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November 21

Rigid motions of the unit sphere in 3-space

Hello reference desk. I'm looking at the rigid motions of a unit sphere in 3-space, and there are some things I'm unclear of.

I know that any rigid motion can be thought of as either a rotation followed by a translation, or a translation followed by a rotation. It seems reasonable to think that rigid motions of the sphere can also be characterised by a rotation about an axis through the sphere's centre, then a translation. I can picture this easily enough - if we rotate the sphere around an arbitrary axis A, we can get the sphere to the same location and orientation by first rotating the sphere about an axis parallel to A, but passing through the sphere's centre, to get the sphere to the right orientation, then translating to get the sphere to have the correct centre. I'm having difficulty proving this rigorously, however. I can't even see if we are rotating in the second case by the same angle as the first, in general (obviously we are if it is by 180 degrees).
I'm also not sure how to represent a rotation about an axis that does not pass through the origin (e.g. an axis passing through the centre of a sphere not centred at the origin) - the Euler axis/angle method, or even just the DCM 3x3 matrix, require the axis to pass through the origin. I want to be able to carry such a rotation out by translating the sphere's centre to the origin, carrying out the rotation (i.e. rotate around the translated axis by the angle we want), then undo the translation. Again, I can see this conceptually, but can't prove it rigorously. We're basically conjugating a rotation by a translation, and wanting this to bring the sphere to the same location/orientation as the original rotation...but I can't see how to write this argument down. Is this really simple, and I just can't see it? Obviously I want $\mathbf {x}$ to be sent by rotation about an axis passing through sphere centre $\mathbf {v}$ by some angle to be sent to $N(\mathbf {x} -\mathbf {v} )+\mathbf {v}$ , where $N$ is the rotation by the same angle, about an axis parallel to the original but passing through the origin, but I can't see how I can set up an equation, as the original rotation cannot be expressed as a 3x3 matrix.

Am I over-complicating this? I'm just having a hard time getting my head around this - I've just never really thought about how to represent rotations whose axes do not pass through the origin. Thanks in advance, Icthyos (talk) 16:45, 21 November 2009 (UTC)[reply]

I'd have a look at SO(3) for a straightforward version and Quaternions and spatial rotation for something I rather like. Translating to the origin and back again is fairly straightforward. Dmcq (talk) 17:57, 21 November 2009 (UTC)[reply]

Thanks for the speedy reply. I'm actually using unit quaternions to represent the rotations - I agree, it's a nice, and at first surprising, representation. I'll try to rephrase my issue a bit more succinctly: in the SO(3) article, it talks about representing the axis of rotation as a unit vector,

\mathbf {n}

,so we we think of the axis along the directed line segment from the origin to the point

\mathbf {n}

. My problem is, how do I represent a rotation around an axis that does not pass through the origin? I'm probably just not understanding something, since it says in the aforementioned article: Every nontrivial proper rotation in 3 dimensions fixes a unique 1-dimensional linear subspace of $\mathbf {R} ^{3}$ which is called the axis of rotation - obviously, an axis that does not pass through the origin is not a linear subspace. Do we need to undergo a a change of coordinate system, then undo this change - essentially what we are doing when we translate? As a concrete example, how would I represent the rotation around the axis passing through (0,0,1) and (1,0,0) by 90 degrees? Thanks, Icthyos (talk) 18:47, 21 November 2009 (UTC)[reply]

When I read "rigid motions of the sphere", I assume you're talking about maps that take the sphere into itself. Any such map is fairly clearly a rotation about an axis through the origin. Are you using the term in a wider sense? Algebraist 18:50, 21 November 2009 (UTC)[reply]

Technically, I'm working with the groupoid of rigid motions of the sphere, in which information about the location of the sphere is included in each element - we only have a partial multiplication, since two elements can only be composed if the motion of the first takes the sphere to the position at which the second requires it to be. If we represent an element by

(\mathbf {v} ,\phi )

where

\mathbf {v}

is the centre of the sphere and

\phi

is the rigid motion we are going to apply, then

(\mathbf {v'} ,\phi ')(\mathbf {v} ,\phi )=(\mathbf {v} ,\phi '\phi )

but is only defined if

\phi (\mathbf {v} )=\mathbf {v'}

. So, I might conceivably want an element which rotates the sphere with centre (5,0,0) around the axis through (0,0,1) and (1,0,0), and I'm not sure how to represent this. Icthyos (talk) 19:04, 21 November 2009 (UTC)[reply]

Using SO(3) if you add an extra 1 at the end of coordinates so they become (x,y,z,1) and then use a 4×4 unit matrix with a,b,c,1 in the last column then you translate the coordinates to (x+a,y+b,z+c,1). I had a quick look but didn't see the trick here and it's only really worth it because hardware often deals with that size nicely. A rotation round an arbitrary axis is then a translate, rotate, translate back again. Dmcq (talk) 19:39, 21 November 2009 (UTC)[reply]

See this. It explains how it works, and how you can describe any affine transformation in this way.--Leon (talk) 19:45, 21 November 2009 (UTC)[reply]

Wow, thanks - that's really helpful. That should simplify things for me, at least when composing elements. I'm still looking for a proof though that a rotation about an arbitrary axis has the same result as translate-rotate-'untranslate' - I can't think what notation to use, even though the picture is so clear in my head. Icthyos (talk) 23:03, 21 November 2009 (UTC)[reply]

There's no need to worry. Why not choose your coordinate system so that the origin is the centre of the sphere?
The Lie group of Euclidean transformations acting on R³ is given by SO(3,R) ⋉ R³ where ⋉ denotes the semi-direct product. SO(3,R) gives the rotations and R³ gives the translations. We can embed SO(3,R) ⋉ R³ in the general linear group GL(4,R) as follows: Let M ∈ SO(3,R) and τ ∈ R, then we have

{\mbox{SO}}(3,\mathbf {R} )\ltimes \mathbf {R} \ni (M,\tau )\mapsto \left({\begin{array}{c|c}1&0\\\hline \tau &M\end{array}}\right)\in {\mbox{GL}}(4,\mathbf {R} ).

So, to compute the composition of two Euclidean transformations, say (M,τ) with (N,t), we simply calculate the (block) matrix product:

(N,t)\cdot (M,\tau )\mapsto \left({\begin{array}{c|c}1&0\\\hline t&N\end{array}}\right)\left({\begin{array}{c|c}1&0\\\hline \tau &M\end{array}}\right)=\left({\begin{array}{c|c}1&0\\\hline t+N\tau &NM\end{array}}\right)

and so it follows that for M, N ∈ SO(3,R) and τ, t ∈ R we have (N,t)·(M,τ) = (NM,t + Nτ). ~~ Dr Dec (Talk) ~~ 14:18, 22 November 2009 (UTC)[reply]

A number with all digits random is random

Suppose (X_i) is a sequence of independent random variables, each variable in discrete uniform distribution from {0,1}. How to prove that $\sum _{i}X_{i}2^{-i}$ has uniform continuous distribution on [0,1]? 212.87.13.71 (talk) 19:02, 21 November 2009 (UTC)[reply]

Looks like homework. Want to tell us how far you've gotten, and where you got stuck? --Trovatore (talk) 20:13, 21 November 2009 (UTC)[reply]

This recent post may give you a hint. --pma (talk) 18:16, 22 November 2009 (UTC)[reply]

Integral

Hi. Could someone suggest to me how to determine $\int {\frac {e^{-2x^{2}}}{x}}$ ? I've tried doing it via integration by parts but that leads nowhere, I considered writing the exponential as sinhs and coshs but that didn't help and I tried rewriting it slightly so that the exponential was multiplied by the derivate of teh index so I could integrate that but again, nothing useful. Thanks 131.111.216.150 (talk) 19:54, 21 November 2009 (UTC)[reply]

Exponential integral 131.111.248.99 (talk) 20:10, 21 November 2009 (UTC)[reply]

Let's see:

\int {\frac {e^{-2w^{2}}}{w}}\,dw={\frac {1}{2}}\int {\frac {e^{-2w^{2}}}{-2w^{2}}}\,(-4w\,dw)={\frac {1}{2}}\int {\frac {e^{u}}{u}}\,du

and we'd like the bounds on the last integral to go from −∞ to x in order that this fit into the topic of the article titled exponential integral. But −2w² = x seems to rule out x > 0. Michael Hardy (talk) 05:42, 22 November 2009 (UTC)[reply]