Let $\mathbf r = \left[ \begin{matrix} r & \phi \end{matrix} \right]^\top \; $be some curvilinear coordinates, with corresponding unit base column vectors $\hat {\mathbf h}_r \; $and $\hat {\mathbf h}_\phi . \; $Let $\frac{ \mathrm d s}{\mathrm d r} = \sqrt{ \left[ \begin{matrix} \frac{ \partial x}{\partial r} & \frac{ \partial y}{\partial r} \end{matrix} \right] \left[ \begin{matrix} \frac{ \partial x}{\partial r} & \frac{ \partial y}{\partial r} \end{matrix} \right]^\top } \; $and let $\frac{ \mathrm d s}{\mathrm d \phi} \; $be defined accordingly. Then $\frac{\mathrm d \phi}{ \mathrm d s} = \left( \frac{ \mathrm d s}{\mathrm d \phi} \right)^{-1} . \; $Let $\frac{ \mathrm d s}{\mathrm d \phi} \equiv h_\phi \; $and let $h_r\; $be defined accordingly. Let $f \; $be some scalar valued function of the variables $r \; $and $\phi . \; $Then the gradient of $f \; $is $$ \nabla f = \left[ \begin{matrix} \hat {\mathbf h}_r & \hat {\mathbf h}_\phi \end{matrix} \right] \left[ \begin{matrix} \frac{ \partial f}{\partial r} \, \frac{ \mathrm d r}{\mathrm d s} \\ \frac{ \partial f}{\partial \phi} \, \frac{ \mathrm d \phi}{\mathrm d s} \end{matrix} \right] $$

By definition: $\hat {\mathbf h}_\phi = \left[ \begin{matrix} \frac{ \partial x}{\partial \phi} & \frac{ \partial y}{\partial \phi} \end{matrix} \right]^\top \div h_\phi \; $and there is a similar definition for $\hat {\mathbf h}_r \, . \; $Let $F_r \; $and $F_\phi \; $be scalar valued functions of $r \; $and $\phi \; $such that $\mathbf F \; $is a vector field with the following definition: $$ \mathbf F = \left[ \begin{matrix} \hat {\mathbf h}_r & \hat {\mathbf h}_\phi \end{matrix} \right] \, \left[ \begin{matrix} F_r & F_\phi \end{matrix} \right]^\top $$

It follows that the divergence of $\mathbf F \; $is $$\begin{align} \nabla \cdot \mathbf F &= \nabla \cdot \left[ \begin{matrix} \hat {\mathbf h}_r & \hat {\mathbf h}_\phi \end{matrix} \right] \, \left[ \begin{matrix} F_r & F_\phi \end{matrix} \right]^\top \\ &= \frac{\partial}{\partial x} \left \{ \frac{F_r}{h_r} \, \frac{ \partial x}{\partial r} + \frac{F_\phi}{h_\phi} \, \frac{ \partial x}{\partial \phi} \right \} + \frac{\partial}{\partial y} \left \{ \frac{F_r}{h_r} \, \frac{ \partial y}{\partial r} + \frac{F_\phi}{h_\phi} \, \frac{ \partial y}{\partial \phi} \right \} \end{align}$$

The following vector calculus operations are quite generally valid, and will soon be applied. Consider the arbitrary, scalar valued dummy variables $g \; $ and $u \; $.

$$\begin{align} \frac{\partial g}{\partial x} &= \frac{\mathrm d \, g}{\mathrm d \, \mathbf r} \frac{\partial \mathbf r }{\partial x} \\ \frac{\partial \mathbf r }{\partial u} &= \left[ \begin{matrix} \frac{\partial r}{\partial u} & \frac{\partial \phi}{\partial u} \end{matrix} \right]^\top = \left[ \begin{matrix} \left( \frac{\partial u}{\partial r} \right)^{-1} & \left( \frac{\partial u}{\partial \phi} \right) ^{-1} \end{matrix} \right]^\top \end{align}$$

More particularly, let $F_r \; $have the following definition: $$ F_r = \frac{\partial f }{\partial r} \div h_r $$ and let $F_\phi \; $ be defined similarly. Then the Laplace operator of $f \; $in this system of curvilinear coordinates is: $$\begin{align} \Delta f &= \nabla \cdot \nabla f \\ &= \frac{\mathrm d}{\mathrm d \mathbf r} \left \{ \frac{\partial x}{\partial r} \, \frac{ \partial f}{\partial r} \div h_r^2 + \frac{\partial x}{\partial \phi} \, \frac{ \partial f}{\partial \phi} \div h_\phi^2 \right \} \frac{\partial \mathbf r}{\partial x} + \frac{\mathrm d}{\mathrm d \mathbf r} \left \{ \frac{\partial y}{\partial r} \, \frac{ \partial f}{\partial r} \div h_r^2 + \frac{\partial y}{\partial \phi} \, \frac{ \partial f}{\partial \phi} \div h_\phi^2 \right \} \frac{\partial \mathbf r}{\partial y} \end{align}$$

The foregoing analysis is quite general as I have not specified the particular functions $x (r, \phi) \; $and $y (r, \phi). \; $From now on, let $r \; $and $\phi \; $be standard polar coordinates such that $x = r \cos \phi \; $and $y = r \sin \phi. \; $When the formula $\Delta f \; $that I derived gets applied, then it consistently returns the following:

$$ \Delta f = 2 \cdot \left( \frac{1}{r} \, \frac{\partial f}{\partial r} + \frac{\partial^2 f}{\partial r^2} + \frac{1}{r^2} \, \frac{\partial^2 f}{\partial \phi^2} \right) $$

What I have derived deviates from the true laplacian for this coordinate system by a factor of two. I have shown in considerable detail how I have produced this formula. If somebody could please indicate where the error lies, then I would will be very thankful.


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