Q28P - Introduction to Electrodynamics

Question

Prove that the curl of a gradient is always zero. Check it for function(b) in Pro b. 1.11.

Step-by-Step Solution

Answer

The curl of gradient of a function is always zero, has been proven. The divergence of curl of function $f(x,y,z)=x^2{y}^2{z}^4,\nabla \times (\nabla f)$ is 0.

Step 1: Define the laplacian

The gradient of a function is $\mathit{v}$ is defined as$\mathbf{\nabla }\mathit{v}$ . The $\nabla$ operator is defined as  .

 $\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k.$

Compute the gradient of function $v$ , as

$$\begin{gathered} \nabla \times (\nabla v)=\nabla \times \left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(v\right) \\ =\frac{\partial v}{\partial x}i+\frac{\partial v}{\partial y}j+\frac{\partial v}{\partial z}k \end{gathered}$$
 

Compute curl of gradient of function $v$ .

$$\begin{gathered} \nabla \times (\nabla v)=\nabla \times \left(\frac{\partial v}{\partial x}i+\frac{\partial v}{\partial y}j+\frac{\partial v}{\partial z}k\right) \\ =\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}& \frac{\partial v}{\partial z}\end{array}\right| \\ =i\left(\frac{{\partial }^{2}v}{\partial y\partial z}-\frac{{\partial }^{2}v}{\partial z\partial y}\right)-j\left(\frac{{\partial }^{2}v}{\partial z\partial x}-\frac{{\partial }^{2}v}{\partial x\partial z}\right)+ki\left(\frac{{\partial }^{2}v}{\partial x\partial y}-\frac{{\partial }^{2}v}{\partial y\partial x}\right) \end{gathered}$$

Compute curl of gradient of function 0.

Step 2: Compute the curl of gradient of function

The function is given as $f(x,y,z)=x^2{y}^3{z}^4$ is computed as follows:

 

Compute the gradient of function $f(x,y,z)=x^2{y}^3{z}^4,$as

$$\begin{gathered} \nabla f(x,y,z)=\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(x^2{y}^3{z}^4\right) \\ =\frac{\partial \left(x^2{y}^3{z}^4\right)}{\partial x}i+\frac{\partial \left(x^2{y}^3{z}^4\right)}{\partial y}j+\frac{\partial \left(x^2{y}^3{z}^4\right)}{\partial z}k \\ =2x{y}^3{z}^{4}i+3x^2{y}^2{z}^4+j4x^2{y}^3{z}^3 \end{gathered}$$

Compute curl of gradient of function $f(x,y,z)$

$$\begin{gathered} \nabla \times (\nabla f)=\nabla \times \left(\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j+\frac{\partial f}{\partial z}k\right) \\ =\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 2x{y}^3{z}^4& 3x^2{y}^2{z}^4& 4x^2{y}^3{z}^3\end{array}\right| \\ =i(12x^2{y}^2{z}^3-12x^2{y}^2{z}^3)-j(8x{y}^3{z}^3-8x{y}^3{z}^3)+k(6x{y}^2{z}^4-6x{y}^2{z}^4) \\ =0 \end{gathered}$$

Thus, curl of gradient is always 0.