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jango_IN_DTOWN

  • one year ago

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  1. jango_IN_DTOWN
    • one year ago
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  2. jango_IN_DTOWN
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    @ganeshie8

  3. jango_IN_DTOWN
    • one year ago
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    @texaschic101

  4. jango_IN_DTOWN
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    @Zarkon

  5. jango_IN_DTOWN
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    @Michele_Laino

  6. Michele_Laino
    • one year ago
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    yes the set of rotation of the \(x,y-\)plane is a group, for example, consider this rotation around the \(z-\)axis \[\left\{ {\begin{array}{*{20}{c}} {x' = x\cos \alpha + y\sin \alpha } \\ {y' = - x\sin \alpha + y\cos \alpha } \\ {z' = z} \end{array}} \right.\]

  7. Michele_Laino
    • one year ago
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    or, in a more compact form: \[\left( {\begin{array}{*{20}{c}} {x'} \\ {y'} \\ {z'} \end{array}} \right) = {R_\alpha }\;\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) \Rightarrow R = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)\]

  8. jango_IN_DTOWN
    • one year ago
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    HI... The group is already defined .. And the rotation is taken about (0,0)

  9. Michele_Laino
    • one year ago
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    even if the elements of a general group of rotation need not commute, namely: \[{R_i}{R_j} \ne {R_j}{R_i}\]

  10. Michele_Laino
    • one year ago
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    In your case, please try to verify the axioms of a group, using my formulas above

  11. Michele_Laino
    • one year ago
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    if you prefer you can consider the case when z=0, namely a rotation of the \(x,y-\)plane

  12. jango_IN_DTOWN
    • one year ago
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    Yes... It will be a subgroup then. What about the cyclic part?

  13. Michele_Laino
    • one year ago
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    I'm thinking....

  14. Michele_Laino
    • one year ago
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    I think that the rotations of a polygon, around the origin are a cyclic group

  15. jango_IN_DTOWN
    • one year ago
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    For a cyclic group we need a generator.. In this case the angles are takes from the set of all real numbers...

  16. Michele_Laino
    • one year ago
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    I know, nevertheless if we can find a cyclic group with respect to our group is a subgroup, we have the thesis

  17. jango_IN_DTOWN
    • one year ago
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    You mean isomorphic to our group?

  18. Michele_Laino
    • one year ago
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    I mean if \(R_1\) is a cyclic group, such that \(R\) is included in \(R_1\) then also \(R\) is cyclic

  19. jango_IN_DTOWN
    • one year ago
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    I am not getting th cyclic part

  20. Michele_Laino
    • one year ago
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    whe we are in quantum mechanics we can speak about the generator of rotations, and if we consider the rotation around the \(z-\)axis, then such generator is the \(z-\)component of angular momentum of our mechanical system: \(J_z\)

  21. Michele_Laino
    • one year ago
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    when*

  22. jango_IN_DTOWN
    • one year ago
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    I mean the group mentioned in the question.. HOw is it having a generator, that part I am not getting

  23. Michele_Laino
    • one year ago
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    for example, the \(n-\)th roots of unity are a cyclic group

  24. jango_IN_DTOWN
    • one year ago
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    Yeah I know that its a cyclic group, I meant the given group is cyclic or not

  25. Michele_Laino
    • one year ago
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    and as a generator I can take any one of that roots, of course the order of such group is \(n\) Yes I think so, nevertheless I'm not able to justify it

  26. Michele_Laino
    • one year ago
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    If we consider the set of symmetry rotations, then we have a cyclic group, since such group is isomorphic to the group of integers modulo n, where \(2\pi/n\) is the littlest rotation

  27. Michele_Laino
    • one year ago
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    so, if we define an infinitesimal rotation, namely \(\delta \theta\), also our group is a cyclic group

  28. jango_IN_DTOWN
    • one year ago
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    Then how can we represent any rotation which is less than it but greater than 0?

  29. Michele_Laino
    • one year ago
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    good question

  30. Michele_Laino
    • one year ago
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    I consider the matrix: \[R = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right)\]

  31. jango_IN_DTOWN
    • one year ago
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    ok

  32. Michele_Laino
    • one year ago
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    Using the induction principle, we can write: \[{R^n} = \left( {\begin{array}{*{20}{c}} {\cos \left( {n\alpha } \right)}&{\sin \left( {n\alpha } \right)} \\ { - \sin \left( {n\alpha } \right)}&{\cos \left( {n\alpha } \right)} \end{array}} \right)\]

  33. Michele_Laino
    • one year ago
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    since not all real numbers are in the form \(n\alpha\) because the real line is a vector space on the re4al set, then our group is not cyclic

  34. Michele_Laino
    • one year ago
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    real* set

  35. jango_IN_DTOWN
    • one year ago
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    OK thanks...:)

  36. Michele_Laino
    • one year ago
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    :)

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