I've found the following example in a vector calculus book: the divergence of the vector field $\vec F(x,y,z) = x\vec i + y\vec j - z \vec k$ in spherical coordinates is

$$ \nabla \cdot \vec F(\rho,\phi,\theta) = -3\cos(2\phi). $$ I understood all the passages of the book and I've found the same result. But why the following argument does not holds true: in cartesian coordinates $\nabla \cdot \vec F(x,y,z)=1$ and then $\nabla \cdot \vec F(\rho,\phi,\theta)=1$? In my mind if $\nabla \vec G(x,y,z)=xy$ (for instance) then $\nabla \vec G(\rho,\phi,\theta) = \rho^2\sin\phi^2 \sin \theta \cos\theta$. Am I missing something?

  • $\begingroup$ Can you show the passages in the book? Because your argument is valid and it's not clear how they got $-3\cos 2\phi$ $\endgroup$ – Vasily Mitch Feb 1 at 17:15
  • $\begingroup$ Ok, the idea is to use the formula for divergence in the spherical coordinates. The orthogonal basis in spherical coordinates is $T_\rho = (\sin \phi \cos\theta,\sin\phi\sin\theta,\cos\phi), T_\phi = (\cos\phi\cos\theta, \cos\phi\sin\theta,-\cos\phi), T_\theta = (-\sin\phi\sin\theta,\sin\phi\cos\theta,0)$. We have to write $\vec F = F_1 T_\rho + F_2 T_\phi + F_3+T_\theta$. Notice that $\vec F(x,y,z) = \rho\sin\phi\cos\theta \vec i + \rho\sin\phi\sin\theta \vec j + \rho\cos\phi \vec k$. So are able to compute $F_1 = \vec F \cdot \vec T_\rho$ (analogously $F_2$ and $F_3$). (continue) $\endgroup$ – user85353 Feb 1 at 17:24
  • $\begingroup$ These calculations leads to: $F_1=-\rho\cos(2\phi), F_2=F_3=0$. Now we put directly in the formula of divergence and we get the answer. $\endgroup$ – user85353 Feb 1 at 17:26
  • $\begingroup$ Another example of the book calculates the Laplacian in spherical coordinates of the function $f(x,y,z) = x^2+y^2-z^2$. The book says that the answer isn't $1$.. for me the same argument can be used. I'm so confused... $\endgroup$ – user85353 Feb 1 at 17:30
  • $\begingroup$ Could you show a scan of the calculations in the book? $\endgroup$ – md2perpe Feb 1 at 17:36

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