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Converting a unit vector from cartesian to cylindrical

  1. May 19, 2015 #1
    Hi, I was wondering if anyone could help with a vector question that I have.

    If I have a unit vector defined in cartesian co-ordinates as p= (0,1,0) how would I go about converting this vector to a cylindrical geometry.

    I understand that I will probably need to use p_r=sqrt(px^2+py^2) and p_theta=arctan(py/px) but also want to convert the vector to a cylindrical geometry and i don't think it can be as simple as this

    thanks
     
  2. jcsd
  3. May 19, 2015 #2

    BiGyElLoWhAt

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    You need to use the sine and cosine definitions. I'll draw a picture.
     
  4. May 19, 2015 #3
    thank you
     
  5. May 19, 2015 #4

    BiGyElLoWhAt

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    Look at the definitions, and see if you can figure out what to do. If not, post back and we'll take it from there.
     

    Attached Files:

  6. May 19, 2015 #5

    BiGyElLoWhAt

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    By the way, welcome to physics forums vector problems.
     
  7. May 19, 2015 #6
    thank you very much for the welcome :)

    just to clarify in case this changes things. The vector i am working with is a direction vector rather than a position vector. i.e. it describes the orientation of a particle with preferred orientation (0,1,0).

    Will the same theory apply, thanks for the help. This is really outside my wheel house
     
  8. May 19, 2015 #7

    BiGyElLoWhAt

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    Yea, it doesn't matter, as far as math is concerned, a vector is a vector. And you're welcome for the welcome. The definition I provided in the image holds true for any position on that circle.
     
  9. May 19, 2015 #8
    from the picture it would seem that p_polar=(r,theta,z)=(1,pi/2/0) but this just seems too simple.

    I would have thought i would have needed to use some sort of transformation matrix to transform from cartesian to polar e.g. p_r=cos(theta)p_x+sin(theta)p_y, p_theta=-sin(theta)p_x+cos(theta)p_y but is this instead only for a flow with multiple components.

    Thanks again
     
  10. May 19, 2015 #9

    BiGyElLoWhAt

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    You can use a transformation matrix, but it's also equivalent to a system of equations relating the coordinates. Since y is independant of x as well as for z in any order, the matrix is unnecessary. Also, what you have is correct. (1,pi/2,0) is the correct set of coordinates for (0,1,0).
     
  11. May 19, 2015 #10
    so if i had a more general point in cartesian co-ordinates; p_cart=(px,py,pz) to express this in a cylindrical geometry do I just do p_polar=(sqrt(px^2+py^2),arctan(py/px),pz) or will I need a transformation matrix in this case

    thanks
     
  12. May 19, 2015 #11

    BiGyElLoWhAt

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    nope, that's good.
     
  13. May 19, 2015 #12
    ok thats great, thank you very much for your help
     
  14. May 19, 2015 #13

    BiGyElLoWhAt

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    No problemo, Glad to help.
     
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