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Unit vectors

  1. Nov 19, 2011 #1
    can i get some help in how i can convert from cartesian to cylindrical and spherical unit vectors and vice versa ???? thank you
     
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  3. Nov 19, 2011 #2

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  4. Nov 19, 2011 #3
    i mean with proofs because i already read that website page
     
  5. Nov 19, 2011 #4

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    Hmm, well, you can basically read them directly off the drawing of the coordinates.

    Is there any one in particular for which you would like an explanation?
     
  6. Nov 19, 2011 #5
    yes
    in the definition of unit vectors how did we get the cartesian coordinates in terms of cylinderical coordinates
     
  7. Nov 19, 2011 #6

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    Do you mean:
    [tex]
    \begin{matrix}
    \mathbf{\hat x} & = & \cos\phi\boldsymbol{\hat \rho}-\sin\phi\boldsymbol{\hat\phi} \\
    \mathbf{\hat y} & = & \sin\phi\boldsymbol{\hat \rho}+\cos\phi\boldsymbol{\hat\phi} \\
    \mathbf{\hat z} & = & \mathbf{\hat z}
    \end{matrix}
    [/tex]
    ?
     
  8. Nov 19, 2011 #7
    yes please because we know how the opposite happens
     
  9. Nov 19, 2011 #8

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    The only interesting ones are [itex]\mathbf{\hat x}[/itex] and [itex]\mathbf{\hat y}[/itex].
    So we're actually talking about 2-dimensional polar coordinates.


    Here's one way to derive the unit vectors.

    [itex]\boldsymbol{\hat \rho}[/itex] corresponds in this case with the vector from the origin to a point on the unit circle at angle [itex]\phi[/itex].

    The vector [itex]\boldsymbol{\hat \rho}[/itex] can be seen as the result of a rotation of [itex]\mathbf{\hat x}[/itex] over an angle of [itex]\phi[/itex].

    Or in reverse, the vector [itex]\mathbf{\hat x}[/itex] is the result of a rotation of [itex]\boldsymbol{\hat \rho}[/itex] over an angle of [itex]-\phi[/itex].
    Similarly the vector [itex]\mathbf{\hat y}[/itex] is the result of a rotation of [itex]\boldsymbol{\hat \phi}[/itex] over an angle of [itex]-\phi[/itex].

    We need the rotation matrix for an angle [itex]\phi[/itex] is to get what we want:
    [tex]R = \begin{pmatrix}\cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{pmatrix}[/tex]

    Multiply [itex]R[/itex] with [itex]\boldsymbol{\hat \rho}[/itex] and [itex]\boldsymbol{\hat \phi}[/itex] and the result rolls out:
    [tex]
    \begin{matrix}
    \mathbf{\hat x} & = & \cos\phi\boldsymbol{\hat \rho}-\sin\phi\boldsymbol{\hat\phi} \\
    \mathbf{\hat y} & = & \sin\phi\boldsymbol{\hat \rho}+\cos\phi\boldsymbol{\hat\phi}
    \end{matrix}
    [/tex]
     
  10. Nov 20, 2011 #9
    i know how raw and phy are in terms of x hat and y hat how did you change that to the last line im still not getting the idea??
     
  11. Nov 20, 2011 #10

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    How about this:

    Apparently you already know:
    [tex]
    \begin{matrix}
    \boldsymbol{\hat \rho} & = & \cos\phi\mathbf{\hat x}+\sin\phi \mathbf{\hat y}\\
    \boldsymbol{\hat\phi} & = & -\sin\phi\mathbf{\hat x}+\cos\phi \mathbf{\hat y}
    \end{matrix}
    [/tex]

    Which is:
    [tex]\begin{pmatrix} \boldsymbol{\hat \rho} \\ \boldsymbol{\hat\phi} \end{pmatrix} = \begin{pmatrix}
    \cos \phi & \sin \phi \\
    -\sin \phi & \cos \phi \end{pmatrix}
    \begin{pmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \end{pmatrix}
    [/tex]

    The inverse is:

    [tex]\begin{pmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \end{pmatrix} = \begin{pmatrix}
    \cos \phi & -\sin \phi \\
    \sin \phi & \cos \phi \end{pmatrix}
    \begin{pmatrix} \boldsymbol{\hat \rho} \\ \boldsymbol{\hat\phi} \end{pmatrix}
    [/tex]
     
  12. Nov 20, 2011 #11
    ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh now i get it okay okay thank you alotttttttttttttttttttt
     
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