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Transform a vector from Cartesian to Cylindrical coordinates

  1. Sep 4, 2008 #1
    1. The problem statement, all variables and given/known data

    Transform the vector below from Cartesian to Cylindrical coordinates:

    [tex]Q\,=\,\frac{\sqrt{x^2\,+\,y^2}}{\sqrt{x^2\,+\,y^2\,+\,z^2}}\,\hat{x}\,-\,\frac{y\,z}{x^2\,+\,y^2\,+\,z^2}\,\hat{z}[/tex]



    2. Relevant equations

    Use these equations:

    [tex]A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A}\,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi[/tex]

    [tex]A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A}\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi[/tex]

    [tex]A_z\,=\,A_z[/tex]

    [tex]x\,=\,\rho\,cos\,\phi[/tex]

    [tex]y\,=\,\rho\,sin\,\phi[/tex]



    3. The attempt at a solution

    [tex]Q_{\rho}\,=\,\frac{\sqrt{\left(\rho\,cos\,\phi\right)^2\,+\,\left(\rho\,\sin\,\phi\right)^2}}{\sqrt{\left(\rho\,cos\,\phi\right)^2\,+\,\left(\rho\,\sin\,\phi\right)^2\,+\,z^2}}\,cos\,\phi[/tex]

    [tex]Q_{\rho}\,=\,\frac{\rho}{\sqrt{\rho^2\,+\,z^2}}\,cos\,\phi[/tex]

    [tex]Q_{\phi}\,=\,-\,\frac{\rho}{\sqrt{\rho^2\,+\,z^2}}\,sin\,\phi[/tex]

    [tex]Q_z\,=\,-\,\frac{z\,\rho\,sin\,\phi}{\rho^2\,+\,z^2}[/tex]




    Does the above look correct?
     
    Last edited: Sep 4, 2008
  2. jcsd
  3. Sep 5, 2008 #2

    tiny-tim

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    Hi VinnyCee! :smile:
    No … these are the equations for an ordinary rotation of the x and y axes through an angle phi.
    hmm … you want polar coordinates of a vector along the x-axis … so what will its phi coordinate be? :smile:
     
  4. Sep 5, 2008 #3
    Really? The equations for transforming from Cartesian to Cylindrical were given in the notes for the class in a matrix form and then solved out to a formula, the one below.

    Here is a more detailed version for the [itex]\rho[/itex] coordinate transformation only:

    [tex]A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A} \,=\,\left(\hat{\rho}\,\cdot\,\hat{x}\right)\,A_x\,+\,\left(\hat{\rho}\,\cdot\,\hat{y}\right)\,A_y\,+\,\left(\hat{\rho}\,\cdot\,\hat{z}\right)\,A_z\,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi[/tex]

    Is that incorrect? If so, what is the correct way to transform from Cartesian to Cylindrical?


    I think the [itex]\phi[/itex] coordinate would be zero in this case since there is no y-component, right?
     
  5. Sep 5, 2008 #4

    Redbelly98

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    Here you've included the Ax cos(phi) term, but forgot the Ay sin(phi) term
     
  6. Sep 5, 2008 #5
    [itex]A_y[/itex] is zero, so it need not be included, right?
     
  7. Sep 5, 2008 #6

    Redbelly98

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    Oops, I was confusing Ay and Az. My bad.

    Your results from post #1 look okay to me. My understanding is the dot product of the vector Q with the coordinate unit-vector is the component of Q in that coordinate direction, which is exactly what you did.

    On the other hand I'm hesitant to disagree with tiny-tim, since he knows this stuff pretty well also.
     
  8. Sep 5, 2008 #7

    tiny-tim

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    That's right! for x > 0 and y = 0, phi must be zero. :smile:
    No!

    A three-dimensional vector needs three coordinates …

    if one coordinate is 0, you have to say so!
    No, that is correct … the other one is wrong.
     
  9. Sep 5, 2008 #8

    Redbelly98

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    Isn't that the same expression as in post #1 for Aρ ?
     
    Last edited: Sep 5, 2008
  10. Sep 5, 2008 #9

    tiny-tim

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    ah … I meant the phi-coordinate [tex]A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A}\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi[/tex] is wrong. :wink:
     
  11. Sep 5, 2008 #10

    Redbelly98

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    Um, if we use the convention that φ's unit vector points counterclockwise, and φ is taken counterclockwise from the positive x-axis, then I agree with VinnyCee's formula.
     
  12. Sep 5, 2008 #11
    So, does that mean that the original post contains the correct answer? Or is the [itex]Q_{\phi}[/itex] variable still going to be zero even though the formulas given produce the expression I wrote in the first post?

    The expression written from a matrix in class is:

    [tex]A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A} \,=\,\left(\hat{\phi}\,\cdot\,\hat{x}\right)\,A_x\,+\,\left(\hat{\phi}\,\cdot\,\hat{y}\right)\,A_y\,+\,\left(\hat{\phi}\,\cdot\,\hat{z}\right)\,A_z\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi[/tex]

    Since...

    [tex]\hat{x}\,\cdot\,\hat{\phi}\,=\,-\,sin\,\phi[/tex]

    [tex]\hat{y}\,\cdot\,\hat{\phi}\,=\,cos\,\phi[/tex]

    [tex]\hat{z}\,\cdot\,\hat{\phi}\,=\,0[/tex]

    Right?
     
    Last edited: Sep 5, 2008
  13. Sep 6, 2008 #12

    tiny-tim

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    sorry!

    Hi VinnyCee and Redbelly98! :smile:

    Yes, I completely mis-read the original question.

    Your original answer was totally right.

    Please ignore my previous posts.

    Sorry. :redface:
     
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