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Chapter 3: Problem 11

a) Let \(u_{k l}^{i j}\) be a tensor of the type indicated. Show that each of the following is a tensor: \(w_{k}^{j}=u_{k i}^{i j}, v_{k}^{i}=u_{k j}^{i j}\) b) Prove generally that contraction of a tensor yields a tensor.

Short Answer

Expert verified
Based on the given step by step solution, provide a short answer to the problem: In this problem, we are given a tensor \(u_{k l}^{i j}\) and are required to show that the following quantities, \(w_{k}^{j}=u_{k i}^{i j}\) and \(v_{k}^{i}=u_{k j}^{i j}\), are also tensors. Additionally, we need to prove that contraction of a tensor yields a tensor. By applying a linear transformation on the given tensor and the derived expressions, and using the transformation rule for tensors, we showed that the transformed tensors have the same structure as the original tensors, thus proving that \(w_{k}^{j}\) and \(v_{k}^{i}\) are indeed tensors. In general, the contraction operation involves summing over one lower index and one upper index, as seen in the case of tensors \(w_{k}^{j}\) and \(v_{k}^{i}\). Therefore, we can generally conclude that the contraction of a tensor yields a tensor.

Step by step solution

01

Transform the given tensor

We will start by applying a transformation on the given tensor \(u_{k l}^{i j}\). We shall apply a linear transformation represented by a transformation matrix \(L\). The transformed tensor will be given as follows: \(u^\prime_{k^\prime l^\prime}^{i^\prime j^\prime}=L^i_{i^\prime} L^j_{j^\prime} L^k_{k^\prime} L^l_{l^\prime} u_{k l}^{i j}\) Notice that we applied the transformation matrix \(L\) to all indices of the tensor \(u_{k l}^{i j}\).
02

Transform the tensor \(w_{k}^{j}\) and \(v_{k}^{i}\)

For the two derived expressions, we also apply the same linear transformation: a) For \(w_{k}^{j}\), the transformed tensor will be given as: \(w^\prime_{k^\prime}^{j^\prime}=L^j_{j^\prime} L^k_{k^\prime} w_{k}^{j}\) And substituting the expression for \(w_{k}^{j}\), we get: \(w^\prime_{k^\prime}^{j^\prime}=L^j_{j^\prime} L^k_{k^\prime} u_{k i}^{i j}\) b) For \(v_{k}^{i}\), the transformed tensor will be given as: \(v^\prime_{k^\prime}^{i^\prime}=L^i_{i^\prime} L^k_{k^\prime} v_{k}^{i}\) And substituting the expression for \(v_{k}^{i}\), we get: \(v^\prime_{k^\prime}^{i^\prime}=L^i_{i^\prime} L^k_{k^\prime} u_{k j}^{i j}\)
03

Show that the transformed tensors have the same structure as the original tensors

Now let's make use of the transformation rule for tensors: a) For the transformed tensor \(w^\prime_{k^\prime}^{j^\prime}\), we can rewrite the expression as: \(w^\prime_{k^\prime}^{j^\prime}=L^j_{j^\prime} L^k_{k^\prime} u_{k i}^{i j}\) Now use the transformation rule for \(u_{k i}^{i j}\) by summing over its indices: \(w^\prime_{k^\prime}^{j^\prime}=L^j_{j^\prime} L^k_{k^\prime} (L^i_{i^\prime}L^k_{k^\prime}L^l_{l^\prime}u_{k l}^{i j})\) It is clear that the transformed expression for \(w_{k}^{j}\) follows the same structure as the original. Thus, \(w_{k}^{j}\) is a tensor. b) For the transformed tensor \(v^\prime_{k^\prime}^{i^\prime}\), we can rewrite the expression as: \(v^\prime_{k^\prime}^{i^\prime}=L^i_{i^\prime} L^k_{k^\prime} u_{k j}^{i j}\) Now use the transformation rule for \(u_{k j}^{i j}\) by summing over its indices: \(v^\prime_{k^\prime}^{i^\prime}=L^i_{i^\prime} L^k_{k^\prime} (L^i_{i^\prime}L^j_{j^\prime}L^k_{k^\prime}L^l_{l^\prime}u_{k l}^{i j})\) It is clear that the transformed expression for \(v_{k}^{i}\) also follows the same structure as the original. Thus, \(v_{k}^{i}\) is a tensor.
04

Contraction of a tensor yields a tensor

In general, the contraction of a tensor is achieved by summing over one lower index and one upper index. In the case of the tensors \(w_{k}^{j}\) and \(v_{k}^{i}\), we did this by summing over indices \(i\) and \(j\) respectively. Since the transformed tensors follow the same structure as the original tensors and the operation of contraction involved summing over one lower index and one upper index for each case, we can generally conclude that the contraction of a tensor yields a tensor.

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Key Concepts

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

Tensor Transformation
Understanding tensor transformation is essential for grasping the manipulation of tensors in higher-dimensional spaces. A tensor can be thought of as a mathematical object that extends the concept of scalars, vectors, and matrices to more complex relationships. When a tensor undergoes a transformation, essentially, each of its components is converted from one coordinate system to another. This process is governed by certain rules that maintain the geometric and physical properties of the tensor.

For example, in our exercise, the tensor transformation rule is applied to each component of the tensor using a transformation matrix, which provides a systematic way to carry out this transition. The critical takeaway here for students is that the transformation of tensor components is not arbitrary; it adheres to a structured law that maintains the tensor's intrinsic characteristics.
Linear Transformation
A linear transformation is a mapping between two vector spaces that preserves the operations of addition and scalar multiplication. The effects of a linear transformation can be captured through a transformation matrix. In context with the given exercise, when we apply a linear transformation to a tensor, we are essentially using the transformation matrix to map each index of the tensor to a new coordinate system while preserving the linearity of the tensor.

To ensure a correct understanding, students should recognize that linear transformations are pivotal in maintaining the structure of a tensor during the transformation. The linear characteristic implies that the new tensor, after transformation, must exhibit a direct correspondence to the original tensor, with its components adjusted linearly according to the transformation matrix used.
Transformation Matrix
The transformation matrix is a powerful tool in tensor analysis. It is a square matrix that acts as a bridge to convert the components of a tensor from one basis to another. In the provided exercise, the matrix denoted by 'L' performs the reshaping of our tensor's indices through multiplication. It's critical for students to comprehend the role of the transformation matrix in the legitimacy of a tensor as a geometric entity, unaffected by the choice of coordinate system.

Highlighting the importance of this concept, students should recognize the transformation matrix as a foundational aspect of tensor algebra, with its role in altering tensors executed in an organized and consistent manner, as demonstrated in the given exercise.
Tensor Analysis
Tensor analysis is the field that encompasses the study and application of tensors. It goes beyond simple operations to include concepts such as transformation, contraction, and invariance under coordinate change. In the context of the exercise, tensor analysis is crucial as it informs us how to properly execute operations like contraction that reduce a tensor's order but preserve its nature as a tensor. It means that the result of the contraction, as shown in our steps with tensors w_{k}^{j} and v_{k}^{i}, is still a valid tensor in its right.

It's imperative for students to realize that tensor analysis covers these transformations and operations, ensuring the consistency and legitimacy of manipulations applied to tensors. Understanding these principles allows students to process and handle higher-order tensors with the same confidence as they do with vectors and matrices.

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Most popular questions from this chapter

Show that the force field \((3.4)\) is the gradient of the scalar $$ f=\log \frac{\sqrt{(x-1)^{2}+y^{2}}}{\sqrt{(x+1)^{2}+y^{2}}} . $$

In \(E^{2}\) let \(\left(\xi^{1}, \xi^{2}\right)\) be standard coordinates and let \(\left(x^{1}, x^{2}\right)\) be new coordinates given by \(x^{1}=3 \xi^{1}+2 \xi^{2}, x^{2}=4 \xi^{1}+3 \xi^{2}\). Find the \(\left(x^{i}\right)\) components of the following tensors, for which the components in standard coordinates are given. a) \(u_{i}\), where \(U_{1}=\xi^{1} \xi^{2}, U_{2}=\xi^{1}-\xi^{2}\). b) \(v^{i}\), where \(V^{1}=\xi^{1} \cos \xi^{2}, V^{2}=\xi^{1} \sin \xi^{2}\). c) \(w_{i j}\), where \(W_{11}=0, W_{12}=\xi^{1} \xi^{2}, W_{21}=-\xi^{1} \xi^{2}, W_{22}=0\). d) \(z_{j}^{i}\), where \(Z_{1}^{1}=\xi^{1}+\xi^{2}, Z_{2}^{1}=Z_{1}^{2}=3 \xi^{1}+2 \xi^{2}, Z_{2}^{2}=\xi^{1}-\xi^{2}\).

Prove the identity (cf. Problem 12): curl \((\mathbf{u} \times \mathbf{v})=\mathbf{u} \operatorname{div} \mathbf{v}-\mathbf{v} \operatorname{div} \mathbf{u}+(\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{u}-(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{v}\)

Prove the identity (cf. Problem 12): \(\operatorname{grad}(\mathbf{u} \cdot \mathbf{v})=(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{v}+(\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{u}+(\mathbf{u} \times \operatorname{curl} \mathbf{v})+(\mathbf{v} \times \operatorname{curl} \mathbf{u})\).

Prove the following identities: a) \(\operatorname{div}[\mathbf{u} \times(\mathbf{v} \times \mathbf{w})]=(\mathbf{u} \cdot \mathbf{w}) \operatorname{div} \mathbf{v}-(\mathbf{u} \cdot \mathbf{v}) \operatorname{div} \mathbf{w}+\operatorname{grad}(\mathbf{u} \cdot \mathbf{w}) \cdot \mathbf{v}-\operatorname{grad}(\mathbf{u} \cdot \mathbf{v}) \cdot \mathbf{w}\), b) \(\operatorname{div}(\operatorname{grad} f \times f \operatorname{grad} g)=0\), c) \(\operatorname{curl}(\operatorname{curl} \mathbf{v}+\operatorname{grad} f)=\) curl curl \(\mathbf{v}\), d) \(\nabla^{2} f=\operatorname{div}(\operatorname{curl} \mathbf{v}+\operatorname{grad} f)\). These should be established by means of the identities already found in this chapter and not by expanding into components.
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