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In the book "Gravitation" by Misner, Thorne and Wheeler, exercise 1.1 on page 44 reads:

Show that the Gaussian curvature $R$ of the surface of a cylinder is zero by showing that geodesics on that surface (unroll!) suffer no geodesic deviation. Give an independent argument for the same conclusion by employing the formula $R = 1/\rho_1 \rho_2$, where $\rho_1$ and $\rho_2$ are the principal radii of curvature at the point in question with respect to the enveloping Euclidean three-dimensional space.

I completed the first part of the exercise by "unrolling" the cylinder. This gives you a flat surface. Therefore, two initially parallel worldlines will remain parallel and the seperation, $\xi$ will be a non-zero constant:

$$\xi = k$$

So,

$$\frac{\textrm{d}^2 \xi}{\textrm{d}s^2} = 0$$

Now, substituting these values into the "equation of geodesic deviation" (Eqn. 1.6 on pg 30):

$$\frac{\textrm{d}^2 \xi}{\textrm{d}s^2} + R\xi = 0$$

We get:

$$Rk = 0$$

And since $k \neq 0$, then:

$$R=0$$

However, the problem I have is with the second part: Show that $R=0$ using $R = 1/\rho_1 \rho_2$. To me, it seems like either $\rho_1$ or $\rho_2$ must equal $\infty$ but I'm not sure why.

Is it because, say $\rho_1$ refers to the radius of curvature of the cross-section of the cylinder and $\rho_2$ refers to curvature of the tube itself and so, since it is straight, has an infinite radius? Is this effectively saying a cylinder is a torus with infinite radius?

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  • $\begingroup$ You are correct; a standard problem from an introduction to differential geometry. $\endgroup$ – Peter Diehr Jun 28 '18 at 20:10

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