In the usual approach to the study of Cartesian tensors the system is considered fixed and the coordinate axes are rotated. The transformation matrix used is therefore that for components relative to rotated coordinate axes. An alternative view is that of taking the coordinate axes as fixed and

Question

In the usual approach to the study of Cartesian tensors the system is considered fixed and the coordinate axes are rotated. The transformation matrix used is therefore that for components relative to rotated coordinate axes. An alternative view is that of taking the coordinate axes as fixed and rotating the components of the system; this is equivalent to reversing the signs of all rotation angles.

Using this alternative view, determine the matrices representing (a) a positive rotation of [latex]\pi[/latex]/4 about the x-axis and (b) a rotation of −[latex]\pi[/latex]/4 about the y-axis. Determine the initial vector r which, when subjected to (a) followed by (b), finishes at (3, 2, 1).

Accepted Answer

The normal notation for the two rotation matrices would be

[latex]A=\left(\begin{array}{ccc}1 & 0 & 0 \0 & \cos heta & \sin heta \0 & -\sin heta & \cos heta\end{array} ight) ext { and } B=\left(\begin{array}{ccc}\cos \phi & 0 & \sin \phi \0 & 1 & 0 \-\sin \phi & 0 & \cos \phi\end{array} ight),[/latex]

with [latex] heta=\phi=\pi / 4[/latex].

In the alternative view (denoted by ”) they would have the same forms but with [latex] heta=\phi=-\pi / 4[/latex], namely

[latex] ext{A"} =\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}\sqrt{2} & 0 & 0 \0 & 1 & -1 \0 & 1 & 1\end{array} ight) ext { and } ext{B"} =\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}1 & 0 & -1 \0 & \sqrt{2} & 0 \1 & 0 & 1\end{array} ight) .[/latex]

The matrix representing (a) followed by (b) in this alternative view is thus

[latex]\begin{aligned} ext { B"A" } & =\frac{1}{2}\left(\begin{array}{ccc}1 & 0 & -1 \0 & \sqrt{2} & 0 \1 & 0 & 1\end{array} ight)\left(\begin{array}{ccc}\sqrt{2} & 0 & 0 \0 & 1 & -1 \0 & 1 & 1\end{array} ight) \& =\frac{1}{2}\left(\begin{array}{ccc}\sqrt{2} & -1 & -1 \0 & \sqrt{2} & -\sqrt{2} \\sqrt{2} & 1 & 1\end{array} ight) .\end{aligned}[/latex]

The required point is the solution of

[latex]\frac{1}{2}\left(\begin{array}{ccc}\sqrt{2} & -1 & -1 \0 & \sqrt{2} & -\sqrt{2} \\sqrt{2} & 1 & 1\end{array} ight)\left(\begin{array}{l}x \y \z\end{array} ight)=\left(\begin{array}{l}3 \2 \1\end{array} ight) .[/latex]

Using the fact that B”A” is orthogonal, and therefore its inverse is simply its transpose, this can be solved directly as

[latex]\left(\begin{array}{l}x \y \z\end{array} ight)=\frac{1}{2}\left(\begin{array}{ccc}\sqrt{2} & 0 & \sqrt{2} \-1 & \sqrt{2} & 1 \-1 & -\sqrt{2} & 1\end{array} ight)\left(\begin{array}{l}3 \2 \1\end{array} ight)=\left(\begin{array}{c}2 \sqrt{2} \-1+\sqrt{2} \-1-\sqrt{2}\end{array} ight) .[/latex]

As a partial check, we compute [latex]\left| \mathbf{r} _{ ext {initial }} ight|^2=8+(3-2 \sqrt{2})+(3+2 \sqrt{2})[/latex]= 14 =3² + 2² + 1² = [latex]\left| \mathbf{r} _{ ext {final }} ight|^2[/latex], i.e. the length of the vector is unchanged by the rotations, as it should be.

Question 26.3: In the usual approach to the study of Cartesian tensors the ...

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