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Symmetry in Landen's transformation in elliptic integrals

  1. Jul 29, 2010 #1
    http://en.wikipedia.org/wiki/Landen%27s_transformation" [Broken]

    Since both the expressions [tex]a_1[/tex] and [tex]b_1[/tex] in Landen's transformation are selected in an arbitrary manner, is it all right to define [tex]a_1[/tex] with the geometric mean and [tex]b_1[/tex] with the arithmetic mean, instead of as given in the above link? I think so, but what is your opinion?

    Irrelevant background: In the 1975 Journal of Colloid and Interface Science paper by the Russians Derjaguin, Muller and Toporov, the authors attempt to take long range van der Waals type adhesion into account during contact deformation. They use the Landau Lifgarbagez formulation (Vol 7, Theory of elasticity) for the deformation profile of two symmetric solids, under loading and uniform contact, for describing the deformed surface profile. They end up with an elliptic integral (Eq. 6, in the paper) which they magically transform to the regular complete elliptic int. of the first kind form. I am trying to re-derive their solution and it appears that the symmetry argument is important to keep the value of the elliptic modulus under 1. Otherwise, they should end up with a diverging series solution to the integral.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 29, 2010 #2
    That looks true to me. Using the first transformation in the citation:

    [tex]\begin{aligned} \int_0^{\pi/2}\frac{dt}{\sqrt{a^2\cos^2(t)+b^2\sin^2(t)}}&=\int_0^{\infty}\frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}\\
    &=\int_0^{\infty}\frac{dx}{\sqrt{(x^2+A^2)(x^2+B^2)}};\quad A=\sqrt{ab}, B=\frac{a+b}{2}\\
  4. Jul 30, 2010 #3
    Thanks Jack. That's how I chose to consider it as well. And it certainly is true when a, b and A, B are defined over the real domain. But if we examine the definition over a complex field, the transformation seems to stretch things a bit. But nevertheless, elliptics are usually defined exclusively over the in the real space and that fits my case. Thanks!
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