Suppose $\Phi$ is a function of $r, \theta$ and $\phi$. If I want to derive the Laplacian for this function, I would assume that..

$$\nabla ^2 \Phi = \nabla \cdot \nabla \Phi$$

And as, in spherical:

$$\nabla = \frac{\partial}{\partial r} \hat r + \frac{1}{r} \frac{\partial}{\partial \theta} \hat \theta + \frac{1}{r \sin \ \theta} \frac{\partial}{\partial \phi} \hat \phi$$

Which implies

$$\nabla \Phi= \frac{\partial \Phi}{\partial r} \hat r + \frac{1}{r} \frac{\partial \Phi}{\partial \theta} \hat \theta + \frac{1}{r \sin \ \theta} \frac{\partial \Phi}{\partial \phi} \hat \phi$$

Thus, the Laplacian would seem to me to be:

$$\large\begin{bmatrix} \frac{\partial}{\partial r} \\ \frac{1}{r} \frac{\partial}{\partial \theta} \\ \frac{1}{r \sin \ \theta} \frac{\partial}{\partial \phi} \\ \end{bmatrix} \cdot \begin{bmatrix} \frac{\partial \Phi}{\partial r} \\ \frac{1}{r} \frac{\partial \Phi}{\partial \theta} \\ \frac{1}{r \sin \ \theta} \frac{\partial \Phi}{\partial \phi} \\ \end{bmatrix}$$

Which will not give me the form for the Laplacian in spherical coordinates in my lecture notes or the internet. Where do I have it wrong?

  • 3
    $\begingroup$ This is a point in which the notation $\nabla\cdot$ for the divergence is slightly misleading. Don't interpret it as a true scalar product, it is not, just a mnemonic. $\endgroup$ Nov 25 '18 at 16:48
  • $\begingroup$ Well... that is indeed misleading. What am I supposed to do instead? $\endgroup$
    – sangstar
    Nov 25 '18 at 16:52
  • $\begingroup$ See youtube.com/watch?v=AOT719oYmt0 The issue is that when you take the derivatives of $\hat r, \hat\theta, \hat \phi$, as opposed to the Cartesian case, those are not zero $\endgroup$
    – Andrei
    Nov 25 '18 at 16:53

The Laplacian is the divergence of the gradient. And the divergence in spherical coordinates is: $$\nabla\cdot \mathbf A = {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over r\sin\theta}{\partial \over \partial \theta} \left( A_\theta\sin\theta \right) + {1 \over r\sin\theta}{\partial A_\varphi \over \partial \varphi}$$

Now substitute the $\nabla\Phi$ that you already have for $\mathbf A$.

That leaves the question how we got that formula for the divergence. You can find the derivation here: Nabla in spherical coordinates. It explains it in terms of how divergence is defined.


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