Let \(\lambda_{i}^{2}(i=1,2,3)\) be the eigenvalues of the tensor \(B=E \cdot E^{t}\). (a) Using the fact that \(\lambda_{i}^{2}(i=1,2,3)\) are the solutions of the eigenvalue equation \(\operatorname{det}(B-\) \(\left.\lambda^{2} I\right)=0\) ( \(I\) being a unit tensor), prove the following relations $$ \begin{aligned} \lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2} &=B_{\alpha \alpha}=E_{\alpha \beta}^{2} \\ \lambda_{1}^{2} \lambda_{2}^{2} \lambda_{3}^{2} &=\operatorname{det}(B) \end{aligned} $$ (b) Prove the relation $$ \lambda_{1}^{-2}+\lambda_{2}^{-2}+\lambda_{3}^{-2}=\left(B^{-1}\right)_{\alpha \alpha} $$ (c) Prove that the free energy of deformation of an incompressible isotropic material can be written as a function of \(\operatorname{Tr} B\) and \(\operatorname{Tr} B^{-1}\).

Step by step solution

01

Understand the Given Problem

Let \( \lambda_i^2(i=1,2,3) \) be the eigenvalues of the tensor \( B = E \cdot E^t \). We need to use \( \lambda_i^2(i=1,2,3) \) which are solutions of the eigenvalue equation: \( \operatorname{det}(B - \lambda^2 I) = 0 \), to prove given relations for parts (a), (b), and (c).

02

Eigenvalue Equation

The eigenvalues of \( B \) (i.e., \( \lambda_1^2, \lambda_2^2, \lambda_3^2 \)) satisfy the characteristic equation: \( \operatorname{det}(B - \lambda^2 I) = 0 \). Here, \( I \) is the unit tensor.

03

Proving Relation (a)

To prove \( \lambda_1^2 + \lambda_2^2 + \lambda_3^2 = B_\alpha \alpha = E_\alpha \beta^2 \): The sum of eigenvalues of matrix \( B \) is equal to the trace of \( B \): \( \lambda_1^2 + \lambda_2^2 + \lambda_3^2 = \operatorname{Tr}(B). \) The trace of \( B \) is the sum of its diagonal elements, which are the sums of squares of elements of \( E: \ \operatorname{Tr}(B) = E_\alpha \beta^2 \). Thus, \( \lambda_1^2 + \lambda_2^2 + \lambda_3^2 = B_\alpha \alpha \).

04

Prove the Product of Eigenvalues Relation

To prove \( \lambda_1^2 \lambda_2^2 \lambda_3^2 = \operatorname{det}(B) \) The product of the eigenvalues of \( B \) is equal to the determinant of \( B: \ \det(B) = \lambda_1^2 \lambda_2^2 \lambda_3^2 \). Thus, the second relation is proven.

05

Proving Relation (b)

To prove \( \lambda_1^{-2} + \lambda_2^{-2} + \lambda_3^{-2} = (B^{-1})_\alpha \alpha \): The eigenvalues of \( B^{-1} \) are \( \lambda_i^{-2} \). The trace of \( B^{-1} \) is the sum of its eigenvalues: \( \operatorname{Tr}(B^{-1}) = \lambda_1^{-2} + \lambda_2^{-2} + \lambda_3^{-2} \). Therefore, \( \lambda_1^{-2} + \lambda_2^{-2} + \lambda_3^{-2} = (B^{-1})_\alpha \alpha \).

06

Free Energy Expression

To prove that the free energy of deformation of an incompressible isotropic material can be written as a function of \( \operatorname{Tr}(B) \) and \( \operatorname{Tr}(B^{-1}) \): In an incompressible isotropic material, the free energy of deformation \( W \) is a function of the principal invariants of \( B \) and \( B^{-1} \. \ \operatorname{Tr}(B) \) and \( \operatorname{Tr}(B^{-1}) \) are the principal invariants. Since \lambda_1^2, \lambda_2^2,\beta^2 \ are the eigenvalues, \( W \) can be written in terms of \( \operatorname{Tr}(B) \) and \( \operatorname{Tr}(B^{-1}) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eigenvalues

Eigenvalues are special numbers that provide insights into the properties of a matrix or tensor. If you have a matrix or tensor, its eigenvalues are those numbers that satisfy the equation: \ \( Ax = \lambda x \) \ Here, \ \(A\) is your matrix or tensor, \ \(x\) is a vector, and \ \(\lambda\) is the eigenvalue. In other words, the matrix or tensor \ \(A\) can be transformed into a scaled version of itself by \ \(\lambda\). Understanding eigenvalues helps in various applications such as stability analysis and solving differential equations. In soft matter physics, eigenvalues can describe the behavior of deformations and stresses.

Tensor Analysis

Tensor analysis is a field in mathematics focusing on transformations that generalize vectors and matrices. Tensors represent multi-dimensional data and can range from vectors (one-dimensional) to matrices (two-dimensional) and beyond. In soft matter physics, tensor analysis is crucial for understanding complex properties, such as stress and strain in materials. Tensors describe how deformations and forces distribute in a material, making it essential for understanding material behavior under various conditions. Since tensors can have higher dimensions, operations like addition, multiplication, and finding eigenvalues require careful mathematical handling. This ensures accurate representation and analysis of physical phenomena.

Characteristic Equation

The characteristic equation is a key concept for finding eigenvalues. For a given tensor \(B\) and eigenvalue \(\lambda^2\), the characteristic equation is: \ \( \operatorname{det}(B - \lambda^2 I) = 0 \) \ Here, \ \(I\) is the identity tensor. Solving this equation yields the eigenvalues \ \(\lambda_i^2\) of tensor \ \(B\). This equation tells us that when we subtract \(\lambda^2 I\) from \(B\), the resulting matrix has a determinant of zero. This is important because the determinant zero condition suggests the presence of non-trivial solutions for eigenvectors. Understanding the characteristic equation is crucial for determining eigenvalues, which in turn helps in analyzing the physical properties of the tensor \(B\).

Trace of a Tensor

The trace of a tensor is the sum of its diagonal elements. For a tensor \(B\), it is often denoted as \(\operatorname{Tr}(B)\). For a 3x3 tensor \(B\), the trace is: \ \( \operatorname{Tr}(B) = B_{11} + B_{22} + B_{33} \) \ In the context of eigenvalues, the sum of the eigenvalues \( \lambda_i^2 \) is equal to the trace of tensor \(B\): \ \( \lambda_1^2 + \lambda_2^2 + \lambda_3^2 = \operatorname{Tr}(B) \) \ This relationship helps in simplifying problems and finding solutions quickly. For instance, it can describe the total energy or deformation applied to a material. In soft matter physics, knowing the trace provides insight into the material's overall behavior under given forces.

Determinant of a Tensor

The determinant of a tensor is a special scalar value that encapsulates several features of the tensor. For a tensor \(B\), the determinant is often represented as \(\det(B)\). It provides information on the volume scaling factor by which the tensor \(B\) transforms space. When dealing with eigenvalues, the product of the eigenvalues \(\lambda_i^2\) is equal to the determinant of \(B\): \ \( \lambda_1^2 \lambda_2^2 \lambda_3^2 = \operatorname{det}(B) \) \ This relation simplifies the computation and helps in understanding how the tensor changes space. In soft matter physics, especially during deformation analysis, the determinant can describe how much a material is stretched or compressed. It also assists in calculating physical properties like stress and strain types.