Timeline for continuous R^2xR^2xR^2/E^+(2) -> R^3 injection?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2013 at 0:39 | comment | added | Gil Bor | Actually, there is more then a sign change. The map I wrote is, in your notation, $(x,y,z) = (2s_1 s_2 cos\theta, 2s_1 s_2 sin\theta, s_1^2-s_2^2)$, i.e. there is an extra factor of 2 in the $x$ and $y$ coordinates, but not in the $z$ coordinate. This has the following effect: $p:\mathbb{C}^2\to\mathbb{R}^3$ maps the sphere $|v_1|^2+|v_2|^2=r^2$ onto the sphere $x^2+y^2+z^2=r^4.$ | |
| Jan 6, 2013 at 12:41 | vote | accept | Leon Avery | ||
| Jan 6, 2013 at 12:41 | |||||
| Jan 6, 2013 at 12:40 | comment | added | Leon Avery | In fact, this turns out to be identical (except for sign changes) to the map I worked out as a result of Eric Wofsey's answer. Let's call $s_i = |v_i|$ and $\theta = arg(v_2/v_1)$. Then it turns out to be $(x,y,z) = (s_1 s_2 cos\theta, s_1 s_2 sin\theta, s_2^2-s_1^2)$. This makes more physical sense if I use square roots: $(\sqrt{s_1 s_2} cos\theta, \sqrt{s_1 s_2} sin\theta, s_2-s_1)$, which, if z is a smooth function of time and the sampling interval is small, are roughly mean speed, radial acceleration, and tangential acceleration. | |
| Jan 6, 2013 at 2:32 | history | answered | Gil Bor | CC BY-SA 3.0 |