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Deriving the Spherical Unit Vectors

  1. Sep 7, 2006 #1
    Does anyone know how to derive the spherical unit vectors in the cartesian basis? Or a good link that might show how its done?

    I would also like to see it done for the cylindrical coordinates. I have tried to do it, especially for the spherical case, but i can only get r-hat.

    It would be a great help!
    Last edited: Sep 7, 2006
  2. jcsd
  3. Sep 8, 2006 #2
    Well, I have written an article about it (its not typed though). If you could wait for a week or so, I can type it, put it on PlanetMath, publish it and then give you the link (I can't do it now because my Final exams are approaching soon).

    Anyways, the basic idea is as follows:

    You know that the conversion from rectangular to spherical coordinates goes like this:
    [tex]x = r\sin(\theta)\cos(\phi), \qquad y = r\sin(\theta)\sin(\phi), \qquad z = r\cos(\theta)[/tex]

    Now we can represent any point in space in terms of the position vector [tex]\vec{R} = x\ha{i} + y\hat{j} + z\hat{k}[/tex] or in spherical coordinates as:
    [tex]\vec{R} = r\sin(\theta)\cos(\phi)\hat{i} + r\sin(\theta)\sin(\phi)\hat{j} + r\cos(\theta)\hat{k}[/tex].

    So far so good! Now, what does it mean for something to be a unit vector?? Take for example the vector [tex]\hat{r}[/tex]. What's special about it? Its not really its length because all unit vectors have the length of 1. That's BORING! BUT its DIRECTION is special, because that is what seperates one unit vector from another. Our unit vector [tex]\hat{r}[/tex], for example, points in the direction where the coordinate [tex]r[/tex] increases while the rest of the coordinates are held constant.

    Well, this is the same thing as taking the PARTAL derivative of the position vector [tex]\vec{R}[/tex] with respect to [tex]r[/tex] while holding [tex]\theta[/tex] and [tex]\phi [/tex] constant (think about it, it'll make sense).

    So therefore:
    [tex]\hat{r} = \frac{\partial \vec{R}}{\partial r} \qquad \hat{\theta} = \frac{\partial\vec{R}}{\partial\theta} \qquad \hat{\phi} = \frac{\partial\vec{R}}{\partial\phi}[/tex]

    Warning: What I have given you are not exactly unit vectors. You still have to divide them by their respective lengths but they do have the right direction which is the more important thing.

    edit: Here's is the more accurate forumulation:

    [tex]\mathbf{ \hat{r} = \frac{\frac{\partial \vec{R}}{\partial r}}{\Big| \frac{\partial \vec{R}}{\partial r} \Big|} \qquad \hat{\theta} = \frac{\frac{\partial\vec{R}}{\partial\theta}}{\Big| \frac{\partial\vec{R}}{\partial\theta} \Big|} \qquad \hat{\phi} = \frac{\frac{\partial\vec{R}}{\partial\phi}}{\Big| \frac{\partial\vec{R}}{\partial\phi} \Big|} }[/tex]

    You can derive the unit vectors for cylindrical coordinates in a similar way by changing the position vecotor [tex]\vec{R}[/tex] accordingly.

    Hope this helps. :smile:
    Last edited: Sep 8, 2006
  4. Sep 19, 2006 #3
    BTW, we can write the above expression more concisely as:

    [tex] \hat{r} = \frac{\frac{\partial \vec{R}}{\partial r}}{h_r} \qquad \hat{\theta} = \frac{\frac{\partial\vec{R}}{\partial\theta}}{h_{\theta}} \qquad \hat{\phi} = \frac{\frac{\partial\vec{R}}{\partial\phi}}{h_{\phi}}[/tex]


    [tex]h_r = \Big| \frac{\partial \vec{R}}{\partial r} \Big|, \quad h_{\theta} = \Big| \frac{\partial\vec{R}}{\partial\theta} \Big|, \quad h_{\phi} = \Big| \frac{\partial\vec{R}}{\partial\phi} \Big|[/tex]

    and [itex]h_r, h_{\theta}, h_{\phi}[/itex] are collectively know as scale factors or metric coefficients of the spherical coordinate system. They are also denoted as [itex]h_1, h_2, h_3[/itex] or simply as [itex]h_i[/itex] sometimes (especially in General relativity) when working with any arbitrary curvilinear coordinate system.
    Last edited: Sep 19, 2006
  5. Dec 22, 2006 #4
    Last edited by a moderator: Dec 9, 2013
  6. May 25, 2011 #5
    Thank you sooooooo much for this.
    Simple but the idea of direction made it so easy!
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