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Delta amplitude and nabla amplitude

  1. Mar 1, 2014 #1
    Delta amplitude and "nabla amplitude"

    Why all jacobi theory and all ellipitc integrals is based in ##\Delta(\theta) = \sqrt{1-m \sin(\theta)^2}## ?

    You already think that this definition is just midle of history, cause' you can define other elementar function: [tex]\nabla(\theta) = \sqrt{1-m \cos(\theta)^2}[/tex] So, a "nabla amplitude", will imply in more interesting definitions:

    if: [tex]\int_{0}^{\phi}\frac{d\theta}{\sqrt{1-m \sin(\theta)^2}}=u[/tex] thus: [tex]\int_{0}^{\phi}\frac{d\theta}{\sqrt{1-m \cos(\theta)^2}}=v[/tex] and so: [tex]\\ u=\int \frac{d\phi}{\Delta(\phi)} \;\;\;\Rightarrow \;\;\;\frac{du}{d\phi}=\frac{1}{\Delta(\phi)} \;\;\;\Rightarrow \;\;\;\frac{d\phi}{du} = \Delta(\phi) = dn(u) \;\;\;\Rightarrow \;\;\; \phi = \int dn(u) du \\ \\ \\ \\ v=\int \frac{d\phi}{\nabla(\phi)} \;\;\;\Rightarrow \;\;\;\frac{dv}{d\phi}=\frac{1}{\nabla(\phi)} \;\;\;\Rightarrow \;\;\;\frac{d\phi}{dv} = \nabla(\phi) = qn(v) \;\;\;\Rightarrow \;\;\; \phi = \int qn(v)dv[/tex]
    And the implications continues... So, why the elliptic integrals are based only in the root square of a sine, why not exist definition based in the root square of a cosine too?
     
  2. jcsd
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