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Euclidean group E(n) elements

  1. Jan 31, 2014 #1

    ChrisVer

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    Well I am not sure if this thread belongs here or in mathematics/groups but since it also has to do with physics, I think SR would be the correct place.
    An element of the Euclidean group [itex]E(n)[/itex] can be written in the form [itex](O,\vec{b})[/itex] which acts:
    [itex] \vec{x} \rightharpoondown O\vec{x}+\vec{b}[/itex]
    With [itex]O \in O(n)[/itex] and [itex] \vec{b} \in R^{n}[/itex]
    This would mean that the vector [itex]\vec{x}[/itex] would be rotated by some angle and then translated by a vector.

    Now I'm having a certain problem. Since it's a group, the multiplication of two of its elements should be an element itself.
    this I write:
    [itex](O_{2},\vec{b_{2}})(O_{1},\vec{b_{1}})\vec{x}[/itex]
    giving:
    [itex](O_{2},\vec{b_{2}}) (O_{1}\vec{x}+\vec{b_{1}})[/itex]

    From here I followed two paths which I think the first gives a wrong answer, while I'm pretty sure the 2nd gives the correct one... However I don't understand what's their difference.

    wrong path
    [itex](O_{2},\vec{b_{2}}) (O_{1}\vec{x}+\vec{b_{1}})=(O_{2},\vec{b_{2}}) O_{1}\vec{x}+(O_{2},\vec{b_{2}}) \vec{b_{1}}[/itex]
    which gives:
    [itex]O_{2} O_{1}\vec{x}+\vec{b_{2}} +O_{2} \vec{b_{1}} + \vec{b_{2}}[/itex]
    see the two times of [itex]\vec{b_{2}}[/itex] appearing

    correct path
    I write that:
    [itex](O_{2},\vec{b_{2}}) (O_{1}\vec{x}+\vec{b_{1}})=(O_{2},\vec{b_{2}}) \vec{x_{2}}[/itex]
    So I get:
    [itex]O_{2}\vec{x_{2}}+\vec{b_{2}}[/itex]
    and reentering the definition of [itex]x_{2}=O_{1}\vec{x}+\vec{b_{1}}[/itex] I get:
    [itex]O_{2}O_{1}\vec{x}+O_{2}\vec{b_{1}}+\vec{b_{2}}[/itex]

    So here we have [itex]\vec{b_{2}}[/itex] only once....
    I was able to distinguish the correct from the wrong due to physical imaging (by double rotations and translations), however I don't understand (or more precisely see) what's the difference (and so the wrong in the 1st case) between the two approaches...
     
    Last edited: Jan 31, 2014
  2. jcsd
  3. Jan 31, 2014 #2

    Bill_K

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    In method 1 you're trying to use the distributive law where it does not apply. You're saying Tb(x + y) = (x + b) + (y + b), whereas it should be just Tb(x + y) = x + y + b.
     
  4. Jan 31, 2014 #3

    BruceW

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    In other words, the transformation is not linear. f(x+y) not equal to f(x) + f(y) in general.
     
  5. Feb 1, 2014 #4

    ChrisVer

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    I think the problem of non-linearity is in the translations part... that's kinda funny, thanks...
     
  6. Feb 1, 2014 #5

    BruceW

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    yep. it is pretty interesting. try looking up affine transformation, that should give more info on this topic.
     
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