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I Convert from rectangular to Spherical Coordinates

  1. Oct 5, 2017 #1
    How do you convert this to Spherical Components?

    Spherical Convention = (radial, azimuthal, polar)

    ##\vec r = |\vec r| * \cos{(\theta)} * \sin{(\phi)} * \hat x +|\vec r| * \sin{(\theta)} * \sin{(\phi)} * \hat y +|\vec r| * \cos{(\phi)} * \hat z##

    Is this correct?

    ##\vec r =\sqrt{(x^2 + y^2 + z^2)} * \hat r + \arctan{(\frac{y}{x})} * \hat \theta + \arccos{(\frac{z}{r})} * \hat \phi##
     
  2. jcsd
  3. Oct 5, 2017 #2

    Orodruin

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    No, it is not correct. Please show your work.
     
  4. Oct 5, 2017 #3
    Cartisian Coordinates
    ##\vec r = x * \hat x + y * \hat y + z * \hat z##
    ##x = |\vec r| \cos{\theta} \sin{\phi}##
    ##y = |\vec r| \sin{\theta} \sin{\phi}##
    ##z = |\vec r| \cos{\phi}##
    Spherical Coordinates
    ##r = \sqrt{(x^2 + y^2 + z^2)}##
    ##\theta = \arctan{(\frac{y}{x})}##
    ##\phi = \arccos{(\frac{z}{r})}##
    ##\vec r = r \hat r + \theta \hat \theta + \phi \hat \phi##
    or is it
    ##\vec r = r \hat r + \frac{\theta}{r\sin{\phi}} \hat \theta +\frac{\phi}{r} \hat \phi##
     
  5. Oct 5, 2017 #4

    Orodruin

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    Neither of those is correct. You are even adding vectors with different physical dimension ... I suggest you express the Cartesian basis vectors in terms of the spherical basis vectors and start from the Cartesian expression for the position vector.
     
  6. Oct 5, 2017 #5
    What would be the expression in terms of Spherical Coordinates
    ##\vec r = r(x,y,z) \hat r + \theta(x,y,z) \hat \theta + \phi(x,y,z) \hat \phi##
    ##r(x,y,z)=?##
    ##\theta(x,y,z)=?##
    ##\phi(x,y,z)=?##
     
  7. Oct 5, 2017 #6

    phyzguy

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    Why don't you try drawing it out? As you said, in Cartesian Coordinates the position vector is [itex]\vec r = x * \hat x + y * \hat y + z * \hat z[/itex]. When you draw this out, in what direction is it pointing in spherical coordinates? How much of it is pointing along the r, θ, and φ unit vectors?
     
  8. Oct 5, 2017 #7
    If I draw it out the Spherical Coordinates are:
    ##r = \sqrt{(x^2 + y^2 + z^2)}##
    ##\theta = \arctan{(\frac{y}{x})}##
    ##\phi = \arccos{(\frac{z}{r})}##
    ##\vec r = r \hat r + \theta \hat \theta + \phi \hat \phi##
    ##\hat \theta## and ##\hat \phi## are along the arcs.
     
  9. Oct 5, 2017 #8

    phyzguy

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    Right. So if I decompose the Cartesian position vector into the spherical coordinate unit vectors, how much of the Cartesian position vector points along these arcs?
     
  10. Oct 5, 2017 #9
    Is correct then?
     
  11. Oct 5, 2017 #10
    ##r = \sqrt{(x^2 + y^2 + z^2)}##
    ##\theta = \arctan{(\frac{y}{x})}##
    ##\phi = \arccos{(\frac{z}{r})}##
    ##\vec r = r \hat r + \theta \hat \theta + \phi \hat \phi##
    is correct then?
     
  12. Oct 5, 2017 #11

    phyzguy

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    No. You didn't answer my question.
     
  13. Oct 5, 2017 #12
    ##\arctan{(\frac{y}{x})}## points in the ##\hat \theta## direction.
    ##\arccos{(\frac{z}{r})}## points in the ##\hat \phi## direction.
     
  14. Oct 5, 2017 #13

    Orodruin

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    A scalar cannot "point" in any direction. You were asked to decompose the Cartesian basis vectors into the spherical ones in post #4 and to make the drawing with the position vector relative to a point in spherical coordinates twice in posts #6 and #8. I suggest that you follow either or both of these suggestions.
     
  15. Oct 5, 2017 #14
    Uncertain how to convert x,y,z to ##(\hat r, \hat \theta, \hat \theta)##. Can you give a hint?
     
  16. Oct 5, 2017 #15

    Orodruin

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    Are you familiar with how to express the spherical basis vectors in Cartesian coordinates?
     
  17. Oct 5, 2017 #16
    Look above. Post #1.
     
  18. Oct 5, 2017 #17

    Orodruin

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    You have not done so above.
     
  19. Oct 5, 2017 #18

    phyzguy

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    Have you drawn out the vectors yet? If so, can you post a picture?
     
  20. Oct 5, 2017 #19
    $$\vec{r}=r\hat{r}$$
     
  21. Oct 6, 2017 #20
    Attached is a drawing. r is in a 3d box or x,y,z. On the xy plane starting on x-axis is ##\theta## starting on the arc is ##\hat \theta## as a unit vector. On the z-xy plane directly down from r to the xy plane. Starting from the z-axis to r to the xy plane is ##\phi## starting on the arc ##\hat \phi## as a unit vector.
     

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    Last edited: Oct 6, 2017
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