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The transformation of coordinate from $(x, y, z)$ to $(x', y', z')$ is given by $$ x' = a_{11}x + a_{12}y + a_{13}z\\ y' = a_{21}x + a_{22}y + a_{23}z\\ z' = a_{31}x + a_{32}y + a_{33}z$$ If $(\hat i , \hat j, \hat k)$ and $(\hat i_1 , \hat j_1, \hat k_1)$ be unit vectors in two coordinates. Equating $ \vec r = \vec r'$ I got $$\begin{bmatrix} \hat i\\ \hat j\\ \hat k \end{bmatrix} = \begin{bmatrix} a_{11} & a_{21} & a_{31}\\ a_{12} & a_{22} & a_{32}\\ a_{13} & a_{23} & a_{33} \end{bmatrix} \times \begin{bmatrix} \hat i_1\\ \hat j_1\\ \hat k_1 \end{bmatrix}$$ The help manual says $$\begin{bmatrix} \hat i_1\\ \hat j_1\\ \hat k_1 \end{bmatrix} = \begin{bmatrix} a_{11} & a_{21} & a_{31}\\ a_{12} & a_{22} & a_{32}\\ a_{13} & a_{23} & a_{33} \end{bmatrix} \times \begin{bmatrix} \hat i\\ \hat j\\ \hat k \end{bmatrix}$$ I don't know if the manual is wrong or I am wrong. Please help.

The original problem is states as

Show that the operator $\nabla$ remains invariant passing from one rectangular Cartesian system of axes to another.

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1 Answer 1

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Which $(i,j,k)$ corresponds to which $(x,y,z)$? In general, if $A$ is a matrix whose row vectors are the basis vectors for coordinates $(i,j,k)$ and $B$ is a matrix whose row vectors are the basis vectors for coordinates $(i_1,j_1,k_1)$ then if they span the same space, there is an invertible 3x3 matrix $R$ so that $A=RB$ and for $\alpha$ in the basis given by $A$, its image $\beta$ in the basis given by $B$ is $β=R^Tα$.

This makes sense intuitively since $R$ expresses the vectors for $A$ as a linear combination of the vectors in $B$, so anything in terms of $A$ multiplied by $R$ will give you the vector as a linear combination in terms of $B$ i.e. under the basis given by $B$. So if,

$$\begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix} $$

Then correspondingly by the explanation above:

$$ \begin{bmatrix} \hat{i} \\ \hat{j} \\ \hat{k} \end{bmatrix} = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}^T \begin{bmatrix} \hat{i}_1 \\ \hat{j}_1 \\ \hat{k}_1 \end{bmatrix} = \begin{bmatrix}a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{bmatrix} \begin{bmatrix} \hat{i}_1 \\ \hat{j}_1 \\ \hat{k}_1 \end{bmatrix}. $$

So unless I'm making the same mistake you are, I believe you're right.

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If you clarify your question I might be able to help more. –  MGN Jun 29 '12 at 0:31
    
$ ( \hat i, \hat j, \hat k) $ corresponds to $ (x,y,z)$ and $ ( \hat i_1, \hat j_1, \hat k_) $ corresponds to $ (x',y',z')$ –  Santosh Linkha Jun 29 '12 at 12:02

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