The length of a path on a sphere (in spherical coordinates)

[ks_wann]

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$\int_{\theta_1}^{\theta_2} \sqrt{1+\sin^2(\theta) \phi'^2(\theta)}d\theta$,
where it is assumed $\phi(\theta)$, and start coordinates are $(\theta_1,\phi_1)$and $(\theta_2, \phi_2)$.

I've tried starting by letting ds be a short path segment, such that
$ds=R \sqrt{ d\theta^2+ d\phi^2}$ and $d\phi=\frac{d\phi}{d\theta}d\theta=\phi'(\theta) d\theta$

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?

 

[voko]

If a curve is given by x(t), y(t), z(t), where t is a parameter, how is its length computed?

 

[ks_wann]

I'd use

$s=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}$

So I tried inserting x, y and z as spherical coordinates, and having θ as the parameter, in the end I end up with $\sqrt{R^2}$. That's after I take the derivative with respect to θ.

I'm not sure I got your hint.

 

[Goddar]

ds^2 = dx^2 + dy^2 + dz^2
To translate this into spherical coordinates in the form:
ds^2 = a*dr^2 + b*dθ^2 + c*dϕ^2,
you need the Jacobian matrix for the transformation, which will give you a, b and c. The result is rather well known, but if you're not familiar with the technique, i would recommend you look it up. Your problem is straightforward from there.

 

[voko]

I'd use

$s=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}$

So I tried inserting x, y and z as spherical coordinates, and having θ as the parameter, in the end I end up with $\sqrt{R^2}$. That's after I take the derivative with respect to θ.

I'm not sure I got your hint.

Your approach is correct in principle, so there must be an error in details. Unless you show all the steps, I can't help.

 

[Chestermiller]

It's a matter of geometry. As you approach the poles, the lines of constant longitude get closer together, and as you approach the equator, they get further apart. You need to draw a diagram to see why geometrically there is a factor of sin²θ.

 

[ks_wann]

ds^2 = dx^2 + dy^2 + dz^2
To translate this into spherical coordinates in the form:
ds^2 = a*dr^2 + b*dθ^2 + c*dϕ^2,
you need the Jacobian matrix for the transformation, which will give you a, b and c. The result is rather well known, but if you're not familiar with the technique, i would recommend you look it up. Your problem is straightforward from there.

I've looked it up, but it puzzles me how I go from $ds^2=dx^2+dy^2+dz^2$ when dx, dy and dz are added together. I mean, if it was $ds^2=dx^2 dy^2 dz^2$ it would be straight forward for me.

abs[∂(x,y,z)/∂(r,θ,phi)] * dr dθ dphi

Can I simply transform each on its own? dx, dy and dz?

 

[ks_wann]

Your approach is correct in principle, so there must be an error in details. Unless you show all the steps, I can't help.

With $\theta$ as the parameter, I insert $x=r\sin\theta\cos\phi$, $y=r\sin\theta\sin\phi$ and $z=r\cos\theta$.

This would give me
$\sqrt{(\frac{d}{d\theta}r\sin\theta\cos\phi)^2+(\frac{d}{d\theta}r\sin\theta\sin\phi)^2+(\frac{d}{d\theta}r\cos\theta)^2}$

From there I end up with

$\sqrt{r^2(\cos^2\theta\cos^2\phi+cos^2\theta\sin^2\phi+\sin^2\theta)}$

Using identities it reduces to $\sqrt{r^2}$..

 

[voko]

Do you understand that, for example, $$ \frac {d} {d\theta} \sin \theta \cos \phi = \cos \theta \cos \phi - \sin^2 \theta \phi'(\theta) $$

 

[ks_wann]

Do you understand that, for example, $$ \frac {d} {d\theta} \sin \theta \cos \phi = \cos \theta \cos \phi - \sin^2 \theta \phi'(\theta) $$

No, I can't figure out the last term on the right.

 

[dreamLord]

Voko, is your 2nd term correct? In my opinion it ought to be " -sin(theta)*sin(phi)*(dphi/dtheta)"

 

[voko]

Darn, brain still half asleep :)

Indeed that should have been $$ \cos \theta \cos \phi - \sin \theta \sin \phi \phi'(\theta) $$

 

[ks_wann]

I see what you did there. So I should treat $\phi(\theta)$ properly in my derivatives.

 

[ks_wann]

Well, yes that was fairly easy. Thanks a lot. :)

 

[Chestermiller]

Here's a simpler way. Draw a circle. Draw the z-axis vertically through the center of the circle. Pick a point P on the upper half of the circumference. Draw a radius from the center of the circle to the point on the circumference. Label the angle θ between the radius and the z axis. Draw a horizontal line from the point P to the z axis. The length of this line is R sin θ. So you see how this length is related to longitudinal distance around the z axis?