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

Short Answer

Expert verified
The components of the tensor product (\(\mathbf{U} \otimes \mathbf{T}\)) are indeed the direct products of the components of the original tensors \(\mathbf{U}\) and \(\mathbf{T}\). This was to be shown and follows directly from the definition of a tensor product.

Step by step solution

01

Understand the Tensor Product

The tensor product of two tensors \(\mathbf{U}\) and \(\mathbf{T}\) is another tensor, denoted as \(\mathbf{U} \otimes \mathbf{T}\). It is defined in such a way that its components in an arbitrary basis are the products of the components of \(\mathbf{U}\) and \(\mathbf{T}\) in the same basis.
02

Write Down the Equation

The task now is to write down the equation we want to prove, which is: \[ (\mathbf{U} \otimes \mathbf{T})_{j_{1} \ldots j_{s+l}}^{i_{1} \ldots i_{r+k}}=U_{j_{1} \ldots j_{j}}^{i_{1} \ldots i_{r}} T_{j_{s+1} \ldots j_{s+l}}^{i_{r+1} \ldots i_{r+k}} \]
03

Break Down the Left-Hand Side

The left-hand side of the equation represents the components of the tensor product \(\mathbf{U} \otimes \mathbf{T}\).
04

Break Down the Right-Hand Side

The right-hand side of the equation represents the product of the components of the tensors \(\mathbf{U}\) and \(\mathbf{T}\)
05

Conclude the Proof

As per the definition of the tensor product, the components of the tensor product are indeed the direct products of the components of the original tensors. Hence, the equation reflects exactly this property and does not need further operations or changes. The equation holds, as it is exactly the definition of the tensor product. Hence, the task is already completed by the definition of a tensor product.

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Key Concepts

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

Tensor Components
Understanding tensor components is fundamental in grappling with tensors, particularly in physics and mathematics.

A tensor can be viewed as a generalization of scalars (zero-order tensors), vectors (first-order tensors), and matrices (second-order tensors) to an arbitrary number of dimensions. The components of a tensor represent its entries in a particular basis. For instance, a vector in a 3-dimensional space has three components, which could represent directions along the x, y, and z-axis.

The power of tensor components lies in their transformation properties. When you change the basis (for example, by rotating your coordinate system), the components of a tensor transform in a specific way that preserves the geometric or physical quantities they represent.

The tensor product operation combines two tensors to form a new tensor with each component resulting from multiplying the corresponding components of the initial tensors. In mathematical physics, where tensors often represent physical quantities like stress or strain, being able to calculate and understand their components is pivotal for solving problems related to elasticity, fluid dynamics, and general relativity.
Mathematical Physics
Mathematical physics intertwines the rigor and methods of mathematics with the empirical and theoretical aspects of physics. It's an area of study that involves developing new mathematical structures and tools, like tensors, to describe physical systems.

Tensors, in particular, serve as a linchpin in this field by providing a concise mathematical language for formulating and solving problems in different areas of physics, including quantum mechanics, electrodynamics, and relativity theory.

Within this crossover domain, tensor algebra and calculus become essential as practitioners work to describe complex interactions and fields. Understanding tensor products, for example, is crucial when dealing with quantum states in quantum mechanics, where the state space of a composite system is described by the tensor product of the state spaces of its parts.
Tensors in Physics
Tensors are woven into the very fabric of physics, serving as a mathematical representation of physical quantities that are independent of the chosen coordinate system.

For example, in the theory of relativity, the spacetime interval, stress-energy tensor, and electromagnetic field tensor are central constructs. These tensors convey properties of spacetime and energy that remain consistent across all frames of reference, a core principle of relativistic physics.

In classical mechanics, tensors describe stresses and strains within materials, while in fluid dynamics, they represent various properties of fluid flow, including viscosity and vorticity. The practice of breaking down tensors into their components allows physicists to extract tangible information, like the magnitude and direction of forces within a material, or the rate of energy transfer in a system.
Tensor Algebra
Tensor algebra is an extension of vector algebra to tensors of any rank. It is the branch of mathematics that deals with operations on tensors, which includes addition, scalar multiplication, and tensor products among other operations.

Just as vectors can be added together or multiplied by scalars to create new vectors, tensors can also be combined in various ways to form new tensors. The tensor product is one such operation that combines two tensors to create a tensor of higher rank. The beauty of tensor algebra is in its ability to describe these complex operations in a structured and consistent way.

The rules governing tensor components during multiplication (as shown in the original exercise and solution) ensure that the resulting tensor product behaves in a way that is consistent with both the algebraic and geometric interpretation of tensors. Mastery of tensor algebra is indispensable for students and professionals dealing with multidimensional physical problems.

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Most popular questions from this chapter

Let \(\left\\{\mathbf{e}_{i}\right\\}_{i=1}^{N}\) be a \(g\) -orthonormal basis of \(\mathcal{V}\). Let \(\boldsymbol{\eta}\) be the matrix with elements \(\eta_{i j}\), which is the matrix of \(\mathbf{g}\) in this orthonormal basis. Let \(\left\\{\mathbf{v}_{j}\right\\}_{j=1}^{N}\) be another (not necessarily orthonormal) basis of \(\mathcal{V}\) with a transformation matrix \(\mathrm{R}\), i.e., \(\mathbf{v}_{i}=r_{i}^{j} \mathbf{e}_{j}\). (a) Using G to denote the matrix of \(\mathbf{g}\) in \(\left\\{\mathbf{v}_{j}\right\\}_{j=1}^{N}\), show that $$ \operatorname{det} \mathbf{G}=\operatorname{det} \boldsymbol{\eta}(\operatorname{det} \mathbf{R})^{2}=(-1)^{v}(\operatorname{det} \mathbf{R})^{2} $$ In particular, the sign of this determinant is invariant. Why is det \(\mathrm{G}\) not equal to \(\operatorname{det} \eta ?\) Is there any conflict with the statement that the determinant is basis-independent? (b) Let \(\boldsymbol{\mu}\) be the volume element related to \(\mathbf{g}\), and let \(|G|=|\operatorname{det} G|\). Show that if \(\left\\{\mathbf{v}_{j}\right\\}_{j=1}^{N}\) is positively oriented relative to \(\boldsymbol{\mu}\), then $$ \boldsymbol{\mu}=|G|^{1 / 2} \mathbf{v}_{1} \wedge \mathbf{v}_{2} \wedge \cdots \wedge \mathbf{v}_{N} $$.

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?

What is the tensor product of \(\mathbf{A}=2 \mathbf{e}_{x}-\mathbf{e}_{y}+3 \mathbf{e}_{z}\) with itself?

Let \(\mathbf{A} \in \Lambda^{2}(\mathcal{V})\) with components \(A^{i j}\). Show that \(\mathbf{A} \wedge \mathbf{A}=0\) if and only if \(A^{i j} A^{k l}-A^{i k} A^{j l}+A^{i l} A^{j k}=0\) for all \(i, j, k, l\) in any basis.

Starting with the definition of the permutation tensor \(\delta_{j_{1} j_{2} \ldots j_{N}}^{i_{1} i_{2} \ldots i_{N}}\), and writing the wedge product in terms of the antisymmetrized tensor product, show that $$ \delta_{j_{1} j_{2} \ldots j_{N}}^{i_{1} i_{2} \ldots i_{N}}=\sum_{\pi} \epsilon_{\pi\left(j_{1}\right) \pi\left(j_{2}\right) \ldots \pi\left(j_{N}\right)} \delta_{\pi\left(j_{1}\right)}^{i_{1}} \delta_{\pi\left(j_{2}\right)}^{i_{2}} \cdots \delta_{\pi\left(j_{N}\right)}^{i_{N}} $$.

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