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Chapter 22: Problem 15

Sind \(e_{i}\) die Komponenten eines Einheitsvektors und \(v_{i}\) jene eines beliebigen Vektors des \(\mathbb{R}^{3}\), so gilt die Identit盲t $$ v_{i}=v_{j} e_{j} e_{i}+\varepsilon_{i j k} e_{j} \varepsilon_{k \operatorname{lm}} v_{l} e_{m-} $$ Begr眉nden Sie diese Identit盲t. Was bedeutet sie in Vektorform?

Short Answer

Expert verified
Question: Prove the identity $v_i = v_j e_j e_i + \varepsilon_{ijk} e_j \varepsilon_{klm} v_l e_m$ for a unit vector and an arbitrary vector in 鈩澛�, and express its meaning in vector form. Answer: The given identity represents a geometric decomposition of an arbitrary vector in 鈩澛� in terms of projections along unit vectors and rotations. The proof involves simplifying dot product and cross product expressions using the properties of the Levi-Civita symbol and Kronecker delta. In vector form, the identity is given by $\vec{v} = (\vec{v} \cdot \vec{e_i})\vec{e_i} + (\vec{e_i} \times \vec{e_j}) \times \vec{v}$.

Step by step solution

01

Define the terms

First, let's define the terms in the identity. - \(e_i\) are the components of a unit vector. - \(v_i\) are the components of an arbitrary vector in 鈩澛�. - \(\varepsilon_{ijk}\) is the Levi-Civita symbol, defined as: \(\varepsilon_ {ijk} = \begin{cases} 1, & \text{if } (i, j, k) \text{ is an even permutation of } (1, 2, 3)\\ -1, & \text{if } (i, j, k) \text{ is an odd permutation of } (1, 2, 3)\\ 0, & \text{if any indices are repeated}\end{cases}\)
02

Prove the identity

Now let's prove the given identity: $$ v_i = v_j e_j e_i + \varepsilon_{ijk} e_j \varepsilon_{klm} v_l e_m $$ Since \(e_i\) are components of a unit vector, we have \(e_ie_i = 1\). We can start with the definition of the dot product of two vectors: $$ v_i = \vec{v} \cdot \vec{e_i} = (\vec{v} \cdot \vec{e_j})\vec{e_i} = v_j e_j e_i $$ Now, we will work on the second term in the identity: $$ \varepsilon_{ijk} e_j \varepsilon_{klm} v_l e_m = \varepsilon_{ijk} e_j (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}) v_l e_m $$ Using the Kronecker delta properties, we can simplify this expression: $$ \delta_{il}\delta_{jm} v_l e_m - \delta_{im}\delta_{jl} v_l e_m = v_i e_j e_j - v_j e_i e_j $$ Combining the two terms, we have: $$ v_i = v_j e_j e_i + \varepsilon_{ijk} e_j \varepsilon_{klm} v_l e_m = v_i e_j e_j $$ Since \(e_j e_j = 1\), we can conclude that the identity is true.
03

Interpret the identity in vector form

Converting the given identity into vector form, we have: $$ \vec{v} = (\vec{v} \cdot \vec{e_i})\vec{e_i} + (\vec{e_i} \times \vec{e_j}) \times \vec{v} $$ This identity states that any vector can be decomposed into two parts: 1. The projection of the vector onto the unit vector (i.e., the component of the vector along the unit vector's direction). 2. The cross product of the unit vectors with the result again crossed with the given arbitrary vector, representing a rotation in space. Thus, the given identity is a geometric representation of vectors in 鈩澛� in terms of unit vectors and their combinations.

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Key Concepts

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

Unit Vector
A unit vector is a vector that has a magnitude of 1. It is often used to specify a direction without assigning it a magnitude. Unit vectors are helpful in defining the directionality of another vector when only the angle or orientation needs to be considered. In mathematical notation, if \( \vec{e} \) is a unit vector, then the magnitude \( \|\vec{e}\| = 1 \). Unit vectors are commonly denoted by symbols with a hat, such as \( \hat{e} \).

Unit vectors play a critical role in vector decomposition, especially when you need to find components of a vector along a specified direction. By projecting any vector onto a unit vector, you can isolate how much of the vector lies in that particular direction. This is the basis for many vector calculations in physics and engineering.
Cross Product
The cross product is an operation on two vectors in three-dimensional space, resulting in a new vector that is perpendicular to both original vectors. If \( \vec{a} \) and \( \vec{b} \) are two vectors, the cross product \( \vec{a} \times \vec{b} \) can be found using the formula:\[\vec{c} = \vec{a} \times \vec{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\]

The magnitude of the cross product can be described as \( \|\vec{a} \times \vec{b}\| = \|\vec{a}\|\|\vec{b}\|\sin(\theta) \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \). The direction of the resulting vector is determined by the right-hand rule. The cross product is particularly useful in determining the rotational effect and torque when analyzing forces in physics.

It is essential in applications involving vector decomposition, as it allows for the separation of vector components that are parallel versus those that are orthogonal to a given vector or axis.
Levi-Civita Symbol
The Levi-Civita symbol, \( \varepsilon_{ijk} \), is used in three-dimensional vector calculus to facilitate cross-product calculations and other vector operations. It is defined as follows:
  • \( \varepsilon_{ijk} = 1 \) if \( (i, j, k) \) is an even permutation of \( (1, 2, 3) \)
  • \( \varepsilon_{ijk} = -1 \) if \( (i, j, k) \) is an odd permutation of \( (1, 2, 3) \)
  • \( \varepsilon_{ijk} = 0 \) if any indices are repeated
This symbol is crucial when we need to express cross products and determinants in a succinct, algebraic form. It effectively encodes the orientation of a three-dimensional coordinate system (typically right-handed).

In vector decomposition, the Levi-Civita symbol is used to express elements of a vector perpendicular to two given vectors, ultimately leading to an easier manipulation of multi-index summations and simplifying complex vector identities.
Projection of Vectors
Projection involves projecting one vector onto another to find how much of one vector falls in the direction of another. It is an operation that results in a scalar or a vector. For vectors \( \vec{v} \) and \( \vec{e} \) (the latter being a unit vector), the projection of \( \vec{v} \) onto \( \vec{e} \) is given by:\[\text{proj}_{\vec{e}} \vec{v} = (\vec{v} \cdot \vec{e})\vec{e}\]

This formula highlights two components:
  • \( \vec{v} \cdot \vec{e} \) gives the magnitude of \( \vec{v} \) in the direction of \( \vec{e} \)
  • Multiplying by \( \vec{e} \) brings the vector back to having direction along \( \vec{e} \)
Projection is a key concept in vector decomposition as it allows us to break down a vector into components that are parallel and perpendicular to a specified axis or another vector. Recognizing the vector's ability to split into these components is particularly useful in physics for resolving forces and motions into their parts.

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Most popular questions from this chapter

Zerlegen Sie den Tensor \(t_{i j}\) mit $$ \left(t_{i j}\right)=\left(\begin{array}{ccc} 3 & 2 & 1 \\ 0 & -4 & 3 \\ 7 & 11 & -5 \end{array}\right) $$ in seinen symmetrischen Anteil \(s_{i j}\) und seinen alternierenden Anteil \(a_{i j}\) und berechnen Sie den Vektor \(d_{j}=\frac{1}{2} \varepsilon_{i j k} t_{i k}\)

Berechnen Sie den Projektionstensor \(n_{i j}\) zum Einheitsvektor \(\widehat{n}=\frac{1}{\sqrt{6}}(1,-1,2)^{T}\) und zerlegen Sie den Vektor \(v=(1,1,1)^{T}\) in zwei orthogonale Komponenten \(v^{\prime}\) und \(v^{\prime \prime}\), wobei \(v^{\prime}\) zu \(\hat{n}\) parallel ist.

Berechnen Sie im \(\mathbb{R}^{3}\) die nachstehend angef眉hrten Ausdr眉cke, die alle das Kronecker-Delta enthalten: $$ \begin{array}{cccc} \delta_{i i}, & \delta_{i j} \delta_{j k}, & \delta_{i j} \delta_{j i}, & \delta_{i j} \delta_{j k} \delta_{k i} . \\ \hline \end{array} $$

Beweisen Sie die Grassmann-Identit盲t von S. 714 $$ u \times(v \times w)=(u \cdot w) v-(u \cdot v) w $$ unter Benutzung der Tensordarstellung (22.12) von Vektorprodukten.

Wir wechseln von der Basis \(B=\left(\boldsymbol{b}_{1}, \boldsymbol{b}_{2}, \boldsymbol{b}_{3}\right) \mathrm{mit}\) dem kovarianten Metriktensor \(\left(g_{i j}\right)=\operatorname{diag}(1,2,1)\) zur Basis \(\bar{B}=\left(\bar{b}_{1}, \bar{b}_{2}, \bar{b}_{3}\right) \mathrm{mit}\) $$ \bar{b}_{1}=b_{1}, \bar{b}_{2}=b_{1}+b_{2}, \bar{b}_{3}=b_{1}+b_{2}+b_{3} $$ Wie sehen die neuen Komponenten \(\bar{t}_{i j}\) des Tensors \(t_{i j}\) mit $$ \left(t_{i j}\right)=\left(\begin{array}{rrr} 1 & -1 & 2 \\ 2 & 2 & -1 \\ -1 & -2 & 1 \end{array}\right) $$ aus? Berechnen Sie ferner die neuen Komponenten \(\bar{r}_{j}^{i}\), und zwar einerseits im neuen Koordinatensystem durch Hinaufziehen, andererseits aus den zugeh枚rigen alten Komponenten nach dem jeweiligen Transformationsgesetz.
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