Rotation

Definition 3..3  

For the vector field in space $\boldsymbol{F} = F_{1}\boldsymbol{i} + F_{2}\boldsymbol{j} + F_{3}\boldsymbol{k}$, we define the curl $of \boldsymbol{F}$ or $\rm {curl} \boldsymbol{F}$ as follows.．

$$\rm {curl} \boldsymbol{F} = \nabla \times \boldsymbol{F} = \left(... ...al F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}\right)\boldsymbol{k}$$

More formally,

$$\nabla \times \boldsymbol{F} = \left\vert\begin{array}{ccc} \bol... ...rac{\partial }{\partial z}\\ F_{1} & F_{2} & F_{3} \end{array} \right\vert$$

Example 3..19  

Find the curl of $${\boldsymbol{F} = z\:\boldsymbol{i} + x^2\:\boldsymbol{j} + 2y\:\boldsymbol{k}}$$．

Answer

$${\rm curl} \boldsymbol{F} = \left\vert\begin{array}{ccc} \boldsy... ... \end{array} \right\vert = 2\boldsymbol{i} + \boldsymbol{j} + 2x\boldsymbol{k}$$

It is not so difficult to find the rotation of the vector field, but it is difficult to understand what the rotation of the vector field is. So let's understand what the rotation of a vector field is by considering the following example.

For $\boldsymbol{F} = (F_{1}, F_{2})$，we study how much quadrilateral ABCD can be rotated, where quadrilateral ABCD are given by A $(x_{A},y_{A})$, B $(x_{A}+\Delta x, y_{A})$, C $(x_{A}+\Delta x, y_{A}+\Delta y)$, D $(x_{A}, y_{A}+\Delta y)$

First，the horizontal component at point A $(x_{A},y_{A})$ is $F_{1}(x_{A},y_{A})$． The horizontal component at point D $(x_{A}, y_{A}+\Delta y)$ is

$$F_{1}(x_{A},y_{A}) + \frac{\partial F_{1}}{\partial y}(x_{A},y_{A})\Delta y$$

The difference between these two values, that is, the difference between the horizontal components $${\frac{\partial F_{1}}{\partial y}(x_{A}, y_{A}) \Delta y}$$ is positive, the quadrilateral ABCD rotates clockwise． Also, the difference in the vertical components at points A and B, $${\frac{\partial F_{2}}{\partial x}(x_{A},y_{A})\Delta x}$$ is positive，the quadrilateral ABCD rotates counter clockwise．Thus

$$\Vert{\rm curl} \boldsymbol{F}\Vert = \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}$$

is the force of rotation of the quadrilateral by ${\boldsymbol{F}}$ and the direction of the force is orthogonal to the quadrilateral $${\boldsymbol{k}(\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y})}$$. This is the name curl comes from．Thus, for $\nabla \times \boldsymbol{F} = 0$, the vector field $\boldsymbol{F}$ becomes no votex. ．

Example 3..20  

If $\boldsymbol{F}$ is conservative， show that $${\nabla \times \boldsymbol{F} = {\bf0}}$$．

Answer Since $\boldsymbol{F}$ is conservative， there exists $f$ so that $\boldsymbol{F} =\nabla f$．Now we find ${\rm curl} \boldsymbol{F}$. Then

$$\nabla \times \boldsymbol{F} = \nabla \times \nabla f = \left\ver... ...} + (f_{xz} - f_{zx})\boldsymbol{j} + (f_{yx} - f_{xy})\boldsymbol{k} = {\bf0}$$

Theorem 3..2  

For any scalar field $\phi$ and any vector field $\boldsymbol{A}$, show that

$$\nabla \times (\nabla \phi) = {\bf0},\hskip 1cm \nabla\cdot(\nabla \times \boldsymbol{A}) = 0$$

Example 3..21  

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k},\ r = \vert\boldsymbol{r}\vert$ and ${\bf\omega}$ be a constant vector, prove the followings.．

$$(1)\ \nabla \times \boldsymbol{r} = {\bf0}\hskip 1cm (2)\ \nabla \times ({\bf\omega} \times \boldsymbol{r}) = 2{\bf\omega}$$

(1) $\nabla \times \boldsymbol{r} = \left\vert\begin{array}{ccc} \boldsymbol{i} & \b... ...\partial y}{\partial y} - \frac{\partial x}{\partial y})\boldsymbol{k} = {\bf0}$

(2) Let ${\bf w} = w_{1}\boldsymbol{i} + w_{2}\boldsymbol{j} + w_{3}\boldsymbol{k}$. Then

$${\bf w} \times \boldsymbol{r} = \left\vert\begin{array}{ccc} \bol... ...ldsymbol{i} + (w_{3}x - w_{1}z)\boldsymbol{j} + (w_{1}y - w_{2}x)\boldsymbol{k}$$

Thus，

$$\nabla \times ({\bf w} \times \boldsymbol{r}) = \left\vert\begin{... ...= 2w_{1}\boldsymbol{i} + 2w_{2}\boldsymbol{j} + 2w_{3}\boldsymbol{k} = 2{\bf w}$$

Scalar potential

If vector field $\boldsymbol{A}$ has the potential $\phi$，then $\boldsymbol{A} = -\nabla \phi$. Thus by the theorem 3.2， $\nabla \times \boldsymbol{A} = - \nabla \times (\nabla \phi) = {\bf0}$．How about the converse?．

Theorem 3..3  

If the vector field $\boldsymbol{A}$ on all spacesatisfies $\nabla \times \boldsymbol{A} = {\bf0}$，then the vector field $\boldsymbol{A}$ has a potential

Example 3..22  

Find $${\int_{C}((2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:\boldsymbol{k}) \cdot d\boldsymbol{r}}$$. for
(1) $C$ is a close curve
(2) $C$ is a curve connecting from the point ${\rm P}(1,0,-1)$ to the point ${\rm Q}(2,-1,3)$.

Answer (1)

$$\nabla \times (2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:... ...y} & \frac{\partial}{\partial z}\\ 2x+yz & zx & xy \end{array}\right\vert = 0$$

Then

$$\oint_{C}((2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:\boldsymbol{k}) \cdot d\boldsymbol{r} = 0$$

(2) Since $2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:\boldsymbol{k}$ has a scalar potential, we find the scalar potential.

$$(2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:\boldsymbol{k}... ...\partial y}\:\boldsymbol{j} + \frac{\partial \phi}{\partial z}\:\boldsymbol{k})$$

implies $\phi = -(x^2 + xyz) + C$．Note that

$$d\boldsymbol{r} = \frac{d\boldsymbol{r}}{dt}dt = (\frac{dx}{dt}\:\boldsymbol{i} + \frac{dy}{dt}\:\boldsymbol{j} + \frac{dz}{dt}\:\boldsymbol{k})dt$$

Then，

$$\int_{C}((2x+yz)\:\boldsymbol{i} + zx\:\boldsymbol{j} + xy\:\boldsymbol{k}) \cdot d\boldsymbol{r} = \int_{C}(-\nabla \phi \cdot(\frac{dx}{dt}\:\boldsymbol{i} + \frac{dy}{dt}\:\boldsymbol{j} + \frac{dz}{dt}\:\boldsymbol{k}))dt$$

$$= \int_{{\rm P}}^{{\rm Q}}(-d \phi) = -\phi]_{P}^{Q} = (x^2 + xyz)]_{P}^{Q} = 4-6 - 1 = -3$$

Vector potential

For the vector field $\boldsymbol{A}$, if there exists the vector field ${\bf p}$ satisfying

$$\boldsymbol{A} = \nabla \times {\bf p}$$

then we say that the vector field $\boldsymbol{A}$ has vector potential${\bf p}$．Here，if the vector field $\boldsymbol{A}$ has the vector potential ${\bf p}$, then by the theorem3.2， $\nabla \cdot(\nabla \times {\bf p}) = 0$. we have $\nabla \cdot\boldsymbol{A} = 0$．How about the converse?．

Theorem 3..4  

The vector field $\boldsymbol{A}$ defined in all space，if $\nabla \cdot\boldsymbol{A} = 0$，then the vector field $\boldsymbol{A}$ has a vector potential．

To solve the exercise，we introduce a nes symbol. Formal inner product of $\boldsymbol{A} = a_{1}\boldsymbol{i} + a_{2}\boldsymbol{j} + a_{3}\boldsymbol{k}$ and $\nabla$ is an operator such as

$$\boldsymbol{A} \cdot\nabla = a_{1}\frac{\partial}{\partial x} + a_{2}\frac{\partial}{\partial y} + a_{3}\frac{\partial}{\partial z}$$

When this is applied to the scalar field $\phi$ and the vector field $\boldsymbol{A}$，we can write

$$(\boldsymbol{A}\cdot\nabla)\phi = a_{1}\frac{\partial \phi}{\partial x} + a_{2}\frac{\partial \phi}... ...ial y} + a_{3}\frac{\partial \phi}{\partial z} = \boldsymbol{A}\cdot\nabla \phi$$

$$(\boldsymbol{A}\cdot\nabla)\boldsymbol{A} = a_{1}\frac{\partial \boldsymbol{A}}{\partial x} + a_{2}\frac{\par... ...boldsymbol{A}}{\partial y} + a_{3}\frac{\partial \phi}{\partial \boldsymbol{A}}$$

The formula

$$\nabla(\boldsymbol{A} \cdot\boldsymbol{B}) = (\boldsymbol{B}\cdot... ...la \times \boldsymbol{A}) + \boldsymbol{A}\times (\nabla \times \boldsymbol{B})$$

and it's proof．

To process an expression containing the operator $\nabla$，put $\nabla$ as $\frac{\partial}{\partial x}\boldsymbol{i} + \frac{\partial}{\partial y}\boldsymbol{j} + \frac{\partial}{\partial z}{\bf z}$. Then excute $\nabla$. After that, processing is performed using the scalar triple product and vector triple product learned in vector algebra．

$$\boldsymbol{A} \times (\nabla \times \boldsymbol{B}) = \boldsymbol{A} \times \left(\boldsymbol{i} \times \frac{\partial ... ...{\partial y} + \boldsymbol{k} \frac{\partial \boldsymbol{B}}{\partial z}\right)$$

$$= (A \cdot\frac{\partial \boldsymbol{B}}{\partial x})\boldsymbol{i}... ... (\boldsymbol{A} \cdot\boldsymbol{k})\frac{\partial \boldsymbol{B}}{\partial z}$$

In the above equation interchange $\boldsymbol{A}$ and $\boldsymbol{B}$. Then

$$\boldsymbol{B} \times (\nabla \times \boldsymbol{A}) = (B \cdot\f... ... (\boldsymbol{B} \cdot\boldsymbol{k})\frac{\partial \boldsymbol{A}}{\partial z}$$

Adding these two equations, we have

$$\boldsymbol{A} \times (\nabla \times \boldsymbol{B}) + \boldsymbol{B} \times (\nabla \times \boldsymbol{A})$$

$$= \nabla(\boldsymbol{A}\cdot\boldsymbol{B}) - (B \cdot\nabla)\boldsymbol{A} - (A \cdot\nabla)\boldsymbol{B}$$

Therefore，

$$\nabla(\boldsymbol{A} \cdot\boldsymbol{B}) = (\boldsymbol{B}\cdot... ...la \times \boldsymbol{A}) + \boldsymbol{A}\times (\nabla \times \boldsymbol{B})$$

Exercise3.6

1.

Let $\boldsymbol{A} = 2xz^2 \boldsymbol{i} - yz\boldsymbol{j} + 3xz^3\boldsymbol{k}, \phi = x^2 yz$．Find the followings.．

(1) $\nabla \times \boldsymbol{A}$

(2) $\nabla \times (\phi \boldsymbol{A})$

(3) $\nabla \times (\nabla \times \boldsymbol{A})$

2.

Find $a,b,c$ so that $${{\bf V} = (x + 2y + az)\boldsymbol{i} + (bx - 3y -z)\boldsymbol{j} + (4x + cy + 2z)\boldsymbol{k}}$$ satisfies $\nabla \times {\bf V} = {\bf0}$.

3.

If $\nabla \times \boldsymbol{A} = {\bf0},\ \nabla \times \boldsymbol{B} = {\bf0}$，then prove that $\nabla \cdot(\boldsymbol{A} \times \boldsymbol{B}) = 0$．

4.

Let $\boldsymbol{C}$ be a constant vector．Then prove the following equation for arbitray vecot field $\boldsymbol{A}, \boldsymbol{B}$.

(1) $\nabla(\boldsymbol{C}\cdot\boldsymbol{A}) = (\boldsymbol{C}\cdot\nabla)\boldsymbol{A} + \boldsymbol{C} \times (\nabla \times \boldsymbol{A})$

(2) $\nabla \cdot(\boldsymbol{C} \times \boldsymbol{A}) = -\boldsymbol{C} \cdot(\nabla \times \boldsymbol{A})$

(3) $\nabla \times (\boldsymbol{C} \times \boldsymbol{A}) = \boldsymbol{C} (\nabla \cdot\boldsymbol{A}) - (\boldsymbol{C} \cdot\nabla)\boldsymbol{A}$

5.

For an arbitrary vector field $\boldsymbol{A}$, prove the following．

$$(\boldsymbol{A}\cdot\nabla)\boldsymbol{A} = \frac{1}{2}\nabla \vert\boldsymbol{A}\vert^2 - \boldsymbol{A} \times (\nabla \times \boldsymbol{A})$$

6.

Let $\rho$ and $p$ be scalar fields．If $\rho \boldsymbol{F} = \nabla p$，then prove that $\boldsymbol{F}\cdot(\nabla \times \boldsymbol{F}) = 0$．