Curl and Gauss Divergence Theorem

Curl

If \(\vec F \) represents a vector field the quantity

$$\hat i \times \frac {\partial \vec F}{\partial x}+ \hat j \times \frac {\partial \vec F}{\partial y} + \hat k \times \frac {\partial \vec F}{\partial z}, $$

Is also a vector, is called curl of \(\vec F \) and is written as

\begin{align*} \text {Curl}\: \vec F &= \hat i \times \frac {\partial \vec F}{\partial x}+ \hat j \times \frac {\partial \vec F}{\partial y} + \hat k \times \frac {\partial \vec F}{\partial z} \\ &= \hat i \frac {\partial}{\partial x}\times \vec F + \hat j \frac {\partial }{\partial y}\times \vec F + \hat k \frac {\partial }{\partial z}\times \vec F \\ &= \left (\hat i \frac {\partial}{\partial x} + \hat j \frac {\partial }{\partial }+ \hat k \frac {\partial}{\partial z}\right )\times \vec F = \nabla \times \vec F \end{align*}

If \(\nabla \times \vec F = 0\) then \(\vec F\) is irrotational vector. Now let \(\vec F = \hat i F_x + \hat j F_y + \hat k F_z \) then

\(\text {Curl}\: \vec F =\)

\begin{matrix} \hat i &\hat j &\hat k \\ \frac {\delta }{\delta x} &\frac {\delta }{\delta y} &\frac {\delta }{\delta z} \\ F_x &F_y &F_z \end{matrix}

If \(\nabla \times \vec A = 0\) then there is a \(\phi \) such that \(\vec A = \nabla \phi \)

Second derivatives of vectors functions or fields

1. \(\nabla .(\nabla \phi \) ; divergence of a gradient \(\phi = \nabla ^2 \phi =\) a scalar field.
2. \(\nabla \times (\nabla \phi \); curl of a gradient \(\phi = 0 \)
3. \(\nabla (\nabla .\vec F \); gradient of divergence \(\vec F\) is a vector field
4. \(\nabla .(\nabla \times \vec F \); divergence of a curl \(\vec F = 0\)
5. \(\nabla \times (\nabla \times \vec F \); curl of curl \(\vec F = \nabla (\nabla . \vec F) - \nabla ^2 \vec F\)
6. \((\nabla . \nabla )\vec F = \nabla ^2 \vec F\) is a vector field.

Where, \(\nabla ^2 = \left (\frac {\delta ^2}{\delta x^2} + \frac {\delta ^2}{\delta y^2} + \frac {\delta }{\delta z^2}\right )\) is referred as Laplacian operator.

Interpretation of curl of \(\vec F\):

If a rigid body is in motion, the curl of its linear velocity at any point gives twice its angular velocity. Let us consider a rigid body rotating with angular velocity \(\vec \omega \) about an axis OA, O being a fixed point in the body.

Let \(\vec r\) be the position vector of any point P on the body. Draw perpendicular PQ from P to PA. then linear velocity \(\vec V\) of P due to circular motion.

\begin{align*} |\vec V|&=\vec {\omega }QP = \vec {\omega } r\sin \theta = |\vec {\omega } \times \vec r | \\ \vec v &= \vec omega \times \vec r \\ \text {where} \vec r &=x \hat i + y \hat j + z\hat k \\ \vec {\omega} &= \omega _1 \hat I + \omega _2 \hat j + \omega _3 \hat k \\ \text {But}\\: \text {curl}\: \vec {\nu } &=(\nabla \times v) = \nabla \times (\vec {\omega} \times \vec r) \\ \end{align*}

\begin{align*} &= \nabla \times \left |\begin{matrix} \hat I &\hat j &\hat k \\ \omega _1 &\omega _2 &\omega_3 \\ x &y &z \end{matrix}\right |\end{align*}

\begin{align*} &= \nabla \times [(\omega _2z -\omega _3y)\hat i + (\omega _3 x -\omega _1z) \hat j + (\omega _1y - \omega _2x)\hat k] \\ &= \left | \begin{matrix}\hat I &\hat j &\hat k \\ \frac {\delta}{\delta x} &\frac {\delta}{\delta y} &\frac {\delta}{\delta z}\\ \omega_2 z - \omega_3y &\omega_3 x -\omega _1z &\omega_1z -\omega_2x \end{matrix} \right |\\ &= 2[\omega _1\hat i + \omega _2 \hat j + \omega _3 \hat k]\\ &= 2\widehat {\omega }\end{align*}

It proves the statement that if a rigid body is in motion, the curl of its linear velocity at any point gives twice its angular velocity.

Gauss Divergence Theorem

Gauss’ divergence law gives relation between surface integral and volume integral for any vector function.

“Gauss’ divergence law states that normal surface integral of a any vector function \(\vec F(x, y, z)\) over the boundary of closed surface S is equal to the volume integral of the divergence of the vector over the volume V enclosed by the surface.”

Symbolically,

$$\iint \vec F.d\vec s = \iiint (\nabla .\vec F)dV $$

In Russian texts Gauss’ theorem is called Ostrogradski’s theorem.

Proof:

To prove Gauss divergence law assume small parallelepiped of dimensions \(\delta x,\: \delta y \: \text {and}\: \delta z\) along axes X, Y and Z respectively as shown in the figure. Let be any vector point function at the centre \(P\:(x, y, z)\) of parallelepiped.\(\vec F_x, \vec F_y , \vec F_z\) are the orthogonal components of vector point function along 3-axes respectively. i.e.

$$\vec F = \hat i\vec F_x + \hat j\vec F_y + \hat k \vec F_z $$

Now at any point on face ABCD, normal vector component along Y-axis is \(\vec F\left (y - \frac {dy}{2} \right )\)

Taylor’s theorem is expressed as,

$$F(x + h) = F(x) + h\frac {\partial F(x,y)}{\partial x}+ \frac {h^2}{2!}\frac {\partial ^2F(x,y)}{\partial x^2}+ \dots $$

Using Taylor’s theorem and neglecting higher power, normal component of vector F from face ABCD along Y-axis can be written as,

\begin{align*} &= F_y - \frac {dy}{2}\frac {\partial F_y}{\partial y} \\ &= F_y - \frac {\partial F_y}{\partial y}\frac {dy}{2}, \dots (i) \end{align*}

Flux is defined as the rate of flow of property per unit area. So, the total flux of vector point function through the surface S can be given as.

\begin{align*} &= \left (F_y - \frac {\partial F_y}{\partial y}\frac {dy}{2} \right )dS_y \\ &= \left (F_y - \frac {\partial F_y}{\partial y}\frac {dy}{2} \right )dxdz \\ \end{align*}

Similarly, the flux of vector point function through the face A’B’C’D’ can be given as,

\begin{align*} &= \left (F_y + \frac {\partial F_y}{\partial y}\frac {dy}{2} \right )dxdz \\ \end{align*}

Thus volume of the flux passing through faces ABCD and A’B’C’D’ of the parallelepiped along the direction of Y becomes,

\begin{align*} &= \left (F_y +\frac {\partial F_y}{\partial y}\frac {dy}{2} \right )dxdz - \left (F_y - \frac {\partial F_y}{\partial y}\frac {dy}{2} \right )dxdz \\&= \frac {\partial F_y}{\partial y} dxdydz,\:\dots (2) \end{align*}

Thus, obeying the law of symmetry, volume of the flux passing through other faces of the parallelepiped becomes,

Along X-axis,

$$= \frac {\partial F_x}{\partial x}dxdydz, \dots (3)$$

Along Z-axis,

$$= \frac {\partial F_z}{\partial z}dxdydz, \dots (4)$$

Thus, total volume of flux flowing away per second from the parallelepiped can be obtained by adding equation (2), (3) and (4) which becomes as

\begin{align*} &= \frac {\partial F_x}{\partial x}dxdydz + \frac {\partial F_y}{\partial y}dxdydz + \frac {\partial F_z}{\partial z}dxdydz \\ &= \left (\frac {\partial F_x}{\partial x} + \frac {\partial F_y}{\partial y} + \frac {\partial F_z}{\partial z}\right )dxdydz \\ &= \left (\hat i \frac {\partial }{\partial x} + \hat j \frac {\partial }{\partial y} + \hat k \frac {\partial }{\partial z}\right )(\hat i F_x + \hat j F_y + \hat k F_z)dxdydz \\ \end{align*}

Since, volume of the small parallelepiped is

$$ dV = dxdydz $$

So, total volume of flux flowing away per second from the parallelepiped is

\begin{align*} &= \left (\hat i \frac {\partial }{\partial x}+ \hat j \frac {\partial }{\partial y} + \hat k \frac {\partial }{\partial z}\right )(\hat i F_x + \hat j F_y + \hat k F_z)dV \\ &= (\nabla .\vec F)dV, \dots (5) \\ \end{align*}

Also, total volume flux passing away of the surface of the parallelepiped per second is

$$ \phi = \iint \vec F.d\vec s, \dots (6) $$

From equations (5) and (6), it becomes

$$\iint \vec F. d\vec s = \iiint (\nabla .\vec F)dV, \dots (7)$$

Following are some examples in which Gauss’ divergence law is applicable:

1. The magnitude of river’s current, i.e. the amount of water that flows through a cross-section of the river each second.
2. The amount of sunlight that lands on a patch of ground each second is also kind.

Bibliography

P.B. Adhikari, Bhoj Raj Gautam, Lekha Nath Adhikari. A Textbook of Physics. kathmandu: Sukunda Pustak Bhawan, 2011.

Jha, V. K.; 'Lecture title'; Elementary Vector Analysis; St. Xavier's College, Kathmandu; 2016.