5 ms·
At least in some formulations (e.g. using bivectors and geometric algebra) you can treat a spatial translation as being a rotation about a point at infinity.
by antognini 4y ago
At least in some formulations (e.g. using bivectors and geometric algebra) you can treat a spatial translation as being a rotation about a point at infinity.
- ThrowawayTestr 4y agoIs that like saying sin(x)=x?
- joe__f 4y agoYou mean, for small x, sin(x) ~= x? I've been thinking for a while and I can't see a way that it's like that
- ThrowawayTestr 4y agoLike if you zoom in on a circle close enough it looks like a straight line.
- topaz0 4y agoThis is basically it, yes, along with cos(x) = 1. If the axis of rotation is much further than the distance that you move through upon rotation, the angle of rotation is very close to zero, so the approximations are very good. In the limit that the axis of rotation is infinitely far away, it is exact.
- hoseja 4y agoCoincidentally, a recent relevant XKCD: https://xkcd.com/2761/ https://xkcd.com/2761/
- afiori 4y agoIt is more like saying 1/∞ = 0. "Translations are rotations around infinity" is what you get if you compactify the real plane with projective geometry, 1/∞ = 0 is what you get when you close the complex field with an holomorphic-friendly point at infinity.
- gowld 4y agosort of... A line is a circle (more precisely, a cline ("circle-or-line") with infinite radius and center on the "horizon" (In 1D, the horizon is the 2 points at +/infinity. In 2-D, the horizon an infinite circular boundary around the plane. In 3-D, it's an infinite circular sphere). The center's position determines the slope/angle of the line. For a translational "rotation" in 3D, the axis of rotation is the line that is tangent to the infinite "celestial" sphere and perpendicular to the line of translational motion.
- mananaysiempre 4y ago(A line at infinity, of course, as rotations are about axes, not points, and in three dimensions an axis is a line.) As long as dimension n = 3 and so [dimension of R^n] n = n(n-1)/2 [dimension of so(n)]. Not to disregard these low-dimensional isomorphisms; there’s quite a few of them (they run out at about n = 10 IIRC), each one different from the next, and most are important in one way or another. But expressing momenta and angular momenta in those terms gives the impression that their existence is due to the (numerous and sometimes very complex) particulars of three-dimensional space, which it isn’t.
- rcme 4y agoYou can rotate around a line, it doesn't necessarily need to be an axis.
- mananaysiempre 4y agoI don’t really see your point, sorry? In arbitrary dimension, you rotate in an oriented plane (aka a bivector up to a positive constant) around a fixed point, if anything; I meant to call that an “axis” by abusing the terminology a bit (it’s an oriented line in three dimensions and a point plus an orientation, i.e. either “clockwise” or “counterclockwise”, in two). Not all isometries with a fixed point will be “rotations” in that sense, though—some will be products of them, as in / cos u sin u 0 0 \ | -sin u cos u 0 0 | | 0 0 cos v sin v | \ 0 0 -sin v cos v / with non-boring values of u and v.
- deleted 4y ago[deleted]
- danbruc 4y agoRotations of a point happen in a plane around some point in the plane, not around a line or axis. In two dimensions there is no line perpendicular to the plane to rotate around, in four and higher dimensions there is no unique plane perpendicular to any given line. It is only in three dimensions that rotating in a plane and around a line coincide.