Gradient of a Scalar Function in Curvilinear Coordinates
Theorem
In a curvilinear coordinate system, the gradient of a scalar function $f=f(q_{1},q_{2},q_{3})$ is as follows.
$$ \nabla f= \frac{1}{h_{1}}\frac{ \partial f }{ \partial q_{1} } \hat{\mathbf{q}}_{1} + \frac{1}{h_{2}}\frac{ \partial f }{ \partial q _{2}}\hat{\mathbf{q}}_{2}+\frac{1}{h_{3}}\frac{ \partial f }{ \partial q_{3} } \hat{\mathbf{q}}_{3}=\sum \limits _{i=1} ^{3}\frac{1}{h_{i}}\frac{ \partial f}{ \partial q_{i}}\hat{\mathbf{q}}_{i} $$
$h_{i}$ is the scale factor.
Formulas
Cartesian Coordinate System:
$$ h_{1}=h_{2}=h_{3}=1 $$
$$ \nabla f= \frac{\partial f}{\partial x}\mathbf{\hat{\mathbf{x}} }+ \frac{\partial f}{\partial y}\mathbf{\hat{\mathbf{y}}} + \frac{\partial f}{\partial z}\mathbf{\hat{\mathbf{z}}} $$
Cylindrical Coordinate System:
$$ h_{1}=1,\quad h_{2}=\rho,\quad h_{3}=1 $$
$$ \nabla f = \frac{\partial f}{\partial \rho}\boldsymbol{\hat \rho} + \frac{1}{\rho}\frac{\partial f}{\partial \phi}\boldsymbol{\hat \phi} + \frac{\partial f}{\partial z}\mathbf{\hat{\mathbf{z}}} $$
Spherical Coordinate System:
$$ h_{1}=1,\quad h_{2}=r\quad, h_{3}=r\sin\theta $$
$$ \nabla f= \frac{\partial f}{\partial r} \mathbf{\hat r} + \frac{1}{r}\frac{\partial f}{\partial \theta} \boldsymbol{\hat \theta} + \frac{1}{r\sin\theta}\frac{\partial f}{\partial \phi}\boldsymbol{\hat \phi} $$
Derivation
In the three-dimensional Cartesian coordinate system, $\mathbf{a}$, which satisfies the following equation, is defined as the gradient of $f$, and is denoted as $\nabla f$.
$$ d f =\mathbf{a} \cdot d\mathbf{r} $$
The same definition is applied to any curvilinear coordinate system. The total differential of $f$ is as follows.
$$ d f = \frac{ \partial f}{ \partial q_{1} }dq_{1}+\frac{ \partial f}{ \partial q_{2}}dq_{2}+\frac{ \partial f}{ \partial q_{3}}dq_{3} $$
In a curvilinear coordinate system, the infinitesimal change in the position vector $\mathbf{r}$ is as follows.
$$ d\mathbf{r}=h_{1}dq_{1}\hat{\mathbf{q}}_{1}+h_{2}dq_{2}\hat{\mathbf{q}}_{2}+h_{3}dq_{3}\hat{\mathbf{q}}_{3} $$
Now, what we seek is $\mathbf{a}$ that satisfies the equation below.
$$ \begin{equation} df=\mathbf{a} \cdot d\mathbf{r} \end{equation} $$
Given $\mathbf{a}=a_{1}\hat{\mathbf{q}}_{1}+a_{2}\hat{\mathbf{q}}_{2}+a_{3}\hat{\mathbf{q}}_{3}$, then $(1)$ is as follows.
$$ \frac{ \partial f}{ \partial q_{1} }dq_{1}+\frac{ \partial f}{ \partial q_{2}}dq_{2}+\frac{ \partial f}{ \partial q_{3}}dq_{3} = a_{1}h_{1}dq_{1}+a_{2}h_{2}dq_{2}+a_{3}h_{3}dq_{3} $$
Therefore, given $a_{i}=\dfrac{1 }{h_{i}}\dfrac{ \partial f}{ \partial q_{i} }$, the following holds true.
$$ \quad \mathbf{a}=\frac{1 }{h_{1}}\frac{ \partial f}{ \partial q_{1} }\hat{\mathbf{q}}_{1}+\frac{1 }{h_{2}}\frac{ \partial f}{ \partial q_{2} }\hat{\mathbf{q}}_{2}+\frac{1 }{h_{3}}\frac{ \partial f}{ \partial q_{3} }\hat{\mathbf{q}}_{3} $$
Now, the vector $\mathbf{a}$ defined as the gradient of $f$ and denoted as $\nabla f$.
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