Explain why all geodesics on a sphere are arcs of great circ

[whatisreality]

Homework Statement

Show that any geodesic with constant $\theta$ lies on the equator of a sphere, with the north pole being on the $\theta = 0$ line. Hence explain why all geodesics on a sphere will be arcs of a great circle.

Homework Equations

The Attempt at a Solution

I've had a go at it, but I'm wondering if my reasoning on the second part is correct.

Previously in the question I showed that $\frac{d^2\theta}{d\sigma^2} = k^2 \frac{cos(\theta)}{sin^3{\theta}}$. For constant theta, $\frac{d^2\theta}{d\sigma^2} = 0$ so we require $cos(\theta)=0$, and therefore that $\theta = \frac{m\pi}{2} \pm m\pi$. If I take $\theta$ to be measured from the north pole, then for any $m$ the solution will lie on the equator.

Great circles are circles formed by the intersection of a plane containing the center of the sphere with the sphere itself. A plane through the centre of the sphere which also intersects with all points on the equator is a great circle. And the geodesics with $\theta = \frac{m\pi}{2} \pm m\pi$ are arcs of this circle.

Here's where it gets a bit sketchy... my thoughts are that the axes are defined arbitrarily. I could do some sort of co-ordinate transformation and repeat the calculation, with the same result, that geodesics on the new equator are arcs of a great circle. The equator would have moved, but there's nothing special about where the equator is. The above reasoning will still hold whatever direction I point the axes in, as long as the origin is the centre of the sphere. That means that whatever great circle I choose, the geodesics will be arcs of it. Because they might be arcs of constant theta in the new co-ordinate systems, but transforming back to the original system they'd be geodesics that don't necessarily have constant theta.

I'm really unsure about this and I'm not explaining it well at all. Is that reasoning correct or wildly wrong? Any help is very much appreciated, thank you!

 

[andrewkirk]

It will be easier to help you if you define your terms. What are $\theta, \sigma,k$?

Also, what definition of geodesic are you using? There is more than one. In most cases they are equivalent, but any proof will need to use the definition you've been given.

I also note that the question seems to presuppose that it is possible for a geodesic to have constant $\theta$, whereas typically that is something that would have to be proved, not just assumed. Of course, one needs to know the definition of geodesic and the meaning of $\theta$ in order to approach that.

 

[whatisreality]

It will be easier to help you if you define your terms. What are $\theta, \sigma,k$?

Also, what definition of geodesic are you using? There is more than one. In most cases they are equivalent, but any proof will need to use the definition you've been given.

I also note that the question seems to presuppose that it is possible for a geodesic to have constant $\theta$, whereas typically that is something that would have to be proved, not just assumed. Of course, one needs to know the definition of geodesic and the meaning of $\theta$ in order to approach that.

A geodesic has been defined as the straightest possible path, and we were given the geodesic equation:
$\frac{d}{d\sigma} (g_{\mu\nu}(x) \frac{dx^{\nu}}{d\sigma}) = \frac{1}{2} \frac{\partial g_{\rho\sigma}}{\partial x^{\mu}} \frac{dx^{\rho}}{d\sigma}\frac{dx^{\sigma}}{d\sigma}$.

k is a constant (of integration) and the definition of theta in the OP is as it was defined to us. Rephrased, theta = 0 is the z axis and where the +ve z axis intersects the sphere, that's the north pole.

 

[andrewkirk]

Thank you, that makes it clearer. Now, when you write

Previously in the question I showed that $\frac{d^2\theta}{d\sigma^2} = k^2 \frac{cos(\theta)}{sin^3{\theta}}$.

do you mean that you showed that that must be true for any geodesic on the sphere?

If so, and your proof of that equation is valid, then your first three paragraphs are correct and you can conclude that any geodesic with constant $\theta$ is a great circle.

Also, your reasoning about changing the coordinate systems is valid. That is, if there is any coordinate system in which a given geodesic has constant $\theta$ then the geodesic lies on the equator of that system, and hence is part of a great circle in any coordinate system.

What remains to be proven is that for any geodesic on a sphere, there is some spherical coordinate system in which it has constant $\theta$. It is not clear to me that that would be easy to prove.