is the line. You draw a line from P to a focus, it intersects the conjugate diameter parallel to the tangent of P, it intersects that at point E.
That's how E is defined. So DX is the conjugate diameter parallel to the tangent at P. Ok? So far, now look at the fact that CS and HC are equal to one another. This is an ellipse, it's symmetric. If they're equal to one another, EI and ES have to be equal to one another.
Because HI is parallel to this conjugate diameter. When you combine that you end up with EP being the same, this distance EP is the same as the semi major axis. Because EP, when you do it algebraically is simply PS plus PI divided by two. It is easy to do that.
But it's a striking feature of ellipse that nobody had noticed before. Probably cuz nobody was looking at lines drawn from a point to a focus the way Newton was. But that's sig-, it's crucial to do that. Now that he's got that QR up there versus PV. Now this PV if you see, it's this little distance.
It is the same as EP to PC. Again, I have to find points on here. EP to PC, that's gonna be similar triangles. Let me say why, PV is this distance, QR is the same as PX. So, this triangle PXV, is a part of this larger triangle PEF, and therefore they are similar triangles and that's what gives me the second line.
Third line, QX squared is to QT squared, as EP squared is to PF squared. That's again, similar triangles, but it's this time because of the relationship of the angles, this is a right triangle, this is a right triangle and this is perpendicular to the hypotenuse here. And what's the one I want to complete that?
This is perpendicular to the hypotenuse there, so they're similar triangles. Okay now I simply write down. And how you discover these proofs I don't know. I've learned over the years to read the proofs. It's very rare that I can discover them. Latus rectum times QR over QT squared.
And you now substitute all these substitutions and you get this very, very long expression for that. And now you start simplifying using things up here, you reduce it down to here and you ask what happens as point R approaches point P at that juncture. This PC and QV, and GV, and QX square, the whole thing collapses to unity.
And therefore, the force is proportional to one over the latus rectum times XP squared. Now notice something there, it's not just SP squared. It's not just one over R squared. It's one over R squared is how the force has to vary within any one. Any one ellipse, that's the result that's significant here.
But the latus rectum is showing up to as a scaling factor. Bigger the latus rectum, the greater the force. Is that right? No, the less the force, cuz it's inverse. That would be significant in the Principia, it would be a little bit significant in what follows, but not greatly so.
The latus rectum is of course a constant. So therefore the force doesn't vary as the latus rectum within any one ellipse. That just becomes irrelevant. It's gonna be relevant when we compare ellipses one to another. Fair enough? So these proofs, as I say, they're hard to discover but they're fundamentally just using facts from Apollonius and similar triangles and setting up ratios that will work.
And when I say I don't know how to discover them, it means I can't anticipate the right ratios. Why? Because I've learned mathematics symbolically and to me I just want to play with the symbols, and you can't do it that way. You've gotta visualize what it is your trying to get and I don't know how Newton did that, but I wasn't trained what he was trained to do.