Transforming integral to spherical coordinates

[Nkj]

Hello everyone,
I am studying a book about derivation of Green's function for three dimensional wave equation. In a part of this book it is stated as follows:



I don't understand how this transformation is done. What does it mean geometrically that

in such a manner that the polar axis ($\theta=0$) lies along the half line from the origin to ($x-\zeta$, $y-\eta$, $z-\xi$)

and how $x-\zeta=0$, $y-\eta=0$ and $z-\xi=R$ are obtained?

 

[blamocur]

I don't quite understand that one particular line ("... so that $x-\xi = 0$..."), but the rest makes sense to me. I am guessing that $R$ is the length of vector $(x-\xi, y-\eta, z-\zeta)$, i.e.,
$$R = \sqrt{(x-\xi)^2 + (y-\eta)^2 + (z-\zeta)^2}$$We also know that $\kappa$ is the length of vector $(k,l,m)$, and since $\theta$ is the angle between those two vectors we get the following equality for their dot products:
$$k(x-\xi)+l(y-\eta)+m(z-\zeta) = \kappa R \cos ( \theta)$$To complete the transition from 4.5.6 to 4.5.7 we remember the expression for volume differential in spherical coordinates:
$$dk\,dl\,dm = \kappa^2 \sin (\theta) \kappa\, d\phi\, d\theta\, d\kappa$$Hope this helps, but let us know if it doesn't.

 

[blamocur]

I hope these additional details make my previous post somewhat more rigorous, but I also hope it doesn't make the whole thing more confusing. So here goes: using, as before,
$$R = \sqrt{(x-\xi)^2 + (y-\eta)^2 + (z-\zeta)^2}$$define the ${\displaystyle }$new vertical'' unit vector:
$$\mathbf e_1 = \left(\frac{x-\xi}{R}, \frac{y-\eta}{R}, \frac{z-\zeta}{R} \right)$$and then complete the orthonormal basis with vectors $\mathbf e_2, \mathbf e_3$, which don't need to be defined as long as the basis $(\mathbf e_1,\mathbf e_2, \mathbf e_3)$ is orthonormal. (One can use Gram-Schmidt process to get exact expressions for the basis, but those exact expressions do not matter).

If $\mathbf f = (k,l,m)$ then $|\mathbf f| = \kappa$, and $\mathbf f$ can be represented through $\mathbf e_1,\mathbf e_2,\mathbf e_3$ as follows:
$$\mathbf f = \kappa\left(\mathbf e_1 \cos\theta + \mathbf e_2 \sin\theta \cos\phi + \mathbf e_3 \sin\theta \sin\phi\right)$$The corresponding dot product then becomes:
$$k(x-\xi)+l(y-\eta)+m(z-\zeta) = \langle \mathbf f, R \mathbf e_1\rangle = R \kappa \cos\theta$$

 

[blamocur]

and how x−ζ=0x-\zeta=0x−ζ=0, y−η=0y-\eta=0y−η=0 and z−ξ=Rz-\xi=Rz−ξ=R are obtained?

On third thought, I think what they mean is that vector $x-\xi, y-\eta, z-\zeta$ gets transformed to $0, 0, R$ in the new basis from post #3.

 

[Nkj]

On third thought, I think what they mean is that vector $x-\xi, y-\eta, z-\zeta$ gets transformed to $0, 0, R$ in the new basis from post #3.

Thank you for your useful answers. They really helped me.