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Chapter 3: Problem 25

Show that if \(\mathbf{A}\) is a \(\left(\begin{array}{c}2 \\\ 0\end{array}\right)\) tensor and \(\mathbf{B}\) a \(\left(\begin{array}{c}0 \\\ 2\end{array}\right)\) tensor, then $$ A^{\alpha \beta} B_{\alpha \beta} $$ is frame invariant, i.e. a scalar

Short Answer

Expert verified
The expression \( A^{\alpha \beta} B_{\alpha \beta} \) is a frame-invariant scalar, as demonstrated by contraction.

Step by step solution

01

Understand the Question

We need to determine if the expression \( A^{\alpha \beta} B_{\alpha \beta} \) results in a scalar, irrespective of the coordinate system in which it is evaluated.
02

Define Tensor Types

Identify the types of tensors involved. Here, \( \mathbf{A} \) is a \((2,0)\) tensor, meaning it has two contravariant indices \( \alpha \) and \( \beta \). \( \mathbf{B} \) is a \((0,2)\) tensor with two covariant indices \( \alpha \) and \( \beta \).
03

Perform Contraction

Evaluate the expression \( A^{\alpha \beta} B_{\alpha \beta} \) by performing contraction between the indices. This results in a summation over all indices, with contravariant indices paired with covariant indices, leading to a scalar quantity.
04

Use Index Transformation

Consider how tensors transform: \( A'^{\mu u} = \frac{\partial x'^\mu}{\partial x^\alpha} \frac{\partial x'^u}{\partial x^\beta} A^{\alpha \beta} \) and \( B'_{\mu u} = \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^u} B_{\alpha \beta} \). When contracted as in \( A'^{\mu u} B'_{\mu u} \), all transformation coefficients cancel out.
05

Conclusion of Invariance

Post-contraction, after cancelling of transformation terms, the result \( A^{\alpha \beta} B_{\alpha \beta} \) remains unchanged regardless of the reference frame. Therefore, it is a scalar.

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Key Concepts

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

Tensor Contraction and Its Role in Frame Invariance
Tensor contraction is a mathematical operation that simplifies tensors by summing over paired indices, transforming them into lower-rank tensors or scalars. In this exercise, tensor contraction transforms the expression involving a (2,0) tensor \( \mathbf{A} \) and a (0,2) tensor \( \mathbf{B} \). The expression \( A^{\alpha \beta} B_{\alpha \beta} \) is evaluated by summing the products of corresponding contravariant and covariant components.
  • Each index \( \alpha \) in \( A^{\alpha \beta} \) has a counterpart in \( B_{\alpha \beta} \), creating pairs \( \alpha \alpha \) and \( \beta \beta \) that are summed over, effectively 'contracting' them.

  • This contraction results in a single numerical value, often called a scalar, which holds true across all coordinate transformations.
  • The key result of this process is that the contracted expression remains invariant under change of basis. Thus, tensor contraction is a core operation in ensuring quantities maintain their physical meaning, irrespective of the chosen reference frame.
Understanding (2,0) Tensors
A (2,0) tensor, as in the case of \( \mathbf{A} \), is a two-dimensional tensor with contravariant indices. These indices, typically represented with superscripts, are not dependent on the metric of the space in which they operate. This makes them suitable for transformations that require adaptability and flexibility.
  • Contravariant Nature: Contravariant indices transform with the basis vectors of the space. This means that when you change coordinate bases, the components of the tensor change inversely with respect to the new basis.

  • Directionality: In geometrical terms, (2,0) tensors can be envisioned as defining planes or surfaces in space, dictated by the two directions specified by the indices.
  • Applications: These tensors are often used to represent physical quantities such as fluxes and fields that need to be described in any coordinate system without losing their inherent characteristics.
Appreciating (0,2) Tensors
A (0,2) tensor, such as \( \mathbf{B} \), is characterized by two covariant indices. Unlike their contravariant counterparts, covariant indices, marked by subscripts, scale with the metric of the space. This relationship is essential for accurately depicting interactions and quantities rooted in physical contexts.
  • Covariant Nature: Covariant indices are aligned with the inner product in the space, meaning they transform in concert with the coordinate changes. They usually relate to concepts like gradients and forces.

  • Dual Concept: (0,2) tensors operate on vectors within the space, allowing the definition of interactions or measurements, much like the concept of potential energy is measured between points.
  • Real-world Usage: These tensors play pivotal roles in formulating mathematical models that describe how changes in one context affect the structure and behavior within another, such as in elasticity and electromagnetic theory.

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Most popular questions from this chapter

Prove, by geometric or algebraic arguments, that \(\mathrm{d} f\) is normal to surfaces of constant \(f\).

Let \(S\) be the two-dimensional plane \(x=0\) in three-dimensional Euclidean space. Let \(\tilde{n} \neq 0\) be a normal one-form to \(S\). (a) Show that if \(\vec{V}\) is a vector which is not tangent to \(S\), then \(\tilde{n}(\vec{V}) \neq 0 .\) (b) Show that if \(\tilde{n}(\vec{V})>0\), then \(\tilde{n}(\vec{W})>0\) for any \(\vec{W}\), which points toward the same side of \(S\) as \(\vec{V}\) does (i.e. any \(\vec{W}\) whose \(x\) components has the same sign as \(V^{x}\) ). (c) Show that any normal to \(S\) is a multiple of \(\tilde{n}\). (d) Generalize these statements to an arbitrary three-dimensional surface in fourdimensional spacetime.

(a) Suppose \(\mathbf{A}\) is an antisymmetric \(\left(\begin{array}{l}2 \\\ 0\end{array}\right)\) tensor. Show that \(\left\\{A_{\alpha \beta}\right\\}\), obtained by lowering indices by using the metric tensor, are components of an antisymmetric \(\left(\begin{array}{l}0 \\ 2\end{array}\right)\) tensor. (b) Suppose \(V^{\alpha}=W^{\alpha}\). Prove that \(V_{\alpha}=W_{\alpha}\).

Prove that tensor differentiation obeys the Leibniz (product) rule: $$ \nabla(\mathbf{A} \otimes \mathbf{B})=(\nabla \mathbf{A}) \otimes \mathbf{B}+\mathbf{A} \otimes \nabla \mathbf{B} $$

Consider a timelike unit four-vector \(\vec{U}\), and the tensor \(\mathbf{P}\) whose components are given by $$ P_{\mu \nu}=\eta_{\mu v}+U_{\mu} U_{\nu} $$ (a) Show that \(\mathbf{P}\) is a projection operator that projects an arbitrary vector \(\vec{V}\) into one orthogonal to \(\vec{U}\). That is, show that the vector \(\vec{V}_{\perp}\) whose components are $$ V_{\perp}^{\alpha}=P^{\alpha}{ }_{\beta} V^{\beta}=\left(\eta_{\beta}^{\alpha}+U^{\alpha} U_{\beta}\right) V^{\beta} $$ is (i) orthogonal to \(\vec{U}\), and (ii) unaffected by \(\mathbf{P}\) : $$ V_{\perp \perp}^{\alpha}:=P^{\alpha}{ }_{\beta} V_{\perp}^{\beta}=V_{\perp}^{\alpha} $$ (b) Show that for an arbitrary non-null vector \(\vec{q}\), the tensor that projects orthogonally to it has components $$ \eta_{\mu v}-q_{\mu} q_{v} /\left(q^{\alpha} q_{\alpha}\right) $$ How does this fail for null vectors? How does this relate to the definition of \(\mathbf{P} ?\) (c) Show that \(\mathbf{P}\) defined above is the metric tensor for vectors perpendicular to \(\vec{U}\) : $$ \begin{aligned} \mathbf{P}\left(\vec{V}_{\perp}, \vec{W}_{\perp}\right) &=\mathbf{g}\left(\vec{V}_{\perp}, \vec{W}_{\perp}\right) \\ &=\vec{V}_{\perp} \cdot \vec{W}_{\perp} \end{aligned} $$
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