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Converting a DISPLACEMENT vector in Cartesian to Cylindrical Corrdinates!

  1. Nov 11, 2012 #1
    Hi all,

    I was wondering how would one go about converting a displacement vector in cartesian coordinates to cylindrical. I am getting a bit confused on how to deal with the unit vectors; converting a point in space is a simple task, but when it's a vector it just confuses me.

    I am specifically working on describing a displacement vector along the axis of symmetry (z-axis) of a arbitrary helix i.e. the displacement vector from the z-axis to any point on the helix:

    Describing the Helix in Cartesian:

    {x=Rcosθ, y=Rsinθ, z=(h/2∏)θ}; h is the pitch of the helix, or height of one revolution of coil

    Describing the Helix in Cylindrical:

    {r=R, ∅=θ, z=(h/2∏)θ}



    In Cartesian I have the displacement vector as...

    (x-x')i + (y-y')j + (z-z')k =...

    ...(x-Rcos[θ])i + (y-Rsin[θ])j + (z-((h/2∏)θ))k=...; apply condition of being on z axis {0,0,z}

    ...(0-Rcos[θ])i + (0-Rsin[θ])j + (z-((h/2∏)θ))k=

    ...(-Rcos[θ])i + (-Rsin[θ])j + (z-((h/2∏)θ))k

    Now, converting to cylindrical (or so I think):

    {(-Rcos[θ])[itex]^{2}[/itex]+(-Rsin[θ])[itex]^{2}[/itex]}[itex]\hat{r}[/itex]=(R)[itex]\hat{r}[/itex]...

    ArcSin[y/r][itex]\hat{θ}[/itex]=ArcSin[(([itex]\frac{-Rsinθ}{R}[/itex])]=-θ[itex]\hat{θ}[/itex]

    (z-((h/2∏)θ))[itex]\hat{z}[/itex]
     
    Last edited: Nov 11, 2012
  2. jcsd
  3. Nov 11, 2012 #2

    chiro

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    Hey jasonpatel.

    A vector should be considered a point with the magnitude of the point and the direction of the point since vectors are invariant under any kind of translation and so if you translate the vector so that the tail is at the origin, you get the head which corresponds to the point.

    So converting the point from one system to another will give you the vector provided you have continuity in the space and the metric, and also provided you have a bijection between all points from one co-ordinate system to another.
     
  4. Nov 11, 2012 #3
    Hey Chiro,

    thanks for the reply, but I think I am over thinking this problem of mine. So, in my confused state its hard for me to really grasp what you have said. Is there anyway you can check my work to see if it is correct?
     
  5. Nov 11, 2012 #4

    chiro

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    For these kind of calculations, I like to back up the answer with a computer calculation through something like Maple.

    One thing I did notice as possible error is that your ArcSin should be Arctan since Rsin(theta)/Rcos(theta) = tan(theta) = y/x so theta = Arctan(y/x) where you adjust it for the right quadrant.

    If I wanted to check this thoroughly I would calculate the metric tensor or calculate all the partial derivatives and then derive the result this way and Maple can do this for you without the possibility of error if the code is right.

    But regardless of that, I think that if the ArcSin is changed to ArcTan(y/x) = phi, R = SQRT(x^2 + y^2), and z involves a function of theta (which is an adjusted value of ArcTan(y/x)) then it should be right if you are going from <x,y,z> to <r,phi,z_bar> (but again double check yourself with the partial derivatives which can be calculated from Maple or by hand).
     
  6. Nov 11, 2012 #5
    Awesome, thanks for the help Chiro! I will do as you mentioned with the ArcTan but I believe the result will be the same.

    Also, Im not exactly sure what you mean by "metric" so Ill have to do some research into that and see if I can clear it up. Hope you dont mind me coming back to ask you a few more questions :)
     
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