Can Geodesic Deviation Prove the Gaussian Curvature of a Cylinder is Zero?

[MathematicalPhysicist]

The question is:"Show that the Guassian curvature R of the surface of a cylinder is zero by showing that geodesics on that surface suffer no geodesic deviation.
Give an independent argument for the same conclusion by employing the formula
$$R=\frac{1}{\rho_1 \rho_2}$$ where $$\rho_1$$ and $$\rho_2$$ are the principal radii of curvature at the point in question wrt the enveloping euclidean 3-dimensional space."

Now if I write down the deviation geodesic equation I get:
$$\frac{d^2\chi}{ds^2}+R\chi=0$$ where chi is the distance between geodesics, now because the cylinder has a quasi rectangular shape, the geodesics which start parallel stay parallel thus there is no geodesic deviation, and R=0 cause $$\frac{d^2\chi}{ds^2}=0$$ and xsi is linear wrt s chi=as+b so R=0.
Is this just plain mambo jambo from my behalf or there's something genuine here?

I am not sure about the second argument, which point is in question here?

Any hints are appreciated.

 

[MathematicalPhysicist]

Anyone?

 

[MathematicalPhysicist]

Any hints?

 

[MathematicalPhysicist]

Can someone move my post to advanced physics HW?, perhaps there my post will get my attention it deserves, or not.

 

[Sam Park]

The question is:"Show that the Guassian curvature R of the surface of a cylinder is zero by showing that geodesics on that surface suffer no geodesic deviation." Now if I write down the deviation geodesic equation I get:
$$\frac{d^2\chi}{ds^2}+R\chi=0$$ where chi is the distance between geodesics, now because the cylinder has a quasi rectangular shape, the geodesics which start parallel stay parallel thus there is no geodesic deviation, and R=0 cause $$\frac{d^2\chi}{ds^2}=0$$ and xsi is linear wrt s chi=as+b so R=0. Is this just plain mambo jambo from my behalf or there's something genuine here?

It's a bit mumbo-jumboish, because you've essentially just re-stated the premise of the question. MTW ask you to show that geodesics on the surface of a cylinder suffer no geodesic deviation, and your answer is "because the cylinder has a quasi rectangular shape, the geodesics which start parallel stay parallel thus there is no geodesic deviation". It isn't clear what you mean when you say "the cylinder has a quasi rectangular shape", nor have you explained why geodesics on a surface with such a shape suffer no geodesic deviation.

Maybe what you mean is that we can "unroll" a cylinder and lay it flat on a table, without changing any of the internal surface distances, so it is a metrically flat surface, but this is really a third argument for flatness. It depends on what kind of answer you want. Most likely MTW expect you to determine the geodesic equations for a cylindrical surface and show explicitly that there is no geodesic deviation. That would be "something genuine".

[MTW also ask:] "Give an independent argument for the same conclusion by employing the formula
$$R=\frac{1}{\rho_1 \rho_2}$$ where $$\rho_1$$ and $$\rho_2$$ are the principal radii of curvature at the point in question wrt the enveloping euclidean 3-dimensional space." I am not sure about the second argument, which point is in question here?

Well, the minimum radius of a cylindrical surface is the usual raduis, whereas the radius perpendicular to that is infinite, and 1/infinity = 0, so the Gaussian curvature of the surface is zero.

 

[MathematicalPhysicist]

Thank you for your post, I thought this thread of mine will get lost.

Anyway, back to topic, so I gather from your reply that I need to find parametric equation of the cylinder and from there show that the deviation is null, correct?

Well finding this parametric coordinates of the cylinder wouldn't be a problem.

If I have more questions I will ask them later one, thanks again for your reply.