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ephedyn
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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?
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