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Show that any contravariant tensor of rank two can be written as the sum of a symmetric tensor and an antisymmetric tensor. Can this be generalized to tensors of arbitrary rank?

Short Answer

Expert verified
A contravariant tensor of rank two can indeed be represented as the sum of a symmetric tensor and an antisymmetric tensor. This decomposition is a fundamental property of tensors and can be extended to tensors of any rank.

Step by step solution

01

Define the tensors

Consider a tensor \( T^{ij} \) of rank two. A tensor is symmetric if it remains the same when its indices are swapped, \( T^{ij} = T^{ji} \). It is antisymmetric if it changes sign when its indices are swapped, \( T^{ij} = -T^{ji} \).
02

Decompose the tensor

The tensor \( T^{ij} \) can be expressed as the sum of two new tensors, \( S^{ij} \) and \( A^{ij} \), where \( S^{ij} \) is symmetric and \( A^{ij} \) is antisymmetric. This is written as \( T^{ij} = S^{ij} + A^{ij} \).
03

Define the symmetric and antisymmetric tensors

The symmetric tensor \( S^{ij} \) can be defined as \( S^{ij} = \frac{1}{2}(T^{ij} + T^{ji}) \) and the antisymmetric tensor \( A^{ij} \) can be defined as \( A^{ij} = \frac{1}{2}(T^{ij} - T^{ji}) \). Note the change of indices and sign in these definitions according to the properties of symmetric and antisymmetric tensors.
04

Verify the decompositon

Upon substituting \( S^{ij} \) and \( A^{ij} \) into the equation from step 2, we get \( T^{ij} = \frac{1}{2}(T^{ij} + T^{ji}) + \frac{1}{2}(T^{ij} - T^{ji}) \). Simplifying gives \( T^{ij} = T^{ij} \), verifying our decomposition.
05

Discuss generalization

The decomposition in the manner stated above is indeed true for tensors of any rank. It is a fundamental property of tensors.

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Key Concepts

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

Symmetric Tensor
A symmetric tensor is a special type of tensor that exhibits a distinct behavior with respect to its indices.
By definition, a tensor is called symmetric when swapping any pair of its indices doesn’t change its value.
For a rank-two tensor, like a matrix, if we have the elements arranged as \( T^{ij} \) and swapping these indices \( T^{ij} = T^{ji} \), the tensor is symmetric.
In simpler terms, the element at position \((i, j)\) is equal to the element at position \((j, i)\).
  • A symmetric tensor will have its elements mirrored across the main diagonal.
  • This property makes symmetric tensors particularly useful in physics and engineering as they can represent quantities like stress and strain, where directionality is mirrored.
A symmetric tensor can be decomposed from any general tensor by taking the average of each pair of entries and its swapped counterpart.
This decomposition is expressed as: \( S^{ij} = \frac{1}{2}(T^{ij} + T^{ji}) \).
This ensures that the resulting tensor retains its symmetry across the diagonal.
Antisymmetric Tensor
Antisymmetric tensors, conversely, have a property that starkly contrasts with symmetric tensors.
An antisymmetric tensor changes sign whenever two of its indices are swapped.
For a rank-two tensor: \( T^{ij} = -T^{ji} \).
  • If you exchange any pair of indices in an antisymmetric tensor, the value becomes negative.
  • The diagonal elements of an antisymmetric tensor are always zero because \( T^{ii} = -T^{ii} \), and the only number equal to its own negative is zero.
The antisymmetric part of a general tensor \( T^{ij} \), can be derived by the formula: \( A^{ij} = \frac{1}{2}(T^{ij} - T^{ji}) \).
This expression ensures that the resulting tensor changes sign with an index swap, preserving antisymmetry.
Antisymmetric tensors are essential in physics, particularly in electromagnetism and fluid mechanics, where they represent rotational characteristics.
Contravariant Tensor
Tensors are versatile mathematical entities used to generalize scalar, vector, and more complex relations across spaces.
A contravariant tensor, specifically, transforms a certain way under coordinate transformations.
  • In simple terms, the components of a contravariant tensor change in opposition to the change in the coordinate grid. This means if the coordinate grid scales up, the tensor's components scale down, and vice versa.
  • Mathematically, if \( x^i \) are the original coordinates and \( \bar{x}^i \) are the new coordinates, the components of a contravariant tensor transform as: \( T^{i'} = \frac{\partial x^{i'}}{\partial x^i} T^i \).
Understanding contravariant tensors is crucial when dealing with spaces where direction and magnitude both matter, such as in physics and advanced calculus.
They lay the foundation for more profound operations and transformations, essential when tensors interact with complex spaces and systems.
Whether decomposing into symmetric and antisymmetric components or transforming across coordinate grids, contravariant tensors maintain their fundamental properties throughout.

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Most popular questions from this chapter

Let \(V\) be a finite-dimensional vector space and \(\omega \in \Lambda^{2}(\mathcal{V}) .\) Suppose there exist a pair of vectors \(\mathbf{e}_{1}, \mathbf{e}_{1}^{\prime} \in \mathcal{V}\) such that \(\omega\left(\mathbf{e}_{1}, \mathbf{e}_{1}^{\prime}\right) \neq 0\). Let \(\mathcal{P}_{1}\) be the plane spanned by \(\mathbf{e}_{1}\) and \(\mathbf{e}_{1}^{\prime}\), and \(\mathcal{V}_{1}\) the \(\boldsymbol{\omega}\) -orthogonal complement of \(\mathcal{P}_{1}\). Show that \(\mathcal{V}_{1} \cap \mathcal{P}_{1}=0\), and that \(\mathbf{v}-\boldsymbol{\omega}\left(\mathbf{v}, \mathbf{e}_{1}^{\prime}\right) \mathbf{e}_{1}+\boldsymbol{\omega}\left(\mathbf{v}, \mathbf{e}_{1}\right) \mathbf{e}_{1}^{\prime}\) is in \(\mathcal{V}_{1}\).

Show that the components of a tensor product are the products of the components of the factors: $$ (\mathbf{U} \otimes \mathbf{T})_{j_{1} \ldots j_{s+l}}^{i_{1} \ldots i_{r+k}}=U_{j_{1} \ldots j_{j}}^{i_{1} \ldots i_{r}} T_{j_{s+1} \ldots j_{s+l}}^{i_{r+1} \ldots i_{r+k}} $$.

Suppose that \(\mathcal{V}\) is a symplectic vector space and \(\mathbf{v}, \mathbf{v}^{\prime} \in \mathcal{V}\) are expressed in a canonical basis of \(\mathcal{V}\) with coefficients \(\left\\{x_{i}, y_{i}, z_{i}\right\\}\) and \(\left\\{x_{i}^{\prime}, y_{i}^{\prime}, z_{i}^{\prime}\right\\}\). Show that $$ \omega\left(\mathbf{v}, \mathbf{v}^{\prime}\right)=\sum_{i=1}^{n}\left(x_{i} y_{i}^{\prime}-x_{i}^{\prime} y_{i}\right) . $$

Show the following vector identities, using the definition of cross products in terms of \(\epsilon_{i j k}\) (a) \(\quad \mathbf{A} \times \mathbf{A}=0\). (b) \(\boldsymbol{\nabla} \cdot(\mathbf{A} \times \mathbf{B})=(\boldsymbol{\nabla} \times \mathbf{A}) \cdot \mathbf{B}-(\boldsymbol{\nabla} \times \mathbf{B}) \cdot \mathbf{A} .\) (c) \(\boldsymbol{\nabla} \times(\mathbf{A} \times \mathbf{B})=(\mathbf{B} \cdot \boldsymbol{\nabla}) \mathbf{A}+\mathbf{A}(\boldsymbol{\nabla} \cdot \mathbf{B})-(\mathbf{A} \cdot \boldsymbol{\nabla}) \mathbf{B}-\mathbf{B}(\boldsymbol{\nabla} \cdot \mathbf{A})\) (d) \(\boldsymbol{\nabla} \times(\boldsymbol{\nabla} \times \mathbf{A})=\boldsymbol{\nabla}(\boldsymbol{\nabla} \cdot \mathbf{A})-\nabla^{2} \mathbf{A} .\)

Using the exterior product show whether the following three vectors are linearly dependent or independent: $$ \begin{array}{l} \mathbf{v}_{1}=2 \mathbf{e}_{1}-\mathbf{e}_{2}+3 \mathbf{e}_{3}-\mathbf{e}_{4} \\\ \mathbf{v}_{2}=-\mathbf{e}_{1}+3 \mathbf{e}_{2}-2 \mathbf{e}_{4}, \\ \mathbf{v}_{3}=3 \mathbf{e}_{1}+2 \mathbf{e}_{2}-4 \mathbf{e}_{3}+\mathbf{e}_{4} . \end{array} $$

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