Timeline for How to convince a student without calculus that great circles are geodesics in a sphere?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 27, 2023 at 12:59 | comment | added | Mikhail Katz | Fully guilty of an esprit d'escalier I furnished a proof here. | |
| Aug 24, 2023 at 17:29 | comment | added | David E Speyer | That's good! Thanks! @MikhailKatz | |
| Aug 24, 2023 at 12:30 | comment | added | Mikhail Katz | Well, if you are willing to admit that the path is piecewise-geodesic, then consider the triple $x_{i-1}, x_i, x_{i+1}$, drop a perpendicular from $x_i$ to the point $a$ on the geodesic arc $x_{i-1}x_{i+1}$. Then by the theorem of cosines for the right-angle triangle, the arc $x_{i-1}a$ is shorter than $x_{i-1}x_i$ and similarly for the other one. Then one argues inductively. | |
| Aug 24, 2023 at 10:57 | comment | added | David E Speyer | I don't think so? The arc length of a path is the limit of the lengths of the best piecewise linear approximations to that path, but it is not the limit of the best right angled piecewise linear approximations. See math.stackexchange.com/questions/12906/… . @MikhailKatz | |
| Aug 24, 2023 at 6:13 | comment | added | Mikhail Katz | If one can reduce this somehow to only using right-angle spherical triangles, the argument might look simpler. | |
| Aug 24, 2023 at 1:01 | history | answered | David E Speyer | CC BY-SA 4.0 |