Question

Consider a two-dimensional Cartesian coordinate system in which a rank 2 contravariant tensor is represented by the matrix \[ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \] The coordinate system is rotated anticlockwise by an acute angle $\theta$ with the origin fixed. Which one of the following matrices represents the tensor in the new coordinate system?

• A. \[ \begin{pmatrix} 0 & \cos 2\theta \\ -\sin 2\theta & 0 \end{pmatrix} \]
• B. \[ \begin{pmatrix} \sin 2\theta & \cos 2\theta \\ \cos 2\theta & -\sin 2\theta \end{pmatrix} \]
• C. \[ \begin{pmatrix} \sin 2\theta & -\cos 2\theta \\ \cos 2\theta & \sin 2\theta \end{pmatrix} \]
• D. \[ \begin{pmatrix} \sin 2\theta & 0 \\ -\cos 2\theta & 0 \end{pmatrix} \]

Hint:

When rotating a tensor in a 2D coordinate system, the new components are found by applying the transformation matrix to the original tensor using matrix multiplication.

Answer 1

The Correct Option is B

- The transformation matrix for rotating a tensor by an angle $\theta$ is given by: \[ R = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \] - To apply the rotation to the original matrix, we use the transformation rule for tensor rotation: \[ T' = R^{-1} T R \] where $R$ is the rotation matrix and $T$ is the tensor matrix. - For the given tensor: \[ T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \] - After applying the rotation matrix and calculating, we get: \[ T' = \begin{pmatrix} \sin 2\theta & \cos 2\theta \\ \cos 2\theta & -\sin 2\theta \end{pmatrix} \]