Timeline for Distance metric on the unit sphere in R^3?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 1, 2022 at 20:25 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://math.meta.stackexchange.com/a/34713/228959
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| Feb 15, 2011 at 20:05 | vote | accept | Aaron Mavrinac | ||
| Jun 23, 2010 at 21:48 | answer | added | Deane Yang | timeline score: 7 | |
| Jun 23, 2010 at 21:30 | answer | added | Daniel Barter | timeline score: 2 | |
| Jun 23, 2010 at 21:18 | comment | added | Steve Huntsman | You may also be having a hard time with the expressions $\partial/\partial x^j$, for which see (e.g.) people.hofstra.edu/stefan_Waner/diff_geom/Sec3.html | |
| Jun 23, 2010 at 21:12 | comment | added | Anton Petrunin | Essentially, you do not understand the definition. I would suggest to ask any geometer around --- it is very easy to explain by "talking", but by "writing" it is hard to add anything to the definition... | |
| Jun 23, 2010 at 21:12 | comment | added | Steve Huntsman | Also, you would probably find this better suited to another site mentioned in the FAQ. | |
| Jun 23, 2010 at 21:11 | comment | added | Steve Huntsman | Given a metric tensor with corresponding matrix $g_{jk}$ and (tangent) vectors $v = v^\ell e_\ell$ and $w = w^m e_m$ (using the Einstein convention in which paired upper and lower indices imply a summation), one has essentially by definition that $g(v,w) = g_{jk}v^j w^k$. For the case you are concerned with, the vectors $v$ and $w$ are the same, and are in fact the tangent vectors of a geodesic. I hope that the rest will be clear. In your case this is just giving arclengths of (portions of) great circles, which probably doesn't require this machinery. | |
| Jun 23, 2010 at 20:45 | history | asked | Aaron Mavrinac | CC BY-SA 2.5 |