Question

P Derive the expression (9.11)for curl V in the following way. Show that $\mathbf{\nabla }{\mathbf{x}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{e}}_{\mathbf{1}}}{{\mathbf{h}}_{\mathbf{1}}}$ and $\mathbf{\nabla }\mathbf{\times }\mathbf{(}\mathbf{\nabla }{\mathit{x}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\mathbf{\nabla }\mathbf{\times }\mathbf{\left(}\frac{{\mathbf{e}}_{\mathbf{1}}}{{\mathbf{h}}_{\mathbf{1}}}\mathbf{\right)}\mathbf{=}\mathbf{0}$ . Write V in the form $\mathit{V}\mathbf{=}\frac{{\mathbf{e}}_{\mathbf{1}}}{{\mathbf{h}}_{\mathbf{1}}}\mathbf{\left(}{\mathit{h}}_{\mathbf{1}}{\mathit{V}}_{\mathbf{1}}\mathbf{\right)}\mathbf{+}\frac{{\mathbf{e}}^{\mathbf{2}}}{{\mathbf{h}}_{\mathbf{2}}}\left(h_2{V}_2\right)\mathbf{+}\frac{{\mathbf{e}}_{\mathbf{3}}}{{\mathbf{h}}_{\mathbf{3}}}\left(h_3{V}_3\right)$and use vector identities from Chapter 6 to complete the derivation.

Short answer

The value of V is $\nabla \times V=\sum _{jik}\frac{1}{h_j{h}_i{h}_k}{\in }_{jik}\frac{\partial }{\partial xj}\left(h_i{V}_i\right)h_k{\stackrel{⏜}{e}}_k$.

Step-by-step solution

Given Information

A vector V .

Definition of curl.

The curl at a given place in the field is represented by a vector whose length and direction correspond to the greatest circulation's magnitude and axis.

Find the value.

Formula states that$\nabla u={\sum }_i\frac{1}{h_i}\frac{\partial u}{\partial x_i}{\stackrel{⏜}{e}}_i$.

Let $u=x_k$,the equation becomes as follows.

$\begin{array}{l}\nabla x_k=\sum _i\frac{1}{h}{\delta }_{ki}{\stackrel{⏜}{e}}_i\\ =\frac{1}{h_k}{\stackrel{⏜}{e}}_k\end{array}$

The curl of the gradient is zero, equation becomes as follows.

$\nabla \times \left(\frac{1}{h_k}{\stackrel{⏜}{e}}_k\right)=0$

Evaluate curl of vector field V.

$\begin{array}{l}V=\sum _i{h}_i{V}_i+\frac{{\stackrel{⏜}{e}}_i}{h_i}\\ \nabla +V=\underset{i}{\sum \nabla }\times \left(h_i{V}_i\frac{{\stackrel{⏜}{e}}_i}{h_i}\right)\end{array}$

Vector field identity states that $\nabla \times \left(\varphi v\right)=\varphi \nabla \times v+\nabla \varphi \times v$

Apply the identity in the equation, the equation becomes as follows.

$$\begin{gathered} \nabla \times V=\sum _i\left\{h_i{V}_i\nabla \times \left(\frac{{\stackrel{⏜}{e}}_i}{h_i}\right)+\nabla \left(h_i{V}_i\right)\times \frac{{\stackrel{⏜}{e}}_i}{h_i}\right\} \\ =\sum _i\nabla \left(h_i{V}_i\right)\times \left(\frac{{\stackrel{⏜}{e}}_i}{h_i}\right)+\nabla \left(h_i{V}_i\right)=\sum _i\left(h_i{V}_i\right)\frac{{\stackrel{⏜}{e}}_j}{h_j} \\ =\sum _i\frac{1}{h_j{h}_i}\frac{\partial }{\partial x_j}\left(h_i{V}_i\right){\stackrel{⏜}{e}}_j\times {\stackrel{⏜}{e}}_i;{\stackrel{⏜}{e}}_j\times {\stackrel{⏜}{e}}_i=\sum _k{\in }_{jik}{\stackrel{⏜}{e}}_k \\ =\sum _{jik}\frac{1}{h_j{h}_i{h}_k}{\in }_{jik}\left(h_i{V}_i\right)h_k{\stackrel{⏜}{e}}_k \end{gathered}$$

Hence, the value of V is$\sum _{jik}\frac{1}{h_j{h}_i{h}_k}{\in }_{jik}\left(h_i{V}_i\right)h_k{\stackrel{⏜}{e}}_k$.