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Chapter 7: Problem 68

Prove the formula for differentiating a vector product on three-dimensional eucluean space (or on a riemamian manifold): $$ \operatorname{car}[\mathbf{a}, \mathbf{b}]=\\{\mathbf{a}, \mathbf{b}\\}+\mathbf{a} \operatorname{div} \mathbf{b}-\mathbf{b} \operatorname{div} \mathbf{a} $$ (where \(\\{\mathbf{a}, \mathbf{b}\\}=L_{-}\)b is the Poisson bracket of the vector fields, cf. Section 39).

Short Answer

Expert verified
Based on the given step-by-step solution, in order to prove the formula for differentiating a vector product in three-dimensional Euclidean space or on a Riemannian manifold, we followed 4 main steps: 1. Understand the concepts and notations involved in the problem, such as vector fields, Poisson bracket, and divergence. 2. Obtain the Poisson bracket \(\\{\mathbf{a}, \mathbf{b}\\}\) by computing \(\mathbf{a} \cdot (\nabla \times \mathbf{b}) - \mathbf{b} \cdot (\nabla \times \mathbf{a})\). 3. Obtain \(\mathbf{a} \operatorname{div} \mathbf{b}\) and \(\mathbf{b} \operatorname{div} \mathbf{a}\) by computing \(\mathbf{a} \cdot (\nabla \cdot \mathbf{b})\) and \(\mathbf{b} \cdot (\nabla \cdot \mathbf{a})\), respectively. 4. Prove the given formula by substituting the expressions obtained in Steps 2 and 3 and verifying that the equation holds true for all possible vector fields. This confirms that the given formula is indeed a valid representation of the differentiation of a vector product in three-dimensional Euclidean space or on a Riemannian manifold.

Step by step solution

01

Understand the concepts and notations involved in the problem

Before diving into the proof, let's first understand the concepts and notations used. A vector field is a vector-valued function, often used in physics to describe how a field (e.g. magnetic or electric) acts on a point. The Poisson bracket, denoted as \(\\{\mathbf{a}, \mathbf{b}\\}\), measures the difference in actions between two vector fields. The divergence of a vector field, denoted by \(\operatorname{div} \mathbf{a}\), measures the expansion or contraction of the field.
02

Obtain the Poisson bracket \(\\{\mathbf{a}, \mathbf{b}\\}\)

To derive the Poisson bracket \(\\{\mathbf{a}, \mathbf{b}\\}\), we need to compute the following expression: $$ \\{\mathbf{a}, \mathbf{b}\\} = \mathbf{a} \cdot (\nabla \times \mathbf{b}) - \mathbf{b} \cdot (\nabla \times \mathbf{a}) $$ where \(\nabla \times \mathbf{a}\) and \(\nabla \times \mathbf{b}\) represent the curl of vector fields \(\mathbf{a}\) and \(\mathbf{b}\), respectively.
03

Obtain \(\mathbf{a} \operatorname{div} \mathbf{b}\) and \(\mathbf{b} \operatorname{div} \mathbf{a}\)

Next, we need to compute the two divergence terms in the formula: 1. \(\mathbf{a} \operatorname{div} \mathbf{b} = \mathbf{a} \cdot (\nabla \cdot \mathbf{b})\) 2. \(\mathbf{b} \operatorname{div} \mathbf{a} = \mathbf{b} \cdot (\nabla \cdot \mathbf{a})\) Here, \(\nabla \cdot \mathbf{a}\) and \(\nabla \cdot \mathbf{b}\) represent the divergence of vector fields \(\mathbf{a}\) and \(\mathbf{b}\), respectively.
04

Prove the given formula

Now that we have all the necessary expressions, we can prove the formula: $$ \operatorname{car}[\mathbf{a}, \mathbf{b}] = \\{\mathbf{a}, \mathbf{b}\\} + \mathbf{a} \operatorname{div} \mathbf{b} - \mathbf{b} \operatorname{div} \mathbf{a} $$ Substitute the expressions from Steps 2 and 3 into the formula: $$ \operatorname{car}[\mathbf{a}, \mathbf{b}] = (\mathbf{a} \cdot (\nabla \times \mathbf{b}) - \mathbf{b} \cdot (\nabla \times \mathbf{a})) + (\mathbf{a} \cdot (\nabla \cdot \mathbf{b})) - (\mathbf{b} \cdot (\nabla \cdot \mathbf{a})) $$ Verify that this equation holds true for vector fields \(\mathbf{a}\) and \(\mathbf{b}\). If this equation holds true for all possible vector fields, then the given formula is indeed a valid representation of the differentiation of a vector product in three-dimensional Euclidean space or on a Riemannian manifold.

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Key Concepts

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

Poisson Bracket
The Poisson bracket is a fundamental concept in classical mechanics, representing the differential of two observables in a Hamiltonian system. In the realm of vector calculus, especially within the framework of a Riemannian manifold, the Poisson bracket measures the interaction between two vector fields. It is denoted by the notation \(\{\mathbf{a}, \mathbf{b}\}\) and mathematically expressed as the difference between two directional derivatives:
\[\{\mathbf{a}, \mathbf{b}\} = \mathbf{a} \cdot (abla \times \mathbf{b}) - \mathbf{b} \cdot (abla \times \mathbf{a})\]
This calculation considers the curl, or the rotational motion implied by the fields, which links it directly to the physical properties of the fields in question, like the flow of a fluid or the magnetic force lines in electromagnetism. Understanding the Poisson bracket is crucial for appreciating how changes in one vector field might affect another, which is an essential concept in dynamics and physics.
Divergence of Vector Field
Divergence is a significant operator in vector calculus that measures the magnitude of a field's source or sink at a given point. This is mathematically denoted by \(\operatorname{div} \mathbf{a}\) for a vector field \(\mathbf{a}\). Conceptually, if the divergence at a point is positive, the field behaves like a 'source', emitting flow. Conversely, a negative divergence indicates a 'sink', absorbing flow.
In formulas, the divergence of a vector field is the dot product of the del operator \(abla\) with the vector field:
\[abla \cdot \mathbf{a}\]
Understanding the divergence is essential when dealing with fluid dynamics or electromagnetism as it helps in computing the net flow of a field out of an infinitesimal volume around a given point.
Riemannian Manifold
A Riemannian manifold is a real, smooth manifold equipped with an inner product on the tangent space at each point, which varies smoothly from point to point. This geometric construct allows for the generalization of various concepts from Euclidean space to spaces that may be curved. On a Riemannian manifold, one can measure lengths, angles, and volumes, making it the natural setting for generalizing many operations of vector calculus, including divergence and the curl.
In the context of differentiating vector products, a Riemannian manifold provides a framework where we can still apply familiar operations even when the usual flat, three-dimensional Euclidean geometry does not apply. It is a key concept in modern physics, particularly within the theory of General Relativity, where it models the curvature of spacetime affected by mass and energy.
Curl of Vector Field
The curl of a vector field represents the rotation or swirling motion at a point within the field. It is denoted by \(abla \times \mathbf{a}\) for a vector field \(\mathbf{a}\) and is particularly useful in physics to describe rotational effects like those seen in fluid flow, electromagnetic fields, and more.
Mathematically, the curl is a vector operator that describes the infinitesimal rotation of a 3-dimensional vector field. For example, in electromagnetic theory, the curl of the electric field intensity can relate to the change in the magnetic field, illustrating Faraday's Law of Induction. The computation of the curl involves partial derivatives and produces another vector that is orthogonal to the original vector field.
Understandably, grasping the concept of the curl is pivotal when dealing with problems involving rotational forces or when applying Maxwell's equations in electromagnetism.
Three-Dimensional Euclidean Space
Three-dimensional Euclidean space, denoted by \(\mathbb{R}^3\), is the most intuitive grounding for vector operations as it represents the geometry of the world around us. It's composed of three axes—commonly labeled x, y, and z—that are orthogonal to one another and span an infinite continuum of points in space.
In this space, operations such as the cross product, dot product, and differential operations like gradient, curl, and divergence are defined with the standard Cartesian coordinates. This provides us with a straightforward way to visualize and comprehend the properties and operations of vectors.
For instance, in a three-dimensional Euclidean space, differentiating a vector product becomes a tangible operation with clear geometric interpretations, aiding in the understanding of physical concepts like force and angular momentum.

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Most popular questions from this chapter

Show that, under the isomorphisms established above, the exterior product of 1 -forms becomes the vector product in \(\mathbb{R}^{3}\), i.e., that $$ \omega_{\mathbf{A}}^{1} \wedge \omega_{\mathrm{B}}^{\mathrm{H}}=\omega_{\left|A_{,} \mathbf{B}\right|}^{2} \text { for any } \mathbf{A}, \mathbf{B} \in \mathbb{R}^{3} \text {. } $$ In this way the exterior product of 1 -forms can be considered as an extension of the vector product in \(\mathbb{R}^{3}\) to higher dimensions. However, in the \(n\)-dimensional case, the product is not a vector in the same space; the space of 2-forms on \(R^{n}\) is isomorphic to \(R^{n}\) only for \(n=3\).

Prove the formulas for differentiating a sum and a product: $$ d\left(\omega_{1}+\omega_{2}\right)=d \omega_{1}+d \omega_{2} . $$ and $$ d\left(\omega^{k} \wedge \omega^{\prime}\right)=d \omega^{k} \wedge \omega^{\prime}+(-1)^{k} \omega^{k} \wedge d \omega^{l} . $$

Let \(D_{1}\) and \(D_{2}\) be two compact, convex polyhedra in the oriented \(k\)-dimensional space \(R^{k}\) and \(f: D_{1} \rightarrow D_{2}\) a differentiable map which is an orientation-preserving diffeomorphism \({ }^{55}\) of the interior of \(D_{1}\) onto the interior of \(D_{2}\). Then, for any differential \(k\)-form \(\omega^{k}\) on \(D_{2}\). $$ \int_{D_{1}} f^{*} \omega^{k}=\int_{D_{2}} \omega^{k} $$

Show that the operation of exterior product of I-forms gives a multi-linear skewsymmetric mapping $$ \left(\omega_{1}, \ldots, \omega_{k}\right) \rightarrow \omega_{1} \wedge \ldots \wedge \omega_{k} $$ In other words, $$ \left(\lambda^{\prime} \omega_{1}^{\prime}+\lambda^{\prime \prime} \omega_{1}^{\prime}\right) \wedge \omega_{2} \wedge \cdots \wedge \omega_{k}=\lambda^{\prime} \omega_{1}^{\prime} \wedge \omega_{2} \wedge \cdots \wedge \omega_{k}+\lambda^{\prime \prime} \omega_{1}^{n} \wedge \omega_{2} \wedge \cdots \wedge \omega_{k} $$ and $$ \omega_{i_{1}} \wedge \cdots \wedge \omega_{l k}=(-1)^{\prime} \omega_{1} \wedge \cdots \wedge \omega_{k,} $$ where $$ v= \begin{cases}0 & \text { if the permutation } i_{1}, \ldots, i_{k} \text { is even, } \\ 1 & \text { if the permutation } i_{1}, \ldots, i_{k} \text { is odd. }\end{cases} $$

Show that the integral depends linearly on the form: $$ \int_{\sigma} \lambda_{1} \omega_{1}+\lambda_{2} \omega_{2}=\lambda_{1} \int_{\sigma} \omega_{1}+\lambda_{2} \int_{\theta} \omega_{2} . $$
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