Chapter 10: Q2P (page 528)

P Derive the expression (9.11)for curl V in the following way. Show that $\nabla x_{1} = \frac{e_{1}}{h_{1}}$ and $\nabla \times (\nabla x_{1}) = \nabla \times (\frac{e_{1}}{h_{1}}) = 0$ . Write V in the form $V = \frac{e_{1}}{h_{1}} (h_{1} V_{1}) + \frac{e^{2}}{h_{2}} (h_{2} V_{2}) + \frac{e_{3}}{h_{3}} (h_{3} V_{3})$ and use vector identities from Chapter 6 to complete the derivation.

Short Answer

Expert verified

The value of V is $\nabla \times V = \sum_{j i k} \frac{1}{h_{j} h_{i} h_{k}} \in_{j i k} \frac{\partial}{\partial x j} (h_{i} V_{i}) h_{k} {\overset{⏜}{e}}_{k}$ .

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 $\nabla u = \sum_{i} \frac{1}{h_{i}} \frac{\partial u}{\partial x_{i}} {\overset{⏜}{e}}_{i}$ .

Let $u = x_{k}$ ,the equation becomes as follows.

$\begin{array}{l} \nabla x_{k} = \sum_{i} \frac{1}{h} δ_{k i} {\overset{⏜}{e}}_{i} \\ = \frac{1}{h_{k}} {\overset{⏜}{e}}_{k} \end{array}$

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

$\nabla \times (\frac{1}{h_{k}} {\overset{⏜}{e}}_{k}) = 0$

Evaluate curl of vector field V.

$\begin{array}{l} V = \sum_{i} h_{i} V_{i} + \frac{{\overset{⏜}{e}}_{i}}{h_{i}} \\ \nabla + V = \underset{i}{\sum \nabla} \times (h_{i} V_{i} \frac{{\overset{⏜}{e}}_{i}}{h_{i}}) \end{array}$

Vector field identity states that $\nabla \times (ϕ v) = ϕ \nabla \times v + \nabla ϕ \times v$

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

$\nabla \times V = \sum_{i} \{h_{i} V_{i} \nabla \times (\frac{{\overset{⏜}{e}}_{i}}{h_{i}}) + \nabla (h_{i} V_{i}) \times \frac{{\overset{⏜}{e}}_{i}}{h_{i}}\} = \sum_{i} \nabla (h_{i} V_{i}) \times (\frac{{\overset{⏜}{e}}_{i}}{h_{i}}) + \nabla (h_{i} V_{i}) = \sum_{i} (h_{i} V_{i}) \frac{{\overset{⏜}{e}}_{j}}{h_{j}} = \sum_{i} \frac{1}{h_{j} h_{i}} \frac{\partial}{\partial x_{j}} (h_{i} V_{i}) {\overset{⏜}{e}}_{j} \times {\overset{⏜}{e}}_{i}; {\overset{⏜}{e}}_{j} \times {\overset{⏜}{e}}_{i} = \sum_{k} \in_{j i k} {\overset{⏜}{e}}_{k} = \sum_{j i k} \frac{1}{h_{j} h_{i} h_{k}} \in_{j i k} (h_{i} V_{i}) h_{k} {\overset{⏜}{e}}_{k}$

Hence, the value of V is $\sum_{j i k} \frac{1}{h_{j} h_{i} h_{k}} \in_{j i k} (h_{i} V_{i}) h_{k} {\overset{⏜}{e}}_{k}$ .

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Short Answer

Step by step solution

Given Information

Definition of curl.

Find the value.

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P Derive the expression (9.11)for curl V in the following way. Show that ∇x1=e1h1 and ∇×(∇x1)=∇×(e1h1)=0 . Write V in the form V=e1h1(h1V1)+e2h2(h2V2)+e3h3(h3V3)and use vector identities from Chapter 6 to complete the derivation.

Short Answer

Step by step solution

Given Information

Definition of curl.

Find the value.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Oscillations

Engineering Physics

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Electrostatics

Study anywhere. Anytime. Across all devices.