# Delta amplitude and nabla amplitude

1. Mar 1, 2014

### Jhenrique

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: $$\nabla(\theta) = \sqrt{1-m \cos(\theta)^2}$$ So, a "nabla amplitude", will imply in more interesting definitions:

if: $$\int_{0}^{\phi}\frac{d\theta}{\sqrt{1-m \sin(\theta)^2}}=u$$ thus: $$\int_{0}^{\phi}\frac{d\theta}{\sqrt{1-m \cos(\theta)^2}}=v$$ and so: $$\\ u=\int \frac{d\phi}{\Delta(\phi)} \;\;\;\Rightarrow \;\;\;\frac{du}{d\phi}=\frac{1}{\Delta(\phi)} \;\;\;\Rightarrow \;\;\;\frac{d\phi}{du} = \Delta(\phi) = dn(u) \;\;\;\Rightarrow \;\;\; \phi = \int dn(u) du \\ \\ \\ \\ v=\int \frac{d\phi}{\nabla(\phi)} \;\;\;\Rightarrow \;\;\;\frac{dv}{d\phi}=\frac{1}{\nabla(\phi)} \;\;\;\Rightarrow \;\;\;\frac{d\phi}{dv} = \nabla(\phi) = qn(v) \;\;\;\Rightarrow \;\;\; \phi = \int qn(v)dv$$
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?