Geodesic of Sphere in Spherical Polar Coordinates (Taylor's Classical Mechanics)

[fehilz]

Homework Statement

"The shortest path between two point on a curved surface, such as the surface of a sphere is called a geodesic. To find a geodesic, one has to first set up an integral that gives the length of a path on the surface in question. This will always be similar to the integral (6.2) but may be more complicated (depending on the nature of the surface) and may involve different coordinates than x and y. To illustrate this, use spherical polar coordinates (r, \theta,\phi ) to show that the length of a path joining two points on a sphere of radius R is

L=R\int\sqrt{1+sin^2\theta\phi'(\theta)^2}d\theta (Don't know how to do it on latex but the integral is between \theta1 and \theta2)

if (\theta1,\phi1) and (\theta2,\phi2) specify two points and we assume that the path is expressed as \phi=\phi(\theta)."

Homework Equations

x=rsin(ϕ)cos(θ)
y=rsin(ϕ)sin(θ)
z=rcos(ϕ)

ds=\sqrt{dx^2+dy^2+dz^2}

The Attempt at a Solution

I'm unsure of how much this question is asking for. I was able to quickly work out the solution after looking up the line element for the surface of a sphere in spherical polar and using that in place of the Cartesian form of ds.

But then I was wondering if whether the question was asking me to derive the line element. It doesn't seem likely since this is one of the * questions which are supposed to be the easiest and I've already completed the ** questions with no difficulty.

In any case, I worked on deriving the line element for the sake of it and got stuck. I found the differentials for x,y and z (dx, dy and dz) and put them into the equation for ds. What do I do from here? Do I tediously expand the brackets involving three terms or is there something that I'm missing?

Thanks in advance.

 

[Oxvillian]

Probably you can just do this intuitively instead of diving into the algebra - just write down the formula for a distance element along a line of longitude on the sphere, and then along a line of latitude. Use Pythagoras to put them together and you're home and dry

 

[vela]

I think you need to adjust your expectations of what constitutes "tedious." When you expand, you'll get only 7 terms. Two of them will obviously cancel, and you can simplify the rest with one or two lines of algebra.

 

[andrien]

your distance formula will not hold here.