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Cross and dot product of two vectors in non-orthogonal coordinate

  1. Jul 3, 2014 #1
    Hi everyone,
    I have to find out how to do cross and dot product for two vectors in non-orthogonal coordinate system.
    thanks
     
  2. jcsd
  3. Jul 3, 2014 #2
    You could always use the [itex]\vec{u}\cdot\vec{v}=|\vec{u}|\ |\vec{v}|\cos\theta[/itex] and [itex]\vec{u}\times\vec{v}=|\vec{u}|\ |\vec{v}|\sin\theta \ \hat{n}[/itex] definitions. In general the dot and cross product are independent of coordinate system.
     
  4. Jul 3, 2014 #3
    Dear jk86,
    I don't think so. Consider a non-orthogonal coordinate system like in which angle between any two axis in less than 90 degree. and two vector along x and y axis [1 0 0] & [0 1 0], then the normal cross product is [0 0 1] which is along z-direction but for this coordinate system, z is not perpendicular to x and y axis. and you know cross product of two vector should be perpendicular to both vector.
     
  5. Jul 3, 2014 #4
    Ah, OK I'm sorry I should have read your post more carefully. If you are calculating the dot product of [itex]\vec{a}\cdot\vec{b}[/itex], you can expand each in terms of its contravariant components. As an example, define a coordinate system (u,v,w) via the Cartesian coordinates (x,y,z) using some relations:
    [tex]
    \begin{align}
    x &= u + v\\
    y &= u - v\\
    z &= 3uv + 2w
    \end{align}
    [/tex]
    If the basis vectors for the non-orthogonal (u,v,w) coordinate system are [itex]\vec{e}_u,\vec{e}_v,\vec{e}_{w}[/itex] (and they are [itex]\hat{e}_x,\hat{e}_y,\hat{e}_z[/itex] for the Cartesian basis) then you can write a general vector [itex]\vec{r}=x\hat{e}_x + y\hat{e}_y+z\hat{e}_z=(u+v)\hat{e}_x + (u-v)\hat{e}_y + (3uv + 2w)\hat{e}_z[/itex]. You can then find the non-orthogonal basis vectors by:
    [tex]
    \begin{align}
    \vec{e}_u &= \frac{\partial \vec{r}}{\partial u} = \hat{e}_x + \hat{e}_y + 3v\hat{e}_z\\
    \vec{e}_v &= \frac{\partial \vec{r}}{\partial u} = \hat{e}_x - \hat{e}_y + 3u\hat{e}_z\\
    \vec{e}_w &= \frac{\partial \vec{r}}{\partial u} = 2\hat{e}_z
    \end{align}
    [/tex]
    You can verify that the example is indeed non-orthogonal by computing dot products such as [itex]\vec{e}_u\cdot\vec{e}_w = (3v\hat{e}_z)\cdot(2\hat{e}_z)=6v[/itex]. To compute more general dot products, and make all this simpler, you should first find the metric tensor:
    [tex]
    g_{ij}\equiv \vec{e}_i\cdot\vec{e}_j = \begin{bmatrix}2+9v^2 & 9uv & 6v\\ 9uv & 2+9u^2 & 6u\\ 6v & 6u & 4\end{bmatrix}
    [/tex]
    where i,j refer to [itex]u,v,w[/itex] basis indices. Then for some vectors [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex], you get [itex]\vec{a}\cdot\vec{b}=(\sum_i a^i\vec{e}_i)\cdot (\sum_j b^j\vec{e}_j)=\sum_{ij}g_{ij}a^ib^j[/itex]. You are then simply sticking a matrix [itex]g_{ij}[/itex] in between the vectors---a matrix which is diagonal in an orthogonal coordinate system. As for the cross product you should be able to do something similar using the orthonormal basis definition [itex][\vec{a}\times\vec{b}]_i = \epsilon_{ijk}\vec{a}^j\vec{b}^k[/itex]. I think it just becomes [itex][\vec{a}\times\vec{b}]_i = g^{ij}\epsilon_{jkl}a^{k}b^{l}[/itex], where [itex]\epsilon_{jkl}[/itex] is the Levi-Civita symbol.
     
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