- #1
ks_wann
- 14
- 0
So, I'm to show that in spherical coordinates, the length of a given path on a sphere of radius R is given by:
L= R[itex]\int_{\theta_1}^{\theta_2} \sqrt{1+\sin^2(\theta) \phi'^2(\theta)}d\theta[/itex],
where it is assumed [itex]\phi(\theta)[/itex], and start coordinates are [itex](\theta_1,\phi_1) [/itex]and [itex](\theta_2, \phi_2)[/itex].
I've tried starting by letting ds be a short path segment, such that
[itex]ds=R \sqrt{ d\theta^2+ d\phi^2} [/itex] and [itex]d\phi=\frac{d\phi}{d\theta}d\theta=\phi'(\theta) d\theta[/itex]
But this is obviously wrong, since I'll be missing a factor of sin^2(θ). If I draw the situation, it's also clear that I'll need the factor.
Is it just a sphercial coordinate concept I'm missing, or am I way off?
L= R[itex]\int_{\theta_1}^{\theta_2} \sqrt{1+\sin^2(\theta) \phi'^2(\theta)}d\theta[/itex],
where it is assumed [itex]\phi(\theta)[/itex], and start coordinates are [itex](\theta_1,\phi_1) [/itex]and [itex](\theta_2, \phi_2)[/itex].
I've tried starting by letting ds be a short path segment, such that
[itex]ds=R \sqrt{ d\theta^2+ d\phi^2} [/itex] and [itex]d\phi=\frac{d\phi}{d\theta}d\theta=\phi'(\theta) d\theta[/itex]
But this is obviously wrong, since I'll be missing a factor of sin^2(θ). If I draw the situation, it's also clear that I'll need the factor.
Is it just a sphercial coordinate concept I'm missing, or am I way off?