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Chapter 3: Problem 3

(a) Prove, by writing out all the terms, the validity of the following $$ \tilde{p}\left(A^{\alpha} \vec{e}_{\alpha}\right)=A^{\alpha} \tilde{p}\left(\vec{e}_{\alpha}\right) $$ (b) Let the components of \(\tilde{p}\) be \((-1,1,2,0)\), those of \(\vec{A}\) be \((2,1,0,-1)\) and those of \(\vec{B}\) be \((0,2,0,0)\). Find (i) \(\tilde{p}(\vec{A})\); (ii) \(\tilde{p}(\vec{B})\); (iii) \(\tilde{p}(\vec{A}-3 \vec{B})\); (iv) \(\tilde{p}(\vec{A})-3 \tilde{p}(\vec{B})\).

Short Answer

Expert verified
Part (a): The expression is valid for a linear function. Part (b): (i) -1, (ii) 2, (iii) -7, (iv) -7.

Step by step solution

01

Understand the Expression

We need to prove \( \tilde{p}\left(A^{\alpha} \vec{e}_{\alpha}\right)=A^{\alpha} \tilde{p}\left(\vec{e}_{\alpha}\right) \). The set \( \{\vec{e}_{\alpha}\} \) represents a basis, and \( A^{\alpha} \) are the components of a vector \( \vec{A} \). The tilde notation indicates a function that takes a vector and returns a scalar, often a type of projection or linear form.
02

Expand the Left Side

Expand the left side: \( \tilde{p}(A^{\alpha} \vec{e}_{\alpha}) = \tilde{p}\left(A^{1} \vec{e}_{1} + A^{2} \vec{e}_{2} + \ldots + A^{n} \vec{e}_{n}\right) \). If \( \tilde{p} \) is linear, it can be written as: \( A^{1} \tilde{p}(\vec{e}_{1}) + A^{2} \tilde{p}(\vec{e}_{2}) + \ldots + A^{n} \tilde{p}(\vec{e}_{n}) \).
03

Expand the Right Side

Write out the right side: \( A^{\alpha} \tilde{p}(\vec{e}_{\alpha}) = A^{1} \tilde{p}(\vec{e}_{1}) + A^{2} \tilde{p}(\vec{e}_{2}) + \ldots + A^{n} \tilde{p}(\vec{e}_{n}) \). This shows that the right side is the same as the expanded left side if \( \tilde{p} \) is linear.
04

Conclusion for Part (a)

Since both the expanded forms of left and right side are equal, we have \( \tilde{p}\left(A^{\alpha} \vec{e}_{\alpha}\right)=A^{\alpha} \tilde{p}\left(\vec{e}_{\alpha}\right) \) is valid for a linear function \( \tilde{p} \).
05

Compute \( \tilde{p}(\vec{A}) \)

Given the components of \( \tilde{p} \) are \((-1,1,2,0)\) and of \( \vec{A} \) are \((2,1,0,-1)\), calculate using the dot product: \( \tilde{p}(\vec{A}) = (-1)\times 2 + 1\times 1 + 2\times 0 + 0\times (-1) = -2 + 1 + 0 + 0 = -1 \).
06

Compute \( \tilde{p}(\vec{B}) \)

Given the components of \( \vec{B} \) are \((0,2,0,0)\), calculate: \( \tilde{p}(\vec{B}) = (-1)\times 0 + 1\times 2 + 2\times 0 + 0\times 0 = 0 + 2 + 0 + 0 = 2 \).
07

Compute \( \tilde{p}(\vec{A} - 3 \vec{B}) \)

First find \( \vec{A} - 3 \vec{B} \): It is \((2, 1, 0, -1) - 3(0, 2, 0, 0) = (2, 1, 0, -1) - (0, 6, 0, 0) = (2, -5, 0, -1)\). Then calculate \( \tilde{p}(\vec{A} - 3 \vec{B}) = (-1)\times 2 + 1\times (-5) + 2\times 0 + 0\times (-1) = -2 - 5 + 0 + 0 = -7 \).
08

Compute \( \tilde{p}(\vec{A}) - 3 \tilde{p}(\vec{B}) \)

Using previous results, \( \tilde{p}(\vec{A}) = -1 \) and \( \tilde{p}(\vec{B}) = 2 \). Calculate: \( \tilde{p}(\vec{A}) - 3 \tilde{p}(\vec{B}) = -1 - 3 \times 2 = -1 - 6 = -7 \).
09

Verify Conclusion for Part (b)

By comparing steps 7 and 8, we see that the two calculations give the same result \(-7\), which confirms the linearity and distribution properties worked correctly.

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Key Concepts

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

Linear Transformations
A linear transformation is a powerful concept in linear algebra. It is essentially a function between vector spaces that preserves the operations of vector addition and scalar multiplication. To check if a function \(f\) is a linear transformation, it must satisfy two main conditions:
  • Additivity: \(f(\vec{u} + \vec{v}) = f(\vec{u}) + f(\vec{v})\)
  • Homogeneity: \(f(c\vec{u}) = c f(\vec{u})\), where \(c\) is a scalar.
When a linear transformation is applied to a vector, it results in another vector within a possibly different vector space. For instance, if \(\tilde{p}\) is a linear transformation and \(\vec{v}\) is a vector, applying \(\tilde{p}\) to \(\vec{v}\) does not change the nature of \(\tilde{p}\) being a linear transformation, as it retains the structure and properties of linearity. In exercises like the one above, verifying linearity involves checking these properties, often by expanding expressions and demonstrating that they produce consistent results.
Vector Spaces
Vector spaces form the foundational framework within which concepts like linear transformations and the dot product are defined. A vector space over a field is a set of vectors that can be added together and multiplied by scalars. For example, consider a set \( V \) with operations addition \(+\) and scalar multiplication \(\cdot\) that satisfies rules such as:
  • There is an additive identity (zero vector): every vector \( \vec{v} \) has an additive inverse \( -\vec{v} \) such that \( \vec{v} + (-\vec{v}) = \vec{0} \).
  • Scalar multiplication is distributive with respect to vector addition and field addition.
  • For any scalars \( a, b \) and vector \( \vec{v} \), \((a + b)\cdot\vec{v} = a\cdot\vec{v} + b\cdot\vec{v} \).
  • Vectors can be multiplied by 1, resulting in themselves: \(1\cdot\vec{v} = \vec{v}\).
Many real-world problems and computational tasks rely on these properties because they guarantee consistent and reliable results. Understanding vector spaces allows one to grasp more advanced topics like basis, dimension, and more complex transformations vital to linear algebra applications.
Dot Product
The dot product, also known as the scalar product, is a fundamental operation in linear algebra that takes two vectors and returns a single scalar value. It is computed by multiplying corresponding components of the vectors and then summing those products. Given two vectors \( \vec{a} = (a_1, a_2, .., a_n) \) and \( \vec{b} = (b_1, b_2, .., b_n) \), the dot product is defined as:
\[ \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + ... + a_n b_n \]
This operation has several useful properties:
  • It is commutative: \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\).
  • It distributes over vector addition: \(\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}\).
  • It is bilinear, meaning it is linear in each of its arguments.
  • If the dot product of two vectors is zero, the vectors are orthogonal (perpendicular).
In problems like the one above, the dot product is often used to compute projections and for various geometric interpretations, such as measuring angles between vectors. Its simplicity and utility make it a cornerstone operation in vector calculus and physics.

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Most popular questions from this chapter

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.

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 \(\vec{U}\) : $$ \begin{aligned} \mathbf{P}\left(\vec{V}_{\perp}, \vec{W}_{\perp}\right) &=\mathbf{g}\left(\vec{V}_{\perp}, \vec{W}_{\perp}\right) \\ &=\vec{V}_{\perp} \cdot \vec{W}_{\perp} \end{aligned} $$

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

Consider the coordinates \(u=t-x, v=t+x\) in Minkowski space. (a) Define \(\vec{e}_{u}\) to be the vector connecting the events with coordinates \(\\{u=1\), \(v=0, y=0, z=0\\}\) and \(\\{u=0, v=0, y=0, z=0\\}\), and analogously for \(\vec{e}_{v}\) Show that \(\vec{e}_{u}=\left(\vec{e}_{t}-\vec{e}_{x}\right) / 2, \vec{e}_{v}=\left(\vec{e}_{t}+\vec{e}_{x}\right) / 2\), and draw \(\vec{e}_{u}\) and \(\vec{e}_{v}\) in a spacetime diagram of the \(t-x\) plane. (b) Show that \(\left\\{\vec{e}_{u}, \vec{e}_{v}, \vec{e}_{y}, \vec{e}_{z}\right\\}\) are a basis for vectors in Minkowski space. (c) Find the components of the metric tensor on this basis. (d) Show that \(\vec{e}_{u}\) and \(\vec{e}_{v}\) are null and not orthogonal. (They are called a null basis for the \(t-x\) plane.) (e) Compute the four one-forms \(\tilde{\mathrm{d}} u, \tilde{\mathrm{d}} v, \mathbf{g}\left(\vec{e}_{u}, \quad\right), \mathbf{g}\left(\vec{e}_{v}, \quad\right)\) in terms of \(\tilde{\mathrm{d}} t\) and \(\tilde{\mathrm{d}} x\).

Given the following vectors in \(\mathcal{O}\) : $$ \vec{A} \rightarrow{\mathcal{O}}(2,1,1,0), \vec{B} \rightarrow{\mathcal{O}}(1,2,0,0), \vec{C} \rightarrow{\mathcal{O}}(0,0,1,1), \vec{D} \underset{\vec{O}}(-3,2,0,0) $$ (a) show that they are linearly independent; (b) find the components of \(\tilde{p}\) if $$ \tilde{p}(\vec{A})=1, \tilde{p}(\vec{B})=-1, \tilde{p}(\vec{C})=-1, \tilde{p}(\vec{D})=0 $$ (c) find the value of \(\tilde{p}(\vec{E})\) for $$ \vec{E} \rightarrow(1,1,0,0) $$ (d) determine whether the one-forms \(\tilde{p}, \tilde{q}, \tilde{r}\), and \(\tilde{s}\) are linearly independent if \(\tilde{q}(\vec{A})=\) $$ \begin{aligned} &\tilde{q}(\vec{B})=0, \quad \tilde{q}(\vec{C})=1, \quad \tilde{q}(\vec{D})=-1, \quad \tilde{r}(\vec{A})=2, \quad \tilde{r}(\vec{B})=\tilde{r}(\vec{C})=\tilde{r}(\vec{D})=0, \quad \tilde{s}(\vec{A})= \\ &-1 . \tilde{s}(\vec{B})=-1, \tilde{s}(\vec{C})=\tilde{s}(\vec{D})=0 \end{aligned} $$
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