Path joining 2 points on a sphere

[ephedyn]

Homework Statement

Using spherical coordinates, show that the length of a path joining two points on a sphere is

$L=\int_{\theta_{1}}^{\theta_{2}}\sqrt{1+\sin^{2}\theta\phi'^{2}}d\theta$

Homework Equations

$x=r\cos\theta\sin\phi$
$y=r\sin\theta\sin\phi$
$z=r\cos\phi$

The Attempt at a Solution

The distance between 2 neighboring points on the path is
$ds=\sqrt{dx^{2}+dy^{2}+dz^{2}}$

and
$dx=\dfrac{dx}{d\theta}d\theta$
$dy=\dfrac{dy}{d\theta}d\theta$
$dz=\dfrac{dz}{d\theta}d\theta$

Consider
$\dfrac{1}{r}\dfrac{dx}{d\theta}=\cos\theta\cos\phi\phi'\left(\theta\right)-\sin\theta\sin\phi$
$\dfrac{1}{r}\dfrac{dy}{d\theta}=\sin\theta\cos\phi\phi'\left(\theta\right)+\sin\phi\cos\theta$
$\dfrac{1}{r}\dfrac{dz}{d\theta}=-\sin\phi\phi'\left(\theta\right)$
$\left(\dfrac{1}{r}\dfrac{dx}{d\theta}\right)^{2}=\cos^{2}\theta\cos^{2}\phi\phi'{}^{2}-2\sin\theta\cos\theta\sin\phi\cos\phi\phi'+\sin^{2}\theta\sin^{2}\phi$

$\left(\dfrac{1}{r}\dfrac{dy}{d\theta}\right)^{2}=\sin^{2}\theta\cos^{2}\phi\phi'^{2}+2\sin\theta\cos\theta\sin\phi\cos\phi\phi+\sin^{2}\phi\cos^{2}\theta$

$\left(\dfrac{1}{r}\dfrac{dz}{d\theta}\right)^{2}=\sin^{2}\phi\phi'^{2}$

Summing, $dy^{2}$ and $dx^{2}$, the term $2\sin\theta\cos\theta\sin\phi\cos\phi\phi$ cancels out, leaving
$L={\displaystyle r\int_{\theta_{1}}^{\theta_{2}}\sqrt{\cos^{2}\theta\cos^{2}\phi\phi'^{2}+\sin^{2}\theta\cos^{2}\phi\phi'^{2}+\sin^{2}\theta\sin^{2}\phi+\sin^{2}\phi\cos^{2}\theta+\sin^{2}\phi\phi'{}^{2}}d\theta}$

$L=r\int_{\theta_{1}}^{\theta_{2}}\sqrt{\phi'^{2}+\sin^{2}\phi}d\theta$

Any idea where I went wrong?

 

[gabbagabbahey]

Homework Statement

Using spherical coordinates, show that the length of a path joining two points on a sphere is

$L=\int_{\theta_{1}}^{\theta_{2}}\sqrt{1+\sin^{2}\theta\phi'^{2}}d\theta$

The Attempt at a Solution

$L=r\int_{\theta_{1}}^{\theta_{2}}\sqrt{\phi'^{2}+\sin^{2}\phi}d\theta$

Any idea where I went wrong?

Looks to me like you are just using a different naming convention for the angles than the author of the problem statement. Try it again using $\theta$ as the polar angle and $\phi$ as the azimuthal angle and you should get the desired result (assuming the missing factor of $r$ was just a typo).

 

[tiny-tim]

hi ephedyn!

(have a theta: θ and a phi: φ and a square-root: √ and an integral: ∫ )

as gabbagabbahey says …

√(1 + sin²θ (dφ/dθ)²) dθ

= √(1 + sin²θ (dφ/dθ)²) dθ/dφ dφ

= √((dθ/dφ)² + sin²θ) dφ