Results 1 to 5 of 5

Math Help - Rigid motions

  1. #1
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2009
    Posts
    54
    Quote Originally Posted by Pinkk View Post
    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 fis 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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2009
    Posts
    54
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rigid Motions
    Posted in the Advanced Algebra Forum
    Replies: 16
    Last Post: November 9th 2011, 12:01 PM
  2. Replies: 3
    Last Post: November 8th 2010, 06:16 PM
  3. two rigid motions agree on three non-colinear points
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 2nd 2010, 08:50 AM
  4. group of motions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 26th 2009, 01:05 PM
  5. [SOLVED] Concerning Rigid Motions
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: February 27th 2008, 08:05 PM

Search Tags


/mathhelpforum @mathhelpforum