Prove that a great circle is a geodesic

[miew]

Homework Statement

L = R $$\int \sqrt{1+ sin^2 \theta \phi ' ^ 2} d\theta$$ from theta 1 to theta 2

Using this result, prove that the geodesic (shortest path) between two given points on a sphere is a great circle. [Hint: The integrand f($$\phi,\phi',\theta$$) in the result is independent of phi so the Euler-Lagrange equation reduces to $$\partial f$$/$$\partial \phi '$$ = c, a constant. This gives you $$\phi '$$ as a function of theta. You can avoid doing the final integral by the following trick: There is no loss of generality in choosing your z axis to pass through the point 1. Show that with this choice the constant c is necessarily zero and describe the corresponding geodesics.

Homework Equations

The Attempt at a Solution

Doing some calculations, I ended up getting

$$\phi ' ^ 2$$ - A sin^2 $$\theta$$ $$\phi ' ^ 2$$= A^2. The hint tells me that I can take theta as 0, and so I get that phi_prime = A.

I integrate that, but I don't get anything that looks like a circle.

 

[Dickfore]

your integrand function does not depend on $\phi$ explicitly. That is why you can immediately perform one integration of the Euler's equation:

$$\frac{\partial F}{\partial \phi'} = C_{1}$$

Finding this partial derivative gives:

$$\frac{R \, \sin^{2}{(\theta)} \, \phi'}{\sqrt{1 + \sin^{2}{(\theta)} (\phi')^{2}}} = C_{1}$$

You can solve this equation by a parametric trick:

$$\sin{(\theta)} \, \phi' = \tan{p}$$

which reduces the equation to:

$$\sin^{2}{(\theta)} \, \phi' \, \cos{p} = C_{1}$$

where I redefined the arbitrary integrating constant $C_{1} \rightarrow R \, C_{1}$.

These equations give:

$$\sin{(\theta)} \, \cos{p} = C_{1} \, \cot{p} \Rightarrow \sin{(\theta)} = \frac{C_{1}}{\sin{p}}$$

and

$$\phi' \eqiuv \frac{d \phi}{d \theta} = \frac{\sin^{2}{p}}{C_{1} \, \cos{p}}$$

You need to find $\phi(p)$ and identify the equation of a great circle in spherical coordinates.