Rigid motions

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February 25th 2010, 01:21 PM
Pinkk

Rigid motions

If $f$ is any rigid motion and $\tau$ is any translation, identity the rigid motion $\tau \circ f \circ \tau^{-1}$

I thought it was simply the rigid motion $f$ but after messing around a bit it's obviously not, so I have no idea.
February 25th 2010, 06:34 PM
cribby

Quote:

Originally Posted by Pinkk

If $f$ is any rigid motion and $\tau$ is any translation, identity the rigid motion $\tau \circ f \circ \tau^{-1}$

I thought it was simply the rigid motion $f$ but after messing around a bit it's obviously not, so I have no idea.

Its been a while, so I hope someone else chimes in on this to confirm or correct what I'm about to say, but any rigid motion can be expressed as a composition of rotations and translations.

Now, IF I'm right about that, then you can assume that $f$ is either a rotation or a translation because if its a composition, we can take it apart. For instance, if $f=r \circ t$ where $r$ is some rotation and $t$ is some translation, we can see that $\tau \circ f \circ \tau ^{-1} = \tau \circ (r \circ t) \circ \tau ^{-1} = \tau \circ r \circ (\tau ^{-1} \circ \tau) \circ t \circ \tau ^{-1}= (\tau \circ r \circ \tau ^{-1}) \circ (\tau \circ t \circ \tau ^{-1})$ and from that point we can just worry about what happens to either the rotation or translation and then combine them later.

That said, my best guess as to what's being asked in this problem is just to say what type of motion $\tau \circ f \circ \tau ^{-1}$ is, and that's going to depend on what type of motion $f$ itself is.

Try thinking about it that way. You're going to basically answer two questions:

(1) What kind of motion is $\tau \circ f \circ \tau ^{-1}$ when $f$ is a rotation?

(2) What kind of motion is $\tau \circ f \circ \tau ^{-1}$ when $f$ is a translation?

One last hint: Based on my long-winded explanation up there, its either going to be a rotation, a translation, or a composition of a rotation with a translation in each case.
February 25th 2010, 07:53 PM
Pinkk

This is what I have so far:

Consider f = rotation not the identity, so f is the composition of two reflections that intersect. Construct those reflection lines, r and q, so that they intersect two parallel lines, m and n, that, when composed, formed the translation. When we have the composition of Pm * Pn * Pq * Pr * Pn *Pm. Note that Pr* Pn is a rotation as well as Pn * Pq, so now we have Pm * Ra * Rb *Pn, and we know that R * p and p *R are both glide reflections, so in total we have the composition of two glide reflections. Now here's where I'm stuck; do two glide reflections result in another glide reflection?
February 26th 2010, 06:27 AM
cribby

I don't think that is always true. For example, the product/composition of a glide reflection with itself will be a pure translation because the "reflections" effectively cancel each other out (being equal). However, I do believe that for the case when the reflections are not equal, then the result will be another glide reflection.
February 26th 2010, 06:52 AM
Pinkk

I actually got that if they are not parallel it's a rotation. Because we'll have P * T * T * P, which is P* T * P, which is Glide * P, which is T * P * P. If those last two P's are parallel, then it's a translation, if they are not, then we have T * R, which is just another rotation R.