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Can you compute geodesics by treating it as a problem where you want to minimize the length of a curve through two points on a specified surface while having the constraint that the curve must reside on the specified surface?

If so, can you explain how one could do so for a cylinder?

Or is the calculus of variations the only method by which to do so?

I tried setting up the constrained optimization problem for a cylinder but was unable to make any progress, leading me to think that maybe it's impossible to solve the problem without the calculus of variations.

Simon M
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  • Since you can cut the cylinder in a rectangle the geodesics are the straight lines on the rectangle, so why do any calculation? – trula Aug 07 '23 at 13:42
  • Although I appreciate explanations with freshman calculus terms, I appreciate learning new terms and concepts. I can give you some directions on differential geometry regarding Levi Civita procedure of parallel translation on surfaces along its tangent space for each point. Are you interested? – Bruno Lobo Aug 07 '23 at 14:23
  • I thought that a geodesic on a cylinder is either a vertical line, a circle, or a helix? – Simon M Aug 07 '23 at 14:40
  • @BrunoPeixoto yes, please. I'm mostly comfortable with differential geometry but there may be a few things that I'll need to brush up on – Simon M Aug 07 '23 at 14:41
  • @SimonM . Correct. And if you cut a vertical line into that cylinder and unfold it to make it flat what lines do you get ? – Kurt G. Aug 07 '23 at 14:50
  • The geodesics are all helices. Vertical line, circle, helix are special cases on cylinder surface . In particular the helix on a cylinder of constant radius $R$ makes constant angle $\gamma$ to vertical line. [In general for non- cylinders Clairaut's geodesic Law holds, it states product $ R \cdot \sin \gamma $ remains constant, equal to the nearest distance it can get to symmetry axis.] – Narasimham Aug 07 '23 at 14:56
  • @KurtG. The cylinder flattens out to a rectangle and all geodesics look like straight lines on the rectangle. Ok, I get it now. Thank you – Simon M Aug 07 '23 at 15:10
  • As to why one should even do any calculation for the cylinder, wouldn't you need to use numbers in order to find the geodesic through two particular points? I'm not just interested in the general shape of the geodesic but rather the precise equation of the geodesic through two specified points – Simon M Aug 07 '23 at 17:04
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    The geodesics emerges from the Levi-Civitta connection of the tangent curve respective to itself, which is a metric-preserving metric with zero normal component on a manifold. I am not familiar with good textbooks, but I think the following articles may suffice: https://mathworld.wolfram.com/Levi-CivitaConnection.html , https://en.wikipedia.org/wiki/Levi-Civita_connection – Bruno Lobo Aug 07 '23 at 23:25

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The calculus of variations is perhaps the most natural way of defining geodesics on a surface (via extremality of length or energy), but there is an easier definition that is sometimes more direct in specific cases. Thus, if a surface in $\mathbb R^3$ has a regular parametrisation $X(u^1,u^2)$ and a curve $\alpha(s)=(\alpha^1(s),\alpha^2(s))$ is regular, then the resulting curve $\beta(s)=X(\alpha(s))$ will be a geodesic if and only if the second derivative $\beta''(s)$ is normal to the surface. This can be easily shown to be equivalent to the usual equations ${\alpha^k}'' +\Gamma_{ij}^k {\alpha^i}'{\alpha^j}'=0$ for $k=1,2$.

Mikhail Katz
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