Struggling with the following:

Prove the identity

$$ \nabla = e_{r}(e_{r} \cdot \nabla) + e_{\theta}(e_{\theta} \cdot \nabla) + e_{\phi}(e_{\phi} \cdot \nabla).$$

Given the vector fields $F=F_{r}e_{r}+F_{\theta}e_{\theta}+ F_{\phi}e_{\phi}$ show that

$$ \nabla \cdot F=\frac{1}{r^{2}}\frac \partial {{\partial r}}(r^{2}F_r)+\frac{1}{r\sin\theta}\frac \partial {\partial \theta}(\sin\theta F_\theta)+\frac{1}{r\sin\theta}\frac \partial {\partial\phi} $$

Any help will be most appreciated, many thanks.

  • 3
    $\begingroup$ For the first line there isn't really much to prove. For the second you can use the definition of the divergence $\nabla\cdot\vec a=\lim_{V\rightarrow 0}\frac{1}{V}\int_{\partial V} \vec a\,d\vec A$ and replace everything with the spherical expressions or alternatively take $\nabla\cdot \vec a=d_x a_x + d_y a_y + d_z a_z$, replace the derivatives and the kartesian coordinate functions by their spherical counterpart and express it in terms of the spherical unit vectors afterwards. It's a very long a tedious calculation though.... $\endgroup$
    – example
    Apr 2, 2012 at 22:02
  • 1
    $\begingroup$ A Google search "Divergence in spherical coordinates" yields among others the following link that completely addresses the question. $\endgroup$ Apr 3, 2012 at 0:25
  • $\begingroup$ The question can be closed. I think I agree with the first comment. There is nothing to prove about the identity. This is by definition. Coming to second part it is quite straightforward and can be looked into wiki article for the derivation of divergence expression for curvilinear coordinates. $\endgroup$
    – user16409
    Apr 4, 2012 at 23:42
  • $\begingroup$ @user16409: You can flag it and ask the moderators to close it. $\endgroup$
    – draks ...
    Apr 5, 2012 at 7:10


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