Using paraboloidal coordinates

Question

I have the 3-dimensional paraboloidal coordinates

$$s_{\pm}=\sqrt{x^2+y^2+z^2}\pm z$$ $$\phi=ArcTan(y/x)$$ with the inverse transformation $$x=\sqrt{s_+ \cdot s_-}\cdot cos(\phi)$$ $$y=\sqrt{s_+ \cdot s_-}\cdot sin(\phi)$$ $$z=\frac{s_+ - s_-}{2}$$

Questions:

1. Properties in integration: Suppose I have $ I= \int_{0}^{\infty} ds_+ \int_{0}^{\infty} ds_- f(s_+,s_-) \cdot \delta(s_+ - t)$, can I naively apply the Dirac-Delta and get $I=\int_{0}^{\infty} ds_- f(t,s_-)$ ?
2. Properties in differentiation: Suppose I have $D=\frac{\partial}{\partial s_-} e^{2\cdot s_-} \cdot s_+$, can I naively ignore the $s_+$ Term and get $D=2\cdot e^{2\cdot s_-} \cdot s_+$ ?

Every hint will be much appreciated!

Answer 1

What you wrote is

$$ \begin{eqnarray} s_+ &=& \sqrt{x^2+y^2+z^2} + z,\\ s_- &=& \sqrt{x^2+y^2+z^2} - z.\\ \end{eqnarray} $$

As both are positive, we can take the square root, thus

$$ \begin{eqnarray} t_+ &=& \sqrt{\sqrt{x^2+y^2+z^2} + z},\\ t_- &=& \sqrt{\sqrt{x^2+y^2+z^2} - z}.\\ \end{eqnarray} $$

Note that

$$ \begin{eqnarray} t_+ t_- &=& \sqrt{x^2+y^2},\\ \frac{t_+^2 - t_-^2}{2} &=& z, \end{eqnarray} $$

so

$$ \begin{eqnarray} x &=& \cos(\phi) t_+ t_-,\\ y &=& \sin(\phi) t_+ t_-,\\ z &=& \frac{t_+^2 - t_-^2}{2}. \end{eqnarray} $$

These coordinates are known as Paraboloidal coordinates $(\phi,u,v)$.

What you have defined is $(\phi,s_+,s_-) = (\phi,u^2,v^2)$.

We can call then the partly squared paraboloidal or the NicoDean coordinates...