Chapter 9: Problem 14

Write the following using \(\epsilon_{i j k}\) notation and simplify if possible. a. \(\mathbf{C} \times(\mathbf{A} \times(\mathbf{A} \times \mathbf{C}))\) b. \(\nabla \times(\nabla \times \mathbf{A})\) c. \(\nabla \times \nabla \phi\).

Short Answer

Expert verified

The simplified forms of the given expressions in \(\epsilon_{ijk}\) notation are: a. 0, b. \(\nabla(\nabla.\mathbf{A}) - \nabla^2\mathbf{A}\) (cannot be further simplified without knowledge about \(\mathbf{A}\)), c. 0.

Step by step solution

Convert \(\mathbf{C} \times(\mathbf{A} \times(\mathbf{A} \times \mathbf{C}))\) into \(\epsilon_{i j k}\) notation

The vector triple cross product can be converted to \(\epsilon_{i j k}\) notation as follows: \(\mathbf{C} \times(\mathbf{A} \times(\mathbf{A} \times \mathbf{C})) = \epsilon_{ijk}C_j(A_l \epsilon_{l m n} A_m C_n) = 0\). The reason this simplifies to zero is because the term \( A_l \epsilon_{l m n} A_m\) is identically zero, as two indices of epsilon cannot be same.

Convert \(\nabla \times(\nabla \times \mathbf{A})\) into \(\epsilon_{i j k}\) notation

This can be re-written using the vector identity \(\nabla \times(\nabla \times \mathbf{A}) = \nabla(\nabla.\mathbf{A}) - \nabla^2\mathbf{A}\), where \(\nabla.\) is the divergence operator, and \(\nabla^2\) is the Laplacian. This can't further be simplified in \(\epsilon_{ijk}\) notation without knowledge about A.

Convert \(\nabla \times \nabla \phi\) into \(\epsilon_{i j k}\) notation

Consider \(\phi\) as a scalar function. Taking both gradient (\(\nabla\)) and curl (\(\nabla \times\)) of a scalar function will yield the zero vector, hence \(\nabla \times \nabla \phi = 0\), regardless of what \(\phi\) is.