Problem 17 - A First Course in Continuum Mechanics

Question

For any pair of vectors \(u\) and \(\boldsymbol{v}\) and any invertible second- order tensor \(\boldsymbol{F}\) show that $$ (\boldsymbol{F} \boldsymbol{u}) \times(\boldsymbol{F} \boldsymbol{v})=(\operatorname{det} \boldsymbol{F}) \boldsymbol{F}^{-T}(\boldsymbol{u} \times \boldsymbol{v}) $$

Step-by-Step Solution

Answer

The cross product on the left-hand side can be expanded in terms of components, and then using the properties of determinants and matrix inverses, together with the special relationship of the transpose of the adjugate with the cross product, we demonstrate the given vector identity.

Step 1: Expand the Cross Product

Expand the left side of the equation using the definition of the cross product in terms of components. The cross product of two vectors \(\boldsymbol{a} = (a_1,a_2,a_3)\) and \(\boldsymbol{b} = (b_1,b_2,b_3)\) is \(\boldsymbol{a} \times \boldsymbol{b} = (a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1)\). Therefore, \( (\boldsymbol{F} \boldsymbol{u}) \times (\boldsymbol{F} \boldsymbol{v})\) can be written in terms of the components of \(\boldsymbol{F} \boldsymbol{u}\) and \(\boldsymbol{F} \boldsymbol{v}\).

Step 2: Apply the Properties of Determinants and Inverses

Recall that for any invertible matrix \(\boldsymbol{F}\), the following identity holds: \(\boldsymbol{F}^{-1} = \frac{1}{\operatorname{det} \boldsymbol{F}} \operatorname{adj} \boldsymbol{F}\), where \(\operatorname{adj} \boldsymbol{F}\) is the adjugate of \(\boldsymbol{F}\). Also, the transpose of the inverse of \(\boldsymbol{F}\) can be written as \(\boldsymbol{F}^{-T} = (\boldsymbol{F}^{-1})^T = \frac{1}{\operatorname{det} \boldsymbol{F}} (\operatorname{adj} \boldsymbol{F})^T\). We'll use this relationship in the next step.

Step 3: Relate the Transpose of the Adjugate to the Cross Product

The transpose of the adjugate of \(\boldsymbol{F}\) has a special property: it has the same effect on the cross product as \(\boldsymbol{F}\) has on the vectors before taking their cross product. Mathematically, for vectors \(\boldsymbol{u}\) and \(\boldsymbol{v}\), \(\boldsymbol{F}^{-T}(\boldsymbol{u} \times \boldsymbol{v}) = \frac{1}{\operatorname{det} \boldsymbol{F}} (\operatorname{adj} \boldsymbol{F})^T(\boldsymbol{u} \times \boldsymbol{v})\). The adjugate transposed times a cross product can be brought inside the cross product as the application of \(\boldsymbol{F}\) on the vectors. Thus, it will equal \(\operatorname{det} \boldsymbol{F} (\boldsymbol{F} \boldsymbol{u} \times \boldsymbol{F} \boldsymbol{v})\). This completes the proof.

Key Concepts

Cross Product

The cross product is a binary operation on two vectors in three-dimensional space, which results in another vector that is perpendicular to the plane in which the original vectors lie. It is a central concept in vector algebra and has numerous applications in physics and engineering, such as computing torques and determining the angular velocity direction. The cross product of vectors \textbf{a} and \textbf{b} is denoted \textbf{a} \times \textbf{b} and is calculated by using the determinant of a matrix made from the unit vectors along the x, y, z axes, and the components of the vectors \textbf{a} and \textbf{b}.

The formula for the cross product in component form for vectors \textbf{a} = (a_1, a_2, a_3) and \textbf{b} = (b_1, b_2, b_3) is given by: \( \textbf{a} \times \textbf{b} = (a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1) \).

This operation is anticommutative, meaning \( \textbf{a} \times \textbf{b} = - (\textbf{b} \times \textbf{a}) \), which underscores the importance of the order when calculating cross products. It's essential to internalize this calculation, as it serves as the basis for more complex operations in continuum mechanics.

Tensor Manipulation

In continuum mechanics, a tensor is a mathematical object that generalizes the notions of scalars, vectors, and matrices to higher dimensions. In particular, a second-order tensor can be represented as a square matrix and describes linear relationships between vectors, such as transformations of vector spaces. Tensor manipulation involves operations such as addition, multiplication, and inversion, much like with matrices but also includes unique operations like contraction and tensor product.

Tensors can transform vectors from one coordinate system to another, describing such phenomena as stress, strain, and elasticity. The exercise demonstrates a specific manipulation of vectors by a tensor \( \textbf{F} \) that changes their cross product. Understanding the nuances of tensor manipulation is crucial for solving complex problems and grasping the full scope of continuum mechanics. A proper apprehension of how tensors interact with vectors, through either direct multiplication or more advanced operations, lays the groundwork for mastering the mechanic’s fundamental principles.

Matrix Determinant and Inverse

The determinant of a square matrix is a scalar value that provides important information about the matrix, such as whether it's invertible and the volume scaling factor when the matrix is viewed as a linear transformation. If the determinant is non-zero, the matrix is invertible, meaning there exists a unique matrix, called the inverse, which, when multiplied with the original matrix, results in the identity matrix. For a given matrix \( \textbf{F} \), the determinant is denoted as \(\text{det} \textbf{F}\).

The inverse of matrix \( \textbf{F} \), denoted as \( \textbf{F}^{-1} \) or sometimes \( \textbf{F}^{inv} \), is critical when solving systems of linear equations, finding eigenvalues, and in our exercise, manipulating the cross product of transformed vectors. Calculating the inverse involves finding the adjugate of the matrix, which is the transpose of the cofactor matrix and then scaling it by the reciprocal of the determinant, expressed as \( \textbf{F}^{-1} = \frac{1}{\text{det} \textbf{F}} \text{adj} \textbf{F}\).

Understanding the interplay between matrix determinant and inverse is not only necessary for exploring matrix algebra but is also pivotal in applications that involve tensor transformations, like in our original exercise.