Solution of the Wave Equation in General Spherical Coordinates

  • Eugen Skudrzyk


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:
  1. (a)
    that through the two surfaces perpendicular to r:
    $$ - d\tau {{\left[ {div\vec{v}} \right]}_{r}} = {{r}^{2}}\sin \theta d\theta d\phi {{v}_{r}}(r) - {{\left[ {r + dr} \right]}^{2}}\sin \theta d\theta d\phi {{v}_{r}}(r + dr) = {\text{ }} = - \frac{\partial }{\partial }r({{r}^{2}}{{v}_{r}})dr\sin \theta d\theta d\phi ,{\text{ }}gathered $$
  2. (b)
    that through the two surfaces normal to the polar axis θ = const:
    $$ \begin{gathered} - d\tau {[div \vec v]_\theta } = rdrd\phi \sin \theta v\theta (\theta ) - rdrd\phi \sin (\theta + d\theta ){v_\theta }(\theta + d\theta ) = \hfill \\ = - \frac{\partial }{{\partial \theta }}(\sin \theta v\theta (\theta ))rdrd\theta d\phi , \hfill \\ \end{gathered} $$
  3. (c)
    that through the two surfaces at φ and φ + :
    $$ \begin{gathered} - d\tau {[div \vec v]_\phi } = rdrd\theta v\phi (\phi ) - rdrd\theta {v_\phi }(\phi + d\phi ) = \hfill \\ = - \frac{{\partial {v_\phi }}}{{\partial \phi }}rdrd\theta d\phi . \hfill \\ \end{gathered} $$


Wave Equation Spherical Harmonic Sound Field Spherical Wave Legendre Function 
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© Springer-Verlag/Wien 1971

Authors and Affiliations

  • Eugen Skudrzyk
    • 1
  1. 1.Ordnance Research Laboratory and Physics DepartmentThe Pennsylvania State UniversityUniversity ParkUSA

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