Solution of the Wave Equation in General Spherical Coordinates

Abstract

If the solution depends not only on r, but also on the polar angle θ and the azimuth φ, the elementary volume becomes a parallelepiped of length rdθ, of width r sinθ dφ and of height dr as shown in Fig. 1. The wave equation is derived by considering the excess of volume that leaves the elementary volume relative to that entering it. This volume (or mass) of flow consists of three different components:

• (a) that through the two surfaces perpendicular to r:

  $$ - d\tau {{\left[ {div\vec{v}} \right]}_{r}} = {{r}^{2}}\sin \theta d\theta d\phi {{v}_{r}}(r) - {{\left[ {r + dr} \right]}^{2}}\sin \theta d\theta d\phi {{v}_{r}}(r + dr) = {\text{ }} = - \frac{\partial }{\partial }r({{r}^{2}}{{v}_{r}})dr\sin \theta d\theta d\phi ,{\text{ }}gathered $$

  (1)

• (b) that through the two surfaces normal to the polar axis θ = const:

  $$ \begin{gathered} - d\tau {[div \vec v]_\theta } = rdrd\phi \sin \theta v\theta (\theta ) - rdrd\phi \sin (\theta + d\theta ){v_\theta }(\theta + d\theta ) = \hfill \\ = - \frac{\partial }{{\partial \theta }}(\sin \theta v\theta (\theta ))rdrd\theta d\phi , \hfill \\ \end{gathered} $$

  (2)

• (c) that through the two surfaces at φ and φ + dφ:

  $$ \begin{gathered} - d\tau {[div \vec v]_\phi } = rdrd\theta v\phi (\phi ) - rdrd\theta {v_\phi }(\phi + d\phi ) = \hfill \\ = - \frac{{\partial {v_\phi }}}{{\partial \phi }}rdrd\theta d\phi . \hfill \\ \end{gathered} $$

  (3)