Chapter 6, Problem 6.1P

To determine

Length of the point joined on the sphere.

Answer to Problem 6.1P

Length is $L=R{\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}\sqrt{1+{\left(\sin \theta \right)}^2\varphi {\text{'}}^2\left(\theta \right)}}d\theta$

Explanation of Solution

Length of a short segment in spherical co-ordinate system $\left(R,\theta ,\varphi \right)$ is

    ${\left(ds\right)}^2={\left(dR\right)}^2+{\left(Rd\theta \right)}^2+{\left(R\sin \theta d\varphi \right)}^2$

Radius is constant on the surface of the sphere.

That is,

  $dR=0$

Therefore the length of short segment becomes,

    $\begin{array}{c}{\left(ds\right)}^2={\left(Rd\theta \right)}^2+{\left(R\sin \theta d\varphi \right)}^2\\ ds=R\sqrt{{\left(d\theta \right)}^2+{\left(\sin \theta d\varphi \right)}^2}\end{array}$        (I)

Represent $\varphi =\varphi \left(\theta \right)$

Then,

    $\begin{array}{c}d\varphi =\frac{d\varphi }{d\theta }d\theta \\ =\varphi \text{'}\left(\theta \right)d\theta \end{array}$        (II)

Substitute (II) in (I)

    $\begin{array}{c}ds=R\sqrt{{\left(d\theta \right)}^2+{\left(\sin \theta \right)}^2\varphi {\text{'}}^2\left(\theta \right)d{\theta }^2}\\ =R\sqrt{1+{\left(\sin \theta \right)}^2\varphi {\text{'}}^2\left(\theta \right)d{\theta }^2}\end{array}$        (III)

Find the total path length between the two points $\left({\theta }_1,{\varphi }_1\right)$ and $\left({\theta }_2,{\varphi }_2\right)$,

    $\begin{array}{c}L={\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}ds}\\ ={\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}R\sqrt{1+{\left(\sin \theta \right)}^2\varphi {\text{'}}^2\left(\theta \right)d{\theta }^2}}\\ =R{\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}\sqrt{1+{\left(\sin \theta \right)}^2\varphi {\text{'}}^2\left(\theta \right)d{\theta }^2}}\end{array}$

Conclusion:

Length is $L=R{\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}\sqrt{1+{\left(\sin \theta \right)}^2\varphi {\text{'}}^2\left(\theta \right)}}d\theta$