Chapter 7, Problem 55P

(a)

To determine

Prove that the curl of the gradient function $\nabla \times \left(\nabla V\right)=0$ in cylindrical coordinates.

Explanation of Solution

Calculation:

The Del operator in a cylindrical coordinates is expressed as,

$\nabla =\frac{\partial }{\partial \rho }{a}_{\rho }+\frac{1}{\rho }\frac{\partial }{\partial \varphi }{a}_{\varphi }+\frac{\partial }{\partial z}{a}_z$

The function $\nabla V$ is expressed as,

$\nabla V=\left(\frac{\partial V}{\partial \rho }{a}_{\rho }+\frac{1}{\rho }\frac{\partial V}{\partial \varphi }{a}_{\varphi }+\frac{\partial V}{\partial z}{a}_z\right)$

Therefore, the function $\nabla \times \left(\nabla V\right)$ is,

$\begin{array}{c}\nabla \times \left(\nabla V\right)=\frac{1}{\rho }\left|\begin{array}{ccc}{a}_{\rho }& \rho a_{\varphi }& a_z\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ \frac{\partial V}{\partial \rho }& \rho \left(\frac{1}{\rho }\frac{\partial V}{\partial \varphi }\right)& \frac{\partial V}{\partial z}\end{array}\right|\\ =\frac{1}{\rho }\left|\begin{array}{ccc}{a}_{\rho }& \rho a_{\varphi }& a_z\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ \frac{\partial V}{\partial \rho }& \frac{\partial V}{\partial \varphi }& \frac{\partial V}{\partial z}\end{array}\right|\end{array}$

$\nabla \times \left(\nabla V\right)=\frac{1}{\rho }\left[\begin{array}{l}\left(\frac{\partial }{\partial \varphi }\left(\frac{\partial V}{\partial z}\right)-\frac{\partial }{\partial z}\left(\frac{\partial V}{\partial \varphi }\right)\right)a_{\rho }-\left(\frac{\partial }{\partial \rho }\left(\frac{\partial V}{\partial z}\right)-\frac{\partial }{\partial z}\left(\frac{\partial V}{\partial \rho }\right)\right)\rho a_{\varphi }\\ +\left(\frac{\partial }{\partial \rho }\left(\frac{\partial V}{\partial \varphi }\right)-\frac{\partial }{\partial \varphi }\left(\frac{\partial V}{\partial \rho }\right)\right)a_z\end{array}\right]$        (1)

The point of interchange of base values of the partial derivatives gives the same values. That is,

$\begin{array}{l}\frac{\partial }{\partial \varphi }\left(\frac{\partial V}{\partial z}\right)=\frac{\partial }{\partial z}\left(\frac{\partial V}{\partial \varphi }\right)\\ \frac{\partial }{\partial \rho }\left(\frac{\partial V}{\partial z}\right)=\frac{\partial }{\partial z}\left(\frac{\partial V}{\partial \rho }\right)\\ \frac{\partial }{\partial \rho }\left(\frac{\partial V}{\partial \varphi }\right)=\frac{\partial }{\partial \varphi }\left(\frac{\partial V}{\partial \rho }\right)\end{array}$

Therefore, the Equation (1) is re-written as follows,

$\begin{array}{c}\nabla \times \left(\nabla V\right)=\frac{1}{\rho }\left[\begin{array}{l}\left(\frac{\partial }{\partial \varphi }\left(\frac{\partial V}{\partial z}\right)-\frac{\partial }{\partial \varphi }\left(\frac{\partial V}{\partial z}\right)\right)a_{\rho }-\left(\frac{\partial }{\partial \rho }\left(\frac{\partial V}{\partial z}\right)-\frac{\partial }{\partial \rho }\left(\frac{\partial V}{\partial z}\right)\right)\rho a_{\varphi }\\ +\left(\frac{\partial }{\partial \rho }\left(\frac{\partial V}{\partial \varphi }\right)-\frac{\partial }{\partial \rho }\left(\frac{\partial V}{\partial \varphi }\right)\right)a_z\end{array}\right]\\ =\frac{1}{\rho }\left[\left(0\right)a_{\rho }-\left(0\right)\rho a_{\varphi }+\left(0\right)a_z\right]\\ =0\end{array}$

Conclusion:

Thus, the curl of the gradient function $\nabla \times \left(\nabla V\right)=0$ in cylindrical coordinates and it is shown.

(b)

To determine

Prove that the divergence of the curl function $\nabla \cdot \left(\nabla \times A\right)=0$ in cylindrical coordinates.

Explanation of Solution

Calculation:

The function $\nabla \times A$ is expressed as,

$\begin{array}{c}\nabla \times A=\frac{1}{\rho }\left|\begin{array}{ccc}{a}_{\rho }& \rho a_{\varphi }& a_z\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ A_{\rho }& \rho A_{\varphi }& A_z\end{array}\right|\\ =\frac{1}{\rho }\left[\begin{array}{c}\left(\frac{\partial }{\partial \varphi }\left(A_z\right)-\frac{\partial }{\partial z}\left(\rho A_{\varphi }\right)\right)a_{\rho }-\left(\frac{\partial }{\partial \rho }\left(A_z\right)-\frac{\partial }{\partial z}\left(A_{\rho }\right)\right)\rho a_{\varphi }\\ +\left(\frac{\partial }{\partial \rho }\left(\rho A_{\varphi }\right)-\frac{\partial }{\partial \varphi }\left(A_{\rho }\right)\right)a_z\end{array}\right]\\ =\left[\begin{array}{c}\frac{1}{\rho }\left(\frac{\partial A_z}{\partial \varphi }-\frac{\partial \left(\rho A_{\varphi }\right)}{\partial z}\right)a_{\rho }-\frac{1}{\rho }\left(\frac{\partial A_z}{\partial \rho }-\frac{\partial A_{\rho }}{\partial z}\right)\rho a_{\varphi }\\ +\frac{1}{\rho }\left(\frac{\partial \left(\rho A_{\varphi }\right)}{\partial \rho }-\frac{\partial A_{\rho }}{\partial \varphi }\right)a_z\end{array}\right]\\ =\left[\begin{array}{c}\frac{1}{\rho }\left(\frac{\partial A_z}{\partial \varphi }-\frac{\partial \left(\rho A_{\varphi }\right)}{\partial z}\right)a_{\rho }-\left(\frac{\partial A_z}{\partial \rho }-\frac{\partial A_{\rho }}{\partial z}\right)a_{\varphi }\\ +\frac{1}{\rho }\left(\frac{\partial \left(\rho A_{\varphi }\right)}{\partial \rho }-\frac{\partial A_{\rho }}{\partial \varphi }\right)a_z\end{array}\right]\end{array}$

Therefore, the function $\nabla \cdot \left(\nabla \times A\right)$ is,

$\begin{array}{c}\nabla \cdot \left(\nabla \times A\right)=\left(\frac{\partial }{\partial \rho }{a}_{\rho }+\frac{1}{\rho }\frac{\partial }{\partial \varphi }{a}_{\varphi }+\frac{\partial }{\partial z}{a}_z\right)\cdot \left[\begin{array}{c}\frac{1}{\rho }\left(\frac{\partial A_z}{\partial \varphi }-\frac{\partial \left(\rho A_{\varphi }\right)}{\partial z}\right)a_{\rho }-\left(\frac{\partial A_z}{\partial \rho }-\frac{\partial A_{\rho }}{\partial z}\right)a_{\varphi }\\ +\frac{1}{\rho }\left(\frac{\partial \left(\rho A_{\varphi }\right)}{\partial \rho }-\frac{\partial A_{\rho }}{\partial \varphi }\right)a_z\end{array}\right]\\ =\left[\begin{array}{l}\frac{\partial }{\partial \rho }\left(\frac{1}{\rho }\left(\frac{\partial A_z}{\partial \varphi }-\frac{\partial \left(\rho A_{\varphi }\right)}{\partial z}\right)\right)a_{\rho }\cdot a_{\rho }-0+0\\ +0-\left(\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(\frac{\partial A_z}{\partial \rho }-\frac{\partial A_{\rho }}{\partial z}\right)\right)a_{\varphi }\cdot a_{\varphi }+0\\ +0-0+\left(\frac{\partial }{\partial z}\left(\frac{1}{\rho }\left(\frac{\partial \left(\rho A_{\varphi }\right)}{\partial \rho }-\frac{\partial A_{\rho }}{\partial \varphi }\right)\right)\right)a_z\cdot a_z\end{array}\right]\left\{\begin{array}{l}\because a_{\rho }\cdot a_{\varphi }=0\\ a_{\rho }\cdot a_z=0\\ a_{\varphi }\cdot a_z=0\end{array}\right\}\\ =\left[\begin{array}{l}\frac{\partial }{\partial \rho }\left(\frac{1}{\rho }\left(\frac{\partial A_z}{\partial \varphi }-\frac{\partial \left(\rho A_{\varphi }\right)}{\partial z}\right)\right)-\left(\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(\frac{\partial A_z}{\partial \rho }-\frac{\partial A_{\rho }}{\partial z}\right)\right)\\ +\left(\frac{\partial }{\partial z}\left(\frac{1}{\rho }\left(\frac{\partial \left(\rho A_{\varphi }\right)}{\partial \rho }-\frac{\partial A_{\rho }}{\partial \varphi }\right)\right)\right)\end{array}\right]\left\{\begin{array}{l}\because a_{\rho }\cdot a_{\rho }=0\\ a_{\varphi }\cdot a_{\varphi }=0\\ a_z\cdot a_z=0\end{array}\right\}\\ =\left[\begin{array}{l}\frac{\partial }{\partial \rho }\left(\frac{1}{\rho }\frac{\partial A_z}{\partial \varphi }-\frac{\partial A_{\varphi }}{\partial z}\right)-\left(\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(\frac{\partial A_z}{\partial \rho }-\frac{\partial A_{\rho }}{\partial z}\right)\right)\\ +\left(\frac{\partial }{\partial z}\left(\frac{1}{\rho }\frac{\partial \left(\rho A_{\varphi }\right)}{\partial \rho }-\frac{1}{\rho }\frac{\partial A_{\rho }}{\partial \varphi }\right)\right)\end{array}\right]\end{array}$

Simplify the above equation.

$\nabla \cdot \left(\nabla \times A\right)=\left[\begin{array}{l}\frac{\partial }{\partial \rho }\left(\frac{1}{\rho }\frac{\partial A_z}{\partial \varphi }\right)-\frac{\partial }{\partial \rho }\left(\frac{\partial A_{\varphi }}{\partial z}\right)-\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(\frac{\partial A_z}{\partial \rho }\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(\frac{\partial A_{\rho }}{\partial z}\right)\\ +\frac{\partial }{\partial z}\left(\frac{1}{\rho }\frac{\partial \left(\rho A_{\varphi }\right)}{\partial \rho }\right)-\frac{\partial }{\partial z}\left(\frac{1}{\rho }\frac{\partial A_{\rho }}{\partial \varphi }\right)\end{array}\right]$

$\nabla \cdot \left(\nabla \times A\right)=\left[\begin{array}{l}\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\frac{\partial A_z}{\partial \varphi }\right)-\frac{\partial }{\partial \rho }\left(\frac{\partial A_{\varphi }}{\partial z}\right)-\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(\frac{\partial A_z}{\partial \rho }\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(\frac{\partial A_{\rho }}{\partial z}\right)\\ +\frac{\partial }{\partial z}\left(\frac{\partial A_{\varphi }}{\partial \rho }\right)-\frac{1}{\rho }\frac{\partial }{\partial z}\left(\frac{\partial A_{\rho }}{\partial \varphi }\right)\end{array}\right]$        (2)

The point of interchange of base values of the partial derivatives gives the same values. That is,

$\begin{array}{l}\frac{\partial }{\partial \rho }\left(\frac{\partial A_z}{\partial \varphi }\right)=\frac{\partial }{\partial \varphi }\left(\frac{\partial A_z}{\partial \rho }\right)\\ \frac{\partial }{\partial \rho }\left(\frac{\partial A_{\varphi }}{\partial z}\right)=\frac{\partial }{\partial z}\left(\frac{\partial A_{\varphi }}{\partial \rho }\right)\\ \frac{\partial }{\partial \varphi }\left(\frac{\partial A_{\rho }}{\partial z}\right)=\frac{\partial }{\partial z}\left(\frac{\partial A_{\rho }}{\partial \varphi }\right)\end{array}$

Therefore, the Equation (2) is re-written as follows,

$\begin{array}{c}\nabla \cdot \left(\nabla \times A\right)=\left[\begin{array}{l}\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\frac{\partial A_z}{\partial \varphi }\right)-\frac{\partial }{\partial \rho }\left(\frac{\partial A_{\varphi }}{\partial z}\right)-\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\frac{\partial A_z}{\partial \varphi }\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(\frac{\partial A_{\rho }}{\partial z}\right)\\ +\frac{\partial }{\partial \rho }\left(\frac{\partial A_{\varphi }}{\partial z}\right)-\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(\frac{\partial A_{\rho }}{\partial z}\right)\end{array}\right]\\ =0\end{array}$

Conclusion:

Thus, the divergence of the curl function $\nabla \cdot \left(\nabla \times A\right)=0$ in cylindrical coordinates and it is shown.