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#  Tunneling Through the Math Barrier
The Fledgling Physics Student’s Field Guide to Essential Mathematics

W. J. Meese

(May 2018)

_For Kyo, who got the ball rolling (hopefully without too much slipping on my
part)._

## Acknowledgements

There are a few people whom I definitely need to acknowledge and thank as
their help and support made this project possible. First and foremost, I need
to thank the Internet and the spirit of Open Source projects. I definitely
would not have been able to learn nearly as much over the years without
information being available to me in the giant database that it currently is
in. Being able to come up with questions and research and then think over
possible solutions helped make this project possible, and I would be remiss to
ignore this fact. Thus, I would also like to thank everyone who continues to
make the Internet the a great tool for learning that it currently is.

Secondly, I would like to thank Professor Peter Persans (Rensselaer
Polytechnic Institute) for taking the time to edit this work and provide
helpful suggestions to improve its utility. I don’t know if he quite knew what
he was agreeing to by signing on to be my Master’s Project adviser, and so I
really do appreciate all the extra time he had to invent to thoroughly read my
work. Additionally, I would like to thank him for distributing snippets of
this book as it was being written to his Quantum Physics II and Honors Physics
II classes so I might get helpful feedback from those I aim to help. Luckily
for me, Linus Koepfer from the Honors Physics II $\beta$-tested the chapter on
Complex Algebra and brought a bunch of my typos to my attention, and so I
would also like to thank him as well for finding the time during the semester
to read my work.

I would also like to thank my sister, Morgan Meese, D.P.T., for her help with
brainstorming ideas for my title. She also brought up a pretty important point
to me that for a lot of people, this book will be a shock because, as she
says, it is Math Without Numbers. Without this reminder, I definitely would
have forgotten to make mention of how infrequently I use numbers, so that you,
my reader, is not too taken aback by their absence.

Additionally, I’d like to thank my parents, Colleen and Bill Meese, and my
sister, Katee Meese, for their support throughout the semester and for their
recommendations of who my target audience could be. Their thoughts and helpful
discussions, particularly early on, definitely helped me collect all of my
incoherent thoughts out of the ether so that they could be written down into
book form.

Nicholas Smieszek and Erika Nelson also deserve a shout-out for being
incredibly patient friends and roommates while I was almost always on my
computer typing away. I know that I would often forget conversations we would
have because my mind was on this project and my brain only has a finite RAM,
and so I really do appreciate how supportive of me and this project they were
throughout the semester.

Last but not least, I would like to thank Jennifer Freedberg for everything
that she contributed to this project. First off, she was the one to originally
encourage me to propose such a project. As I began to write this book, she
would kindly remind me to cut down on my rambly-ness and get to the point. She
also brainstormed many useful examples or would check my work when I thought I
had a cool way of deriving something. She also had to listen to me blabber
about this book for four months which definitely could not have always been
easy.

## README.tex

When I started my undergrad in 2014, I expected that there would be some
potential barrier for me to overcome between high school and college as far as
physics goes, but I didn’t think that there would be one for math. I thought,
if anything, I’d be ahead of the game since I came into college already
placing out of Calc II. To be clear, coming into college I might have known a
lot of math through the Calc II level, but I did not think about it in the way
a physicist does. The way in which physicists think about things is weird,
like I mean really out there. Or at least that was the way it seemed to me as
a freshman physics student. The first striking thing was that physicists, at
least those I had just met, do not remember a lot of formulas for things. In
fact, they rarely remember any unless they are immediately relevant to the
problem at hand. As I was just coming from the world of AP classes, this
realization was heretical and confusing. But the more shocking thing, even
without remembering all of those formulas, physicists could still smoothly
move from one mathematical idea to the next effortlessly. It was as if I were
watching a physicist speak mathematics fluently. Sure they may forget a word
or two here or there, but it was no different than when I had talked to
someone in the past and inserted an “umm…” or “uhh…” here or there. So it was
pretty clear to me that the way I had learned to think about math was wrong.
The physicists I saw did not work within the confines of the rules I had so
carefully followed to get decent AP scores in high school calculus. They used
what formulas they knew, but instead of working within some predetermined,
College-Board-approved template, they translated their thoughts into
equations.

My goal with this book is to provide some kind of bridge for mathematics
between the high-school-level and college-level for physics students. From my
perspective, our job as physicists is to observe and understand the universe
around us. Unfortunately our universe happens to be pretty complicated, at
least from a mathematical point-of-view. However, a lot of the underlying
physics — the underlying set of rules surrounding how things move and behave —
is usually not too complicated. Sure, when we couple things together,
everything turns disgusting, and we need to turn to really powerful computers
or simulators to get a lot accomplished. But the basic rules are not that bad.
My hope is to provide enough of a conceptual framework for you to learn to
read math as I saw a lot of other physicists do. I plan to cover some of the
topics that tend to trip up a lot of physics students, although I could not
cover these exhaustively. When I found that I should have covered more, but
could not due to finite time constraints, I left a reference to other work
that I think you may find helpful. However, unlike a lot of works, my focus is
to help modify your thinking of how math is used, rather than just pummel you
with algorithms for you to memorize without giving you the proper context for
such algorithms. Whenever you find my explanations long-winded, feel free to
proceed through faster. If you get stuck, you are always free to go back and
see what I might have referenced before.

I am not going to lie to you — I expect a lot out of you as a reader. It’s not
that I expect you be able to do everything out on your own — I hate when
textbooks do that. I will show you the mechanics of a lot of basic
calculations, or argue to you how I might interpret an equation based on its
logical underpinnings. I might also ask you to imagine certain scenarios, or
really push your inherent ability to reason beyond plugging in numbers
whenever they appear111I actually loathe numbers (I am really bad at them),
and so I work almost exclusively with symbols. Luckily, almost everything in
physics (and math, too) is symbolic because numerical values tend to hide
general trends.. That, in some sense, may be harder than just having you do
problems out without any kind of help. I recognize that learning is very hard
(I am still a student after all), and so I wrote a lot of this book as though
I was sitting next to you and explaining it in person. I hope that I can serve
as a guide, or a significant perturbation, so your transition from a high
school understanding of math to a physics-level understanding of math is not
as tough as it otherwise could be.

I really do think that many people with some hard work are more than capable
of understanding most of the intricacies in physics; unfortunately mathematics
can have a tendency to clutter everything, especially for people who may
struggle to read it as any other human language. But once you can use it as
such, it becomes any other human tool designed for us to make sense of
everything else in our lives. This book is long, but I wrote in in mind that
it can serve as a reference guide for you as you move throughout your
undergraduate career — although it honestly will probably be most helpful when
you are a first- and second-year student. Later on, this book may serve as a
memory bank for some interesting references or a place where you can find a
formula or two that you will forget (yes, as a physics major, you will forget
a lot of math; it’s part of the job). If you happen to have a PDF version,
then there are hyperlinks everywhere for you to use. If you have a hardcopy,
then feel free to still try and click on the equation numbers to activate the
hyperlinks. Worth a shot, right?

So, without further ado, let’s get started.

_William “Joe”_  _Meese_ $\sim$ _May 2018_
_M.Sc. Physics, Rensselaer Polytechnic Institute, 2018_
_B.S. Physics & Mathematics, Rensselaer Polytechnic Institute, 2017_

###### Contents

  1. Acknowledgements
  2. README.tex
  3. 1 Vectors
    1. 1.1 Mathematical Objects
    2. 1.2 Numbers, Lists, and Functions
      1. 1.2.1 Real Numbers
      2. 1.2.2 Collections of Real Numbers: Lists and Functions
    3. 1.3 Vector Operations
      1. 1.3.1 The Geometry Behind Vector Addition
      2. 1.3.2 The Geometry Behind Vector Subtraction
      3. 1.3.3 Vector Addition and Coordinate Systems
      4. 1.3.4 Unit Vectors
      5. 1.3.5 Component-wise Addition and Subtraction of Vectors
      6. 1.3.6 Scaling Vectors
      7. 1.3.7 Dot Product
      8. 1.3.8 Cross Product
      9. 1.3.9 Unit Vectors are Ambidextrous (but we like them to be Right-Handed)
      10. 1.3.10 Three-Dimensional Cross Products
      11. 1.3.11 A Note on Linearity
    4. 1.4 Concluding Remarks
  4. 2 Complex Algebra
    1. 2.1 The Lie of Imaginary Numbers
    2. 2.2 Some Important Definitions
    3. 2.3 Conjugates and Magnitudes
    4. 2.4 Cartesian versus Polar Representations
    5. 2.5 Multiplication = Dilation + Rotation
      1. 2.5.1 Roots of Unity
    6. 2.6 Functions of a Complex Variable
      1. 2.6.1 Polynomials
      2. 2.6.2 Complex Exponentiation Revisted
      3. 2.6.3 Complex Logarithms
      4. 2.6.4 Hyperbolic Trigonometry
    7. 2.7 Concluding Remarks
  5. 3 Calculus of a Single Variable
    1. 3.1 Position, Velocity, and Acceleration
    2. 3.2 Continuity
    3. 3.3 Differentiability and Integrability
      1. 3.3.1 Definition of Derivative: Rise Over Run
      2. 3.3.2 Fundamental Theorems of Calculus
      3. 3.3.3 Definition of (Riemann) Integration: Adding Boxes
    4. 3.4 Applications of One-Dimensional Calculus
      1. 3.4.1 Local Extrema
      2. 3.4.2 Taylor Series
      3. 3.4.3 A Couple of Key (Linear) Differential Equations in Physics
    5. 3.5 Concluding Remarks
  6. 4 Fourier Analysis
    1. 4.1 Frequencies versus Wavenumbers
    2. 4.2 Fourier Series as a Linear Combination
      1. 4.2.1 Review of Linear Combinations
      2. 4.2.2 The Fourier Series
      3. 4.2.3 Calculating the Fourier Series
    3. 4.3 Fourier Transforms
      1. 4.3.1 The Case of the Infinite Domain
      2. 4.3.2 The Complex Fourier Series
      3. 4.3.3 The Fourier Transform
      4. 4.3.4 Some Important Examples
    4. 4.4 Concluding Remarks

###### List of Figures

  1. 1.1 We assign real numbers to a number line by determining what length should be 1 unit and then project those lengths onto a line (or vector).
  2. 1.2 A function $f:A\rightarrow B$ takes every single point $a\in A$ and maps it into exactly one point $f(a)=b\in B$. The set of all $f(a)\in B$ is the $\textrm{range}(f)$.
  3. 1.3 Vectors are simply just mathematical objects that describe directions. This definition leads to an interesting way to add them.
  4. 1.4 The the _function_ of aligning the vectors $\vec{v}$ and $\vec{u}$ tail-to-tail leaves the resultant (output) vector $\vec{R}$ ambiguous. Drawing the resultant vector from tip-to-tip could then yield two different answers.
  5. 1.5 If we choose to use the vectors $\vec{x}$ and $\vec{y}$ of known lengths and directions, then we can rewrite vector $\vec{v}$ as the sum of $\vec{x}$ and $\vec{y}$. In this case, $\vec{v}=2.5\vec{x}+1.5\vec{y}$.
  6. 1.6 Here we combine adding vectors tip-to-tail with the components we can measure in our $\hat{x},\hat{y}$ coordinate system.
  7. 1.7 Whenever we have a function that scales all of the components of a vector $\vec{v}$ by the same real number $a$, then we interpret this as $a\vec{v}$. Notice that this leaves Eq. 1.5 unchanged.
  8. 1.8 By the Law of Cosines, we can relate the lengths of three legs of a triangle to their angles. Likewise, we can relate the lengths of these legs to their components. Therefore, we should be able talk about angles in terms of the components of the legs.
  9. 1.9 When we take the dot product of any vector $\vec{v}$ with any unit vector $\hat{u}$, then we obtain $v\cos\theta$ by Eq. 1.21. This situation corresponds geometrically to the diagram above, where the value $v\cos\theta$ is the length of $\vec{v}$ along the $\hat{u}$ direction in the (green) plane spanned by $\vec{v}$ and $\hat{u}$. If $v\cos\theta>1$ ($v\cos\theta<1$), this length will be longer (shorter) than $\hat{u}$.
  10. 1.10 A geometric calculation of the area of the central parallelogram created by $\vec{v}$ and $\vec{u}$, denoted by $|\vec{v}\times\vec{u}|$. The combination of geometry with the idea of vector components leads to a peculiar property of the area: without careful considerations, the resultant area may become negative!
  11. 1.11 A diagram of how we define right-handedness among the Cartesian unit vectors $\\{\hat{x},\hat{y},\hat{z}\\}$. Throughout this diagram, we choose to label each direction differently, but the direction of each vector remains the same in each set of axes. Starting from the leftmost set of axes, we have $\hat{x}\times\hat{y}=\hat{z}$, where the dashed (blue) vector defines the otherwise ambiguous cross product $\hat{z}$. By rotating the labels counter-clockwise and then taking (green vector) $\times$ (red vector) we can define $\hat{z}\times\hat{x}=\hat{y}$ and $\hat{y}\times\hat{z}=\hat{x}$.
  12. 1.12 An illustration of the right-hand rule.
  13. 2.1 If we consider $z=x+iy$, then we can draw it and its conjugate, $z^{\ast}=x-iy$, in the complex plane.
  14. 2.2 When we multiply the complex number $z=r\textrm{e}^{i\theta}$ by $w=\rho\textrm{e}^{i\phi}$, the magnitude of $z$ is dilated by a factor of $\rho$, while the angle of $z$ is rotated by $\mathrm{arg}(w)=\phi$. Thus $|zw|=r\rho$ and $\mathrm{arg}(zw)=\theta+\phi$.
  15. 2.3 In this figure, we multiply each of the following points along the complex unit circle by the imaginary (lateral!) unit $i=\textrm{e}^{\pi i/2}$: $(1,0)$, $(0,i)$, $(-1,0)$, and $(0,-i)$. Starting with $(1,0)$, we have $1\cdot i=i=\textrm{e}^{\pi i/2}$ over the solid red arc. Hence, multiplying $1$ by $i$, rotates $1$ by $\pi/2$ radians ($90^{\mathrm{o}}$). Next, we multiply $i$ by $i$ in the dashed blue arc. But $i^{2}=-1=\textrm{e}^{\pi i/2+\pi i/2}=\textrm{e}^{\pi i}$. Thus, again, multiplying by $i$ rotates our starting complex number $i$ by $\pi/2$ radians. I leave you to confirm the other two rotations for yourself.
  16. 2.4 A few plots of the symmetry in the roots of unity. The orange points are the complex numbers centered around their common origin. The vectors attached to each point is to help with our geometrical interpretation of the complex plane. The regular polygons within each unit circle represent the internal symmetry within the set of the $n^{\mathrm{th}}$-roots of unity.
  17. 2.5 The symmetry in the square-roots of unity is a rotation by $\pi$ radians since the set of roots is identical under this rotation.
  18. 2.6 In the left plot, we define the unit square in the $z$-plane. It is through the $w(z)=z^{2}$ mapping that we connect the unit square to the square-squared region in the $w$-plane in the right plot. The colored curves represent sets of the complex plane that are connected through the $w(z)=z^{2}$ “map.”
  19. 2.7 The unit hyperbola $x^{2}-y^{2}=1$ parameterized by the functions $x=\cosh t$ and $y=\sinh t$. Think of the parameter $t$ as the time it takes for the black point to move from the point $(x,y)=(1,0)$ to the point $(x,y)=(x(t),y(t))$ along the unit hyperbola.
  20. 2.8 Plots of the hyperbolic cosine (Eq. 2.57) and hyperbolic sine (Eq. 2.58) functions over a few real numbers $x$.
  21. 3.1 If the position of an object is known at two times, perhaps at $t$ and $t+\Delta t$, it is possible to describe the change in position from $\vec{r}(t)$ to $\vec{r}(t+\Delta t)$ as $\Delta\vec{r}(t)=\vec{r}(t+\Delta t)-\vec{r}(t)$.
  22. 3.2 If the position of an object is known at three times, perhaps at $t$, $t+\Delta t$, and $t+2\Delta t$, it is possible to describe the change in the change in the position from $\vec{r}(t)$ to $\vec{r}(t+2\Delta t)$ as $\Delta\Delta\vec{r}(t+\Delta t)=\Delta\vec{r}(t+\Delta t)-\Delta\vec{r}(t)$. Since the change in position is a measure of velocity by Eq. 3.2, then this figure shows the existence in the change in velocity, also known as the acceleration.
  23. 3.3 The position, velocity, and acceleration of particle as a function of time $t$ and it meanders from an initial position $\vec{r}_{0}$ to a final position $\vec{r}_{f}$. During the trip, it can have many different velocities and accelerations, whose directions are represented by the red dashed arrows and the green dotted arrows, respectively.
  24. 3.4 A sketch of a function $f$ (purple solid curve) and its corresponding Riemann sum between $x^{\prime}$ and $x$ (green shaded boxes). Here the function has a particular value at each point $x^{\prime}+j\Delta x_{N}$, and that value is multiplied by $\Delta x_{N}$. This is geometrically equivalent of multiplying a base and a height, and thus we have an approximation for the area under a little segment of the function $f(x)$. By adding up all of the little boxes we can then approximate the total area under the curve.
  25. 3.5 A function and its local extrema. When there is a local maximum, the nearby points are lower than it, creating an upside-down cup shape. When there is a local minimum, the nearby points are all above it, creating a rightside-up shape. When there is an inflection point, the nearby points are both above and below it and the concavity changes sign. Notice that these are all local extrema, but they are not global.
  26. 4.1 A few capital sines $\mathrm{sinc}(\omega\tau)=\sin\omega\tau/\omega\tau$ plotted as functions of $\omega$ for varying values of $\tau$. Notice that as $\tau\rightarrow 0$, the whole function $\mathrm{sinc}(\omega\tau)$ gets wider, i.e. the Fourier Transform needs more values of $\omega$ to account for the smaller pulse.

###### List of Tables

  1. 1.1 A summary of the important and general equations derived for vector operations.
  2. 2.1 A summary of the important and general equations in Complex Algebra.
  3. 3.1 A summary of the important and general equations in one dimensional calculus.
  4. 4.1 A bunch of useful and general formulas for Fourier analysis and their text references.

##  Chapter 1 Vectors

Vectors are everywhere in physics. Their utility comes in many different
varieties, from helping us measure position in classical mechanics, to
allowing us to describe the distribution of available momentum states for
electrons in semiconductors, to aiding us as we describe the interactions
between particles and fields. Since they are everywhere, we must tackle them
first. They will probably seem pretty difficult/annoying/infuriating to you,
particularly if you have never dealt with vectors before. But truth be told,
vectors allow us to manipulate and transport a bunch of otherwise cumbersome
information in a nice compact way. For example, in Maxwell’s original work on
electromagnetism, he needed _twenty_ coupled equations to fully capture all of
electricity, magnetism, and light because vectors were not yet a mathematical
tool [charap_2011]. Now we can write his famous equations111To have a full
picture of the dynamics between the fields and any nearby particles, we need a
couple more equations, but this is besides the point.  down rather quickly as

$\displaystyle\nabla\cdot\vec{D}$ $\displaystyle=\rho_{free}$
$\displaystyle\nabla\cdot\vec{B}$ $\displaystyle=0$
$\displaystyle\nabla\times\vec{E}$
$\displaystyle=-\frac{\partial\vec{B}}{\partial t}$
$\displaystyle\nabla\times\vec{H}$
$\displaystyle=\vec{j}_{free}+\frac{\partial\vec{D}}{\partial t}$

The specific details of what each of these equations means are not within the
scope of this book, so I will leave their explanation for your electromagnetic
theory coursework (or you can go to for a fairly quick overview at
[hyperphysics_maxwells_equations]).

The purpose of this chapter is twofold. Firstly, and most importantly, I want
to help you learn how to use vectors for implementation in your study of the
natural world. Another goal I have in mind is to help you think about
mathematical objects, like numbers, sets, or functions, in ways that you may
not have thought of before. In many ways, mathematics is a game. It has
players (you and me), it has pieces (mathematical objects), and it has rules
for how the players and pieces can interact. Hopefully, by the end of this
chapter you will begin to see this for yourself, too.

###  1.1 Mathematical Objects

> Disclaimer: this section will be pretty rudimentary as far as mathematical
> concepts go — we will will only talk about the basis of mathematical
> reasoning, properties of real numbers, and functions. However, the level of
> abstractness is pretty high. In other words, this level of abstractness is
> not commonly taught in most high school math classes, nor is it usually
> taught in introductory calculus courses (although there are exceptions!). I
> understand that the formalism that I use below may be a little difficult to
> get through for a lot of you the first time you see it, so do not worry when
> it looks like we start at a much higher level than you are expecting. This
> section is not absolutely necessary for one to understand vectors; meanwhile
> I think it is worthwhile to peruse so that you can gain some familiarity
> with the idea of the “math game,” and how we establish rules and then use
> those rules to prove mathematical ideas. I chose to start with something as
> elementary as the real numbers partly because they are used to build
> essentially the rest of mathematics, and vectors by extension, but I mostly
> chose them because most of you will be pretty familiar with the real
> numbers. Thus, you will already have some intuition about them that you can
> use as a reference for when I build them from a set of rules. If this
> section is too mathy for you, feel free to proceed to Section 1.3. Later on
> in the book, if I ever use ideas proven in this section, just refer back
> here and brush up on what you need. Otherwise, especially if you are very
> mathematically-inclined222Like more mathematically inclined than a
> physicist. Or even an applied mathematician… So proceed to read this section
> basically if you really like pure logic., sit back and enjoy constructing
> mathematical reasoning and the real numbers!

Before we get going any further, we need to lay down some ground rules. These
rules will help us understand what we are doing, and at worst, will be
something to fall back on if we get lost. The first two rules are the most
fundamental, but they are necessary for us to move forward.

  1. 1.

We are allowed to define new mathematical objects that have a specified set of
properties.

  2. 2.

Mathematical objects can only be used consistently with their given
properties. If we need a new property for our object, we must redefine the
object to incorporate it.

However, this is not a pure mathematics book, so I will not build mathematics
from set theory. I will lean on a lot of your prior knowledge of math (maybe
up through algebra, geometry, and trigonometry), but whenever I will be
talking about math in this much more elementary way, I will warn you.

To illustrate how we might use these rules, let’s say we wanted to invent
Euclidean geometry for the first time. I might say something like, “there
exists an object called a point”. After that statement, the only thing we know
about points — i.e. their only predefined property — is that they are extant.
We conclude then that points, by themselves, are boring. We decide to define a
means for points to interact with each other. We do so by saying “any two
points belong to a common mathematical object called a line that runs through
both”. Now we have a means of talking about two points with respect to one
another, since they are connected by a line. We may move to say that there are
then infinitely many different points on the line since they are all extant
mathematical objects. This would not be quite right, though, as we have not
given our points a property that they can be distinguished. To try to do this,
we would have to redefine our point object and say something along the lines
of “there exists only one object called a point for every location”. We do
start to get into trouble here because we would need to be clear about what a
“location” object is in order for this new definition to make sense. To
proceed with the infinite number of points idea, we would eventually have to
come up with some kind of length and throw in integers somehow to argue that a
line has at least as many points as integers, therefore is comprised of
infinitely many points. The point is that we could build up Euclidean geometry
in our own way, our only limitation is the way in which we define our
gamepieces. But after we have our pieces in order, we continue to play the
game to see if we can beat our previous score.

###  1.2 Numbers, Lists, and Functions

We will build up the idea of vectors in a much less geometric way than many
other people do, mostly because I think that the geometry can sometimes muddle
the concision that vectors bring to the table. Once we have a few rules in
place for vectors, then I will bring up some geometric interpretations for
what we have found.

####  1.2.1 Real Numbers

I am going to assume that you know the difference between types of numbers,
such as integers and irrational numbers, or at least I will assume that you
could type them into a calculator if someone asked you to. What I will quickly
review though, is a few defining properties of how real numbers interact with
each other. The set of all real numbers is denoted by $\mathds{R}$, and
whenever we say “$x$ is in the set of real numbers”, we say write it
mathematically as $x\in\mathds{R}$, where the symbol $\in$ means “in”.

Let’s choose three real numbers, $a$, $b$, and $c$. It does not matter what
these numbers are, as long as they are real. And since they are all real
numbers, we would write $a,b,c\in\mathds{R}$. If you need to at this point,
whip out your calculator and pick your three favorite real numbers and test
the following properties to show yourself that they hold (recall that
subtraction is denoted by $a-b$ or $a+(-b)$ and multiplication is denoted by
either $ab$ or $a\cdot b$).

  1.   2. 1.

Addition is commutative:
$a+b=b+a\in\mathds{R}$

  3. 2.

Addition is associative:
$(a+b)+c=a+(b+c)\in\mathds{R}$

  4. 3.

There exists an identity operator for addition, namely $0\in\mathds{R}$ such
that
$a+0=0+a=a$.

  5. 4.

There exists an additive inverse for every $a\in\mathds{R}$, denoted by $-a$
such that
$a+(-a)=(-a)+a=0$.

  6. 5.

Multiplication is commutative:
$ab=ba\in\mathds{R}$

  7. 6.

Multiplication is associative:
$(ab)c=a(bc)\in\mathds{R}$

  8. 7.

There exists an identity operator for multiplication, namely $1\in\mathds{R}$
such that
$a\cdot 1=1\cdot a=a$.

  9. 8.

There exists an multiplicative inverse for every $a\neq 0\in\mathds{R}$,
denoted by $\frac{1}{a}$ such that
$a\cdot\frac{1}{a}=\frac{1}{a}\cdot a=1$.

  10. 9.

There exists a distributive property between addition and multiplication,
namely

$a(b+c)=ab+ac\in\mathds{R}$

These properties, although written in perhaps a rather abstract way, are our
initial rules for the real numbers. They outline everything that we are
allowed to do with real numbers as far as calculations are concerned. These
rules tell us how to add, subtract, multiply, and divide real numbers. They
also tell us that we will always get a real number back by adding,
subtracting, multiplying, or dividing other real numbers. This is important
because it is sometimes more important just for us to know what type of object
we are dealing with, rather than knowing the specific value for the object.
Thus, we will always know that when we are combining real numbers according to
the 9 rules above, our final result will always be just another real number.

Real numbers also have the following _Ordering Property_ : if
$a,b\in\mathds{R}$, then only one of the following statements is true. Either
$a-b<0$, $-a+b<0$, or $a=b$. These could equivalently be written as $a<b$,
$b<a$, or $a=b$. This ordering property essentially says that some real
numbers have lower values than other real numbers ($a<b$ or $b<a$), unless the
two numbers we are talking about are the same ($a=b$). Furthermore, there are
two rules governing how addition and multiplication work with ordering. For
example,

$\displaystyle 1.$ $\displaystyle\textrm{ If }b<c\textrm{, then }a+b<a+c.$
$\displaystyle 2.$ $\displaystyle\textrm{ If }0<a,b\textrm{, then }0<ab.$

Using these rules, it can be _proven_ that $0<1$. This may sound silly at
first, but remember, we only defined 1 and 0 as the multiplicative and
additive identities. We did not define them as _something_ and _nothing_ as is
usually done. To show that this is true, we must actually show a few other
things first (these things that facilitate the execution of proof are called
lemmas).

  1. 1.

If $0<a$, then $-a<0$ and if $a<0$ then $0<-a$.
To prove the first part, we note that $0=-a+a$. But $0<a$, so by the first
rule $-a+0<-a+a=0$. Thus, $-a+0=-a<0$. I recommend you use the same line of
reasoning to prove the second part.

  2. 2.

$-ab=-(ab)$.
Consider $-ab+ab=(-a+a)b=0b=0$. We conclude then that since $-ab+ab=0$, then
it must be true that $-ab$ is the the additive inverse of $ab$, otherwise
denoted by $-(ab)$. Hence, $-ab=-(ab)$. (This shows that our rules for real
numbers allow us to permute the negative sign around products.)

  3. 3.

$-(-a)=a$.
This proof may be a little confusing because of the negative signs (they
confuse me, at least). Since $(-a)\in\mathds{R}$ whenever $a\in\mathds{R}$,
then $(-a)+-(-a)=0$. But by definition, $(-a)+a=0$. Thus,

$-(-a)=-(-a)+0=-(-a)+(-a)+a=0+a=a$

. We conclude that $-(-a)=a$.

  4. 4.

$1^{2}=1$.
Given that $1\in\mathds{R}$, then we apply the multiplicative identity to it
to get $1=1\cdot 1=1^{2}$.

  5. 5.

$(-a)(-b)=ab$.
We use the second lemma above to move the negative on $a$ to the outside:
$(-a)(-b)=-[a(-b)]$. Then we apply it again to get
$-[a(-b)]=-[(-b)a]=(-)(-)ba=-[-(ab)]$. By the third lemma, $-[-(ab)]=ab$.

Finally, we have enough lemma-backup to proceed. We will do so by showing that
if $a\in\mathds{R}$, then $a^{2}>0$ if $a\neq 0$. We start with any real
number $a\neq 0$, so there are two cases ($0<a$ or $-a<0$). If $0<a$, then by
the multiplication ordering rule, $0<aa=a^{2}$. Next, if $a<0$, then by the
first lemma, $0<-a$. Thus, $0<(-a)(-a)$. By the fifth lemma,
$(-a)(-a)=aa=a^{2}$, so $0<(-a)(-a)=a^{2}$, proving that when $a\neq 0$, then
$0<a^{2}$ for all other real numbers. Now we choose $a=1$, since by the fourth
lemma, $1^{2}=1$. Therefore, $0<1^{2}=1$, completing our proof.

Again, this little exercise may have seemed silly because we have always been
told that $0<1$, or equivalently $1>0$. But all of that was based on numbers
being representations of physical things. Now we have showed that this holds
for our new abstract set of real numbers, given our rules for them. Hence, our
numbers no longer have to represent sets of things for them to have any
meaning to us. They now can stand on their own. Furthermore this allows us to
interpret our abstract real numbers geometrically so that we can regain the
physical intuition we all had about numbers before reading this book. First
off, we can define something of unit length, as is shown in Fig. 1.1. Then we
call some point $0$ and measure out to a point $1$ using the unit length we
have. But since we know that $1>0$, then the position of $1$ relative to $0$
shows us which direction represents an increase in our abstract real numbers.
From here, we can assign these numbers to represent whatever we want, whether
it be money, distance, energy, complex numbers, complex quantum states, et
cetera. The power we have now obtained from this set of rules for real numbers
is that we control how and when we use the real numbers to help us in
explaining physical phenomena, rather than having to wait until we understand
how a number line can somehow be morphed into something that may not be
strictly geometrical.

FIG. 1.1: We assign real numbers to a number line by determining what length
should be 1 unit and then project those lengths onto a line (or vector).

The final important property that I want to address is that the real numbers
are considered mathematically _complete_. This essentially means that for any
$a,b\in\mathds{R}$, there exists a real number $x\in\mathds{R}$ such that
$a\leq x\leq b$. The implications here are actually quite surprising. This
means that the real numbers form a continuum, meaning we can zoom in between
any two points on a number line indefinitely. There will always be another
real number there no matter how far in we zoom. This property is essential for
continuum mathematics and physics overall. If the reals were not complete,
then every once in a while we might measure a length and fail because our
ruler could suddenly run into a gap in the real numbers!

####  1.2.2 Collections of Real Numbers: Lists and Functions

Sometimes it is important for us to connect numbers to each other. For
example, let’s say we have money and want to buy things. To determine how many
things we could buy with our money, we might divide our money by how much it
costs per thing. For example, to use a case that is probably fairly relatable,
if you only have $100 and a textbook for Physics 1 costs $200, then you can
buy exactly $100/$200 per textbook which is exactly half of a Physics 1
textbook. What we have done though is we have implicitly created a list of
real numbers, $(\textrm{money},\textrm{textbooks})$, and we created a function
to join the numbers together in the list. Using this example, we come up with
a couple more gamepieces to use and we define them below.

  * •

The Cartesian Product of $n$ sets of real numbers, formally written as
$\mathds{R}^{n}=\mathds{R}\times\mathds{R}\times\dots\times\mathds{R}$, is the
set of all $n$-tuples, where any element $a\in\mathds{R}^{n}$ is written as
$a=(a_{1},a_{2},\dots,a_{n})$.

  * •

Consider any two sets of objects, $A$ and $B$. A function (or mapping) $f$
from $A$ into $B$, written $f:A\rightarrow B$, relates the two sets $A$ and
$B$ such that every object $a\in A$ is assigned a unique object $b\in B$,
denoted by $f(a)=b$. The set $A$ is called the domain of $f$ while the set of
all $f(a)\in B$ is called the range of $f$333Note that the range of $f$ is not
necessarily all of $B$. If $\textrm{range}(f)=B$, then we say that $f$ is an
onto function, meaning it is a mapping from $A$ onto $B$ instead of just $A$
into $B$. .

The Cartesian Product allows us to group collections of real numbers together,
like we did before with money and textbooks with
$\mathds{R}^{2}=\mathds{R}\times\mathds{R}$, and the function allows us to
find how many textbooks $b=f(a)\in B$ we can buy with our money $a\in A$. A
schematic of what a function looks like is given in Fig. 1.2. Now there are
MANY special properties of functions, and we will only cover a few throughout
the course of this book. Whenever we must use them, I will define/derive them
if they are relatively easy, but an in-depth study about the nature and
structure of functions will ultimately distract us from physics, and so anyone
interested in learning about these topics should either add a pure mathematics
major or reference one of the following [wolfram_function, rudin_1976,
wikipedia_functions]. For now, the key take away from functions is that every
function has an input and it will give us exactly one output.

FIG. 1.2: A function $f:A\rightarrow B$ takes every single point $a\in A$ and
maps it into exactly one point $f(a)=b\in B$. The set of all $f(a)\in B$ is
the $\textrm{range}(f)$.

###  1.3 Vector Operations

Now that we have developed a little bit of prerequisite mathematical analysis,
we will proceed with our discussion of vectors. For right now, we will define
a vector simply as a length in a particular “direction,” like the thing drawn
in Fig. 1.3a, and we denote a vector with a little arrow on top, like
$\vec{v}$. Many other texts write vectors as $\boldsymbol{v}$, but since bold-
faced type is hard for most people to write by hand, I will stick to arrows.

(a) A vector $\vec{v}$ defined solely as a direction.

(b) When two vectors, or directions, $\vec{v}$ and $\vec{u}$ are added
together, they form a third vector $\vec{v}+\vec{u}$.

(c) When two vectors $\vec{u}$ and $\vec{v}$ are added together, they form a
third vector $\vec{u}+\vec{v}$.

(d) The superposition of either addition method.

FIG. 1.3: Vectors are simply just mathematical objects that describe
directions. This definition leads to an interesting way to add them.

####  1.3.1 The Geometry Behind Vector Addition

For right now, the important property of vectors is that they have directions.
This means that they represent, for example, the arrow that connects your eyes
and these words. Since vectors only have directions, they do not belong to any
particular points in whatever space they live in444The term “direction” is not
clearly defined yet, but for right now, just consider it a ray-segment instead
of a line-segment.. Now let’s say that two vectors exist, namely $\vec{v}$ and
$\vec{u}$. These vectors (directions) are shown in the left-hand side of the
equality in Fig. 1.3b. Since these vectors have defined lengths and
directions, but are allowed to be translated in space, we are free to align
$\vec{v}$ and $\vec{u}$, as shown in the right-hand side of Fig. 1.3b such
that the tip (arrowhead) of $\vec{v}$ touches the tail (not arrowhead) of
$\vec{u}$. This is equivalent to looking at these words, and then following
the arrow between them and something else, until you end up looking at
something else. The end results is that you ended up at the something else
even though you started at those words. Since we started somewhere and ended
up elsewhere, we will call the final result $\vec{v}+\vec{u}$ which is the
resultant vector drawn from the tail of $\vec{v}$ to the tip of $\vec{u}$ when
$\vec{v}$ and $\vec{u}$ are aligned tip-to-tail.

But if this is a summation, it is natural for us to ask whether vector
addition is commutative like regular old real number addition. For example,
does something like $3+4=4+3$ hold for vectors? To check this, we look at
$\vec{u}+\vec{v}$ by translating $\vec{v}$ such that it is aligned with
$\vec{u}$ tip-to-tail, as is shown in Fig. 1.3c. If we compare the direction
of the vector $\vec{v}+\vec{u}$ from Fig. 1.3b with that of the vector
$\vec{u}+\vec{v}$ from Fig. 1.3c, we can start to see that they look very
similar, if not identical. To see that the directions are the same, and
further that the vector sum is indeed commutative with respect to this
definition, then we superimpose the right-hand sides of Fig. 1.3b with Fig.
1.3c to get the parallelogram shown in Fig. 1.3d. By looking at this
parallelogram, it is clear that adding $\vec{u}$ to $\vec{v}$
($\vec{v}+\vec{u}$) tip-to-tail and adding $\vec{v}$ to $\vec{u}$
($\vec{u}+\vec{v})$ leave us with a vector that has the same length and
direction. This discovery allows us to conclude two very important things
about vectors:

  1. 1.

Vector addition is commutative when we add vectors tip-to-tail.

  2. 2.

Two vectors $\vec{A}$ and $\vec{B}$ are equivalent if they have the same
length and direction.

The first point is important because in nature we observe all kinds of
phenomena, such as something’s position and displacement, whose directions are
commutative. That is, the resultant directions due to a sum are independent of
the order in which we add them. For example, if you were to draw a square on
some paper and then follow the sides from one corner to the opposite corner (a
lot like what’s shown in Fig. 1.3d), you’d see that you would end up in that
corner regardless of which direction you first moved along the sides. In other
words, the addition of either sets of sides is commutative. But now we know
definitively that adding vectors tip-to-tail geometrically gives us the result
that we know from our everyday experience. The second point is important
because it gives us a way to specify uniqueness in our vectors. In the
physical world, if we are describing something’s velocity at any particular
point, we intuitively would only need one measurement of it to be complete. We
should not have to name a bunch of different velocities if an object only has
one.

####  1.3.2 The Geometry Behind Vector Subtraction

Before we continue, I want to make note of something important. To define
vector addition, I said that we would add vectors tip-to-tail. But you may
ask, what happens if instead we add them tail-to-tail? Why can’t this also be
vector addition?

This is a very good question, and it actually involves the definition of
_functions_ we used above. However, now we will construct a function that
takes in two vectors arguments and outputs one. By the definition of a
function, there can only be one output for every input in the domain, so we
need our one vector to be unique; thus, the resultant vector of this function
must have a well-defined length and direction. Consider the same two vectors
$\vec{v}$ and $\vec{u}$ from Fig. 1.3b, Fig. 1.3c, and Fig. 1.3d. Now we align
them tail-to-tail as is shown in Fig. 1.4 and draw the resultant vector tip-
to-tip. But which way do we go? Do we draw from $\vec{u}$ to $\vec{v}$ (as is
done in the left-hand side of the figure), or draw from $\vec{v}$ to $\vec{u}$
(as is done in right-hand side)?

FIG. 1.4: The the _function_ of aligning the vectors $\vec{v}$ and $\vec{u}$
tail-to-tail leaves the resultant (output) vector $\vec{R}$ ambiguous. Drawing
the resultant vector from tip-to-tip could then yield two different answers.

Let the function that aligns vectors tail-to-tail be named $\vec{R}$. Since it
aligns vectors tail-to-tail, the function takes in two vectors as inputs. Now
let us say that it has the following possible output vectors

$\displaystyle\vec{R}=\begin{cases}\vec{R}_{1}=\vec{R}(\vec{u},\vec{v}),&\textrm{
if drawn from }\vec{u}\textrm{ to }\vec{v}\\\
\vec{R}_{2}=\vec{R}(\vec{v},\vec{u}),&\textrm{ if drawn from }\vec{v}\textrm{
to }\vec{u}\end{cases}$

It is clear from the geometry in Fig. 1.4 that the lengths of the two outputs
are the same, however the directions are reversed. Thus
$\vec{R}(\vec{u},\vec{v})\neq\vec{R}(\vec{v},\vec{u})$. Hence, this particular
function is NOT commutative, unlike aligning vectors tip-to-tail. therefore,
if we chose this particular function to be vector addition, our answers would
not be commutative, and we would have a hard time describing a lot of physics.

This non-commutative function is reminiscent of subtraction from arithmetic
though. For example, $4-3\neq 3-4$. However, from the properties of the real
numbers above,

$\displaystyle 3-4=-(-3)+(-4)=-(-3+4)=-(4-3),$

or more generally, for any two real numbers $a$ and $b$, $a-b=-(b-a)$ (prove
it). If we consider the length or magnitude of a number as its distance from
$0$ of a number line, then both $a-b$ and $b-a$ have the same length even
though their signs are opposite. By analogy, $\vec{R}_{1}$ and $\vec{R}_{2}$
in Fig. 1.4 both have the same length, however their directions are opposite.
Thus, we use this analogy to define vector subtraction by aligning vectors
tail-to-tail. But just like $4-3+3=4$, or $a-b+b=a$, we need
$\vec{v}-\vec{u}+\vec{u}=\vec{v}$ and $\vec{u}-\vec{v}+\vec{v}=\vec{u}$. In
other words, the tip of the resultant vectors $\vec{R}$ should fall on the
vector being subtracted from and the tail should fall on the vector being
subtracted. This means that

$\displaystyle\vec{R}_{1}$
$\displaystyle=\vec{R}(\vec{u},\vec{v})=\vec{v}-\vec{u},$
$\displaystyle\vec{R}_{2}$
$\displaystyle=\vec{R}(\vec{v},\vec{u})=\vec{u}-\vec{v}.$

Again, by analogy, $a-b=-(b-a)$ for real numbers. By looking at $\vec{R}_{1}$
and $\vec{R}_{2}$ we see that

$\displaystyle\vec{R}_{2}=\vec{u}-\vec{v}=-(\vec{v}-\vec{u})=-\vec{R}_{1}.$

if we allow ourselves to factor out negative signs in an identical way as we
did with real numbers. Since $\vec{R}_{2}$ is in the opposite direction as
$\vec{R}_{1}$, and if we factor out the negative sign like we did above, then
we conclude that multiplying a vector by a negative sign reverses its
direction! This then means that we can eliminate a vector completely by
subtracting it from itself, or by aligning it with a copy of itself tail-to-
tail and then drawing the resultant vector tip-to-tip. Since the two tips will
occupy the same point, the vector will have zero length, and we call this the
null or zero vector, defined by

$\vec{u}-\vec{u}=\vec{u}+(-\vec{u})=\vec{0}.$

To summarize this point, we can subtract a vector $\vec{u}$ from a $\vec{v}$
by aligning the two tail-to-tail and drawing the resultant vector from
$\vec{u}$ to $\vec{v}$. Meanwhile, since multiplying by a negative reverses a
vector’s direction, we can equivalently add $-\vec{u}$ to $\vec{v}$ to get
$\vec{v}-\vec{u}$. Hence, we can reverse the direction of $\vec{u}$ and then
add it tip-to-tail to $\vec{v}$. The ability to subtract vectors is really
helpful in physics, particularly whenever we want to describe the change in a
system. When something changes, for example, position, it moves from one point
to another. The vector that connects the initial and final points would then
represent a displacement, that is, a change in position. But we will talk more
about this point later. Now that we established a visual (geometrical)
representation of how vectors can combine, we move on to study more
complicated systems of vectors.

####  1.3.3 Vector Addition and Coordinate Systems

In physics, the real power of vectors comes from their simplification of
otherwise annoying or intangible systems of coordinates. Sometimes it is
really useful to talk about a problem in terms of three perpendicular
directions (hint hint vectors) — electrons that swim through a solid chunk of
metal, for instance, are described very well with a system of three
perpendicular axes. Other times, however, we need to use a set of coordinates
that have one direction that emanates radially outward from some center and
then two others that are mutually perpendicular with each other and the radial
direction — these spherical coordinates are used in a lot of places, from
electromagnetic or gravitational radiation problems to black holes to proton-
proton collisions (and more). It actually turns out that some incredibly
interesting physics comes out of the study of coordinate systems; for example,
time and space can dilate and contract because of a difference in two
observer’s coordinate systems (thank you, Einstein [einstein_menendez]). So a
thorough understanding of coordinate systems is imperative for a thorough
understanding of the universe, and vectors facilitate such an understanding.

FIG. 1.5: If we choose to use the vectors $\vec{x}$ and $\vec{y}$ of known
lengths and directions, then we can rewrite vector $\vec{v}$ as the sum of
$\vec{x}$ and $\vec{y}$. In this case, $\vec{v}=2.5\vec{x}+1.5\vec{y}$.

Since we have established what vector addition is (remember tip-to-tail555In
case you haven’t caught on, this should be a geometrical mantra of yours.), we
can apply it to define a coordinate system for ourselves. This is because that
any vector in a system of coordinates is represented by a list of lengths in
some predetermined set of directions. But no particular system is universal
over the others — any one that is used to describe a vector works666To get a
better idea of this artifact in the mathematics, draw an arrow on a piece of
paper. Now draw a little $xy$-axis on the side of the arrow. Then rotate the
page and draw another little $xy$-axis. Here, neither axis is necessarily
better than the other one; whichever we choose to use is purely that — a
choice.. Thus, we start by choosing a set of directions that have some
particular kind of meaning to us. As can be seen in Fig. 1.5, we choose
$\vec{x}$ to represent the rightward horizontal direction and $\vec{y}$ to
represent the upward vertical direction (when it comes to constructing our
coordinate system, we align them tail-to-tail). We will call these coordinate-
defining vectors basis vectors. For reasons that will become clearer later and
in the Linear Algebra chapter, we intentionally choose $\vec{x}$ and $\vec{y}$
to be perpendicular. Suppose that we know the lengths of both $\vec{x}$ and
$\vec{y}$ so these vectors are clearly defined. Now whenever we have another
vector $\vec{v}$ living in the same space as $\vec{x}$ and $\vec{y}$, we can
use vector addition to write $\vec{v}$ in terms of $\vec{x}$ and $\vec{y}$.

To do this, we start by extending $\vec{x}$ and $\vec{y}$, as is shown with
the dashed lines in Fig. 1.5. Then we draw lines from the tip (arrowhead) of
$\vec{v}$ onto the extensions of $\vec{x}$ and $\vec{y}$. One of the reasons
why it helps to have perpendicular basis vectors is that these lines that we
drew define two right triangles with $\vec{v}$ as the hypotenuse. All we have
to do now is count how many $\vec{x}$ vectors fit along the horizontal
extension and how many $\vec{y}$ vectors fit along the vertical extension. In
Fig. 1.5, the horizontal extension is $2.5$ $\vec{x}$s long, while the
vertical extension is $1.5$ $\vec{y}$s long. Therefore, we would write777If I
were being 100% truthfull, I would technically need to precisely define what
scalar multiplication is before we multiply vectors by numbers, so for right
now consider it like we have 2.5 US dollars and 1.5 pecan pies. The idea is
that we can assign some numerical amount to some fixed quantity, such as money
and pies. However, the money and pies are not necessarily the same (we would
need some function to connect the two).

$\vec{v}=2.5\vec{x}+1.5\vec{y}$

Furthermore, by exploiting the following trigonometric relationships,

$\displaystyle\sin(\textrm{angle})$ $\displaystyle=\frac{\textrm{opposite
side}}{\textrm{hypotenuse}},$ $\displaystyle\cos(\textrm{angle})$
$\displaystyle=\frac{\textrm{adjacent side}}{\textrm{hypotenuse}},$
$\displaystyle\tan(\textrm{angle})$
$\displaystyle=\frac{\sin(\textrm{angle})}{\cos(\textrm{angle})}=\frac{\textrm{opposite
side}}{\textrm{adjacent side}},$

we have

$\displaystyle\sin\theta$ $\displaystyle=\frac{1.5y}{v},$
$\displaystyle\cos\theta$ $\displaystyle=\frac{2.5x}{v},$
$\displaystyle\tan\theta$ $\displaystyle=\frac{1.5y}{2.5x},$

where $x$, $y$, and $v$ are the lengths of the vectors $\vec{x}$, $\vec{y}$,
and $\vec{v}$, respectively. The tangent relationship is pretty useful,
because if we now think of the direction of the vector $\vec{v}$ as the angle
$\theta$ between it and $\vec{x}$ when aligned tail-to-tail as is drawn in
Fig. 1.5, then we find that the direction is independent of the length $v$.
This is reassuring because we can intuitively reason that we can stretch or
shrink a vector’s length arbitrarily without changing its direction. For
example if you pick a point somewhere near you at eye-level and stare at it,
and then you take a few steps forwards or backwards while still looking at it,
you will see you never really have to turn your head or readjust to keep fixed
on the point888This is assuming you take normal-sized steps of course. If you
wiggle your head too much then this experiment will not help at all.. Hence,
we keep this new definition for the “direction” of a vector.

We can actually learn a couple more things from the trigonometry. First, if
you remember the Pythagorean identity

$\sin^{2}(\textrm{angle})+\cos^{2}(\textrm{angle})=1,$

Then,

$\displaystyle 1$ $\displaystyle=\sin^{2}\theta+\cos^{2}\theta$
$\displaystyle=\left(\frac{1.5y}{v}\right)^{2}+\left(\frac{2.5x}{v}\right)^{2}$
$\displaystyle=\frac{1}{v^{2}}\left[(1.5y)^{2}+(2.5x)^{2}\right].$

By multiplying both sides by $v^{2}$ we find that

$\displaystyle v^{2}=(2.5x)^{2}+(1.5y)^{2}\Rightarrow
v=\sqrt{(2.5x)^{2}+(1.5y)^{2}}.$

which is an expression showing that the length of a vector is independent of
its direction $\theta$ (also notice that is this the geometry version of the
Pythagorean identity, a.k.a the Pythagorean Theorem). Ergo, since a vector is
only defined as an object with a length and a direction, the establishment of
our coordinate system allows us to talk about them in three ways:

$\displaystyle\vec{v}$ $\displaystyle=2.5\vec{x}+1.5\vec{y},$
$\displaystyle\tan\theta$ $\displaystyle=\frac{1.5y}{2.5x},$ (1.1)
$\displaystyle v$ $\displaystyle=\sqrt{(2.5x)^{2}+(1.5y)^{2}}.$ (1.2)

All that we had to do is agree upon a set of perpendicular basis vectors of
known lengths.

####  1.3.4 Unit Vectors

The discussion of coordinates systems relies pretty heavily on the choice of
basis vectors $\vec{x}$ and $\vec{y}$. In order to actually take out a
calculator and get a number for lengths or directions, it is necessary for us
to specify the lengths $x$ and $y$. By looking at Eq. 1.1, we may be motivated
to try and simplify things by making the vectors $\vec{x}$ and $\vec{y}$ have
the same length, i.e. we assume that $y=x$. This would certainly make our
lives easier because Eq. 1.1 would simplify to

$\displaystyle\tan\theta=\frac{1.5x}{2.5x}=\frac{1.5}{2.5}=\frac{3}{5}.$

This is great if we wanted to find a numerical value for $\theta$ since we can
isolate $\theta$ as follows,

$\displaystyle\theta=\arctan\left(\frac{3}{5}\right)\approx 0.540\textrm{
rad}\approx 31.0^{\textrm{o}}.$

I want to emphasize though that this numerical value for $\theta$ only applies
to the vector in Fig. 1.5. We can now reduce Eq. 1.2 as well

$\displaystyle
v=\sqrt{(2.5x)^{2}+(1.5x)^{2}}=\sqrt{x^{2}\left((2.5)^{2}+(1.5)^{2}\right)}=x\,\sqrt{(2.5)^{2}+(1.5)^{2}}.$

It is at this last step though that we realize that to go any further
(numerically) we need to specify what $x$ is.

If we look at the equation above, we have an arbitrary length $x$ and then a
numerical factor. But that numerical factor contains the values 2.5 and 1.5 —
I did not square either intentionally. These numbers are important because
they again represent the number of times the vectors $\vec{x}$ and $\vec{y}$
need to be extended to intersect the lines drawn from the tip of the arrow in
Fig. 1.5. It would be incredibly helpful to be able to describe a vector
exclusively by reading off the set of analogous numbers from any system of
basis vectors. In other words, it would be the most helpful to NOT have to
carry around either factors of $x$ or $y$. However, we are not free to simply
ignore $x$ or $y$ to remain consistent with our definition of vectors since we
make them have lengths. Further, since we use basis vectors to build our
coordinate systems, in order for our coordinate systems to make any sense we
need lengths associated with our basis. Otherwise, again, our work would be
inconsistent. So all that is left is for us to choose $x=y=1$ because $1\cdot
a=a\cdot 1=a$ for all $a\in\mathds{R}$. When we do this, we no longer have to
lug the $x$s and $y$s around our equations because they will always multiply
other numbers without changing their value. And finally would we be able to
write the length of $\vec{v}$ from Fig. 1.5 as

$\displaystyle
v=x\,\sqrt{(2.5)^{2}+(1.5)^{2}}=1\cdot\sqrt{(2.5)^{2}+(1.5)^{2}}=\sqrt{(2.5)^{2}+(1.5)^{2}}\approx
2.92.$

Vectors that have $\textrm{length}=1$ are defined as unit vectors. As we have
seen, their definition and implementation comes about rather naturally when we
use vectors to build a coordinate system. They have many other useful vector
properties that we will discuss more a bit later. For right now, it is very
important to note that unit vectors are so special that physicists and
mathematicians gave them their own “hats”,

$\displaystyle\vec{x}\rightarrow\hat{x}\textrm{ if and only if }x=1.$

and so from now on, whenever I am talking about unit vectors, they will have
the little triangular hat to distinguish them from regular vectors with the
arrows. By extension then, we could write that same vector $\vec{v}$ from Fig.
1.5 as

$\displaystyle\vec{v}=2.5\hat{x}+1.5\hat{y}=v_{x}\hat{x}+v_{y}\hat{y}$

where the symbols $v_{x}$ and $v_{y}$ stand for the $x$ and $y$ components of
the vector, repectively. This notation generalizes quite nicely, too. If our
vector lives in three dimensions, for example, we would write it as

$\displaystyle\vec{v}=v_{x}\hat{x}+v_{y}\hat{y}+v_{z}\hat{z},$ (1.3)

or if our vector lives in $n$-dimensions and we call the $j^{\textrm{th}}$
mutually perpendicular unit vector $\hat{x}_{j}$, then we would write
$\vec{v}$ as

$\displaystyle\vec{v}=v_{1}\hat{x}_{1}+v_{2}\hat{x}_{2}+v_{3}\hat{x}_{3}+\dots+v_{j-1}\hat{x}_{j-1}+v_{j}\hat{x}_{j}+\dots+v_{n-1}\hat{x}_{n-1}+v_{n}\hat{x}_{n},$
(1.4)

where $v_{j}$ is the $j^{\textrm{th}}$ the components of the vector $\vec{v}$.
Using the components of vectors, we can then rewrite the direction and length
formulas given by Eq. 1.1 and Eq. 1.2 as follows

$\displaystyle\tan\theta$ $\displaystyle=\frac{v_{y}}{v_{x}}$ (1.5)
$\displaystyle v$ $\displaystyle=\sqrt{v_{x}^{2}+v_{y}^{2}}$ (1.6)

in two dimensions for any general set of coordinates $\\{v_{x},v_{y}\\}$. In
$n$ dimensions these formulas generalize to

$\displaystyle\tan\theta_{jk}$ $\displaystyle=\frac{v_{k}}{v_{j}}$ (1.7)
$\displaystyle v$
$\displaystyle=\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+\dots+v_{j-1}^{2}+v_{j}^{2}+\dots+v_{n-1}^{2}+v_{n}^{2}}$
(1.8)

where the notation $\theta_{jk}$ means represents the angle measured with
respect to the $j^{\textrm{th}}$ unit vector in the $jk$-plane (notice that if
you substitute in $x$ for $j$ and $y$ for $k$, you get Eq. 1.5 back). I know
this formula may look a little clunky, but it is a way to generalize the
notion of an angle — an object that only exists between two axes $j$ and $k$ —
to any dimensional space. For example, in three dimensions, we can have angles
measured above the $x$-axis in either the $xy$-plane or the $xz$-plane.

Knowing when to switch back and forth between talking about a vector’s
components or its length and direction can get confusing sometimes. In
practice, it is important to remember that although vectors do have
components, those components depend on the coordinate system built from one’s
choice of unit vectors. Even though these components can have different values
in different coordinate systems, the vector should always remain the same. For
example, if we are talking about a vector drawn between the positions of two
points, it does not matter if we are looking directly at it that vector, or we
rotate ourselves, or me move ourselves further from that vector. Its location
is invariant. However, the numerical values we would use to describe it in
each of our coordinate systems would vary wildly. When coordinate systems get
weird, it is useful to stick to thinking of the vector as its own object that
has many different sets of components (or representations). Problems like
these occur in relativity and quantum mechanics. However, when it suffices to
only deal with one coordinate system per problem, like a lot of introductory
physics, it is usually safe to think of the vector as only a set of components
in that one coordinate system.

####  1.3.5 Component-wise Addition and Subtraction of Vectors

Since we have discovered a way to represent vectors in terms of a set of
components in a particular coordinate system, it is useful to go through and
define addition and subtraction in terms of these components. We start with
addition.

Recall that vector addition is a geometric function that aligns two vectors
$\vec{v}$ and $\vec{u}$ tip-to-tail and outputs the vector drawn from the tail
of $\vec{v}$ to the tip of $\vec{u}$. Meanwhile, we can write $\vec{v}$ and
$\vec{u}$ in terms of their $x$ and $y$ components999We are starting in two
dimensions for simplicity. as

$\displaystyle\vec{v}$ $\displaystyle=v_{x}\hat{x}+v_{y}\hat{y}$ (1.9)
$\displaystyle\vec{u}$ $\displaystyle=u_{x}\hat{x}+u_{y}\hat{y}$ (1.10)

once we establish a set of perpendicular basis vectors $\hat{x}$ and
$\hat{y}$. By looking at Fig. 1.6, we can see that there is a rather
convenient way to express vector addition using the components. From the
figure we can conclude that the $x$-component of the sum $\vec{v}+\vec{u}$ is
$(v+x)_{x}=v_{x}+u_{x}$, while the $y$-component of the sum is
$(v+u)_{y}=v_{y}+u_{y}$. Therefore,

$\displaystyle\vec{v}+\vec{u}=(v_{x}+u_{x})\hat{x}+(v_{y}+u_{y})\hat{y}$
(1.11)

Note that by using Eq. 1.11, we can show that component-wise vector addition
is commutative, just like the geometrical version of vector addition.

$\displaystyle\vec{v}+\vec{u}=(v_{x}+u_{x})\hat{x}+(v_{y}+u_{y})\hat{y}=(u_{x}+v_{x})\hat{x}+(u_{y}+v_{y})\hat{y}=\vec{u}+\vec{v}$

Here these equalities hold because $u_{x},u_{y},v_{x},v_{y}\in\mathds{R}$.

FIG. 1.6: Here we combine adding vectors tip-to-tail with the components we
can measure in our $\hat{x},\hat{y}$ coordinate system.

Using the same idea with vector subtraction, the $x$-component of
$\vec{v}-\vec{u}$ is $(v-u)_{x}=v_{x}-u_{x}$ and the $y$-component is
$(v-u)_{y}=v_{y}-u_{y}$. Therefore, in component notation, vector subtraction
becomes

$\displaystyle\vec{v}-\vec{u}=(v_{x}-u_{x})\hat{x}+(v_{y}-u_{y})\hat{y}$
(1.12)

I leave it to you to show that component-wise subtraction is anti-commutative
in agreement with the geometrical version of vector subtraction in Problem
1.3.5.

Show that $\vec{v}-\vec{u}=(\vec{v}-\vec{u})$ using the vector components in
Eq. 1.12. (Hint: factor out a negative sign from each term.)

We can generalize these two dimensional formulas pretty easily. If we have $n$
mutually perpendicular unit vectors, $\\{\hat{x}_{j}\\}_{j=1}^{n}$, then to
get the individual components, we draw the relevant
$\hat{x}_{j}$,$\hat{x}_{k}$ coordinate systems like we did with $j=x$ and
$k=y$ in Fig. 1.6. Then we add tip-to-tail component-wise. Thus, if

$\displaystyle\vec{v}$
$\displaystyle=v_{1}\hat{x}_{1}+v_{2}\hat{x}_{2}+\dots+v_{n}\hat{x}_{n}=\sum_{j=1}^{n}v_{j}\hat{x}_{j},$
$\displaystyle\vec{u}$
$\displaystyle=u_{1}\hat{x}_{1}+u_{2}\hat{x}_{2}+\dots+u_{n}\hat{x}_{n}=\sum_{j=1}^{n}u_{j}\hat{x}_{j},$

then the sum and difference between these vectors will be

$\displaystyle\vec{v}+\vec{u}$
$\displaystyle=(v_{1}+u_{1})\hat{x}_{1}+(v_{2}+u_{2})\hat{x}_{2}+\dots+(v_{n}+u_{n})\hat{x}_{n}=\sum_{j=1}^{n}(v_{j}+u_{j})\hat{x}_{j},$
(1.13) $\displaystyle\vec{v}-\vec{u}$
$\displaystyle=(v_{1}-u_{1})\hat{x}_{1}+(v_{2}-u_{2})\hat{x}_{2}+\dots+(v_{n}-u_{n})\hat{x}_{n}=\sum_{j=1}^{n}(v_{j}-u_{j})\hat{x}_{j}.$
(1.14)

where we have made use of summation notation only to write a potentially
gigantic sum rather concisely101010In summation notation, the dummy index $j$
takes on values of all of the integers between $1$ and $n$, inclusive, inside
the summand, where each time we change a $j$-value, we then add it to all the
terms we had before. For example,  $\sum_{j=1}^{5}j=1+2+3+4+5=15.$ In the sum
above, $n=5$ and the summand is $j$.. As we can see, using components to add
and subtract vectors is a rather powerful tool. Using these vector operations
as inspiration, we will search for useful geometrical functions with vector
components for the remainder of this chapter.

####  1.3.6 Scaling Vectors

In this section, we will study what happens if we set $\vec{u}=\vec{v}$ in the
addition formulas given by Eq. 1.11 and Eq. 1.13. Making a direct
substitution, we have

$\displaystyle\vec{v}+\vec{v}$
$\displaystyle=(v_{x}+v_{x})\hat{x}+(v_{y}+v_{y})\hat{y}=(1+1)v_{x}\hat{x}+(1+1)v_{y}\hat{y}=2v_{x}\hat{x}+2v_{y}\hat{y},$
$\displaystyle\vec{v}+\vec{v}$
$\displaystyle=\sum_{j=1}^{n}(v_{j}+v_{j})\hat{x}_{j}=\sum_{j=1}^{n}(1+1)v_{j}\hat{x}_{j}=\sum_{j=1}^{n}2v_{j}\hat{x}_{j}.$

Hence, adding $\vec{v}$ to itself scales each of the components of $\vec{v}$
by a factor of 2. In regular old real number arithmetic, whenever we add a
number to itself, that is the equivalent of scaling the number by a factor of
2. By analogy, we could factor the 2 out of the equations above to get
$\vec{v}+\vec{v}=2\vec{v}$. Then, if we add $\vec{v}$ to $2\vec{v}$, again, by
analogy, we would get $\vec{v}+2\vec{v}=3\vec{v}$, and so on for all the
integers (prove this to yourself using the same argument for positive
integers, and then use the subtraction formulas for the negative integers.).
But the question remains of whether we can scale a vector by any number even
if that number is not an integer.

One could try to answer this question by adding some series of vectors
together to get non-integer scaling, however this would ultimately require us
to be able to scale vectors by non-integer numbers anyway, and so the logic
would kind of be circular. To fix this, we instead make use of vector
components to help us understand what it means to scale vectors outside of
adding or subtracting the same vector over and over again. We just need two
pieces of information from $\vec{v}+\vec{v}=2\vec{v}$ to generalize. First, we
notice that integer scaling keeps the vector colinear, meaning it leaves Eq.
1.5 and Eq. 1.7 invariant.

$\displaystyle\tan\theta_{2\vec{v}}$
$\displaystyle=\frac{2v_{y}}{2v_{x}}=\frac{v_{y}}{v_{x}}=\tan\theta_{\vec{v}}$
$\displaystyle\tan(\theta_{2\vec{v}})_{jk}$
$\displaystyle=\frac{2v_{k}}{2v_{j}}=\frac{v_{k}}{v_{j}}=\tan(\theta_{\vec{v}})_{jk}$

where the vector subscripts on $\theta$ differentiate between the directions
for either vectors. Further, scaling $\vec{v}$ by $2$ scales the length of the
vector by $2$, as can be seen from Eq. 1.6 and Eq. 1.8:

$\displaystyle v_{2\vec{v}}$
$\displaystyle=\sqrt{(2v_{x})^{2}+(2v_{y})^{2}}=\sqrt{2^{2}\left[(v_{x})^{2}+(v_{y})^{2}\right]}=2\left(\sqrt{v_{x}^{2}+v_{y}^{2}}\right)=2v_{\vec{v}}$
$\displaystyle v_{2\vec{v}}$
$\displaystyle=\sqrt{\sum_{j=1}^{n}(2v_{j})^{2}}=\sqrt{2^{2}\sum_{j=1}^{n}v_{j}^{2}}=2\left(\sqrt{\sum_{j=1}^{n}v_{j}^{2}}\right)=2v_{\vec{v}}$

Therefore, when we end up scaling vectors in a more general sense, we want to
make sure that the scaled vector is also colinear and the length is scaled by
the length (magnitude) of same factor.

FIG. 1.7: Whenever we have a function that scales all of the components of a
vector $\vec{v}$ by the same real number $a$, then we interpret this as
$a\vec{v}$. Notice that this leaves Eq. 1.5 unchanged.

It is important to realize that the reason why the direction and length of a
vector behave as they do under scaling by a factor of $2$ is because each
individual component is scaled by that same factor. Before, this was because
we added $\vec{v}$ to itself tip-to-tail, and the scaling of the components
was a consequence of our use of coordinates. Now, however, we will use an
alternative approach, and seek a function that scales all the components by
the same amount, as is shown in Fig. 1.7. In this figure, we consider a
function $f_{a}$ that takes in vector $\vec{v}$ with a set of component
$\\{v_{x},v_{y}\\}$ and multiplies each component by a real number $a$.
Therefore, under this definition,

$\displaystyle f_{a}:\\{v_{x},v_{y}\\}\rightarrow\\{av_{x},av_{y}\\}.$

Notice we can do this for all $a\in\mathds{R}$ because
$v_{x},v_{y}\in\mathds{R}$, and any two real numbers can be multiplied
together to get another real number, meaning this function is defined
component-wise. This function also keeps the vector $\vec{v}$
colinear111111The only limitation here is that $a\neq 0$ because there is no
multiplicative inverse of 0. In other words, we are only guaranteed to have a
$1/a\in\mathds{R}$ if $a\neq 0$. while scaling the vector’s length.

$\displaystyle\tan\theta_{a\vec{v}}$
$\displaystyle=\frac{av_{y}}{av_{x}}=\frac{v_{y}}{v_{x}}=\tan\theta_{\vec{v}}$
$\displaystyle v_{a\vec{v}}$
$\displaystyle=\sqrt{(av_{x})^{2}+(av_{y})^{2}}=\sqrt{a^{2}\left[(v_{x})^{2}+(v_{y})^{2}\right]}=\sqrt{a^{2}}\left(\sqrt{v_{x}^{2}+v_{y}^{2}}\right)=|a|v_{\vec{v}}.$

We must take care to remember that negative signs reverse the direction of a
vector. This is equivalent to $\tan\theta$ switching between the first and
third quadrants of a Cartesian plane, when both the numerator and denominator
switch sign. However, since the reversed vector falls on the same line, the
resultant vector of this scaling is still colinear. Additionally, since we can
insert $a=2$ into the function above and obtain the same results as we had
before, we conclude that in two dimensions, multiplying a vector’s components
by any real number is equivalent to multiplying the whole vector by the same
factor. In other words, our generalization holds! I leave it to you to use a
similar argument to show that in $n$-dimensions,

$\displaystyle
a\vec{v}=(av_{1})\hat{x}_{1}+(av_{2})\hat{x}_{2}+\dots+(av_{n})\hat{x}_{n}=\sum_{j=1}^{n}(av_{j})\hat{x}_{j},$
(1.15)

is the correct scaling function $f_{a}$ because it also keeps the vector
colinear (all the $\tan\theta$ are invariant) while scaling the length of
$\vec{v}$ by $\sqrt{a^{2}}$.

This picture of vector scaling is actually what gives normal numbers their
name of scalar. Many other texts describe scalars as object with only
magnitude or length. In this context, our scalar is $a$ and the magnitude of
$a$ is $\sqrt{a^{2}}=|a|$. I chose to introduce scalars as a function that act
on vectors because that is all we can do with scalars and vectors. We can
never add a scalar to a vector because our definition of vector addition
involves aligning vectors tip-to-tail. We can never subtract a scalar from a
vector (or vice versa) because vector subtraction involves aligning vectors
tail-to-tail. What we are allowed to do with scalars is multiply them by
vectors. This then either stretches or shrinks the vector, but it always keeps
the vector colinear. It may reverse the vector’s direction, but scalar
multiplication can never change one component without changing all of the
others by the same proportion.

A useful application of vector scaling is in finding a formula to transform
any vector into a unit vector that points in the same direction as the
original. To do this, we recall that a vector is only a unit vector if its
length is one. Thus, if we have a vector $\vec{v}$, we seek a scalar $a$ such
that the length of $a\vec{v}$ is given by

$\displaystyle|a\vec{v}|=\sqrt{\sum_{j=1}^{n}(av_{j})^{2}}=1.$

The use of absolute value bars is used to represent length in analogy with the
absolute value of real numbers used to represent the length or magnitude of
that number. But since the scalar multiplies every term in the sum then we can
factor out the $a^{2}$ inside the radicand.

$\displaystyle|a\vec{v}|=\sqrt{a^{2}\sum_{j=1}^{n}v_{j}^{2}}=|a|\sqrt{\sum_{j=1}^{n}v_{j}^{2}}=1.$

But if we choose our set of coordinates, then we should know all of the
components of that vector $\\{v_{j}\\}$, and those are just numbers, therefore
if we knew their values, we could in principle square them and then add them
and then take their square-root. But ultimately, that is just a number that we
know121212Also, we assume it to be nonzero.. Remember that we want
$|a\vec{v}|=1$ and we want the vectors to point in the same direction, hence
$a>0$. Then we conclude

$\displaystyle a=\frac{1}{\sqrt{\sum\limits_{j=1}^{n}v_{j}^{2}}}=\frac{1}{v},$

where the last equality holds from the length of the $n$-dimensional vector
$\vec{v}$, written as $v=|v|$, and given by Eq. 1.8. Hence, for any nonzero
vector $\vec{v}$, we have

$\displaystyle\hat{v}=\left(\frac{1}{v}\right)\vec{v}=\frac{\vec{v}}{v},$
(1.16)

which is the general way to write any unit vector from a known vector.

####  1.3.7 Dot Product

We return now to a consequence of combining geometry and coordinate systems to
talk about vectors. Specifically, we will look at vector subtraction of
$\vec{u}$ from $\vec{v}$, as is shown in Fig. 1.8. Using Eq. 1.12, we can find
$\vec{v}-\vec{u}$ as

$\displaystyle\vec{v}-\vec{u}=(v_{x}-u_{x})\hat{x}+(v_{y}-u_{y})\hat{y}.$

FIG. 1.8: By the Law of Cosines, we can relate the lengths of three legs of a
triangle to their angles. Likewise, we can relate the lengths of these legs to
their components. Therefore, we should be able talk about angles in terms of
the components of the legs.

Therefore the length of the difference, denoted by absolute value bars just
like as is done for the length of real numbers, is

$\displaystyle|\vec{v}-\vec{u}|=\sqrt{\left(v_{x}-u_{x}\right)^{2}+\left(v_{y}-u_{y}\right)^{2}},$

given Eq. 1.6. Now we employ the Law of Cosines131313For a refresher on the
Law of Cosines, please check out [hyperphys_law_of_cosines,
wikipedia_law_of_cosines]. to write

$\displaystyle|\vec{v}-\vec{u}|=\sqrt{v^{2}+u^{2}-2uv\cos\theta}$

where $\theta$ is the angle between $\vec{v}$ and $\vec{u}$, as is shown in
Fig. 1.8. But these two expressions for $|\vec{v}-\vec{u}|$ must be
equivalent. To proceed, we square both expressions141414We square the square-
roots because square-roots are algebraic nightmares (hard to handle); so it
is, in general, a good idea to get rid of them if you can in any derivation.
and equate them

$\displaystyle\left(v_{x}-u_{x}\right)^{2}+\left(v_{y}-u_{y}\right)^{2}=v^{2}+u^{2}-2uv\cos\theta$

By rearranging terms, we can write the angular part in terms of the squares as

$\displaystyle 2uv\cos\theta$
$\displaystyle=v^{2}+u^{2}-\left(v_{x}-u_{x}\right)^{2}-\left(v_{y}-u_{y}\right)^{2}$
$\displaystyle=\left(\sqrt{v_{x}^{2}+v_{y}^{2}}\right)^{2}+\left(\sqrt{u_{x}^{2}+u_{y}^{2}}\right)^{2}-\left(v_{x}-u_{x}\right)^{2}-\left(v_{y}-u_{y}\right)^{2}$
$\displaystyle=v_{x}^{2}+v_{y}^{2}+u_{x}^{2}+u_{y}^{2}-\left(v_{x}-u_{x}\right)^{2}-\left(v_{y}-u_{y}\right)^{2}.$

If we recall that $(a-b)^{2}=a^{2}+b^{2}-2ab$ when $a,b\in\mathds{R}$, then we
have

$\displaystyle 2uv\cos\theta$
$\displaystyle=v_{x}^{2}+v_{y}^{2}+u_{x}^{2}+u_{y}^{2}-\left(v_{x}^{2}+u_{x}^{2}-2u_{x}v_{x}\right)-\left(v_{y}^{2}+u_{y}^{2}-2u_{y}v_{y}\right)$
$\displaystyle=v_{x}^{2}+v_{y}^{2}+u_{x}^{2}+u_{y}^{2}-v_{x}^{2}-u_{x}^{2}+2u_{x}v_{x}-v_{y}^{2}-u_{y}^{2}+2u_{y}v_{y}$
$\displaystyle=(v_{x}^{2}-v_{x}^{2})+(v_{y}^{2}-v_{y}^{2})+(u_{x}^{2}-u_{x}^{2})+(u_{y}^{2}-u_{y}^{2})+2u_{x}v_{x}+2u_{y}v_{y}$
$\displaystyle=0+0+0+2(u_{x}v_{x}+u_{y}v_{y})$
$\displaystyle=2(u_{x}v_{x}+u_{y}v_{y}).$

After dividing both sides by 2 then we can conclude

$\displaystyle uv\cos\theta=u_{x}v_{x}+u_{y}v_{y}.$ (1.17)

This formula is very powerful because it tells us how the angle between any
two, two dimensional vectors is related to the components of those vectors.
Namely,

$\displaystyle\cos\theta=\frac{u_{x}v_{x}+u_{y}v_{y}}{\sqrt{u_{x}^{2}+u_{y}^{2}}\,\sqrt{v_{x}^{2}+v_{y}^{2}}},$
(1.18)

when we divide both sides by the lengths of $\vec{u}$ and $\vec{v}$. The
quantity in the numerator is pretty important. For example, if it happens to
be the case that $u_{x}v_{x}=-u_{y}v_{y}$, then

$\displaystyle\cos\theta=\frac{u_{x}v_{x}+u_{y}v_{y}}{\sqrt{u_{x}^{2}+u_{y}^{2}}\,\sqrt{v_{x}^{2}+v_{y}^{2}}}=\frac{-u_{y}v_{y}+u_{y}v_{y}}{\sqrt{u_{x}^{2}+u_{y}^{2}}\,\sqrt{v_{x}^{2}+v_{y}^{2}}}=\frac{0}{\sqrt{u_{x}^{2}+u_{y}^{2}}\,\sqrt{v_{x}^{2}+v_{y}^{2}}}=0.$
(1.19)

Then we would conclude that $\theta=\arccos(0)=90^{\textrm{o}}$. This actually
turns out to be a necessary and sufficient condition to tell if two vectors
are perpendicular, or in fancy linear algebra talk, this condition tells us if
two vectors are orthogonal. Because of this seeming utility, we give that
quantity the name of the vector dot product151515Other courses or textbooks
call this quantity a scalar product because it itself is a function that
returns a scalar. However, I think this will probably lead to confusion for
anyone first learning there is a difference between a scalar and a vector..
The two dimensional dot product is defined below

$\displaystyle\vec{u}\cdot\vec{v}=u_{x}v_{x}+u_{y}v_{y}=uv\cos\theta.$ (1.20)

where the little $\cdot$ symbol is explicitly written; hence the name
“dot“product. In $n$-dimensions, the dot product formula becomes

$\displaystyle\vec{u}\cdot\vec{v}=u_{1}v_{1}+u_{2}v_{2}+\dots+u_{n}v_{n}=\sum_{j=1}^{n}u_{j}v_{j}=uv\cos\theta.$
(1.21)

Actually it is pretty straightforward to derive the $n$-dimensional case in
the same way that the two dimensional case was. It is worth it to try it out
in Problem 1.3.7.

In this problem, I will walk you through how to derive the $n$-dimensional dot
product cosine formula. I would recommend adhering to the following steps, but
by all means, if you have a better way to derive it, definitely use it
instead. (a) Define two $n$-dimensional vectors $\vec{u}$ and $\vec{v}$
according to Eq. 1.4. (b) Write out
$|\vec{u}-\vec{v}|=\sqrt{u^{2}+v^{2}-2uv\cos\theta}$ according to the Law of
Cosines. (c) Square both sides. (d) Expand the left-hand side as
$|\vec{u}-\vec{v}|^{2}=(\vec{u}-\vec{v})\cdot(\vec{u}-\vec{v})$ using the
$n$-dimensional dot product (only the components though). (e) Expand the right
hand side’s $u^{2}+v^{2}$ in terms of their components. (f) Compare both side
and divide by any of the 2’s that may appear at the end.

Now again, if the dot product between two nonzero vectors is zero, then they
are mutually orthogonal (perpendicular).

Given that we know $\vec{u}\cdot\vec{v}$, is it possible to determine what
$\vec{v}\cdot\vec{u}$ is? Based off of our geometrical understanding of the
dot product as being a way to measure the angle in between two vectors, it
intuitively makes sense that this angle should be the same in both
$\vec{u}\cdot\vec{v}$ and $\vec{v}\cdot\vec{u}$. This intuition turns out to
be completely correct as the following line of reasoning shows

$\displaystyle\vec{u}\cdot\vec{v}=u_{1}v_{1}+u_{2}v_{2}+\dots+u_{n}v_{n}=v_{1}u_{1}+v_{2}u_{2}+\dots+v_{n}u_{n}=\vec{v}\cdot\vec{u}.$

In other words, the dot product is commutative.

Now that I have made the claim that any two vectors with a vanishing dot
product are orthogonal, we should test it with two vectors that we know to be
perpendicular: $\hat{x}$ and $\hat{y}$. If we write out these unit vectors in
terms of their components, then we have

$\displaystyle\hat{x}$ $\displaystyle=1\hat{x}+0\hat{y},$
$\displaystyle\hat{y}$ $\displaystyle=0\hat{x}+1\hat{y}.$

Since Eq. 1.21 says that the dot product is the sum of the products of all the
individual components of a vector, then

$\displaystyle\hat{x}\cdot\hat{y}=1(0)+0(1)=0.$

We then conclude that the angle between $\hat{x}$ and $\hat{y}$ is indeed
$90^{\textrm{o}}$, which shows that this new vector function is also
consistent with the framework we have built up so far.

FIG. 1.9: When we take the dot product of any vector $\vec{v}$ with any unit
vector $\hat{u}$, then we obtain $v\cos\theta$ by Eq. 1.21. This situation
corresponds geometrically to the diagram above, where the value $v\cos\theta$
is the length of $\vec{v}$ along the $\hat{u}$ direction in the (green) plane
spanned by $\vec{v}$ and $\hat{u}$. If $v\cos\theta>1$ ($v\cos\theta<1$), this
length will be longer (shorter) than $\hat{u}$.

We move on now to something slightly different. Since we have an expression
for the dot product in terms of the lengths of two vectors being dotted, let’s
consider the case where one of the vectors, $\vec{u}$, is actually a unit
vector. Thus $\vec{u}\rightarrow\hat{u}$ and $u\rightarrow 1$ in Eq. 1.21 as
shown below

$\displaystyle\hat{u}\cdot\vec{v}=v\cos\theta.$

(We cannot use the component notation here because we have not specified any
basis vectors for our coordinates, yet.) This problem is drawn in Fig. 1.9. In
the figure, the angle between $\vec{v}$ and $\hat{u}$ is $\theta$. Notice that
this angle is in the plane spanned by the two vectors. It is not necessarily
in the (orange-ish) horizontal plane, nor is it necessarily in either vertical
planes. This would physically correspond to the length of the shadow of
$\vec{v}$ cast on $\hat{u}$ if a light were shined behind it at the unit
vector. I want to emphasize here that we have not established a set of
coordinates yet. In fact, I really don’t even need the black coordinate
vectors to talk about the cosine side of the dot product — I put them in
because it helps me draw a three dimensional picture in only two dimensions.
It turns out that we don’t need components to talk about the dot product
representing directions as it is here because the cosine side of Eq. 1.21 is
derived entirely from coordinate-free geometry (Law of Cosines). Hence, this
length, or projection, of $\vec{u}$ in the $\hat{u}$ direction allows us to
talk about the directionality of $\vec{v}$ without necessarily specifying our
coordinates! Formally, the projection function161616Some authors say that the
projection is a vector, whereas the way I have written it in Eq. 1.22 makes it
scalar. I have chosen not to include the vector form in this chapter to try
and mitigate as much potential confusion as possible. is given by

$\displaystyle\textrm{proj}_{\hat{u}}(\vec{v})=\hat{u}\cdot\vec{v}=v\cos\theta,$
(1.22)

where the function is read as “the projection of $\vec{v}$ onto the unit
vector $\hat{u}$“. In this sense, we are holding the unit vector fixed while
we see how much different $\vec{v}$ vectors point in its direction. If we
recall from Eq. 1.16 that any $n$-dimensional unit vector is

$\displaystyle\hat{u}=\frac{\vec{u}}{u}=\frac{u_{1}\hat{x}_{1}+u_{2}\hat{x}_{2}+\dots+u_{n}\hat{x}_{n}}{\sqrt{u_{1}^{2}+u_{2}^{2}+\dots+u_{n}^{2}}},$

then

$\displaystyle\textrm{proj}_{\hat{u}}(\vec{v})=\left(\frac{\vec{u}}{u}\right)\cdot\vec{v}=\frac{\vec{u}\cdot\vec{v}}{u},$
(1.23)

which then implies the dot product can be written in terms of a projection as

$\displaystyle\vec{u}\cdot\vec{v}=u\,\textrm{proj}_{\hat{u}}(\vec{v}).$ (1.24)

Since the dot product is really just a number $u$ multiplied by
$\textrm{proj}_{\hat{u}}(\vec{v})$, then many people interchangeably refer to
taking a dot product as finding a projection of one vector in the direction of
another.

This comes up all the time in classical mechanics, electromagnetism,
relativity, but probably most interestingly, using the idea of a dot product
to tell us something about a vector’s direction is everywhere in quantum
mechanics. When systems behave quantum mechanically, they are only allowed to
occupy certain states in a many, or infinitely, dimensional Hilbert space, and
we associate each possible state with its own unit vector, or direction in
that space. For example, we can run things like Stern-Gerlach
experiments171717These are really interesting experiments that I highly
recommend checking out the following sources to try and understand them
[gerlach1922experimental, stern_gerlach_exp_revisited]. A user-friendly applet
to experiment with is here [phet_dubson_mckagan_wieman_2011]. to find out that
electrons can only ever be spin-up or spin-down when we measure them, but can
never be both at once. We then would use spin-up as something similar to an
$\hat{x}$ and spin-down as something similar to a $\hat{y}$, since we have
seen that $\hat{x}\cdot\hat{y}=0$, which implies that these vectors have no
projection in the direction of the other. Typically, we normalize state
vectors in quantum mechanics, meaning we make every state vector unit-length
(so they are unit vectors). By doing this, we make a state’s projection along
a particular direction related to the probability of that state of the system
being equivalent to the physical meaning behind that direction. For example,
if we have a state vector $\vec{\psi}$ given by

$\displaystyle\vec{\psi}=\vec{\mathrm{up}},$

then it only is projected in the spin-up direction, meaning the state of the
electron is spin-up. We could also ask what is the probability of this state
being spin-down using the square of the projection operation, Eq. 1.24.

$\displaystyle\mathrm{Probability}=\left[\mathrm{proj}_{\mathrm{down}}(\vec{\psi})\right]^{2}=\left(\vec{\mathrm{down}}\cdot\vec{\mathrm{up}}\right)^{2}=0^{2}=0.$

We are also free to study electrons whose spins may be less clearly defined
for us, but more on that when you get there in a quantum mechanics course. If
you are very eager to read ahead right now, some very thorough introductory
resources on this matter are Townsend’s and Sakurai’s quantum mechanics books
[townsend_2012, sakurai_napolitano_2011].

A more immediate application of the dot product’s ability to select the
projection of one vector onto another vector is in the physical description of
work. In physics, we define work as the amount of force exerted along a
particular distance. The term “along”, though, implies that direction is
involved. If we think about this intuitively, if we push an object so that it
moves in a particular direction, we can only possible associate the object’s
motion with our push. But if we push an object one way, and it totally moves
in an orthogonal direction, then the distance the object moved along our push
is zero. Therefore, we cannot say we were responsible for the object’s motion
at all. But this is is the exact behavior that the dot product describes. As
is shown in Fig. 1.9, the projection selects the amount one vector points in
the direction of another. Then the dot product scales the projection by the
length of the other vector.

An application of the dot-product-as-projection idea comes into play when we
calculate the $\textrm{proj}_{\hat{v}}(\vec{v})$. This function should
intuitively yield $v=|\vec{v}|$ because the projection measures how much a
vector points in the direction of the unit vector. But here the unit vector is
$\hat{v}$, which by Eq. 1.16 points in the same direction as $\vec{v}$. So we
are really just measuring the length of $\vec{v}$ in the direction of
$\vec{v}$, which is exactly the length of $\vec{v}$. This, however, is just
our intuition based on our (wordy) English interpretations of our math. We
need to verify that our work does indeed return the same result with math, not
just with words. By Eq. 1.22, we have

$\displaystyle\textrm{proj}_{\hat{v}}(\vec{v})=\hat{v}\cdot\vec{v}=v\cos\theta=v.$

The last equality holds because the angle between $\hat{v}$ and $\vec{v}$ is
$\theta=0^{\textrm{o}}$. Hence, our projection function does indeed return
$v$, which means that our intuition from before is consistent. Now, however,
if we combine Eq. 1.24 with $\textrm{proj}_{\hat{v}}(\vec{v})$, then we find

$\displaystyle\vec{v}\cdot\vec{v}=v\,\textrm{proj}_{\hat{v}}(\vec{v})=v\cdot
v=v^{2},$

which is again, the projection of $\vec{v}$ onto $\hat{v}$ scaled by the
length of $\vec{v}$. More importantly, this shows that dot products are deeply
related to the lengths of vector, especially because $\vec{v}\cdot\vec{v}$ was
derived without specifying any components. It holds that in general

$\displaystyle v=|v|=\sqrt{\vec{v}\cdot\vec{v}}.$ (1.25)

For the sake of completeness/honesty, this is not the only way to establish
Eq. 1.25. We could have done it using the component form of the dot product,
but I wanted to show this using the projection formula without components to
provide more intuition as to why this relationship exists. The generality that
exists from the dot product, or the more general inner products, is physicists
use to talk about lengths.

####  1.3.8 Cross Product

To finish up this chapter, we just need to talk about one more incredibly
useful (geometrical) vector operation. So far, we have added and subtracted
vectors, scaled vectors, and found a way to multiply vectors so that we can
talk about how much they point in the directions of others. But whatever
happened to multiplying two lengths to get an area?

This also turns out to be a physically interesting quantity, but it must be
distinct from the dot product, since the dot product only scales the length of
one vector by the length of the other in the direction of the first. It is not
explicitly considered as the area of the parallelogram created by two vectors;
hence we cannot conclude that the dot product is in general that particular
area. We seek now to find this area.

FIG. 1.10: A geometric calculation of the area of the central parallelogram
created by $\vec{v}$ and $\vec{u}$, denoted by $|\vec{v}\times\vec{u}|$. The
combination of geometry with the idea of vector components leads to a peculiar
property of the area: without careful considerations, the resultant area may
become negative!

Consider the two vectors $\vec{u}$ and $\vec{v}$ drawn in Fig. 1.10. If we
line them up tail-to-tail, then we see that they span a (purple-ish)
parallelogram. From simple geometry, we remember that the area of this
parallelogram, labeled $|\vec{v}\times\vec{u}|$, is

$\displaystyle|\vec{v}\times\vec{u}|=uv\sin\theta,$ (1.26)

where the quantity $v\sin\theta$ is the “height” in the formula $\textrm{area
}=\textrm{ base }\times\textrm{ height}$. We can also find the area of the
parallelogram though by using the components of $\vec{u}$ and $\vec{v}$ in the
$\hat{x}\hat{y}$ coordinate system. Specifically,

$\displaystyle|\vec{v}\times\vec{u}|$
$\displaystyle=(v_{x}+u_{x})(u_{y}+v_{y})-2a_{1}-2a_{2}-2a_{3},$
$\displaystyle=v_{x}u_{y}+v_{x}v_{y}+u_{x}u_{y}+u_{x}v_{y}-2a_{1}-2a_{2}-2a_{3}.$

where the quantities $a_{1}$, $a_{2}$, and $a_{3}$ correspond to the areas of
each of the non-parallelogram shaded regions in Fig. 1.10. Using the
components in the figure, we determine those areas to be

$\displaystyle a_{1}$ $\displaystyle=\frac{1}{2}v_{x}v_{y}.$ $\displaystyle
a_{2}$ $\displaystyle=u_{x}v_{y},$ $\displaystyle a_{3}$
$\displaystyle=\frac{1}{2}u_{x}u_{y}$

where the area of the triangles are $(\textrm{base }\times\textrm{
height})/2$. Substituting these areas into the previous equation yields

$\displaystyle|\vec{v}\times\vec{u}|$
$\displaystyle=v_{x}u_{y}+v_{x}v_{y}+u_{x}u_{y}+u_{x}v_{y}-v_{x}v_{y}-2u_{x}v_{y}-u_{x}u_{,}$
$\displaystyle=v_{x}u_{y}+(u_{x}v_{y}-2u_{x}v_{y})+(v_{x}v_{y}-v_{x}v_{y})+(u_{x}u_{y}-u_{x}u_{y},$
$\displaystyle=v_{x}u_{y}-u_{x}v_{.}$

Hence we have

$\displaystyle|\vec{v}\times\vec{u}|=uv\sin\theta=v_{x}u_{y}-u_{x}v_{y},$
(1.27)

which is the way to relate the area of the parallelogram to the lengths of the
vector that span it to the components of those vectors. So naturally, we need
to check if this area is commutative, just like we did for the dot product.

$\displaystyle|\vec{u}\times\vec{v}|=u_{x}v_{y}-v_{x}u_{y}=-(v_{x}u_{y}-u_{x}v_{y})=-|\vec{v}\times\vec{u}|.$

I hope you have issues with the equation above. In it, I claim that a
nonnegative area $|\vec{u}\times\vec{v}|$ is equal to a negative area
$-|\vec{v}\times\vec{u}|$. The only way this would hold for arbitrary areas is
if the area of parallelograms is always zero! What happened? Technically,
there isn’t anything actually wrong here besides my sloppy notation. I should
have included absolute value bars on everything to guarantee nonnegativity.
However, leaving it out highlights a problem here with this area. There is a
nontrivial sign change that accompanies the component form of evaluating the
parallelogram’s area. Furthermore this sign change arose when we switched the
order of the vectors being multiplied — a property called anti-commutativity —
and so there is a significant mathematical difference between multiplying the
vectors one way versus the other.

If we think of this sign problem along the lines of vector scaling, then we
would remember that we can completely reverse a vector’s direction by
multiplying it by $-1$. This leads us to think that maybe we could encapsulate
this odd sign behavior if we treated the area of the parallelogram in Fig.
1.10 as a vector instead of a scalar. But if this area is the length of some
vector, what is the vector’s direction? By some dimensional analysis, the area
must have different units than either constituent vector if these vectors are
going to have any physical utility for us. For example, if $\vec{u}$
represents position and $\vec{v}$ represents velocity, then the area would
have dimensions of $\textrm{length }\times\textrm{ speed}$ — a quantity that
is clearly different than the dimensions of either. Additionally, by looking
at Fig. 1.10, the area of the parallelogram is invariant when we rotate the
whole picture by some angle clockwise or counterclockwise. Even though the
constituent vectors change direction under rotations, the magnitude and sign
of the cross product do not as long as the order of the vectors is the same.
These two facts seem to indicate that the cross product behaves differently
than either constituent vector181818This is true in general because the cross
product is an axial vector rather than a polar vector like all of those we
have been talking about up until this point. For more information, see
[axial_vector]. it makes sense that this new vector should be in a direction
that is totally distinct from either constituent vector191919Actually, in a
unit-less space (one where the axes do not have units ascribed to them), the
rotation argument is sufficient to establish the orthogonality of the cross
product.. By Eq. 1.22 and the discussion that follows it, we know that two
vectors are totally distinct if and only if they are orthogonal — that is, the
angle between the vectors is $90^{\textrm{o}}$. But that means our new “area
vector” must be perpendicular to both of its constituent vectors. This idea
holds when we consider that there is a perpendicular line running through Fig.
1.10, and since scaling the vector by $-1$ keeps a vector colinear, the anti-
commutativity of this “area vector” preserves the orthogonality. Thus, we
define the vector cross product. Explicitly, the cross product of the vectors
$\vec{v}$ and $\vec{u}$ is a vector whose magnitude is the area spanned by
both vectors, and whose direction is orthogonal to both vectors.

Hence, we could introduce a new basis vector, let’s call it $\hat{z}$, and use
it to write down the cross product of the vectors $\vec{v}$ and $\vec{u}$ in
Fig. 1.10

$\displaystyle\vec{v}\times\vec{u}=\left(v_{x}\hat{x}+v_{y}\hat{y}\right)\times\left(u_{x}\hat{x}+u_{y}\hat{y}\right)=0\hat{x}+0\hat{y}+(v_{x}u_{y}-v_{y}u_{x})\hat{z},$
(1.28)

where the $\times$ symbol is explicitly written. Now, as long as
$\hat{x}\cdot\hat{z}=\hat{y}\cdot\hat{z}=0$, then this definition of the cross
product will be orthogonal to both $\vec{v}$ and $\vec{u}$ and this vector
will have the length of the parallelogram spanned by the two — I leave it to
you to prove this using Eq. 1.21 for the orthogonality and Eq. 1.8 for the
length. Furthermore, the anti-commutativity of the area calculation is
satisfied for this definition as well. Written explicitly,

$\displaystyle\vec{u}\times\vec{v}$
$\displaystyle=0\hat{x}+0\hat{y}+(u_{x}v_{y}-u_{y}v_{x})\hat{z}$
$\displaystyle=0\hat{x}+0\hat{y}-(u_{y}v_{x}-u_{x}v_{y})\hat{z}$
$\displaystyle=-\left[0\hat{x}+0\hat{y}+(v_{x}u_{y}-v_{y}u_{x})\hat{z}\right]$
$\displaystyle=-\left(\vec{v}\times\vec{u}\right),$ (1.29)

By looking at Eq. 1.28, we can deduce a couple of properties. First, we
consider the cross product of a vector with itself, namely
$\vec{v}\times\vec{v}$. By Eq. 1.28, we have

$\displaystyle\vec{v}\times\vec{v}=0\hat{x}+0\hat{y}+(v_{x}v_{y}-v_{y}v_{x})\hat{z}=0\hat{x}+0\hat{y}+(v_{x}v_{y}-v_{x}v_{y})\hat{z}=0\hat{x}+0\hat{y}+0\hat{z}=\vec{0}.$
(1.30)

Now this normally may seem weird202020It is actually a more general property
of anti-commutative operators., but in terms of our geometrical picture given
by Fig. 1.10, this interpretation makes complete sense: the area of the
parallelogram spanned by $\vec{v}$ and $\vec{v}$ is zero, because the area of
a line is zero! In other words, if the angle between the two vectors $\theta$
aligned tail-to-tail approaches zero (or 180${}^{\textrm{o}}$), the vectors
being crossed span a thinner and thinner parallelogram. Another property that
is important to us to consider is what happens when we scale one vector by a
scalar $a$ and then cross them. In other words, what is
$(a\vec{v})\times\vec{u}$? By Eq. 1.28 and Eq. 1.15, we have

$\displaystyle(a\vec{v})\times\vec{u}$
$\displaystyle=\left(av_{x}\hat{x}+av_{y}\hat{y}\right)\times\left(u_{x}\hat{x}+u_{y}\hat{y}\right)$
$\displaystyle=0\hat{x}+0\hat{y}+(av_{x}u_{y}-av_{y}u_{x})\hat{z}$
$\displaystyle=\hat{x}+0\hat{y}+a(v_{x}u_{y}-v_{y}u_{x})\hat{z}$
$\displaystyle=a\left[\hat{x}+0\hat{y}+(v_{x}u_{y}-v_{y}u_{x})\hat{z}\right]$
$\displaystyle=a\left(\vec{v}\times\vec{u}\right).$ (1.31)

What we conclude from Eq. 1.31 is that scaling one of the vectors being
crossed effectively scales the entire cross product, itself, by that same
exact factor. But this again makes sense in terms of parallelograms because if
we were to stretch or shrink one of the sides of the parallelogram by a factor
of $a$, then we would expect the area to increase or decrease by $|a|$. I
recommend that you show that

$\displaystyle(a\vec{v})\times(b\vec{u})=ab\left(\vec{v}\times\vec{u}\right),$
(1.32)

using a very similar argument (except now both vectors are scaled).

####  1.3.9 Unit Vectors are Ambidextrous (but we like them to be Right-
Handed)

Let’s dive a little deeper into what Eq. 1.28 implies if our coordinate system
be consistent with itself. To do this, we expand the first equality in the
equation just like we would for binomial multiplication:

$\displaystyle\left(v_{x}\hat{x}+v_{y}\hat{y}\right)\times\left(u_{x}\hat{x}+u_{y}\hat{y}\right)=v_{x}\hat{x}\times
u_{x}\hat{x}+v_{x}\hat{x}\times u_{y}\hat{y}+v_{y}\hat{y}\times
u_{x}\hat{x}+v_{y}\hat{y}\times u_{y}\hat{y}$

By Eq. 1.32, we can simplify the equation above by factoring out the scalar
values of the components to get

$\displaystyle\left(v_{x}\hat{x}+v_{y}\hat{y}\right)\times\left(u_{x}\hat{x}+u_{y}\hat{y}\right)=v_{x}u_{x}\left(\hat{x}\times\hat{x}\right)+v_{x}u_{y}\left(\hat{x}\times\hat{y}\right)+v_{y}u_{x}\left(\hat{y}\times\hat{x}\right)+v_{y}u_{y}\left(\hat{y}\times\hat{y}\right)$

By Eq. 1.30, both terms with $\hat{x}\times\hat{x}$ and $\hat{y}\times\hat{y}$
are identically zero (again, because the parallelogram spanned by only a
single vector is really just a line which has vanishing area). Then, by Eq.
1.29, $\hat{y}\times\hat{x}=-\left(\hat{x}\times\hat{y}\right)$. Thus,

$\displaystyle\left(v_{x}\hat{x}+v_{y}\hat{y}\right)\times\left(u_{x}\hat{x}+u_{y}\hat{y}\right)=v_{x}u_{x}\left(\vec{0}\right)+v_{x}u_{y}\left(\hat{x}\times\hat{y}\right)-v_{y}u_{x}\left(\hat{x}\times\hat{y}\right)+v_{y}u_{y}\left(\vec{0}\right).$

Given that $a\vec{0}=\vec{0}$, for every scalar $a\in\mathds{R}$ (prove this
for yourself — remember that the zero vector $\vec{0}$ is really just a set of
components that are all $0$), and $\vec{0}+\vec{v}=\vec{v}$, we conclude

$\displaystyle\left(v_{x}\hat{x}+v_{y}\hat{y}\right)\times\left(u_{x}\hat{x}+u_{y}\hat{y}\right)$
$\displaystyle=\vec{0}+\left(v_{x}u_{y}-v_{y}u_{x}\right)\left(\hat{x}\times\hat{y}\right)+\vec{0}$
$\displaystyle=\left(v_{x}u_{y}-v_{y}u_{x}\right)\left(\hat{x}\times\hat{y}\right)$

Furthermore, since $0\vec{v}=\vec{0}$, it must be true that

$\displaystyle\vec{v}\times\vec{u}=\left(v_{x}\hat{x}+v_{y}\hat{y}\right)\times\left(u_{x}\hat{x}+u_{y}\hat{y}\right)=0\hat{x}+0\hat{y}+\left(v_{x}u_{y}-v_{y}u_{x}\right)\left(\hat{x}\times\hat{y}\right).$

Upon comparing this result with Eq. 1.28, we are forced to conclude that

$\displaystyle\hat{x}\times\hat{y}=\hat{z},$ (1.33)

in order for the cross product to be consistent with the way we write vectors
using components and the basis vectors $\hat{x}$ and $\hat{y}$.

But even though Eq. 1.33 must be true for consistency, it actually does not
tell us anything about the geometrical direction of $\hat{z}$. Remember that
up until now, the cross product is not really geometrically unique — all that
is required up until this point is that the cross product be orthogonal to the
two vectors being crossed. Since there is always a perpendicular line running
through the parallelogram spanned by two vectors212121This is only true is
three and seven dimensions. In other dimensions, a more general object called
the wedge or exterior product must be used in lieu of the cross product., then
the direction of our cross product up until now is still ambiguous. It could
be either one direction along the line or the other and everything cross-
product-wise would still check out. In Fig. 1.10, the two directions would be
either out-of-the-page or into-the-page. This ambiguity leads us to the idea
of handedness in coordinate systems.

FIG. 1.11: A diagram of how we define right-handedness among the Cartesian
unit vectors $\\{\hat{x},\hat{y},\hat{z}\\}$. Throughout this diagram, we
choose to label each direction differently, but the direction of each vector
remains the same in each set of axes. Starting from the leftmost set of axes,
we have $\hat{x}\times\hat{y}=\hat{z}$, where the dashed (blue) vector defines
the otherwise ambiguous cross product $\hat{z}$. By rotating the labels
counter-clockwise and then taking (green vector) $\times$ (red vector) we can
define $\hat{z}\times\hat{x}=\hat{y}$ and $\hat{y}\times\hat{z}=\hat{x}$.

Since there are two options for our $\hat{z}$ in Fig. 1.10 to have, either
out-of-the-page or into-the-page, we should have two options for handedness in
our coordinate systems. In practice, we must choose our coordinates to be
either right-handed or left-handed. In a right-handed coordinate system,
$+\hat{z}$ defined to be out-of-the-page. In a left-handed coordinate system,
$+\hat{z}$ is defined to be into-the-page222222This may not always be true.
Sometimes, $+\hat{z}$ is still out-of-the-page, but in these cases,
$\hat{x}\times\hat{y}=-\hat{z}$, which would point into-the-page. In those
cases, the coordinate system is still left-handed, but the axis labeling
scheme is different than the one I adopt. It heavily depends on the author
because left-handed coordinates are so infrequently used that there really
isn’t a convention established as well as there is for right-handed
coordinates.. The reason why coordinate systems are given handedness is
because three orthogonal unit vectors behave like our thumb, pointer finger,
and middle finger when we orient them to be all mutually orthogonal.

To illustrate this point, consider the set of perpendicular basis vectors
$\\{\hat{x},\hat{y},\hat{z}\\}$ drawn in the leftmost set of axes in Fig.
1.11. Since these are mutually orthogonal and fixed in space, these vectors
are called Cartesian basis vectors232323Not all basis vectors have to be
Cartesian. For example, there exist coordinate systems that have cylindrical
or spherical symmetry as opposed to rectangular-prism symmetry. However, these
curvilinear coordinate systems have basis vectors that are functions of the
Cartesian basis vectors.. By Eq. 1.33, we know that
$\hat{z}=\hat{x}\times\hat{y}$. To remove the ambiguity in the direction of
$\hat{z}$, we choose our system of coordinates to be right-handed so that
$+\hat{z}$ would point out-of-the-page. Now take your pointer finger on your
right hand and align it with the $\hat{x}$-axis. Next take your right middle
finger and align it with the $\hat{y}$-axis. Finally, as you should see, if
you stick your right thumb straight out, then it will be pointing along the
$\hat{z}$ direction. This method for finding the direction of a cross product
of two vectors is called the right-hand rule, and therefore all systems that
abide by it are right-handed. Notice that if you did the exact same procedure
with your left pointer and middle fingers, then your left thumb would point in
the exact opposite direction as your right thumb, and therefore the leftmost
set of coordinates in Fig. 1.11 cannot be left-handed.

I want to emphasize something now: the choice of Roman letters, $x$, $y$, or
$z$, that are used to describe each direction established by our basis vectors
is nothing special. Letters are just letters. We could have easily enough
chosen to use ☉, ☺, and § instead. The important things are the directions
that these labels represent, namely the green solid vector, the red solid
vector, and the blue dashed vector in Fig. 1.11, respectively. So since the
directions are the things that matter, we could really say something like

$\displaystyle(\textrm{green solid direction})\times(\textrm{red solid
direction})=(\textrm{blue dashed direction}),$

as long as we understand that the cross product relationship above is between
the three mutually orthogonal directions sketched in Fig. 1.11. This then
leads us to another property of right-handed coordinate systems: we must keeps
the right-hand rule symmetry in $\hat{z}=\hat{x}\times\hat{y}$ whenever we
change the axis labels in our coordinate system. This is required so that our
geometric and component interpretations of vectors are consistent. Thus, if
the green solid direction is labeled as $\hat{z}$, then if we were to align
our right thumb with the green solid vector in the middle set of axes in the
figure and our right pointer finger with the red solid vector, we would
conclude from the right-hand rule that the red solid direction (our right
pointer finger) would have to be labeled $\hat{x}$ and the blue dashed
direction would have to be labeled $\hat{y}$. Notice that even with our new
labeling-scheme, the directions the vectors point in are identical to the ones
they pointed in before. Then by crossing the colors in the second set of axes,
we conclude that

$\displaystyle\hat{z}\times\hat{x}=\hat{y}.$ (1.34)

By looking to the third set of axes, relabeling the red solid direction as
$\hat{z}$ and aligning our right thumbs with it, and so on, we would see then
that

$\displaystyle\hat{y}\times\hat{z}=\hat{x}.$ (1.35)

Hence we have found the following cross product relationship for the right-
handed Cartesian basis vectors:

$\displaystyle\begin{cases}\hat{x}\times\hat{y}&=\hat{z}\\\
\hat{z}\times\hat{x}&=\hat{y}\\\ \hat{y}\times\hat{z}&=\hat{x}\end{cases}$
(1.36)

Notice that these relationship have a cyclical symmetry; we can permute each
of the basis vectors rightward (the rightmost basis vector then is moved to
the leftmost) and proceed from one relationship to the next. This cyclical
property is used extensively in physics and mathematics, particularly when
dealing with observable quantities in Quantum Mechanics.

I want to take the time to note that a similar cross product relationship
holds for left-handed Cartesian basis vectors. However, it is conventional
that coordinate systems be right-handed in physics, and so I will neither
diagram it nor write the relationships because I risk confusing myself (and
probably you, too). You can totally decide to change your conventions though
if you wanted — although your graders will definitely not like it! The reason
you can is that the into-the-page direction that I arbitrarily decided to NOT
be the $+\hat{z}$ direction is still a physical thing. The physical direction
of does not change whether we decide to use right-handed coordinates or left-
handed ones. The physics is usually the same either way. However, again, the
convention in physics is that coordinate systems are right-handed.

Oddly enough, there are physical processes that are exclusively left-handed.
For example, the weak force governing the particle physics behind nuclear
decays will only act on particles that have a left-handed chirality which is a
measure of the relationship between a particle’s momentum and spin angular
momentum242424If it just so happens that if there were particles that had
right-handed chirality, then they would be undetectable via weak force
interactions. The short version of the reason why this is true is weak force
carriers can only “see” the left-handed particles. The much longer version is
given in [griffiths_2014]. Some of these particles, the sterile neutrinos, are
hypothesized to exist as candidates for dark matter — a bunch of matter in the
universe that is only detectable via gravitational interactions, but
outnumbers regular matter 5-to-1 [nasa_science_dark_matter]! One could argue
now that since there is at least one fundamental physical process that prefers
left-handedness, we physicists should all learn how to use left-handed
coordinate systems so that our physical models have a closer connection to
nature. While that may be a fair argument for some people, the truth of the
matter is that any natural preference for either handedness is so rare in most
physical systems it isn’t apparent at all. This is true for essentially all of
undergraduate and a lot of graduate physics. Therefore, since most of our
written Laws and Theories of Physics, including Maxwell’s Equations written at
the very beginning of this chapter, are written within the framework of a
right-handed coordinate system, the physics community has stuck with the
right-handed coordinate convention.

Before finishing up this section, I want to take the time to talk a little bit
more about the right-hand rule because of how useful it is in finding
directions of cross products. An algorithm that you can use that will never
fail to give you the correct direction of $\vec{v}\times\vec{u}$ for any two
three-dimensional vectors $\vec{v}$ and $\vec{u}$ is as follows:

  1. 1.

Align your right pointer finger with the first vector in the cross product
(remember the order matters) $\vec{v}$.

  2. 2.

Point your right middle finger in the direction of the second vector
$\vec{u}$.

  3. 3.

Stick your right thumb out straight so it is perpendicular to both $\vec{v}$
and $\vec{u}$. This is the direction of $\vec{v}\times\vec{u}$.

This algorithm is summarized in Fig. 1.12.

FIG. 1.12: An illustration of the right-hand rule.

It is imperative to remember that the order of the cross product matters. This
is why I italicized the words “first” and “second” in the algorithm above. I
recommend proving to yourself that $\vec{u}\times\vec{v}$ has the exact
opposite direction of $\vec{v}\times\vec{u}$ using the right-hand rule, just
to make sure I’m telling you the truth about the right-hand rule retaining the
cross product’s antic-commutativity. I also want to note that this is not the
only algorithm that you can use. Just like how we changed which of our fingers
pointed in the direction of the blue dashed vector in Fig. 1.11 to calculate
all of the cross product relations in Eq. 1.36, we could similarly do the same
thing with general three-dimensional vectors. The key is to move from the
first vector to the second to the cross product, and align your first right
finger to the second right finger to your third right finger in a counter-
clockwise rotation. However, for a lot of people who are learning the right-
hand rule for the first time, the more general cyclic properties of cross
products can be overwhelming. So when in doubt, use the algorithm above; since
it is a special case of the more general cyclic properties of the cross
product, it will never fail you. Just please — and this may sound silly at
first but you’d be surprised how many newbies252525This has literally cost me
full letter grades on Physics exams before. On the exam I would write with my
right hand and use my left to find cross products. DO NOT DO THIS. IT IS
WRONG. mess this up when they are first learning the right-hand rule —
remember to use your right (not your left) hand!

####  1.3.10 Three-Dimensional Cross Products

So now that we have a relationship between the cross products of our right-
handed Cartesian basis vectors, we will use them to calculate a general three-
dimensional formula for cross products. Interestingly enough, this formula
does NOT generalize as a cross product in higher dimensions, unlike all of the
other formulas we have derived so far. The behavior of the cross product as a
orthogonal vector whose magnitude is that of a parallelogram only holds in
three dimensions and seven dimensions. There are generalizations called wedge
products, but they are outside the scope of this chapter. Without further ado,
we will find the cross product between two three-dimensional vectors in a
similar way to the how we found Eq. 1.33.

Consider the two vectors $\vec{v}$ and $\vec{u}$ and their right-handed
Cartesian components given below

$\displaystyle\vec{v}$ $\displaystyle=v_{x}\hat{x}+v_{y}\hat{y}+v_{z}\hat{z},$
$\displaystyle\vec{u}$ $\displaystyle=u_{x}\hat{x}+u_{y}\hat{y}+u_{z}\hat{z}.$

Then the cross product $\vec{v}\times\vec{u}$ is given by

$\displaystyle\vec{v}\times\vec{u}$
$\displaystyle=\left(v_{x}\hat{x}+v_{y}\hat{y}+v_{z}\hat{z}\right)\times\left(u_{x}\hat{x}+u_{y}\hat{y}+u_{z}\hat{z}\right),$
$\displaystyle=\;\;v_{x}u_{x}\left(\hat{x}\times\hat{x}\right)+v_{x}u_{y}\left(\hat{x}\times\hat{y}\right)+v_{x}u_{z}\left(\hat{x}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{y}u_{x}\left(\hat{y}\times\hat{x}\right)+v_{y}u_{y}\left(\hat{y}\times\hat{y}\right)+v_{y}u_{z}\left(\hat{y}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{z}u_{x}\left(\hat{z}\times\hat{x}\right)+v_{z}u_{y}\left(\hat{z}\times\hat{y}\right)+v_{z}u_{z}\left(\hat{z}\times\hat{z}\right).$

But since
$\hat{x}\times\hat{x}=\hat{y}\times\hat{y}=\hat{z}\times\hat{z}=\vec{0}$, then
we can simplify the nine terms above to only six.

$\displaystyle\vec{v}\times\vec{u}$
$\displaystyle=\;\;v_{x}u_{x}\left(\hat{x}\times\hat{x}\right)+v_{x}u_{y}\left(\hat{x}\times\hat{y}\right)+v_{x}u_{z}\left(\hat{x}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{y}u_{x}\left(\hat{y}\times\hat{x}\right)+v_{y}u_{y}\left(\hat{y}\times\hat{y}\right)+v_{y}u_{z}\left(\hat{y}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{z}u_{x}\left(\hat{z}\times\hat{x}\right)+v_{z}u_{y}\left(\hat{z}\times\hat{y}\right)+v_{z}u_{z}\left(\hat{z}\times\hat{z}\right),$
$\displaystyle=\;\;v_{x}u_{x}\left(\vec{0}\right)+v_{x}u_{y}\left(\hat{x}\times\hat{y}\right)+v_{x}u_{z}\left(\hat{x}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{y}u_{x}\left(\hat{y}\times\hat{x}\right)+v_{y}u_{y}\left(\vec{0}\right)+v_{y}u_{z}\left(\hat{y}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{z}u_{x}\left(\hat{z}\times\hat{x}\right)+v_{z}u_{y}\left(\hat{z}\times\hat{y}\right)+v_{z}u_{z}\left(\vec{0}\right),$
$\displaystyle=\;\;\;v_{x}u_{y}\left(\hat{x}\times\hat{y}\right)+v_{x}u_{z}\left(\hat{x}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{y}u_{x}\left(\hat{y}\times\hat{x}\right)+v_{y}u_{z}\left(\hat{y}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{z}u_{x}\left(\hat{z}\times\hat{x}\right)+v_{z}u_{y}\left(\hat{z}\times\hat{y}\right).$

We now make use of the anti-commutativity of the cross product to combine
terms like $\hat{x}\times\hat{y}$ and $\hat{y}\times\hat{x}$.

$\displaystyle\vec{v}\times\vec{u}$
$\displaystyle=\;\;\;v_{x}u_{y}\left(\hat{x}\times\hat{y}\right)-v_{x}u_{z}\left(\hat{z}\times\hat{x}\right)$
$\displaystyle\;\;\;-v_{y}u_{x}\left(\hat{x}\times\hat{y}\right)+v_{y}u_{z}\left(\hat{y}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{z}u_{x}\left(\hat{z}\times\hat{x}\right)-v_{z}u_{y}\left(\hat{y}\times\hat{z}\right).$

Notice that some of the $+$ signs became $-$ signs. And now we use the cross
product relations Eq. 1.36:

$\displaystyle\vec{v}\times\vec{u}$
$\displaystyle=\;\;\;v_{x}u_{y}\left(\hat{x}\times\hat{y}\right)-v_{x}u_{z}\left(\hat{z}\times\hat{x}\right)$
$\displaystyle\;\;\;-v_{y}u_{x}\left(\hat{x}\times\hat{y}\right)+v_{y}u_{z}\left(\hat{y}\times\hat{z}\right)$
$\displaystyle\;\;\;+v_{z}u_{x}\left(\hat{z}\times\hat{x}\right)-v_{z}u_{y}\left(\hat{y}\times\hat{z}\right),$
$\displaystyle=\;\;\;v_{x}u_{y}\left(\hat{z}\right)-v_{x}u_{z}\left(\hat{y}\right)$
$\displaystyle\;\;\;-v_{y}u_{x}\left(\hat{z}\right)+v_{y}u_{z}\left(\hat{x}\right)$
$\displaystyle\;\;\;+v_{z}u_{x}\left(\hat{y}\right)-v_{z}u_{y}\left(\hat{x}\right).$

From here, we combine like-terms to obtain the general cross product formula
for three-dimensional vectors:

$\displaystyle\vec{v}\times\vec{u}=\left(v_{y}u_{z}-v_{z}u_{y}\right)\hat{x}+\left(v_{z}u_{x}-v_{x}u_{z}\right)\hat{y}+\left(v_{x}u_{y}-v_{y}u_{x}\right)\hat{z}.$
(1.37)

I leave it to you to show that this cross product is still orthogonal to both
constituent vectors, $\vec{v}$ and $\vec{u}$ (Problem 1.3.10). For
completeness, I want to emphasize that even in three-dimensions,

$\displaystyle\left|\vec{v}\times\vec{u}\right|=vu\sin\theta,$ (1.38)

where $v$ and $u$ are the lengths of the vectors $\vec{v}$ and $\vec{u}$,
respectively, and $\theta$ is the angle between the two lined up tail-to-tail,
and so the full three-dimensional cross product still has a magnitude that is
equal to the parallelogram spanned between $\vec{v}$ and $\vec{u}$.

Use the component form of the vectors $\vec{v}$ and $\vec{u}$ to show both of
the following equations
$\displaystyle\vec{v}\cdot\left(\vec{v}\times\vec{u}\right)$
$\displaystyle=0,$
$\displaystyle\vec{u}\cdot\left(\vec{v}\times\vec{u}\right)$
$\displaystyle=0.$ Since these dot products are zero for both constituent
vectors, then we must conclude that the cross product is indeed orthogonal to
both of its constituent vectors.

####  1.3.11 A Note on Linearity

There is one very useful property that comes up all over the place in physics,
and I actually surreptitiously exploited it in a couple different places in
this chapter. This exceedingly useful property is called linearity, and it is
so useful because it can greatly simplify calculations with vectors, or
derivatives, or electromagnetic waves, or quantum mechanical wavefunctions, et
cetera. Remarkably, this property is also incredibly simple to write down. It
essentially goes something like this: consider any operation $\mathcal{O}$
that acts on two vectors $a\vec{u}$ and $b\vec{v}$, where $a$ and $b$ are just
numbers. Then we define $\mathcal{O}$ to be linear if the following statement
is true:

$\displaystyle\mathcal{O}(a\vec{u}+b\vec{v})=a\mathcal{O}(\vec{u})+b\mathcal{O}(\vec{v}),\;\;\textrm{for
all }a,b\in\mathds{R}.$ (1.39)

In other words, an operation is linear if

  1. 1.

if it is distributative over a sum,

  2. 2.

if it ignores numerical coefficients.

To understand this property, let’s consider two operations defined below

$\displaystyle\mathcal{O}_{1}(\vec{c})$
$\displaystyle=\vec{c}+\vec{r},\;\;\vec{r}\neq\vec{0}$
$\displaystyle\mathcal{O}_{2}(\vec{c})$ $\displaystyle=\gamma\vec{c}.$

These two operations are ones we have already discussed. The first is vector
addition, where the operation takes in a vector $\vec{c}$ and adds
$\vec{r}\neq\vec{0}$ to it. The second is scalar multiplication, where the
operation takes in a vector $\vec{c}$ and stretches or shrinks it by a factor
of $\gamma$. We will now check to see if they fit the definition of linearity.
To do this, we essentially act on $a\vec{u}+b\vec{v}$ with each operator, and
see if we can massage the resulting equation to look like the right-hand side
of Eq. 1.39. We start with $\mathcal{O}_{1}$.

$\displaystyle\mathcal{O}_{1}(a\vec{u}+b\vec{v})=(a\vec{u}+b\vec{v})+\vec{r}\neq
a(\vec{u}+\vec{r})+b(\vec{v}+\vec{r})=a\mathcal{O}_{1}(\vec{u})+b\mathcal{O}_{1}(\vec{v})$

Since the equality does not hold, we conclude vector addition is not linear.
To see why it does not hold, remember based on the definition of linearity,
Eq. 1.39 must hold for all values of $a$ and $b$. Therefore, if we can even
find just one value where the equality does not hold, the operation is not
linear (based on our definition). To do this, let’s take $a=b=1$. Then we
would have

$\displaystyle\mathcal{O}_{1}(a\vec{u}+b\vec{v})=\mathcal{O}_{1}(\vec{u}+\vec{v})=(\vec{u}+\vec{v})+\vec{r}=\vec{u}+\vec{v}+\vec{r}.$

Meanwhile,

$\displaystyle
a\mathcal{O}_{1}(\vec{u})+b\mathcal{O}_{1}(\vec{v})=\mathcal{O}_{1}(\vec{u})+\mathcal{O}_{1}(\vec{v})=(\vec{u}+\vec{r})+(\vec{v}+\vec{r})=\vec{u}+\vec{v}+2\vec{r}.$

If these two equations were equal, then by comparing them, we would find
$\vec{r}=2\vec{r}$, which can only be true if $\vec{r}=\vec{0}$ (prove this to
yourself). But since $\vec{r}\neq\vec{0}$, then this is a contradiction. Hence
we conclude that vector addition cannot be linear.

Now we test scalar multiplication, $\mathcal{O}_{2}$.

$\displaystyle\mathcal{O}_{2}(a\vec{u}+b\vec{v})$
$\displaystyle=\gamma(a\vec{u}+b\vec{v})$
$\displaystyle=\gamma\left[a(u_{1}\hat{x}_{1}+u_{2}\hat{x}_{2}+\dots)+b(v_{1}\hat{x_{1}}+v_{2}\hat{x}_{2}+\dots)\right]$
$\displaystyle=\gamma\left[(au_{1}+bv_{1})\hat{x_{1}}+(au_{2}+bv_{2})\hat{x}_{2}+\dots\right]$
$\displaystyle=\gamma(au_{1}+bv_{1})\hat{x_{1}}+\gamma(au_{2}+bv_{2})\hat{x}_{2}+\dots$
$\displaystyle=(\gamma au_{1}+\gamma bv_{1})\hat{x_{1}}+(\gamma au_{2}+\gamma
bv_{2})\hat{x}_{2}+\dots$ $\displaystyle=\left[a(\gamma
u_{1}\hat{x}_{1}+\gamma
u_{2}\hat{x}_{2}+\dots)+b(v_{1}\gamma\hat{x_{1}}+v_{2}\gamma\hat{x}_{2}+\dots)\right]$
$\displaystyle=a(\gamma\vec{u})+b(\gamma\vec{v})$
$\displaystyle=a\mathcal{O}_{2}(\vec{u})+b\mathcal{O}_{2}(\vec{v}).$

Thus, scalar multiplication is linear! I want to emphasize that it was
necessary to expand out both $\vec{u}$ and $\vec{v}$ in terms of the basis
vectors $\\{\hat{x}_{i}\\}$ because we have only defined scalar multiplication
in terms of multiplying all the components equally by the same factor. Then to
finish the proof, we had to rearrange the multiplication of the scalars $a$,
$b$, and $\gamma$ (which is possible because they are all numbers that have
commutative multiplication rules), and then recombine the components. So the
proof is a little less trivial than it may have initially seemed, but it shows
something pretty important: stretching and shrinking is distributive over a
sum of vectors and totally ignores the numerical coefficients in front.
Remember, this was absolutely not the case for vector addition!

Using the component-based definitions of the dot product and the cross
product, we can see that both of these operations are linear, as well. I will
show the proof of the dot product formula, and I encourage you to show the
cross product case. To begin, we start with the $n$-dimensional dot product
formula given by Eq. 1.21 and set $\vec{v}=a\vec{s}+b\vec{r}$, for any two
vectors $\vec{s}$ and $\vec{r}$. Then we would have

$\displaystyle\vec{u}\cdot(a\vec{s}+b\vec{r})$
$\displaystyle=\sum_{j=1}^{n}u_{j}(as_{j}+br_{j})$
$\displaystyle=u_{1}(as_{1}+br_{1})+u_{2}(as_{2}+br_{2})+\dots$
$\displaystyle=u_{1}as_{1}+u_{1}br_{1}+u_{2}as_{2}+u_{2}br_{2}+\dots$
$\displaystyle=au_{1}s_{1}+bu_{1}r_{1}+au_{2}s_{2}+bu_{2}r_{2}+\dots$
$\displaystyle=a(u_{1}s_{1}+u_{2}s_{2}+\dots)+b(u_{1}r_{1}+u_{2}r_{2}+\dots)$
$\displaystyle=a\sum_{j=1}^{n}u_{j}s_{j}+b\sum_{j=1}^{n}u_{j}r_{j}$
$\displaystyle=a(\vec{u}\cdot\vec{s})+b(\vec{u}\cdot\vec{r}).$

And therefore the dot product is indeed linear because it is distributive and
totally ignores numerical coefficients. As I said before, it is also true that
the cross product is linear, therefore

$\displaystyle\vec{u}\times(a\vec{s}+b\vec{r})=a(\vec{u}\times\vec{s})+b(\vec{u}\times\vec{r}).$

To prove this one, make use of Eq. 1.37. It is actually possible to prove both
of these operations are linear using only geometry (i.e. without the use of
components), but since we have the component machinery in place (and it is
much more straightforward than the geometric proof), I figured it was best to
use it.

The reason why linearity matters in physics is because it allows us to
computer superpositions of quantities. You may have heard of superposition in
terms of wavefunctions and Schrödinger’s famous cat, but superposition is not
strictly a quantum phenomena. It occurs in (linear) wave phenomena, finding
the net force on a particle, summing over the torques on a system, et cetera.
The utility as far as vectors go, comes in from the following example. Suppose
$\vec{u}=u\hat{x}$ and $\vec{v}=v_{x}\hat{x}+v_{y}\hat{y}$. Thus $\vec{u}$
ONLY points in the $\hat{x}$-direction, while $\vec{v}$ points in both the
$\vec{x}$\- and $\hat{y}$-directions. We now take the dot product and cross
product of these two vectors, starting with the dot product.

$\displaystyle\vec{u}\cdot\vec{v}=\vec{u}\cdot(v_{x}\hat{x}+v_{y}\hat{y})=v_{x}\vec{u}\cdot\hat{x}+v_{y}\vec{u}\cdot\hat{y}=uv_{x}(\hat{x}\cdot\hat{x})+uv_{y}(\hat{x}\cdot\hat{y}).$

But since $\hat{x}\cdot\hat{x}=|\hat{x}|^{2}=1$ by Eq. 1.25 and the definition
of a unit vector, and $\vec{x}\cdot\hat{y}=0$ since they are perpendicular
then we have

$\displaystyle\vec{u}\cdot\vec{v}=uv_{x}.$

But since $\vec{u}$ exclusively points in the $\hat{x}$-direction, the
linearity of the dot product selected the parallel component of $\vec{v}$ and
totally ignored the perpendicular components it! Further, the linearity in the
second term shows that it DOES NOT MATTER how big the perpendicular component
of $\vec{v}$ is when dotting it with $\vec{u}$ — it is totally zero by the
unit-vector construction. This idea generalizes well in $n$-dimensions, too,
actually. Whenever vectors are dotted together, only the parallel components
will survive while all of the perpendicular components will always go to zero.

The cross product actually works in the exact opposite way (except only in
three and seven dimensions). If we take these two vectors again and calculate
their cross product, then we have

$\displaystyle\vec{u}\times\vec{v}=\vec{u}\times(v_{x}\hat{x}+v_{y}\hat{y})=v_{x}\vec{u}\times\hat{x}+v_{y}\vec{u}\times\hat{y}=uv_{x}(\hat{x}\times\hat{x})+uv_{y}(\hat{x}\times\hat{y}).$

But now $\hat{x}\times\hat{x}=\vec{0}$ and $\hat{x}\times\hat{y}=\hat{z}$, by
anti-commutativity and Eq. 1.33. Therefore,

$\displaystyle\vec{u}\times\vec{v}=uv_{y}\hat{z}.$

To contrast this result with the dot product, the statement above means that
the linearity of the cross product selects the component of $\vec{v}$ that is
perpendicular to $\vec{u}$ and totally ignores the parallel component.
Furthermore, this property means that it does not matter how large the
parallel component is — all that matters is the size of the perpendicular
part!

###  1.4 Concluding Remarks

In this chapter we covered the basics of vector operations. In the very
beginning, we started the discussion with talking about mathematics as a game
that has a set of rules (the proper name for “rule” is axiom). Although the
first few sections may have initially seemed a little unnecessarily abstract
when they simply were talking about real numbers and functions, I kept them in
because the abstractness is to illustrate how we can formulate familiar
mathematics in terms of this game structure. From here, we moved to defining a
vector geometrically as lengths in a particular direction. Using geometry, we
saw that we can define a commutative function of vectors that we interpret as
vector addition (tip-to-tail!). We then found an anti-commutative function