# The Great Circle et al

1. Aug 29, 2006

### Feles Cestriana

The Great Circle et al....

It is certainly intuitively clear that the geodesics on the sphere are the great circles. But, showing that seems to be a bit tricky...

I tried working out the Christoffel symbols, and then plugging that into the differential equations for the geodesics, and aside from it being a bit of a mess, I'm up against a bit of a wall with the system.

Now, perhaps I just need to grind it a little more, and it will all fall out, but, I can't help but wonder if there is a better way to "show" what the geodesics of a surface are.

I've had the same issue with the cylinder...

Any tips, to just point me in the right direction, I'd be most appreciative.

Thanks,

-FC

2. Aug 29, 2006

### chroot

Staff Emeritus
Wouldn't it be easier to just show that great circles satisfy the geodesic equation?

- Warren

3. Aug 29, 2006

### Feles Cestriana

Hmmm. Yes, I suppose it might be. Thank you. I will try that.

But I think I will have to make some geometric arguments to show that those are all the geodesics.

Thanks for the tip. Is this the typical method of attack for these sorts of problems?

Thanks again.

-FC

4. Aug 29, 2006

### Feles Cestriana

Hmmm... okay, well I'm still having a bit of trouble. Are there any slightly more explicit tips to be had? I'm a bit lost in the symbology, and seem to be unable to calculate an answer here.

Thanks,

-FC

5. Aug 31, 2006

### Feles Cestriana

oooookay....

Well, since no one has given any further assistance, I had to figure this out for myself.

And just in case someone else out there reads this thread in hopes of finding something out, I'll post my findings, so this was not all in vain.

For a 2-dimentional surface in R^3, all geodesics will have second derivatives normal to the surface.

This is the easiest way to deal with geodesics on a surface in R^3.

Note that this doesn't work in higher dimentions.

I hope you all have better luck than I have had trying to figure this stuff out.

-FC