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.(adsbygoogle = window.adsbygoogle || []).push({});

Give an independent argument for the same conclusion by employing the formula

[tex]R=\frac{1}{\rho_1 \rho_2}[/tex] where [tex]\rho_1[/tex] and [tex]\rho_2[/tex] 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:

[tex]\frac{d^2\chi}{ds^2}+R\chi=0[/tex] 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 [tex]\frac{d^2\chi}{ds^2}=0[/tex] 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.

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# Question on MTW (not HW).

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