/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 To show that it takes more than ... [FREE SOLUTION] | 91Ó°ÊÓ

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To show that it takes more than direction and magnitude to make a vector, let's try to define a vector which we'll name squrl \(\mathbf{F}\) by a relation like Eq. 61 with the right-hand side squared: $$ (\operatorname{squrl} \mathbf{F}) \cdot \hat{\mathbf{i}}=\left[\lim _{a_{i} \rightarrow 0} \frac{\int_{C_{i}} \mathbf{F} \cdot d \mathbf{s}}{a_{i}}\right]^{2} $$ Prove that this does not define a vector. (Hint: Consider reversing the direction of f.).

Short Answer

Expert verified
Squaring the integral removes direction dependence, violating vector properties, hence it doesn't define a vector.

Step by step solution

01

Understanding the Expression

The given expression \((\operatorname{squrl} \mathbf{F}) \cdot \hat{\mathbf{i}}=\left[\lim _{a_{i}\rightarrow 0} \frac{\int_{C_{i}} \mathbf{F} \cdot d\mathbf{s}}{a_{i}}\right]^{2}\) describes a component along \(\hat{\mathbf{i}}\) that involves the square of a limit. Here, the notation suggests evaluating a line integral around a closed path \(C_{i}\), with \(a_{i}\) representing some small area.
02

Vector Characteristics

A genuine vector quantity should obey two main conditions: (1) Magnitude and direction are independent of the choice of the coordinate system, and (2) It reverses sign when the direction is reversed.
03

Analyzing Magnitude and Direction

The operation on the right results in a squared limit expression. Squaring removes the sign dependence on the original integral \(\int_{C_{i}} \mathbf{F} \cdot d\mathbf{s}\). This means the magnitude remains the same regardless of the path's orientation, violating the vector's reversal property.
04

Applying the Hint - Reversing the Direction

Consider the case where the path \(C_{i}\) is reversed. For a true vector, reversing the direction of the path should result in reversing the sign of the quantity. Here, \(\int_{C_{i}}\mathbf{F}\cdot d\mathbf{s}\) would become \(-\int_{C_{i}}\mathbf{F}\cdot d\mathbf{s}\), but since the expression is squared, the result does not change.
05

Conclusion

Since reversing the direction of \(\mathbf{F}\) does not change the squared term, \(\operatorname{squrl} \mathbf{F}\) does not change sign. Thus, the directionality criterion for vectors is violated, showing that \(\operatorname{squrl} \mathbf{F}\) cannot be a vector.

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Key Concepts

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

Line Integral
In vector calculus, a line integral is an essential concept that involves the integration of a vector field along a given curve or path. It provides valuable insight into the behavior of vectors by considering both magnitude and direction. Line integrals can evaluate work done by a force field on an object as it moves along a path or calculate fluid flow across a boundary.

The mathematical expression for a line integral of a vector field \(\mathbf{F}\) along a curve \(C\) is \(\int_{C} \mathbf{F} \cdot d\mathbf{s}\). The dot product \(\mathbf{F} \cdot d\mathbf{s}\) implies how much of the vector field aligns with the curve's tangential direction.

For closed paths, the line integral is particularly significant in assessing circulation and flux, linking to concepts like Green's Theorem and Stokes' Theorem. When applied correctly, line integrals help in characterizing dynamic system properties effectively.
Vector Properties
Vectors are not just quantities possessing magnitude; they are entities that also demand direction for a complete description. Two essential properties help us classify a true vector:
  • The magnitude and direction must remain unchanged regardless of any transformations applied to the coordinate system.
  • A vector should reverse its sign if the path or direction is reversed, inherently maintaining its orientation fidelity.
In the context of the exercise, the concept of a vector could be misunderstood by considering expressions like \((\operatorname{squrl} \mathbf{F}) \cdot \hat{\mathbf{i}}\). Squaring this expression fails the second property because squaring erases any direct effects of directional reversal. Thus, despite possessing magnitude, it does not reverse its direction, disqualifying it from being a true vector.
Coordinate System Independence
A crucial characteristic of vectors is their independence from the coordinate systems used to describe them. This means that any proper vector should maintain its properties under any transformations of the coordinate axes, be it rotation, translation, or reflection.

In practical terms, the ability to transform from one coordinate system to another without altering a vector's real-world interpretation is fundamental in physics and engineering. When checking for vector properties, this coordinate independence is vital as it ensures uniformity and applicability across different scenarios and analyses.

When expressions dependent on coordinate choices are mistaken for vectors, like in the exercise's example, the inability to maintain the same magnitude and direction under coordinate adjustments ultimately highlights the misconstruction of assuming such expressions define vectors in any universal sense.

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Most popular questions from this chapter

A thin disk, radius \(3 \mathrm{~cm}\), has a circular hole of radius \(1 \mathrm{~cm}\) in the middle. There is a uniform surface charge of \(-4 \mathrm{esu} / \mathrm{cm}^{2}\) on the disk. (a) What is the potential in statvolts at the center of the hole? (Assume zero potential at infinite distance.) (b) An electron, starting from rest at the center of the hole, moves out along the axis, experiencing no forces except repulsion by the charges on the disk. What velocity does it ultimately attain? (Electron mass \(=9 \times 10^{-28} \mathrm{gm} .\) )

Does the function \(f(x, y)=x^{2}+y^{2}\) satisfy the two-dimensional Laplace's equation? Does the function \(g(x, y)=x^{2}-y^{2}\) ? Sketch the latter function, calculate the gradient at the points \((x=\) \(0, y=1) ;(x=1, y=0) ;(x=0, y=-1) ;\) and \((x=-1, y=\) 0) and indicate by little arrows how these gradient vectors point.

Describe the electric field and the charge distribution that go with the following potential: $$ \begin{array}{ll} \phi=x^{2}+y^{2}+z^{2} \quad \text { for } x^{2}+y^{2}+z^{2}

Calculate the curl and the divergence of each of the following vector fields. If the curl turns out to be zero, try to discover a scalar function \(\phi\) of which the vector field is the gradient: (a) \(F_{x}=x+y ; F_{y}=-x+y ; F_{z}=-2 z\). (b) \(G_{x}=2 y ; G_{y}=2 x+3 z ; G_{z}=3 y\). (c) \(H_{x}=x^{2}-z^{2} ; H_{y}=2 ; H_{z}=2 x z\).

Consider a charge distribution which has the constant density \(\rho\) everywhere inside a cube of edge \(b\) and is zero everywhere outside that cube. Letting the electric potential \(\phi\) be zero at infinite distance from the cube of charge, denote by \(\phi_{0}\) the potential at the center of the cube and \(\phi_{1}\) the potential at a corner of the cube. Determine the ratio \(\phi_{0} / \phi_{1}\). The answer can be found with very little calculation by combining a dimensional argument with superposition. (Think about the potential at the center of a cube with the same charge density and with twice the edge length.)

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