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Newton's Lemma 28

  1. Jul 25, 2004 #1
    I have a problem with Newton’s reasoning in Lemma 28 (“There is no oval figure whose area, cut off by right lines at pleasure, can be universally found by means of equations of any number of finite terms and dimensions.”) It’s right after the start of Section 6 of the Principia, Book 1. I came across this lemma while reading Peter Pesic’s book “Abel’s Proof” about how equations of the fifth degree and higher cannot be solved in radicals. Newton’s Lemma is supposed to be a clear and elegant analog to this notion, going “far beyond the irrationality of pi” as Pesic describes it. (This description takes up the first part of Chapter 4, Spirals and Seahorses, in Pesic’s book.) Here’s a description of Newton’s argument, taken from the latter book:

    “To use the term that Euler later introduced, the area of a circle is transcendental, meaning it cannot be expressed as the root of any equation of finite degree whose coefficients are rational numbers. At one stroke, Newton indicates that such magnitudes exist (because circles exist, and have areas), and also that there are infinitely many of them, since his proof is not restricted to circles but hold for any “oval” curve. His proof is a miracle of simplicity and power, for which he does not even bother to draw a picture or write down a line of algebra. It follows from a single brilliant contrivance. Inside the oval, pick any point whatever; let us call it the pole, P. Now let a straight line come out from that pole and rotate around it at uniform angular speed. Picture a clock hand that makes a complete revolution in one hour. Now imagine a point of light moving along that hand, starting from the pole and moving outward along the hand with speed given by the square of the of the distance from the pole to the point A where the hand intersects the oval.

    Newton has set up a way of measuring the area of the circle, for each hour the hand sweeps through that area, and the moving point keeps track of that area because it is traveling with speed proportional to the area swept out. Here Newton is implicitly using his new calculus of motion, for he knows that, in an infinitesimally short time, the point travels a distance from the pole equal to the area the hand has swept out in that time. However, we don’t need to know anything about calculus in what follows. All that matters is that since, second by second, the moving point is registering the area that the hand sweeps out, we can measure the whole area of the oval merely by waiting until an hour has elapsed, and measuring the distance the moving point has traveled radially outward from the pole. For the two dimensional problem of measuring an area, Newton has substituted a one dimensional problem that gives the same answer: find the length traveled outward by the moving point during an hour.

    The hand moves around uniformly, but the moving point speeds up and slows down in the course of each hour, in proportion to the square of the distance from the pole to the oval at any given moment. Each hour it returns to its initial speed of an hour before. If you were to watch the lighted point (or if you were to open the shutter of your camera and make a long exposure), you would see it move in a spiral, starting at the pole and making “an infinite number of gyrations” as Newton puts it. Now Newton applies a reductio ad absurdum: Suppose that it is possible to describe this spiral (and hence also the area of the oval) by some polynomial equation with a finite number of terms, f(x,y) = 0. For instance, Descartes showed that all the conic sections can be described by equations of the second degree, and those curves can be cut by a straight line no more than two times. Now Newton relies on the fact that an equation of finite degree can have only a finite number of roots, no matter how large. But the spiral in its “infinite number of gyrations” crosses the line an infinite number of times. Thus there should be an infinite number of intersection points, corresponding to an infinite number of roots of the equation. This contradicts our hypothesis that the equation has finite degree, and hence Newton’s conclusion follows: there is no such equation that gives the area of the oval.”

    I have read the original text by Newton in the Principia, and it seems to me that the argument is much the same as Pesic describes it, though I find Newton’s archaic language a bit odd and hard to follow clearly. I understand about the area and the behavior of the spiral within the first hour. Subsequent to that, the lighted point sweeps out an irregular Archimedian spiral with separation equal to the area of the oval. But as the spiral “gyrates” on and on, it calculates not the area of the oval but, at first, twice the area of the oval, then three times the area, then four, etc. The area of the oval is determined within that first hour, and there the straight line intersects the spiral at most twice (trivially at the pole and at the full area point). That the line is allowed to continue sweeping out past the first hour introduces what seems to me a superfluous consideration. If the oval was a square the same steps would apply and it’d be hard to tell from the spiral what the shape of the generating figure was, yet the area of a square is expressible in radicals. It seems to me that Newton is saying that since x+1 and 2x+1 and 3x+1 and 4x+1 have an infinite number of solutions, that ax+b is not expressible in radicals.

    Clearly that’s wrong, and I doubt that Newton and those that followed and Pesic missed this simple point. Since the spiral itself is not equivalent to a formula for the area of the oval but an infinite repeatition of the area, I think the deduced contradiction is invalid. So there must be something I’m overlooking or misunderstood. If anyone can point it out to me, I’d be very grateful. I’m interested in the Pesic book, but have suspended reading it until I clear up this point.

    Again, let me thank the people in this forum in advance for your assistance and insights.
    Last edited: Jul 25, 2004
  2. jcsd
  3. Aug 3, 2004 #2
    No replies after a week? Does that mean no one can explain the flaw in my objection? Oh well, I'll just post this reply to bump the thread once, then let the topic go if it stays quiet.
  4. Aug 3, 2004 #3
    Hi Bob,

    Yes ... sometimes it's difficult to get the balance in posts, of saying too much or (as in lots of cases), not enough.

    There's quite a lot to read on your post and then some supp. to get to the nitty-gritty of your question.

    Your question seems to be asking whether or not Newton could lay claim to a 'proof' that pi is transcendental. Is that right?

    I'd be careful to note that 'rigorous' proofs weren't always quite as rigorous as they might be today. Also, Newton does need heaps of 'interpretation' because, as you pointed out, the language is archaic. So this might be (I don't know) just Pesic's personal opinion of what Newton wrote.
  5. Jul 5, 2005 #4

    I stumbled across this forum when googling for an answer on the same question. I'm reading the German edition of Peter Pesic's book and found that proof pretty disturbing as well.

    Can anyone explain it by now?
  6. Jul 5, 2005 #5
    Well, I have just started reading Principia but as soon as I have seen this post I couldn't restrain myself from checking out. I think I have found a way to explain this.

    Let C be the area of the oval. This is a fixed number.
    Let T be the time it takes for the straight line to come to the point it has begun. This is fixed as well.

    Now, think of an oval and draw a line from a point that is inside the oval. then draw another line. call the area between these two lines A. Then,
    [tex]A=\frac {C*dt}{T} [/tex]

    where dt is a time interval that is indefinetely small. Newton says that the speed of the movable point on the line is the square of the line at the given time. Let's call the line's length r.

    [tex] \frac {ds}{dt}=r^2 [/tex]

    Now replace dt with

    [tex]dt=\frac {A*T}{C} [/tex]

    now we have

    [tex] A=\frac{ds*C}{r^2*T} [/tex].

    From here it is clear that the area swept in an indefinetely small time is proportinal to the place of the movable line on the line. The quote you have given says that it is equal to the area but I could not prove it. I think proportionality is sufficient. I hope it is clear to you that but the time you add all those indefinetely small time intervals "dt" to get T, the point has "described a spiral with an infinite number of gyrations." so it is clear that the lemma 28 is true...

    Correct me if I'm wrong or I misunderstood you
  7. Jul 6, 2005 #6
    Unfortunately not. :confused:

    When T is the time it takes the straight line to come where it has begun (a turn of 360° around the pole) and the movable point is on that line, the points trace consists of two elements: the 360° turn and its own accelerated linear move along the line. So I understand that within T the point has described a spiral with exactly one gyration.

    That's also how I understood Peter Pesic - I haven't read Newton's original argumentation. According to him the infinite spiral is formed when you infinitely continue to turn the straight line and move the point. But that would not happen within T but within x*T whereby x -> oo.

    Under those preconditions I don't understand the validity of the proof. :uhh: In fact I can imagine to apply the same proof to a square and the result would be the same. I'd have to draw the same conclusion for squares as for ovals, which is wrong. The area of a square is not transcendental, "meaning it can be expressed as the root of any equation of finite degree whose coefficients are rational numbers."

    Help me, please!
    Last edited: Jul 6, 2005
  8. Jul 6, 2005 #7
    The hand moves around uniformly, but the moving point speeds up and slows down in the course of each hour, in proportion to the square of the distance from the pole to the oval at any given moment. Each hour it returns to its initial speed of an hour before. If you were to watch the lighted point (or if you were to open the shutter of your camera and make a long exposure), you would see it move in a spiral, starting at the pole and making “an infinite number of gyrations” as Newton puts it...

    this i quoted from above. And kepler's law of areas holds only for oval shapes, not squares. I have used it in what I said.
  9. Jul 6, 2005 #8
    I have read the book from which the quote you quoted again is taken. My problem is that I don't see how this prooves anything. Maybe I'm just not educated enough to understand. :(

    I don't know, can you explain how you can obtain a spiral with infinite gyrations by a point that moves with a (varying but always) finite velocity for a finite time interval T? That's basically the problem I have.

    My understanding is that in order to obtain the infinite spiral newton lets the line (and the point moving along it) infinitly rotate around the pole. But if he would do so, he would also measure the area of the oval not once but infinitely often.

    I'd appreciate any enlightment! Please be patient with me. :blushing:
  10. Jul 6, 2005 #9
    Kaorin: Under those preconditions I don't understand the validity of the proof. In fact I can imagine to apply the same proof to a square and the result would be the same.

    My objection is the same as Kaorin. But as far as the infinite turns of the spiral, this only shows that it has endless roots and so is transcendental. (I did not object to that part of the proof.)
  11. Jul 6, 2005 #10
    Oh right you are! I'm sorry. But I will help you now and I think I got everything right.
    If you have read Newton's original text he says that if we can express the area of the oval by a finite equation we can find the place of the point at a given time. You were right about the gyration part. Think of it this way. We have an equation f that gives us the area by revolving the line around the oval once, that is the number of gyration is one. if we revolve the line twice we have twice the area, that is 2f. by revolving it three times we have three times the area that is 3f. As the number n(1,2,3 in the examples i have given) tends toward infinity nf tends to infinity as well, but it's degree does not change. that if it is a polinomial of degree a, so it will ve no matter how many times you revolve the line. That is to say it has a finite number of roots. The line crosses the spiral the point has described infinitely(because we have an infinite number of gyrations, at last :) )The intersection points are the roots of the area equation we have. This is so because the place of the point on the line is proportional to the area as I have shown. There is a condradiction here. The equation says that there is a finite number of intersection, but the line crosses the spiral infinitely...

    This is how it should be, I believe. what I previously wrote does not make any sense as you have seen :)

    Notice that there have been much discussion about this lemma. Leibniz and Huygens did not accept what newton has written. D.T. Whiteside does not accept it as well. I think it is right.

  12. Jul 6, 2005 #11
    That is not a correct point of view. I have used Kepler's area law to show that the place of the point is proportional to the area. Kepler's Law is not valid for squares.The proof was flowed because I hadn't seen that in order to have an infinite number of gyrations we should revolve the line around the oval infinetly. My last post hopefully settles everything...
  13. Jul 6, 2005 #12
    Ah, many thanks, wisredz. :smile: I think I get Newton's point now.

    I'm afraid I'll have to go with Leibniz and the other sceptics at this point though.
    Maybe I can find some literature on the question to engross my insight and back up or change my opinion.
  14. Jul 6, 2005 #13
    Oh, and now I see that my statement about the square was wrong when Kepler's law is used and it doesn't apply to squares. I wasn't aware of that. I haven't had mathematics or physics since grammar school and this had been some years ago. :uhh:

    I wonder if it could work without using Kepler's law: Could the distance between the point and the pole be proportional to the area of a square anyway? I'll think about it tomorrow. Good night.
  15. Jul 6, 2005 #14
    Hehe anytime... As I have said D.T Whiteside published a counter-example for this in Mathematical Papers of Isaac Newton, if I'm not mistaken. But it's a pain in the neck to find those books. I don't see anything wrong with newton's proof now, but as soon as I see Leibniz's or Whiteside's counter-examples I might change my mind. :)
  16. Jul 6, 2005 #15
    As to the appliance of the law of areas to the square, I don't think it is possible. the law says that in an oval, ellipse precisely, the areas swept in equal times are equal. But what happens the line skips a corner while moving with uniform motion in the square? I don't think it would work
  17. Jul 6, 2005 #16


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    Why would it make an infinite number of gyrations? I've attached a figure of what I see going on. The circle is the oval, the red line is the hand, the thick black curve is the path of the point of light in one hour, and the yellow dot is the point of light.
    When you say "the line," which line are you referring to?
    Again, which line are you talking about? Are you talking about the line that represents the minute hand after 1 hour has passed? If so, are we considering the full "extension" of this line, or just the segment between the pole and the "full area point" as you put it? If we're considering the fully extended line, then it will also pass through the point on the spiral path corresponding to 30 minutes past the hour.

    Attached Files:

  18. Jul 6, 2005 #17
    We are considering the full extension of course. The point is proportional to the area that's what is important for us. Whether it is inside the circle or not doesn't matter...

    And "the" line is the one that moves with uniform speed...
    Last edited: Jul 6, 2005
  19. Jul 6, 2005 #18


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    I think I would make the same objection as the original poster. In one hour, the spiral path traced out will look something like what appears in my figure. Any given line will intersect this path at most 3 times. And is the arc length of the spiral supposed to give us the area, or is the distance between the pole and the "area point" supposed to give it? In the first case, the roots of the equation f(x,y) = 0 don't really tell us anything about the arc length. And if we're looking for the total displacement, I don't see how the number of roots are significant. You can have infinitely many roots to an equation but still have the distance between roots be rational (like y = sin(x/2pi)).
  20. Jul 6, 2005 #19
    As far as I am concerned, we use the point's place on the line that revolves around the pole. I have shown the proportionality thus.

    You said that the line can intersect the spiral at most 3 times. I don't understand this because I think it is at most 1. That is because the spiral is made up with the position of a point that moves on the line. Could you elaborate on that a little?
  21. Jul 6, 2005 #20


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    Use it for what? I don't understand the context of your statement
    Look at my diagram a few posts back. Extend the red line so that it really is a line. Well it already passes through the pole and the "final point" of the spiral. Extending the line will send it through the bottom of the spiral as well. That's three points.
  22. Jul 6, 2005 #21
    I never thought that way but it is not needed. It's like the lines that are in theradar screens. I looked at your diagram, but the process is, I think, like building a circle by using a line segment and revolving it around a point. The two are similar.

    We use the point's place on the line to see how much area the line has swept. Actually the original poster's quotations explain this good enough.
  23. Aug 24, 2006 #22
    Con respecto al mismo problema

    Yo me uno al grupo de los que han leido este libro sobre la prueba de Abel y caido en una profunda confucion cuando se trata de desifrar exactamente lo que dice el Lema 28 de Newton.
    Lo que puedo agregar es que Peter Pesic se equivoca cuando afirma que al cabo de una hora el area de la elipse es igual a la distancia recorrida por el punto de luz, por que lo que sucede es que el area de la elipse y la distancia recorrida estan (con las condiciones dadas por Peter Pesic) en una proporcion de 1/2. Sin embargo este detalle no descalifica la prueba por que muestra que esa distancia (1/2)x es una numero tracendental por lo tanto el area de la elipse x tambien lo es!
    La confucion que tengo es: Que sucede cuando se supone un circulo de radio 1/SQR(2)? Por que dicho circulo tiene area 1!!!! y claramente ese numero no presenta ningun problema en cuando a tracendencia!!!!
    Evidentemente tambien consulte Principia para poder apreciar la prueba original.
    Sorry if i write in Spanish!!
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