Cross products for unit vectors in other coordinate systems

  1. I am a bit confused often when I have to compute cross products in other coordinate systems (non-Cartesian), I can't seem to find any tables for cross products such as "phi X rho." in spherical I think that these unit vectors are considered to be "perpendicular," so would phi X rho just be "+/- theta," in general? Typically when I'm doing problems in E&M it takes me a while to convince myself that my computations are correct in terms of direction and it's just frustrating. On an exam I need to just know what the cross products are quick. My hang up is just that they vary from place-to-place.
     
    Last edited: Feb 27, 2014
  2. jcsd
  3. SteamKing

    SteamKing 9,891
    Staff Emeritus
    Science Advisor
    Homework Helper

    It's not clear why you are trying to calculate cross products in non-cartesian coordinate systems.

    AFAIK, the vector cross product is defined only for cartesian coordinates, and then only for 3-dimensional (and 7-dimensional) coordinates.

    http://en.wikipedia.org/wiki/Cross_product
     
  4. 3-dimensional which would include spherical and cylindrical correct? An example would be a boundary condition problem we had in one of our homeworks (for E&M).

    We were given the B field just inside a spherical shell as [tex]\vec B_{in}=B_o \hat k[/tex] where we were given a surface current density.

    [tex]\vec K=K_o\hat\theta[/tex]

    and then asked to find the field right oustide applying the boundary conditions for B-fields,

    [tex]\vec B_{out}=\mu_o(K_o\hat\theta\times\hat r) + B_o \hat k[/tex]

    You can see that the field just inside is in Cartesian and K is in spherical. I converted everything into Cartesian coordinates by writing out the determinant matrix for [tex]\hat \theta \times \hat r[/tex] that churns out a Cartesian vector and I ended up with the field just outside as, [tex]\vec B_{out}=\mu_o K_o(sin(\phi)\hat i -cos(\phi)\hat j)+B_o \hat k[/tex]

    He did not object to my expression, it's pretty straight forward, but it was more tedious computationally because of the determinant matrix and everything.... I remember he had a different expression though in spherical coordinates and I can't really ask him at the moment I just have a test coming and was thinking it would be nice to deal with vectors more quickly than I do because I don't have a lot of insight when it comes to using these basic vector operations in other coordinate systems.




    I guess my question is, can I compute [tex]\hat \theta \times \hat r[/tex] without going to Cartesian?
     
    Last edited: Feb 27, 2014
  5. [tex]\begin{matrix} \times & \hat{q}_1 & \hat{q}_2 & \hat{q}_3 \\ \hat{q}_1 & \vec{0} & +\hat{q}_3 & -\hat{q}_2 \\ \hat{q}_2 & -\hat{q}_3 & \vec{0} & +\hat{q}_1 \\ \hat{q}_3 & +\hat{q}_2 & -\hat{q}_1 & \vec{0} \\ \end{matrix}[/tex]

    [tex]\begin{matrix} system & \hat{q}_1 & \hat{q}_2 & \hat{q}_3 \\ cartesian & \hat{x} & \hat{y} & \hat{z} \\ cylindrical & \hat{r} & \hat{\theta} & \hat{z} \\ spherical & \hat{\rho} & \hat{\phi} & \hat{\theta} \\ \end{matrix}[/tex]
     
  6. If the coordinate system is orthogonal, then you do it the way you said (i.e., the same as with cartesian coordinates). But, since the unit vectors generally change direction with position, you need to evaluate the cross product using the unit vectors at the point of interest.

    Chet
     
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