Chapter 4, Problem 4.43P

(a)

To determine

For, the central force $F\left(r\right)$, $\nabla \times F\left(r\right)=0$

Answer to Problem 4.43P

For, the central force $F\left(r\right)$, $\nabla \times F\left(r\right)=0$ is proved.

Explanation of Solution

The expression for Central force can be expressed as,

    $\begin{array}{c}F\left(r\right)=\frac{f\left(r\right)}{r}\left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\\ =h\left(r\right)\left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\left(\because h\left(r\right)=\frac{f\left(r\right)}{r}\right)\end{array}$

To avoid difficulty in calculation $h\left(r\right)$ can be considered as $\frac{f\left(r\right)}{r}$, $h\left(r\right)=\frac{f\left(r\right)}{r}$.

Write the expression to calculate $\nabla \times F\left(r\right)$

$\begin{array}{c}\nabla \times F\left(r\right)=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{\text{k}}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ h\left(r\right)x& h\left(r\right)y& h\left(r\right)z\end{array}\right|\\ =\left(z\frac{\partial h\left(r\right)}{\partial y}-y\frac{\partial h\left(r\right)}{\partial z}\right)\widehat{i}+\left(x\frac{\partial h\left(r\right)}{\partial z}-z\frac{\partial h\left(r\right)}{\partial x}\right)\widehat{\text{j}}+\left(y\frac{\partial h\left(r\right)}{\partial x}-x\frac{\partial h\left(r\right)}{\partial y}\right)\widehat{z}\\ =\left(z\left(\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial y}\right)-y\left(\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial z}\right)\right)\widehat{i}+\left(x\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial z}-z\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial x}\right)\widehat{\text{j}}\\ +\left(y\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial x}-x\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial y}\right)\widehat{z}\\ =\left(z\left(\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial y}\right)-y\left(\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial z}\right)\right)\widehat{i}+\left(x\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial z}-z\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial x}\right)\widehat{\text{j}}\\ +\left(y\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial x}-x\frac{\partial h\left(r\right)}{\partial r}\frac{\partial r}{\partial y}\right)\widehat{z}\end{array}$        (I)

Write the expression magnitude of position vector,

$r=\sqrt{x^2+y^2+z^2}$

Differentiate with respect to $x$, $y$, and $z$.

$\begin{array}{c}\frac{\partial r}{\partial x}=\frac{\partial }{\partial x}\left(\sqrt{x^2+y^2+z^2}\right)\\ =\frac{1}{2}\frac{2x}{\left(\sqrt{x^2+y^2+z^2}\right)}\\ =\frac{x}{\left(\sqrt{x^2+y^2+z^2}\right)}\\ =\frac{x}{r}\end{array}$

$\begin{array}{c}\frac{\partial r}{\partial y}=\frac{\partial }{\partial y}\left(\sqrt{x^2+y^2+z^2}\right)\\ =\frac{1}{2}\frac{2y}{\left(\sqrt{x^2+y^2+z^2}\right)}\\ =\frac{y}{\left(\sqrt{x^2+y^2+z^2}\right)}\\ =\frac{y}{r}\end{array}$

$\begin{array}{c}\frac{\partial r}{\partial z}=\frac{\partial }{\partial z}\left(\sqrt{x^2+y^2+z^2}\right)\\ =\frac{1}{2}\frac{2z}{\left(\sqrt{x^2+y^2+z^2}\right)}\\ =\frac{z}{\left(\sqrt{x^2+y^2+z^2}\right)}\\ =\frac{z}{r}\end{array}$

Conclusion:

Substitute $\frac{x}{r}$ for $\frac{\partial r}{\partial x}$, $\frac{y}{r}$ for $\frac{\partial r}{\partial y}$, and $\frac{z}{r}$ for $\frac{\partial r}{\partial z}$ in (I) and solve.

$\begin{array}{c}\nabla \times F\left(r\right)=\left(z\left(\frac{\partial h\left(r\right)}{\partial r}\frac{y}{r}\right)-y\left(\frac{\partial h\left(r\right)}{\partial r}\frac{z}{r}\right)\right)\widehat{i}+\left(x\frac{\partial h\left(r\right)}{\partial r}\frac{z}{r}-z\frac{\partial h\left(r\right)}{\partial r}\frac{x}{r}\right)\widehat{\text{j}}\\ +\left(y\frac{\partial h\left(r\right)}{\partial r}\frac{x}{r}-x\frac{\partial h\left(r\right)}{\partial r}\frac{y}{r}\right)\widehat{z}\\ =0\end{array}$

Therefore, conservative $\nabla \times F\left(r\right)=0$ is proved.

(b)

To determine

To show $\nabla \times F\left(r\right)=0$ using the expression $\nabla \times F\left(r\right)$ for polar.

Answer to Problem 4.43P

$\nabla \times F\left(r\right)=0$ using the expression $\nabla \times F\left(r\right)$ for polar is shown.

Explanation of Solution

Use vector identity for $\nabla \times F\left(r\right)$ in spherical polar coordinates.

    $\begin{array}{c}\nabla \times F\left(r\right)=\left\{\begin{array}{l}\widehat{\text{r}}\frac{1}{r\sin \theta }\left(\frac{\partial }{\partial \theta }\left(\sin \theta f_{\varphi }\right)-\frac{\partial }{\partial \varphi }{f}_{\theta }\right)+\widehat{\theta }\left(\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }{f}_r-\frac{1}{r}\frac{\partial }{\partial r}\left(r{f}_{\varphi }\right)\right)\\ +\widehat{\varphi }\frac{1}{r}\left(\frac{\partial }{\partial r}\left(r{f}_{\varphi }\right)-\frac{\partial }{\partial \varphi }{f}_r\right)\end{array}\right\}\\ =\widehat{\theta }\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }{f}_r-\widehat{\varphi }\frac{1}{r}\frac{\partial }{\partial r}{f}_r\\ =0\end{array}$

Conclusion:

Therefore, conservative $\nabla \times F\left(r\right)=0$ is proved much quicker using spherical polar corodinates.