30 May 2025

Vector Operators in Cartesian, Cylindrical, and Spherical Coordinates

Derivation of gradient, divergence, curl, and the scalar Laplacian in orthogonal coordinates.

bsc semester-ii mathematical-physics coordinate-systems laplacian

Let $(q_1,q_2,q_3)$ be orthogonal coordinates with

\[h_i=\left\lvert\frac{\partial\mathbf r}{\partial q_i}\right\rvert, \qquad \hat{\mathbf e}_i=\frac1{h_i}\frac{\partial\mathbf r}{\partial q_i}.\]

Then

\[d\mathbf r=\sum_i h_i\,dq_i\,\hat{\mathbf e}_i, \qquad dV=h_1h_2h_3\,dq_1dq_2dq_3,\]

and the area normal to $\hat{\mathbf e}_1$ is $dS_1=h_2h_3dq_2dq_3$, with cyclic analogues.

General orthogonal formulas

Comparing $d\phi=\sum_i\phi_{q_i}dq_i$ with $d\phi=\nabla\phi\cdot d\mathbf r$ gives

\[\boxed{\nabla\phi=\sum_i\hat{\mathbf e}_i\frac1{h_i}\frac{\partial\phi}{\partial q_i}}.\]

The net flux through the $q_1$ face pair is $\partial_{q_1}(h_2h_3A_1)dq_1dq_2dq_3$. Summing face pairs and dividing by $dV$ gives

\[\boxed{ \nabla\cdot\mathbf A =\frac1{h_1h_2h_3} \sum_{\rm cyclic}\frac{\partial}{\partial q_1}(h_2h_3A_1)}.\]

The circulation around the $q_1q_2$ face gives

\[(\nabla\times\mathbf A)_3 =\frac1{h_1h_2} \left[\frac{\partial(h_2A_2)}{\partial q_1} -\frac{\partial(h_1A_1)}{\partial q_2}\right],\]

so all components are

\[\boxed{ \nabla\times\mathbf A =\frac1{h_1h_2h_3} \begin{vmatrix} h_1\hat{\mathbf e}_1&h_2\hat{\mathbf e}_2&h_3\hat{\mathbf e}_3\\ \partial_{q_1}&\partial_{q_2}&\partial_{q_3}\\ h_1A_1&h_2A_2&h_3A_3 \end{vmatrix}}.\]

Finally, $\nabla^2\phi=\nabla\cdot\nabla\phi$ gives

\[\boxed{ \nabla^2\phi =\frac1{h_1h_2h_3} \sum_{\rm cyclic}\frac{\partial}{\partial q_1} \left(\frac{h_2h_3}{h_1}\frac{\partial\phi}{\partial q_1}\right)}.\]

Cartesian coordinates

For $(x,y,z)$, $h_x=h_y=h_z=1$:

\[\nabla\phi=\hat{\mathbf x}\phi_x+\hat{\mathbf y}\phi_y+\hat{\mathbf z}\phi_z, \qquad \nabla\cdot\mathbf A=A_{x,x}+A_{y,y}+A_{z,z},\] \[\nabla\times\mathbf A =\hat{\mathbf x}(A_{z,y}-A_{y,z}) +\hat{\mathbf y}(A_{x,z}-A_{z,x}) +\hat{\mathbf z}(A_{y,x}-A_{x,y}),\] \[\boxed{\nabla^2\phi=\phi_{xx}+\phi_{yy}+\phi_{zz}}.\]

Cylindrical coordinates

From

\[\mathbf r(\rho,\varphi,z) =\rho\cos\varphi\,\hat{\mathbf x} +\rho\sin\varphi\,\hat{\mathbf y}+z\hat{\mathbf z},\]

the coordinate tangent vectors are

\[\frac{\partial\mathbf r}{\partial\rho}=\hat{\boldsymbol\rho}, \qquad \frac{\partial\mathbf r}{\partial\varphi}=\rho\hat{\boldsymbol\varphi}, \qquad \frac{\partial\mathbf r}{\partial z}=\hat{\mathbf z}.\]

Therefore

\[d\mathbf r=d\rho\,\hat{\boldsymbol\rho} +\rho d\varphi\,\hat{\boldsymbol\varphi}+dz\,\hat{\mathbf z},\]

so $(h_\rho,h_\varphi,h_z)=(1,\rho,1)$. Therefore

\[\boxed{\nabla\phi =\hat{\boldsymbol\rho}\phi_\rho +\hat{\boldsymbol\varphi}\frac1\rho\phi_\varphi +\hat{\mathbf z}\phi_z},\] \[\boxed{\nabla\cdot\mathbf A =\frac1\rho\frac{\partial(\rho A_\rho)}{\partial\rho} +\frac1\rho\frac{\partial A_\varphi}{\partial\varphi} +\frac{\partial A_z}{\partial z}},\] \[\boxed{ \begin{aligned} \nabla\times\mathbf A ={}&\hat{\boldsymbol\rho}\left(\frac1\rho A_{z,\varphi}-A_{\varphi,z}\right) +\hat{\boldsymbol\varphi}(A_{\rho,z}-A_{z,\rho})\\ &+\hat{\mathbf z}\frac1\rho \left[\frac{\partial(\rho A_\varphi)}{\partial\rho}-A_{\rho,\varphi}\right], \end{aligned}}\] \[\boxed{\nabla^2\phi =\frac1\rho\frac{\partial}{\partial\rho}(\rho\phi_\rho) +\frac1{\rho^2}\phi_{\varphi\varphi}+\phi_{zz}}.\]

Spherical coordinates

Use

\[\mathbf r(r,\theta,\varphi) =r\sin\theta\cos\varphi\,\hat{\mathbf x} +r\sin\theta\sin\varphi\,\hat{\mathbf y} +r\cos\theta\,\hat{\mathbf z}.\]

Direct differentiation gives

\[\left\lvert\frac{\partial\mathbf r}{\partial r}\right\rvert=1, \qquad \left\lvert\frac{\partial\mathbf r}{\partial\theta}\right\rvert=r, \qquad \left\lvert\frac{\partial\mathbf r}{\partial\varphi}\right\rvert=r\sin\theta.\]

Thus

\[d\mathbf r=dr\,\hat{\mathbf r}+r d\theta\,\hat{\boldsymbol\theta} +r\sin\theta d\varphi\,\hat{\boldsymbol\varphi},\]

so $(h_r,h_\theta,h_\varphi)=(1,r,r\sin\theta)$. Hence

\[\boxed{\nabla\phi =\hat{\mathbf r}\phi_r+\hat{\boldsymbol\theta}\frac1r\phi_\theta +\hat{\boldsymbol\varphi}\frac1{r\sin\theta}\phi_\varphi},\] \[\boxed{\nabla\cdot\mathbf A =\frac1{r^2}\frac{\partial(r^2A_r)}{\partial r} +\frac1{r\sin\theta}\frac{\partial(\sin\theta A_\theta)}{\partial\theta} +\frac1{r\sin\theta}\frac{\partial A_\varphi}{\partial\varphi}},\] \[\boxed{ \begin{aligned} \nabla\times\mathbf A ={}&\hat{\mathbf r}\frac1{r\sin\theta} \left[\partial_\theta(\sin\theta A_\varphi)-\partial_\varphi A_\theta\right]\\ &+\hat{\boldsymbol\theta}\frac1r \left[\frac1{\sin\theta}\partial_\varphi A_r-\partial_r(rA_\varphi)\right]\\ &+\hat{\boldsymbol\varphi}\frac1r \left[\partial_r(rA_\theta)-\partial_\theta A_r\right], \end{aligned}}\] \[\boxed{\nabla^2\phi =\frac1{r^2}\partial_r(r^2\phi_r) +\frac1{r^2\sin\theta}\partial_\theta(\sin\theta\phi_\theta) +\frac1{r^2\sin^2\theta}\phi_{\varphi\varphi}}.\]

The scale factors contain the coordinate geometry. Every gradient term has units $[\phi]/{\rm m}$ and every Laplacian term $[\phi]/{\rm m^2}$.

Sample Cartesian-to-cylindrical and Cartesian-to-spherical Laplacian transformations are checked in the Unit I Maxima worksheet; every reported residual is zero.