- #1
ephedyn
- 170
- 1
Homework Statement
Using spherical coordinates, show that the length of a path joining two points on a sphere is
[itex]L=\int_{\theta_{1}}^{\theta_{2}}\sqrt{1+\sin^{2}\theta\phi'^{2}}d\theta[/itex]
Homework Equations
[itex]x=r\cos\theta\sin\phi[/itex]
[itex]y=r\sin\theta\sin\phi[/itex]
[itex]z=r\cos\phi[/itex]
The Attempt at a Solution
The distance between 2 neighboring points on the path is
[itex]ds=\sqrt{dx^{2}+dy^{2}+dz^{2}}[/itex]
and
[itex]dx=\dfrac{dx}{d\theta}d\theta[/itex]
[itex]dy=\dfrac{dy}{d\theta}d\theta[/itex]
[itex]dz=\dfrac{dz}{d\theta}d\theta[/itex]
Consider
[itex]\dfrac{1}{r}\dfrac{dx}{d\theta}=\cos\theta\cos\phi\phi'\left(\theta\right)-\sin\theta\sin\phi[/itex]
[itex]\dfrac{1}{r}\dfrac{dy}{d\theta}=\sin\theta\cos\phi\phi'\left(\theta\right)+\sin\phi\cos\theta[/itex]
[itex]\dfrac{1}{r}\dfrac{dz}{d\theta}=-\sin\phi\phi'\left(\theta\right)[/itex]
[itex]\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[/itex]
[itex]\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[/itex]
[itex]\left(\dfrac{1}{r}\dfrac{dz}{d\theta}\right)^{2}=\sin^{2}\phi\phi'^{2}[/itex]
Summing, [itex]dy^{2}[/itex] and [itex]dx^{2}[/itex], the term [itex]2\sin\theta\cos\theta\sin\phi\cos\phi\phi[/itex] cancels out, leaving
[itex]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}[/itex]
[itex]L=r\int_{\theta_{1}}^{\theta_{2}}\sqrt{\phi'^{2}+\sin^{2}\phi}d\theta[/itex]
Any idea where I went wrong?