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Unit Vectors and Spherical Coordinates

  1. May 19, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\mathbf{r} = rsin(\theta)cos(\phi) \hat x + rsin(\theta)sin(\phi) \hat x + r cos(\theta) \hat z[/tex]
    I am kind of following the description of the process given at http://mathworld.wolfram.com/SphericalCoordinates.html

    I want to find [tex]\hat r[/tex] and I understand everything except:
    Why is [tex]\hat r = \frac{\frac{d\mathbf{r}}{dr} }{|\frac{d\mathbf{r}}{dr}|} [/tex] (why the derivatives)?

    Normally if I were going to find the unit vector I would just say the unit vector u hat = u/|u|
     
    Last edited: May 19, 2008
  2. jcsd
  3. May 19, 2008 #2
    I'm not exactly sure what you're asking, but i have a feeling you want to find the base vectors in spherical. To do so, it might be easier to do so graphically. To find the vectors you need to look at the surfaces that are created when you hold each of r, theta, and phi constant, individually. So, when you hold r constant your surfaces are spheres. to find a unit vector that is normal to the surface, you just use the fact that x = rsin(phi)cos(theta), y = rsin(phi)sin(theta), z = r cos(phi), and then divide the vector by its length. When you hold theta constant you get a sheet that hangs from the z axis. when you hold phi constant you get cones. THe easiest way to find all of them is to find 2 and then do a cross product (in the appropriate order so your sign is correct) on them to find the third. hope this helps some
     
  4. May 19, 2008 #3

    Dick

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    They ARE using u/|u|. But the u's they are applying that to are the vectors pointing in the coordinate directions, the partials dR/dr, dR/dtheta and dR/dphi (where R is the vector r).
     
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