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Derivation of Laplace in spherical co-ordinates
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[QUOTE="vanhees71, post: 4660602, member: 260864"] This is a pretty cumbersome way. The most easy is to use the action principle. The action [tex]A[\phi]=\int \mathrm{d}^3 \vec{x} [(\vec{\nabla} \phi)^2+f \phi][/tex] leads to the equation [tex]\Delta \phi=f.[/tex] Now you write the gradient in terms of spherical coordinates (which is easy to derive by your direct method) [tex]\vec{\nabla} \phi=\vec{e}_r \partial_r \phi+\frac{1}{r} \vec{e}_{\theta} \partial_{\theta} \phi + \frac{1}{r \sin \theta} \vec{e}_{\varphi} \partial_{\varphi} \phi.[/tex] The volume element is [tex]\mathrm{d}^3 {\vec{r}}=\mathrm{d} r \mathrm{d} \theta \mathrm{d} \varphi r^2 \sin \theta.[/tex] This you plug into the action integral and use the Euler-Lagrange equations to derive the field equation in terms of spherical coordinates. This leads to the Laplacian by identifying it with [itex]f[/tex]. [/QUOTE]
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Derivation of Laplace in spherical co-ordinates
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