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Spherical coordinates?

  1. Jun 2, 2006 #1
    What is equivalent to the unit k (vector in cartesian coords) in spherical coordinates? And why?

    z=rcos(t)
     
    Last edited: Jun 2, 2006
  2. jcsd
  3. Jun 2, 2006 #2
    if the vector is in three dimensions, one more variable(of spherical) is required to define j.
     
  4. Jun 2, 2006 #3

    dav2008

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    Last edited by a moderator: Apr 22, 2017
  5. Jun 2, 2006 #4
    I made a mistake which has been corrected, it should be the unit k vector.

    In cartesian, it is (0,0,1). What is it in spherical (0,0,what)?
     
  6. Jun 3, 2006 #5

    arildno

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    For the most common choice of spherical polar coordinates,
    [tex]x=r\sin\phi\cos\theta,y=r\sin\phi\sin\thea,z=r\cos\phi,0\leq{r},0\leq\phi\leq\pi,0\leq\theta\leq{2}\pi[/itex]
    we have the following unit vetors relations:
    [tex]\vec{i}_{r}=\sin\phi(\cos\theta\vec{i}+\sin\theta\vec{j})+\cos\phi\vec{k}[/tex]
    [tex]\vec{i}_{\phi}=\frac{\partial}{\partial\phi}\vec{i}_{r}=\cos\phi(\cos\theta\vec{i}+\sin\theta\vec{j})-\sin\phi\vec{k}[/tex]
    [tex]\vec{i}_{\theta}=\frac{1}{\sin\phi}\frac{\partial}{\partial\theta}\vec{i}_{r}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]
    Solving for the Cartesian unit vectors we gain, in particular:
    [tex]\vec{k}=\cos\phi\vec{i}_{r}-\sin\phi\vec{i}_{\phi}[/tex]
    That is of course equal to the coordinate transformation:
    [tex](0,0,1)\to(\cos\phi,0,\sin\phi)[/tex]
    In order to find the correct expressions for the other two unit Cartesian vectors, utilize the intermediate result:
    [tex]\sin\phi\vec{i}_{r}+\cos\phi\vec{i}_{\phi}=\vec{i}_{\hat{r}}=\cos\theta\vec{i}+\sin\theta\vec{j}[/itex]
    [itex]\vec{i}_{\hat{r}},\vec{i}_{\theta}[/itex] are polar coordinate vectors in the horizontal plane.
     
    Last edited: Jun 3, 2006
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