Problem 7 - A First Course in Continuum Mechanics

Question

Given two vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\), and a second-order tensor \(\boldsymbol{S}\), prove: (a) \(\boldsymbol{S}(\boldsymbol{a} \otimes \boldsymbol{b})=(\boldsymbol{S a}) \otimes \boldsymbol{b}\), (b) \((\boldsymbol{a} \otimes \boldsymbol{b}) \boldsymbol{S}=\boldsymbol{a} \otimes\left(\boldsymbol{S}^{T} \boldsymbol{b}\right)\) (c) \((\boldsymbol{a} \otimes \boldsymbol{b})^{T}=(\boldsymbol{b} \otimes \boldsymbol{a})\) Hint: Recall that two second-order tensors \(\boldsymbol{A}\) and \(\boldsymbol{B}\) are equal if and only if \(\boldsymbol{A} \boldsymbol{v}=\boldsymbol{B} \boldsymbol{v}\) for all \(\boldsymbol{v} \in \mathcal{V}\).

Step-by-Step Solution

Answer

\((a) \boldsymbol{S}(\boldsymbol{a} \otimes \boldsymbol{b})=(\boldsymbol{S}\boldsymbol{a}) \otimes \boldsymbol{b}\), \((b) (\boldsymbol{a} \otimes \boldsymbol{b}) \boldsymbol{S}=\boldsymbol{a} \otimes (\boldsymbol{S}^T \boldsymbol{b})\), \((c) (\boldsymbol{a} \otimes \boldsymbol{b})^T=\boldsymbol{b} \otimes \boldsymbol{a}\).

Step 1: Understand the Tensor Product

The tensor product \(\boldsymbol{a} \otimes \boldsymbol{b}\) creates a second-order tensor such that for any vector \(\boldsymbol{v}\), \(\boldsymbol{a} \otimes \boldsymbol{b}\boldsymbol{v} = (\boldsymbol{b} \cdot \boldsymbol{v})\boldsymbol{a}\). Similarly, for any vector \(\boldsymbol{v}\), we have \(\boldsymbol{v} \cdot (\boldsymbol{a} \otimes \boldsymbol{b}) = (\boldsymbol{a} \cdot \boldsymbol{v})\boldsymbol{b}\).

Step 2: Prove Part (a)

Let \(\boldsymbol{v}\) be an arbitrary vector. Then, \(\boldsymbol{S}(\boldsymbol{a} \otimes \boldsymbol{b})\boldsymbol{v} = \boldsymbol{S}((\boldsymbol{b} \cdot \boldsymbol{v})\boldsymbol{a}) = (\boldsymbol{b} \cdot \boldsymbol{v})(\boldsymbol{S}\boldsymbol{a}) = (\boldsymbol{S}\boldsymbol{a}) \otimes \boldsymbol{b}\boldsymbol{v}\). Since this holds for any vector \(\boldsymbol{v}\), we have proven that \(\boldsymbol{S}(\boldsymbol{a} \otimes \boldsymbol{b})=(\boldsymbol{S}\boldsymbol{a}) \otimes \boldsymbol{b}\).

Step 3: Prove Part (b)

Again, let \(\boldsymbol{v}\) be arbitrary. We have \(\boldsymbol{v} \cdot ((\boldsymbol{a} \otimes \boldsymbol{b})\boldsymbol{S}) = (\boldsymbol{a} \cdot (\boldsymbol{S}^T\boldsymbol{v}))\boldsymbol{b} = \boldsymbol{a} \otimes (\boldsymbol{b} \cdot \boldsymbol{S}^T\boldsymbol{v}) = \boldsymbol{a} \otimes (\boldsymbol{S}\boldsymbol{b} \cdot \boldsymbol{v})\). Thus, since \(\boldsymbol{v}\) is arbitrary, \(\boldsymbol{a} \otimes \boldsymbol{b})\boldsymbol{S} = \boldsymbol{a} \otimes (\boldsymbol{S}^T \boldsymbol{b})\).

Step 4: Prove Part (c)

Considering the definition of the transpose of a tensor, we have that for any vector \(\boldsymbol{v}\), \(\boldsymbol{v} \cdot (\boldsymbol{a} \otimes \boldsymbol{b})^T = (\boldsymbol{a} \otimes \boldsymbol{b})\boldsymbol{v} = (\boldsymbol{b} \cdot \boldsymbol{v})\boldsymbol{a}\), which is equivalent to \(\boldsymbol{v} \cdot (\boldsymbol{b} \otimes \boldsymbol{a}) = (\boldsymbol{a} \cdot \boldsymbol{v})\boldsymbol{b}\). Therefore, \(\boldsymbol{a} \otimes \boldsymbol{b})^T = \boldsymbol{b} \otimes \boldsymbol{a}\), since both produce the same result for any vector \(\boldsymbol{v}\).

Key Concepts

Second-Order Tensor

In the realm of continuum mechanics, a second-order tensor is a mathematical object that can be visualized as a 2D array of numbers with specific transformation properties under rotation or change of basis. It is essentially a linear operator that can act on vectors to yield other vectors, and can be represented by a matrix in a given coordinate system.

Considering vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\), a tensor product \(\boldsymbol{a} \otimes \boldsymbol{b}\) is a second-order tensor. It carries the crucial property of mapping an arbitrary vector to a new vector that is scaled by the dot product of \(\boldsymbol{b}\) with that vector and directed along \(\boldsymbol{a}\). For instance, for any vector \(\boldsymbol{v}\), the tensor product \(\boldsymbol{a} \otimes \boldsymbol{b}\) operates on \(\boldsymbol{v}\) to produce a vector parallel to \(\boldsymbol{a}\) scaled by the factor \(\boldsymbol{b} \cdot \boldsymbol{v}\).

Tensor Transpose

Just like a matrix, a tensor can also be transposed. The tensor transpose is an operation that essentially flips the tensor’s components with respect to its main diagonal when viewed as a matrix. For a second-order tensor, this operation interchanges the action on the input and output spaces.

For any two vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\), the transpose of their tensor product \(\boldsymbol{a} \otimes \boldsymbol{b}\) is denoted by \(\left(\boldsymbol{a} \otimes \boldsymbol{b}\right)^T\) and is equal to \(\boldsymbol{b} \otimes \boldsymbol{a}\). This means that for any vector \(\boldsymbol{v}\), applying the transposed tensor yields the same result as swapping the order of vectors in the tensor product and then applying it to the vector.

Vector and Tensor Operations

When dealing with vector and tensor operations, it’s important to understand how they interact. Multiplying a vector by a tensor or two tensors together involves summing over the paired components to produce a new entity – which could be a vector or another tensor. These operations are subject to rules regarding the shapes of the tensors and vectors involved.

For example, in the context of the exercise, a tensor \(\boldsymbol{S}\) acts on the tensor product of \(\boldsymbol{a}\) and \(\boldsymbol{b}\) to distribute its effect across the resulting composite operation. Similarly, the tensor product acting on the tensor from the right involves transposition to align the dimensions correctly for the operation to make sense.

Proofs in Continuum Mechanics

In continuum mechanics, proofs often involve establishing equality between tensors by showing they behave identically under all possible operations with vectors. This technique emphasizes tensor equality criteria, which states that two tensors are considered equal if they produce the same result when applied to any vector from the space they operate on.

This approach was applied in the three parts of the given exercise. The formulations were proven by manipulating the operations and demonstrating that the tensors on either side of the equation acted identically on arbitrary vectors \(\boldsymbol{v}\). Such methods are common in continuum mechanics, where establishing tensor identities can help express complex physical laws and behaviors in a succinct and mathematically rigorous manner.