3
$\begingroup$

Hi guys,

I have a simple question

Linear movement can be found in 1D, 2D and 3D world objects

Rotation can be found in 2D and 3D world objects.

Now, are there any kind of motion can only be found in 3D world objects? This kind of motion should be as simple as linear movement or rotation, and very basic that other 3D motions can be described based upon.

If there is not, why -

  1. There are only 2 kinds of basic movements?
  2. perhaps there are more advanced movements in 4D or higher dimensions?
Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ The notion of direct isometries formalises your notion of rigid body motions, and it is known for 3 dimensions at least that every direct isometry is the composition of a translation and at most 3 rotations - see, for example, the Wikipedia article about Euler angles. Analogously, in n dimensions, every direct isometry is the composition of a translation and at most n rotations. So — in short, the answer is no. $\endgroup$
    – Zhen Lin
    Commented Dec 22, 2010 at 14:33

1 Answer 1

Reset to default
18
$\begingroup$

I interpret your question to be asking for a description of 1-parameter groups of isometries.

The generic kind of motion in 3 dimensions is a screw-motion, which is translation along an axis combined with rotation about the same axis. Up to changing the speed of a parametrization, translations and rotations in 2 and 3 dimensions are all alike, but screw motions have an invariant, the pitch of the screw: how many complete turns are made with one unit of translation along the axis? Rotation is the limiting case pitch $\rightarrow 0$, while translation is the limiting case pitch $\rightarrow \infty$.

In 4 dimensions, in addition to rotations, translations and screw motions, there are compound rotations, which involve rotating about one 2-plane at one speed while rotating about the perpendicular plane at another speed. In 5 dimensions, there is also a compound screw-motion, which combines translation along a 1-dimensional axis with compound rotation in the perpendicular 4-plane.

The general pattern is similar. In dimension n, you can take any collection of mutually orthogonal 2-planes intersecting at a point and independently rotate each of them, and you can independently combine these rotations with translation in any direction that is a common perpendicular.

Improve this answer
$\endgroup$
4
  • $\begingroup$ One small point, given that the question was phrased in physical terms, is that this answer assumes that the space is euclidean. It would be more realistic to consider particle motion in Minkowski spacetime, in which we are talking about one-parameter subgroups of the Poincaré group instead. The answer there is a little different. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 22, 2010 at 14:56
  • $\begingroup$ +1 Ah, I hadn't thought about this interpretation of the question — I interpreted it as asking whether every rigid body motion consists of some composition of translations and rotations. Another question which could have been asked is the classification of all (direct) isometries up to conjugacy. $\endgroup$
    – Zhen Lin
    Commented Dec 22, 2010 at 15:00
  • $\begingroup$ thanks. I didn't get some part of the answers, but I understand "screw". It all seems so natural now :) $\endgroup$
    – est
    Commented Dec 22, 2010 at 15:03
  • 1
    $\begingroup$ @Zhen Lin: I wasn't sure which of these two questions was intended, so I made my best guess. Direct motions up to conjugacy is similar. One can first look at the derivative $d\phi$ of a direct isometry $\phi$, an element of $SO(n)$, for which there is an orthogonal splitting of $\mathbb E^n$ into the +1 eigenspace, the -1 eigenspace, and even-dimensional characteristic subspaces for some set of rotations. If the +1 eigenspace is trivial, $\phi$ has a fixed point and $\phi$ is a compound rotation. In general, $\phi$ mod $+1$-eigenspace has a fixed point, and it may translate as well as rotate. $\endgroup$
    – Bill Thurston
    Commented Dec 22, 2010 at 17:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .