Question

Show that the transformation equation for a ${\mathbf{2}}^{\mathbf{n}\mathbf{d}}$-rank Cartesian tensor is equivalent to a similarity transformation. Warning hint: Note that the matrix C in Chapter 3 , Section 11 , is the inverse of the matrix A we are using in Chapter 10 (compare$\mathbf{}\mathbf{r}\mathbf{\text{'}}\mathbf{=}\mathbf{Ar}\mathbf{}\mathbf{and}\mathbf{}\mathbf{r}\mathbf{=}\mathbf{Cr}\mathbf{\text{'}}$). Thus a similarity transformation of the matrix T with tensor components${\mathit{T}}_{\mathbf{i}\mathbf{j}}$ is$\mathbf{T}\mathbf{\text{'}}\mathbf{=}{\mathbf{ATA}}^{\mathbf{-}\mathbf{1}}$. Also, see “Tensors and Matrices” in Section 3 and remember that A is orthogonal.

Short answer

The equation has been proven.

Step-by-step solution

Given Information

the $2^{nd}$-rank cartesian tensor.

Definition of a cartesian tensor.

The first rank tensor is just a vector. A tensor of the second rank has nine components (in three dimensions) in every rectangular coordinate system.

Find the value.

Let summation convention be${{\mathrm{T}}_{\mathrm{kl}}}^{\text{'}}={\mathrm{a}}_{\mathrm{ki}}{\mathrm{a}}_{\mathrm{lj}}{\mathrm{T}}_{\mathrm{ij}}.$.

The inverse of the rotation matrix is given below.

$$\begin{gathered} {\mathrm{A}}^{-1}={\mathrm{A}}^{\mathrm{\tau }} \\ {{\mathrm{T}}_{\mathrm{kl}}}^{\text{'}}={\mathrm{a}}_{\mathrm{ki}}{\mathrm{T}}_{\mathrm{ij}}{{\mathrm{a}}_{\mathrm{jl}}}^{\mathrm{\tau }} \end{gathered}$$

Write the above statement in matrix form.

$$\begin{gathered} T^{\text{'}}=AT{A}^{\tau } \\ {T}^{\text{'}}=AT{A}^{-1} \end{gathered}$$

Hence, the equation has been proven.