Vector Calculus¶

\(\require{physics}\)

Del Operators¶

Gradient¶

The gradient in the direction of \(\vec{e_{q_1}}\) is

\[\label{eq:dirgradcurvi} \nabla_{q_i} = \frac{\partial}{\partial s_i} = \frac{1}{h_{q_i}} \frac{\partial}{\partial q_i}\]

thus, the total gradient operator is

\[\label{eq:gradincurvi} \nabla = \sum_i \vec{e_{q_i}} \frac{\partial}{\partial s_i} = \sum_i \vec{e_{q_i}} \frac{1}{h_{q_i}} \frac{\partial}{\partial q_i}\]

Divergence¶

We can define the divergence from the gradient, since

\[\nabla \cdot \vec{A} = \nabla \cdot \left( \sum_j A_j \vec{e_{q_j}} \right)\]

clearly we need \(\nabla \cdot \vec{e_{q_j}}\) to continue; from equation ([eq:dirgradcurvi]),

\[\nabla \cdot q_j = \sum_i \frac{1}{h_{q_i}} \frac{\partial q_j}{\partial q_i} = \sum_i \vec{e_{q_i}} \frac{1}{h_{q_i}} \delta_{ij} = \vec{e_{q_j}} \frac{1}{h_{q_j}}\]

Thus,

\[\vec{e_{q_j}} = h_{q_j} \nabla q_j\]

The basis vectors form a right-handed set, so,

\[e_{q_1} \times e_{q_2} = e_{q_3}\]

and

\[\nabla q_1 \times \nabla q_2 = \frac{\vec{e_{q_3}}}{h_{q_1}h_{q_2}}\]

Considering that \(\curl(\grad) = 0\),

\[\begin{split}\begin{aligned} \nabla \cdot (\nabla_{q_1} \times \nabla_{q_2}) &= \nabla q_2 \cdot (\nabla \times \nabla q_1) - \nabla q_1 \cdot ( \nabla \times \nabla q_2) \\ &= 0\end{aligned}\end{split}\]

\[\nabla \cdot \left( \frac{\vec{e_{q_3}}}{h_{q_1} h_{q_2}} \right) = 0\]

and the same argument applies for cyclic permutations,

\[\nabla \cdot \left( \frac{\vec{e_{q_3}}}{h_{q_1} h_{q_2}} \right) = \nabla \cdot \left( \frac{\vec{e_{q_1}}}{h_{q_2} h_{q_3}} \right) = \nabla \cdot \left( \frac{\vec{e_{q_2}}}{h_{q_3} h_{q_1}} \right) = 0\]

Then returning to

\[\begin{split}\begin{aligned} \nabla \cdot \vec{A} &= \nabla \cdot \left( \sum_j A_j \vec{e_{q_j}} \right) &\\ &= \nabla \cdot \left\{ [h_{q_1}h_{q_2}A_1] \left[ \frac{\vec{e_{q_1}}}{h_{q_2} h_{q_3}} \right] \right. &+& [h_{q_3}h_{q_1}A_2] \left[ \frac{\vec{e_{q_2}}}{h_{q_3} h_{q_1}} \right] \\ &&+&\left.[h_{q_1} h_{q_2}A_3] \left[ \frac{\vec{e_{q_3}}}{h_{q_1} h_{q_2}} \right] \right\} \\ &= \left[ \frac{\vec{e_{q_1}}}{h_{q_2} h_{q_3}} \right] \cdot \nabla [h_{q_1}h_{q_2}A_1] &+&\left[ \frac{\vec{e_{q_2}}}{h_{q_3} h_{q_1}} \right] \cdot \nabla [h_{q_3}h_{q_1}A_2]\\ &&+&\left[ \frac{\vec{e_{q_3}}}{h_{q_1} h_{q_2}} \right]\cdot \nabla [h_{q_1}h_{q_2}A_3]\end{aligned}\end{split}\]

Since

\[\vec{e_{q_i}} \cdot \nabla = \frac{1}{h_{q_i}} \frac{\partial}{\partial q_i}\]

\[\begin{split}\label{eq:divincurvi} \begin{split} \nabla \cdot \vec{A} = \frac{1}{h_{q_1}h_{q_2}h_{q_3}} \left( \frac{\partial}{\partial q_1} (h_{q_2} h_{q_3} A_1) \right. \\ + \left. \frac{\partial}{\partial q_2} (h_{q_3} h_{q_1} A_2) + \frac{\partial}{\partial q_3} (h_{q_2} h_{q_1} A_3) \right) \end{split}\end{split}\]

Curl¶

The curl can be derived from earlier results,

\[\begin{split}\begin{aligned} \nabla \times \vec{A} &=&& \nabla \times \left( \sum_j A_j \vec{e_{q_j}} \right) \\ &=&& \nabla \times \left( \begin{matrix} [h_{q_1}A_1] \qty[\frac{\vec{e_{q_1}}}{h_{q_1}} ] &+ [h_{q_2}A_2] \qty[ \frac{\vec{e_{q_2}}}{h_{q_2}} ] \\ &+ [h_{q_3}A_3] \qty[ \frac{\vec{e_{q_3}}}{h_{q_3}} ] \end{matrix} \right) \\ &= && - \left[ \frac{\vec{e_{q_1}}}{h_{q_1}} \right] \times \nabla (h_{q_1}A_1) - \left[ \frac{\vec{e_{q_2}}}{h_{q_2}} \right] \times \nabla (h_{q_2}A_2) \\ &&& - \left[ \frac{\vec{e_{q_3}}}{h_{q_3}} \right] \times \nabla (h_{q_3}A_3)\end{aligned}\end{split}\]

so,

\[\begin{split}\label{eq:curlincurvi} \nabla \times \vec{A} = \frac{1}{h_{q_1}h_{q_2}h_{q_3}} \begin{vmatrix} h_{q_1} \vec{e_1} & h_{q_2} \vec{e_2} & h_{q_3} \vec{e_{3}} \\ \frac{\partial}{\partial q_1} & \frac{\partial}{\partial q_2} & \frac{\partial}{\partial q_3} \\ h_{q_1} A_1 & h_{q_2} A_2 & h_{q_3} A_3 \end{vmatrix}\end{split}\]

Laplacian¶

\[\begin{split}\label{eq:laplacianincurvi} \begin{split} \nabla^2 = \frac{1}{h_{q_1}h_{q_2}h_{q_3}} \bigg( \pdv{q_1} \qty[ \frac{h_{q_2}h_{q_3}}{h_{q_1}} \pdv{q_1} ] + \pdv{q_2} \qty[ \frac{h_{q_3}h_{q_1}}{h_{q_2}} \pdv{q_2} ] \\ + \pdv{q_3} \qty[ \frac{h_{q_2}h_{q_1}}{h_{q_1}} \pdv{q_3} ] \bigg) \end{split}\end{split}\]

Operations¶

Dot Product¶

\[\label{eq:dotprod} \vec{a} \cdot \vec{b} = a_\theta b_\theta\]

Total Differential¶

\[\label{eq:totaldifferential} \dif{f} = \frac{\partial f}{\partial x_\theta} \dif{x_\theta}\]

Matrix Multiplication¶

\[\label{eq:matmult} a_{ij} = b_{i \theta} c_{\theta j}\]

Cross Product¶

\[\label{eq:crossprod1} [ \vec a \times \vec b]_i = \epsilon_{i \theta \phi} a_{\theta} b_{\phi}\]

\[\label{eq:crossprod2} \vec{a} \times \vec{b} = \epsilon_{i \theta \phi} a_{\theta} b_{\phi} \vec{e_{i}}\]

Here \(\epsilon_{i \theta \phi}\) is the Levi-Civita tensor, and takes values

\[\begin{split}\label{eq:levitcivita} \epsilon_{i \theta \phi} = \left\{ \begin{array}{rl} 0 & \text{if any indices are repeated.} \\ 1 & \text{if indices are a cyclic permutation of } (1,2,3). \\ -1 & \text{if indices are a cyclic permutation of } (1,3,2). \end{array}\right.\end{split}\]

There is a simple relationship between this tensor and the Kronecker delta,

\[\label{eq:deltalevi} \epsilon_{\theta j k } \epsilon_{\theta l m} = \delta_{jl} \delta_{k m} - \delta_{j m} \delta_{l k}\]

Del Operators¶

The gradient is now expressed:

\[ \begin{align}\begin{aligned} \label{eq:grad} \nabla = \vec e_i \frac{\partial }{\partial x_i}\\The divergence:\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \label{eq:divergence} \nabla \cdot \vec{A}(\vec{r}) = \frac{\partial A_i(\vec{r})}{\partial x_i}\\and the curl:\end{aligned}\end{align} \]

\[\label{eq:curl} \nabla \times \vec{A}(\vec{r}) = \epsilon_{ijk} \frac{\partial A_j(\vec{r})}{\partial x_i} \vec{e_k}\]

Consider

\[ \begin{align}\begin{aligned}\left[ \vec{a} \times ( \vec{b} \times \vec{c} ) \right]_m\\using the summation convention,\end{aligned}\end{align} \]

\[\begin{split}\begin{aligned} \left[ \vec{a} \times ( \vec{b} \times \vec{c} ) \right]_m &= \epsilon_{m \alpha \beta} a_{\alpha} \epsilon_{\beta \gamma \delta} b_{\gamma} c_{\delta} \\ &= \epsilon_{\beta m \alpha} \epsilon_{\beta \gamma \delta} a_{\alpha} b_{\gamma} c_{\delta} \\ &= (\delta_{m \delta}\delta_{\alpha \delta} - \delta_{m \delta} \delta_{a \gamma}) a_{\alpha} b_{\gamma} c_{\delta}\\ &= (a_{\alpha}c_{\alpha}) b_m - (a_{\alpha}b_{\alpha}) c_m \\ &= \left[ (\vec a \cdot \vec c) \vec b - (\vec a \cdot \vec b) \vec c \right]_{m} \end{aligned}\end{split}\]

Note,

\[\delta_{i\theta} a_{\theta} = a_i\]

\[\begin{split}\begin{aligned} \nabla \times ( \nabla \phi ) &= \epsilon_{m\alpha \beta} \frac{\partial }{\partial x_a} \frac{\partial \phi}{\partial x_{\beta}}\\ &= 0 \end{aligned}\end{split}\]

This arises because the Levi-Civita tensor is anti-symmetric, but the two partial derivatives are symmetric.

\[\begin{split}\begin{aligned} \vec A \cdot \vec B \times \vec C &= A_{\alpha}\epsilon_{\alpha \beta \gamma} B_{\beta} C_{\gamma} \\ &= C_{\gamma} \epsilon_{\gamma \alpha \beta} A_{\alpha} B_{\beta} \\ &= C_{\gamma} [\vec A \times \vec B]_{\gamma} \\ &= C \cdot [ \vec A \times \vec B] \end{aligned}\end{split}\]