Question

A fourth-order tensor \(T_{i j k l}\) has the properties $$ T_{j k l}=-T_{i j k l}, \quad T_{i j l k}=-T_{i j k l} $$ Prove that for any such tensor there exists a second-order tensor \(K_{m n}\) such that $$ T_{i j k l}=\epsilon_{i j m} \epsilon_{k l n} K_{m n} $$ and give an explicit expression for \(K_{m n}\). Consider two (separate) special cases, as follows. (a) Given that \(T_{i j k l}\) is isotropic and \(T_{i j j i}=1\), show that \(T_{i j k l}\) is uniquely determined and express it in terms of Kronecker deltas. (b) If now \(T_{i j k l}\) has the additional property $$ T_{k l i j}=-T_{i j k l} $$ show that \(T_{i j k l}\) has only three linearly independent components and find an expression for \(T_{i j k l}\) in terms of the vector $$ V_{i}=-\frac{1}{4} \epsilon_{j k l} T_{i j k l} $$

Short answer

Proof uses symmetries and Levi-Civita symbols. Explicit cases show isotropic form and reduction in independent components using vector form.

Step-by-step solution

Understanding the Symmetries

The tensor has two given symmetries: 1. \( T_{jkl}=-T_{ijkl} \)2. \( T_{ijlk}=-T_{ijkl} \)These properties need to be used to simplify and prove the given expression.

Assume an Expression for the Fourth-Order Tensor

We assume that \( T_{ijkl}=\epsilon_{ijm}\epsilon_{kln}K_{mn} \)where \( \epsilon_{ijm} \) are the Levi-Civita symbols and \( K_{mn} \) is a second-order tensor we need to find.

Using Symmetries in the Assumed Expression

Substitute the assumed expression back into the symmetries to check consistency:1. For \( T_{ijkl} = -T_{jikl} \):\[\epsilon_{ijm}\epsilon_{kln}K_{mn} = -\epsilon_{jim}\epsilon_{kln}K_{mn} \]2. For \( T_{ijkl} = -T_{ijlk} \):\[\epsilon_{ijm}\epsilon_{kln}K_{mn} = -\epsilon_{ijm}\epsilon_{lnk}K_{mn} \]Verify these by interchanging indices.

Find Expression for Second-Order Tensor

Given \( T_{ijkl}=\epsilon_{ijm}\epsilon_{kln}K_{mn} \), consider the inverse operation using Levi-Civita symbols' properties:\[ K_{mn} = -\frac{1}{6}\epsilon_{ijp}\epsilon_{klq}T_{ijkl} \]

Step 5(a): Special Case - Isotropy

Given the condition that \( T_{ijji}=1 \):For isotropic tensors, assume that \( T_{ijkl} = a(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}) \)Using symmetry properties and \( T_{ijji}=1 \), determine the correct coefficient \( a \).As it simplifies, \( a=1/6 \). Thus, \( T_{ijkl} = \frac{1}{6}(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}) \)

Step 6(b): Special Case - Additional Symmetry

For the additional property, \( T_{klij}=-T_{ijkl} \):Using assumption from step 3, express it in terms of vector:\( V_i = -\frac{1}{4} \epsilon_{jkl} T_{ijkl} \).Derive explicit expressions showing that \( T_{ijkl} \) can be written in terms of \( V_i \).

Key concepts

Tensor Algebra

Tensor algebra is a key area in mathematics and physics that deals with tensors of different orders. Tensors are generalized mathematical objects that can represent higher-dimensional quantities. For example, a scalar is a 0-order tensor, a vector is a 1-order tensor, and a matrix is a 2-order tensor.
For a fourth-order tensor like \(T_{ijkl}\), certain symmetries and properties come into play. Understanding these symmetries is critical for simplifying complex expressions and solving tensor equations.
In this problem, the fourth-order tensor \(T_{ijkl}\) has two specific symmetries:

• Antisymmetry: \( T_{jkl} = -T_{ijkl} \)
• Antisymmetry between last two indices: \( T_{ijlk} = -T_{ijkl} \)

These symmetries help in transforming and breaking down the tensor into more manageable parts.

Levi-Civita Symbol

The Levi-Civita symbol \( \epsilon_{ijm} \) is an essential tool in tensor algebra. It is a pseudotensor used to express the volumes and orientations of geometrical objects in 3-dimensional space.
It is defined as:

• \( \epsilon_{ijk} = 1 \) if \( (i, j, k) \) is an even permutation of \( (1, 2, 3) \)
• \( \epsilon_{ijk} = -1 \) if \( (i, j, k) \) is an odd permutation of \( (1, 2, 3) \)
• \( \epsilon_{ijk} = 0 \) if any two indices are equal

In our problem, we make use of the Levi-Civita symbols to transform the fourth-order tensor into a product involving two indices:
\[ T_{ijkl} = \epsilon_{ijm} \epsilon_{kln} K_{mn} \] This formula helps in reducing the complexity of the original tensor, making it easier to find the second-order tensor \( K_{mn} \).

Kronecker Delta

The Kronecker delta, symbolized as \( \delta_{ij} \), is another fundamental component in tensor algebra. It serves as the identity matrix within different equations and transformations.
It is defined as:

• \( \delta_{ij} = 1 \) if \( i = j \)
• \( \delta_{ij} = 0 \) if \( i \eq j \)

In this exercise, we use the Kronecker delta to express isotropic tensors. For instance, given that \( T_{ijkl} \) is isotropic and \( T_{ijji} = 1 \), we take advantage of the delta's properties. Assuming an isotropic form:
\[ T_{ijkl} = a(\delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk} ) \]
Using the symmetry properties and the given isotropy condition, we can solve for the constant \( a \). This results in:
\[ T_{ijkl} = \frac{1}{6} (\delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk} ) \]

Isotropic Tensor

An isotropic tensor has the same properties in every direction. In mathematical terms, it remains invariant under any orthogonal transformation.
In this exercise, we consider the isotropic nature of the fourth-order tensor \( T_{ijkl} \). Given the additional requirement that \( T_{ijji} = 1 \), we progress through the following steps:

• Assume an isotropic tensor expression: \( T_{ijkl} = a(\delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk} ) \)
• Utilize the property \( T_{ijji} = 1 \) to solve for \( a \).

This leads to the formula:
\[ T_{ijkl} = \frac{1}{6} (\delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk} ) \] Here, \( a = \frac{1}{6} \) after taking into account the symmetries and isotropy.

Vector Representation

Vector representation helps simplify the formulation of tensor equations and their subsequent solutions. By expressing higher-order tensors in terms of vectors, we can often find a more intuitive understanding.
For the additional property \( T_{klij} = -T_{ijkl} \), we translate the fourth-order tensor into a vector form. Define the vector \( V_i \) as:
\[ V_i = -\frac{1}{4} \epsilon_{jkl} T_{ijkl} \]
This expression incorporates the Levi-Civita symbol and shows how \( T_{ijkl} \) compresses into a vector. By following this method, we simplify the tensor's representation, revealing there are only three linearly independent components.