PART I.1 - Physical Mathematics

BAAABAAA
XQQPQQ βαβαµ
µ
βαβα εδσ == },{2},{ and&&
From First Principles
PART I – PHYSICAL MATHEMATICS
January 2017 – R4.2
Maurice R. TREMBLAY
BAAABAAA
XQQPQQ βαβαµ
µ
βαβα εδσ == },{2},{ and&&
Chapter 1

To Chrisy
If physical mathematics could explain how I found you, got to
know you and Love you still so much today as yesterday…
Then the true equation of our life together must be an infinite
series of happy moments locked in our memories forever…
MRT
2015

The Greek Alphabet
Α, α, a Alpha άλφα
Β, β, b Beta βήτα
Γ, γ, g Gamma γάµµα
∆, δ, d Delta δέλτα
Ε, ε, é Epsilon [épsilon] έψιλον
Ζ, ζ, dz Zeta [dz] ζήτα
Η, η, ê Eta [êta] ήτα
Θ, θ, t Theta [th] θήτα
Ι, ι, i Iota [eeota] ιώτα
Κ, κ, k Kappa κάππα
Λ, λ, l Lambda λάµδα
Μ, µ, m Mu µυ
Ν, ν, n Nu νυ
Ξ, ξ, ks Xi [ks] ξι
Ο, ο, o Omicron όµικρον
Π, π, p Pi πι
Ρ, ρ, r Rho ρώ
Σ, σ, s Sigma σίγµα
Τ, τ, t Tau ταυ
Υ, υ, u Upsilon ύψιλον
Φ,ϕ,φ, ph Phi φι
Χ, χ, kh Chi [kh] χι
Ψ, ψ, ps Psi ψι
Ω, ω, o Omega ωµέγα
These are the equations and phenomenology of the Standard Model of particle physics:
123
110 150120 130 140 GeV100
Super-
symmetry Multiverse
MHiggs
?Theory
……
Mass
3
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This is a Hubble
Space Telescope
image which
features the star
cluster Trumpler 14.
It is one of the
largest gatherings of
hot massive and
bright stars (like our
Sun) in the Milky
Way, the name of
the galaxy our sun is
located in.
This cluster houses
some of the most
luminous stars in
our entire galaxy.
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Mean distance from Earth: 150
million kilometers (149.6××××106 km)
or 8.31 minutes at the speed of
light
Absolute magnitude: 4.8mag
Mean distance from Milky Way
core:
~2.5××××1017 km (26,000-28,000 l-y)
Galactic period: 2.25-2.50××××108 y
Velocity: 217 km/s orbit around the
center of the Galaxy, 20 km/s
relative to average velocity of other
stars in stellar neighborhood.
Completes one revolution in about
225–250 million years.
Rotation velocity at equator:
7,174 km/h
Luminosity: 3.827××××1026 Watts
* 1 Å ==== 1××××10−−−−10 m ==== 0.1 nm (in units wavelength)
The Sun is composed of 73.46 % Hydrogen, 24.85 % Helium and other elements.
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Prolog – The Celestial Bodies
The Sun – SOHO 304 Å *
5

The Sun use the principle of nuclear fusion of hydrogen to produce helium and releases
the energy that causes stars to shine and hydrogen bombs to explode.
Mean diameter: 1.392××××10
6
km
(109 Earth diameters)
Circumference: 4.373××××10
6
km
(342 Earth diameters)
Oblateness: 9××××10
−6
Surface area: 6.09××××10
12
km²
(11,900 Earth’s)
Volume: 1.41××××10
18
km³
(1,300,000 Earth’s)
Mass (M ⋅⋅⋅⋅ ): 1.988 435(27)××××10
30
kg
(~333 Earths)
Surface temperature: 5780K
(1K = °°°°C+273)
Core temperature: ~13.6 MK
(with M being a Mega or 1M = 1,000,000 = ××××106)
* NEVER LOOK DIRECTLY AT THE SUN:
https://en.wikipedia.org/wiki/Sun
† A human eye responds to wavelengths from
about 390 to 700 nm (the 430-770THz frequency)
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The Sun – Visible light *†
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Atmospheric constituents: 78.08% Nitrogen, 20.94% Oxygen, 0.93% Argon, 0.04%
(400 parts per million) carbon dioxide, and water vapor trace which varies with climate.
Average orbital speed: 29.783 km/s
(107,218 km/h)
Satellites: 1 (the Moon)
Mean radius (〈R⊕〉): 6,372.797 km
Mean circumference: 40,041.47 km
Surface area: 510,065,600 km²
Land area: 148,939,100 km²
(29.2 %)
Water area: 361,126,400 km²
(70.8 %)
Volume: 1.083××××10
12
km³
Mass (M⊕): 5.9742××××10
24
kg
Equatorial surface gravity:
9.78 m/s² (1.000 g)
Escape velocity: 1,118.6 m/s
(40,269.6 km/h)
Equatorial rotation velocity:
465.1 m/s (1,674.4 km/h)
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The Earth
7

The Moon is in synchronous rotation – it keeps about the same face turned toward Earth
at all times. The Apollo missions (11 through 17 with the exception of 13) are shown.
Average orbital speed: 1,022 km/s
Equatorial diameter: 3,476.2 km
(0.273 Earth’s)
Surface area: 3.793××××10
7
km²
(0.074 Earth’s)
Volume: 2.1958××××10
10
km³
(0.020 Earth’s)
Mass: 7.347673××××10
22
kg
(0.0123 Earth’s)
Volume: 1.083 207 3××××10
12
km³
Mass (M ): 5.9742××××10
24
kg
Equatorial surface gravity:
1.622 m/s² (0.1654 g)
Escape velocity: 8,568 km/h
(2,000.38 m/s)
Equatorial rotation velocity:
4.627 m/s
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Apollo 17
Apollo 15
Apollo 11
(Landing July 20, 1969)
Apollo 16
Apollo 14
Apollo 12
Sea of Tranquility
Alpha
The Moon
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Introduction
Now that I have pretty much covered all the subject material I set forth to format in Microsoft® Power-
Point® (yes I did have a plan and it used rum and progressive elaboration!) I can now revisit the content
and tell you why I did this. Believe it or not some people have asked! Back when I was a teenager with
modest skills in mathematics (and yet to follow a course on Physics or Chemistry but quite inclined to
studying Astronomy) I found a book on Fluid Mechanics in the basement that belonged to my father (a
metallurgical engineer who had done work on column flotation). I can still remember bringing it back to my
room and my first perusal of it made a significant impact on me because of the appendices – Appendix A
showed page after page of long-winded equations while Appendix B showed the key equations of the
book (in Cartesian and Cylindrical coordinates). Now that I am writing this, I can say that the first thought I
had was: ‘Well, if my Dad can understand this, so can I! ’ Insofar, I graduated from Physics and Engineer-
ing Physics and ended up as a Project Manager in a big telecom company’s Network Engineering group!
So, this presentation is meant for some inquisitive 14 year old (I hope he or she finds this in time!); a lib-
rary rat like I was that is quite astute at finding things out for himself or herself. To help him along I start with
Useful Mathematics as a primer. With that knowledge, even he or she can actually read the rest of the con-
tent provided: my compilation of subjects and material that made a contribution in shaping my way of see-
ing things (c.f., References). While some like it more abstract and geometrical and others more textual than
mathematically inclined, I have had my way in picking, choosing and picturing things and these slides
are the result! In this way the Contents highlight some key discussion topics where I have tried as
best I could to keep the mathematical technicalities and prerequisites within a logical sequence…
But ever since I have wondered how things would be if I managed to format the key mathematical ideas
taken from so many courses and books and made it a necessary prerequisite for a physics novice to dis-
cover the beauty and elegance that PART I – PHYSICAL MATHEMATICS brings to one’s understanding of
the ‘world’ (or ‘universe’ if I add relativity and gravitation into the mix). The problem over the years (25-35)
was that I just kept finding the best stuff in many different books! Eventually, I just started formatting things
letting myself flow with the elaborations telling myself that eventually I would fill all the gaps in learning.
9

Now, as for content, we start with the mathematical topic that occupies a prominent place in applied
mathematics – infinite series! Students of applied sciences meet infinite series in most of the formulas
they use, and it is quite essential that they acquire an intelligent understanding of the concepts under-
lying the subject. Vectors are a key subject whether in its mathematical and technical description or its
elegant and often uncanny way of picturing things such as position, velocity, or whatever has magni-
tude and direction in space. Familiarity with the concepts discussed up to that point is essential to under-
standing the rest. I have spent a great deal of time picturing these in 3D so that the reader becomes
very comfortable with them in preparation for the higher dimensional (e.g., 4D) version – tensors!
Differential equations are also another key technical and crucial mathematical tool for physical mathe-
matics to the point that without differential equations there would be no accessible way to formulate a
relevant and physical solution that describes a repeatable and predictive behavior that can be verified
experimentally. So, formulating and solving differential equations in physics problems is essential to
learning physics and a few examples using appropriate and well known techniques (such as would be
presented to 1-st year physics or engineering students) are reviewed and applied in exacting detail.
Now, many have asked how I did this. The answer is simple… All my years in a fruitful career have
exposedme to Microsoft® Office® products and with time and practice I just came to be adept at format-
ting things in either a technical marketing function or an engineering function. The rest was my under-
standing of physics with a clear path (with many stops along the way) and a determination to complete
this for me alone. So, to you who first browsed your way to this presentation, I hope you will not be
discouraged from its inherent length but on the contrary come back to it again and appreciate it too!
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10
I do have to stress that the content is in no way original except maybe in its presentation as the
Notation to follow outlines. I think this is the originality that is being brought to this material. I have not
seen it so well displayed beforehand and frankly, this is my way of learning so I think that this can even
plant a seed for learning somehow… Overall, this was a labor of love that took some time to complete
(without any financial assistance whatsoever – just a patient wife that was understanding that I had to
do this and complete it for posterity). [See Appendix – all of which are available on www.slideshare.net]

You will notice three primary colors in the slides: Dark Red, Blue and Green. They either
are curves, vectors or variables in Figures, variables or terminology within the text, bold
italic terms (e.g., how something is called – either mathematical or physical) or even to
help in focusing on ‘results’ such as equations boxed with one of these colors:
• Dark Red is good for skimming through and paying attention to the ‘key stuff’;
• Blue is the ‘from first principles’ part (and pretty much the reason for doing this so
anything blue you need to read through to understand the whole thing!) and;
• Green is the ‘physical part’ where the subject matter (i.e., the physics) is highlighted.
Otherwise, these colors just makes for a nicer presentation! Italics are also used to spell
out key terms(e.g.,of variables and/or constants),Bold italics are importantdefinitions.
Since mathematics is a language you will be able to peruse a slide by homing onto the
mathematics which is set in black. Scalars are typically black italic (e.g., the Cartesian
coordinates x, y or z) while vectors are typically black bold (e.g., the position vector r).
This applies to constants, variables, &c. This presentation is key in developing a sense
of ‘reading’ the physics through ‘reading’ equations… It was the reason why I did this –
to be able to peruse the slides and take on the whole whopping content in order to
assimilate a result, equation, complex term or mathematical or physical conclusion.
Notation
“Now you may ask, ‘What is mathematics doing in a physics lecture?’ We have several possible excuses:
first, of course, mathematics is an important tool, but that would only excuse us for giving the formula in
two minutes. On the other hand, in theoretical physics we discover that all our laws can be written in
mathematical form; and that this has a certain simplicity and beauty about it. So, ultimately, in order to
understand nature it may be necessary to have a deeper understanding of mathematical relationships.”
Richard Feynman, Feynman Lectures on Physics, Vol. I, Chap. 22, § 22-1 (1964).
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Contents of 5-Chapter PART I
Useful Mathematics and Infinite Series
Determinants, Minors and Cofactors
Scalars, Vectors, Rules and Products
Direction Cosines and Unit Vectors
Non-uniform Acceleration
Kinematics of a Basketball Shot
Newton’s Laws
Moment of a Vector
Gravitational Attraction
Finite Rotations
Trajectory of a Projectile with Air
Resistance
The Simple Pendulum
The Linear Harmonic Oscillator
The Damped Harmonic Oscillator
General Path Rules
Vector Calculus
Fluid Mechanics
Generalized Coordinates
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The Line Integral
Vector Theorems
Calculus of Variations
Gravitational Potential
Kinematics of Particles
Motion Under a Central Force
Particle Dynamics and Orbits
Space Vehicle Dynamics
Complex Functions
Derivative of a Complex Function
Contour Integrals
Cauchy’s Integral Formula
Calculus of Residues
Fourier Series and Fourier Transforms
Transforms of Derivatives
Matrix Operations
Rotation Transformations
Space Vehicle Motion
Appendix
“The enormous usefulness of mathematics in the natural sciences is something bordering on the
mysterious and that there is no rational explanation for it.” Eugene Wigner, ‘The unreasonable
effectiveness of mathematics in the natural sciences.’ Richard Courant lecture in mathematical sciences
delivered at New York University, May 11, 1959 (1960).
12

Contents
Useful Mathematics and Infinite Series
Newton’s Laws
Moment of a Vector
Gravitational Attraction
Finite Rotations
Trajectory of a Projectile with Air
Resistance
The Simple Pendulum
The Linear Harmonic Oscillator
The Damped Harmonic Oscillator
General Path Rules
Vector Calculus
Fluid Mechanics
Generalized Coordinates
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The Line Integral
Vector Theorems
Calculus of Variations
Gravitational Potential
Kinematics of Particles
Motion Under a Central Force
Particle Dynamics and Orbits
Space Vehicle Dynamics
Complex Functions
Derivative of a Complex Function
Contour Integrals
Cauchy’s Integral Formula
Calculus of Residues
Fourier Series and Fourier Transforms
Transforms of Derivatives
Matrix Operations
Rotation Transformations
Space Vehicle Motion
Appendix
13

The natural numbers are the positive integers such as one, two, three:
Useful Mathematics
Irrational numbers (i.e., the quotient or fraction of one integer with another) have
decimal digits:
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The complex numbers are real numbers with one new number i whose square is −1:
12
−=i
...
7
22
2
3
8
5
2
1
,,,,
is rational but:
LL 82.71828182e141592654.3π == :numberEuler:Pi &
is irrational (note the ellipsis L). The real numbers include the rational numbers and the
irrational numbers; they correspond to all the points on an infinite line called the real line.
...1428.3
7
22
5.1
2
3
625.0
8
5
5.0
2
1
,,,, ====
and zero, 0. Negative numbers are labeled by −1,−2,−3,&c. (et cetera - ‘and so forth’).
The Prime numbers are 2,3,5,7,11,13,17,… and are natural numbers greater than 1 that
have no positive divisors other than 1 and itself. Rational numbers are ratios of integers:
14

Addition is a mathematical operation that represents the operation of adding objects
to a collection (or to the total amount of objects together in a collection). It is signified by
the plus sign (+) and is expressed with an ‘equals’ sign (=) as in 2+3=5 (N.B., this is the
same commutative result as 3+2=5 – N.B., nota bene-‘note well’). When it is written
mathematically it looks like:
The plus sign can also serve as a unary operator that leaves its operand unchanged
(i.e., +x means the same as x – i.e., id est - ‘that is’). This notation may be used when it
is desired to emphasise the ‘positiveness’ of a number, especially when contrasting with
the negative (+5 vs −5 – vs, versus-‘against’).
and so on:
11245633422111 =++=+=+=+ and,,
123333 =+++
A negative number is a real number that is less than zero (i.e., <0). Such numbers
are often used to represent the amount of a loss or absence. Negative numbers are
usually written with a minus sign (−) in front (e.g., −3 represents a negative quantity
with a magnitude of three, and is pronounced ‘minus three’ or ‘negative three’ – e.g.,
exempli gratia-‘for example’ or ‘for instance’).
Subtraction is a mathematical operation that represents the operation of removing
objects from a collection. It is signified by the minus sign (−). 2016
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The division operation 20÷4=5 (inverse of 4×5=20).
Division is another arithmetic operation denoted by (÷).
Specifically, if b times c equals a. Written mathematically:
where b is not zero, then a divided by b equals c, which
means, when written arithmetic mathematically:
cba ×=
Multiplication (often denoted by the cross symbol (×) or by the absence of symbol).
The multiplication of two whole numbers is equivalent to the addition of one of them with
itself as many times as the value of the other one (e.g., 3 multiplied by 4 – often said as
‘3 times 4’) can be calculated by adding 3 copies of the quantity/number 4 together.
Written mathematically, we have:
Here 3 and 4 are the factors and 12 is the product – the result of multiplication – not
addition (c.f., 3+3+3+3=12 – c.f., confer - ‘compare’).
12333343933422111 =+++=×=×=×=× and,,
cba =÷
5420 =÷
For instance (see Figure):
since 4×5=20.
In the expression a÷b=c, a is called the dividend or
numerator, b the divisor or denominator and the result c is
called the quotient.
16

Exercise: Memorize the Base 10 Multiplication Table shown. What are the common
factors and what do they represent - physically? What do the diagonal elements mean?
17

Algebra is a generalized arithmetic in which symbols are used in place of numbers. Al-
gebra thus provides a language in which general relationships can be expressed among
quantities (e.g., a, b, x, y, &c.) whose numeral values need not be known in advance.
The arithmetical operation of addition, subtraction, multiplication, and division have the
same meanings in algebra. The symbols of algebra are normally letters of the alphabet.
cba =+
If we subtract b from a to give the difference d, we would write:
dba =−
Multiplying a and b together to give e may be written in any of these ways:
ebaebaebaeba ===⋅=× ))((
Whenever two algebraic quantities are written together with nothing between them (i.e.,
like ab=e above), it is understood that they are to be multiplied.
Dividing a by b to give the quotient (a fraction) f is usually written (‘≡’ means ‘equivalent’):
f
b
a
ba =≡÷
If we have two quantities a and b and add them to give a sum c, we could write:
but it may sometimes (especially in exponentials) be more convenient to write:
fba =
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In order of priority, parentheses (i.e., (…)) are used first and brackets (i.e., […]) are
used second to show the order in which various operations are to be performed. Thus:
fe
d
cba
=−
+ )(
means that, in order to find f, we are first to add a and b together, then multiply their sum
by c and divide by d, and finally subtract e. If () and [] are used, choose curly brackets {}.
27123 =+x
An equation is simply a statement that a certain quantity is equal to another. Thus 7+2
=9, which contains only numbers, is an arithmetic equation, while:
which contains a symbol as well (i.e., the unknown variable x), is an algebraic equation.
The symbols in an algebraic equation usually cannot have any arbitrary (i.e., unspeci-
fied) values if the equality is to hold. Finding the possible values of these symbols is
called solving the equation. The solution of the latter equation above is:
5=x
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since only when x is 5 is it true that 3x+12=27 (i.e., 3⋅5+12=15+12=27 QED. N.B.,
Latin QED – quod erat demonstrandum – meaning ‘which had to be demonstrated’).
19

In order to solve an equation, a basic principle must be kept in mind: Any operation
performed on one side of an equal sign must also be performed on the other side.
153
122712123
=
−=−+
x
x
To check a solution, we substitute it back in the original equation and see whether the
equality is still true. Thus we can check that x=5 by reducing the original algebraic
equation to an arithmetical one (where the symbol ‘⇒’ means ‘implies’):3x+12=27⇒
(3)(5)+12=27⇒15+12=27⇒27=27.
An equation therefore remains valid when the same quantity, numerical or otherwise,
is added to or subtracted from both sides, or when the same quantity is used to multiply
or divided both sides of the equal sign (=). Other operations, for instance squaring ( ²) or
taking the square root (√ ), also do not alter the equality if the same thing is done to both
sides. As a simple example, to solve 3x+12=27 above, we subtract 12 from both sides:
To complete the solution we divide both sides by 3:
5
3
15
3
3
=
=
x
x
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Two helpful rules follow directly from the principle stated above. The first rule is: Any
term on one side of an equation may be transposed to the other side by changing
its sign. To verify this rule, we subtract b from each side of the equation:
cba =+
by b. The result is:
to obtain:
cba =
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bcabcbba −=⇒−=−+
We see that b has disappeared from the left-hand side and −b is now on the right-hand
side (c.f., the equation 3x+12=27 above, using the same treatment, gave us 3x=27−−−−12).
The second rule is: A quantity which multiplies one side of an equation may be
transposed in order to divide the other side, and vice versa. To verify this rule, we
divide both sides of the equation:
b
c
a
b
c
b
ba
=⇒=
We see that b, a multiplier on the left-hand side, is now a divisor on the right-hand
side (c.f., once again, the equation 3x+12=27 above gave us x=15/3).
21

Exercise: Solve the following equation for x:
7)3(4 =−x
Solution: The above rules are easy to apply here (I did expand on the fractions):
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4
19
4
127
4
12
4
7
4
4
3
4
7
3
4
7
4
7
3
4
7
4
)3(4
7)3(4
=
+
=+=⋅+=+=∴
=−⇒=
−⋅
⇒=−
x
x
x
x
thatso
When each side of an equation consists of a fraction, all we need to do to remove the
fractions we use cross multiplication:
bcad
d
c
b
a
=⇒=
What was originally the denominator (i.e., the lower part) of each fraction now
multiplies the numerator (i.e., the upper part) of the other side of the equation.
22
or if you prefer doing things lazily with a calculator (when the point is ‘learning’ algebra!):
75.4375.1347 =⊕=⊕÷=x

Exercise: Solve the following equation for y:
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2
3
2
5
−
=
+ yy
Solution: First we cross multiply to get rid of the fractions, and then solve in the usual
way:
8
2
16
2
2
162
10635
63105
)2(3)2(5
=
=
=
+=−
+=−
+=−
y
y
y
yy
yy
yy
23
y =8 is the final answer. If you solved this on your own, give yourself a million bucks!

Two rules for multiplying and dividing positive and negative quantities are
straightforward. The first rule is: Perform the indicated operation on the absolute
value of the quantity (e.g., the absolute value of −7 is 7). The second rule is: If the
quantities are both positive or both negative, the result is positive:
Here are a few examples:
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b
a
b
a
b
a
abbaba +=
−
−
==−−= and))(())((
If one quantity is positive and the other negative the result is negative:
2
15
30
70)7)(10(3
2
6
20)5)(4(4
5
20
18)3)(6(
−=
−
−=−−=
−
−=−=
−
−
=−−
and,
,,,
b
a
b
a
b
a
abbaba −=
−
=
−
−=−=− and))(())((
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Exercise: Find the value of:
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yx
yx
z
−
=
Solution: We begin by evaluating xy and x−y, which are:
164)12(
484)12(
−=−−=−
−=⋅−=
yx
yx
when x=−12 and y=4.
Hence:
3
16
48
=
−
−
=
−
=
yx
yx
z
25
Ok. The answer is z =3. Now I have to point out something that you might have realized
by now. If not, well, here’s a thought! Mathematics will give you an answer that is not
based on anyone’s own interpretation! Of course, to find the answer you have to solve
an equation! This reminds me that when I was a college student I always had difficulty
learning how to read French literature books… Why? Because it was subject to the
teacher’s opinion of what you needed to learn as opposed to providing a firm answer!

It is often necessary to multiply a quantity by itself a number of times. This process is
indicated by a superscript number called the exponent, according to the following
scheme:
When we multiply a quantity raised to some particular power (say An) by the same
quantity raised to another power (say Am), the result is that quantity raised to a power
equal to the sum of the original exponents. That is:
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4321
AAAAAAAAAAAAAA =⋅⋅⋅=⋅⋅=⋅= and,,
We read A2 as ‘A squared’ because it is the area of a square of length A on a side;
similarly A3 as ‘A cubed’ because it is the volume of a cube of whose sides is A long.
More generally we speak of An as ‘A to the n-th power’ – A4 is ‘A to the 4-th power’, &c.
)( mnmn
AAA +
=
For example:
75252
AAAA == +
which we can verify directly by writing out the terms:
7
)()( AAAAAAAAAAAAAAA =⋅⋅⋅⋅⋅⋅=⋅⋅⋅⋅⋅⋅
26

From the above result we see that when a quantity raised to a particular power (say
An) is to be multiplied by itself a total of m times, we have:
since:
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mnmn
AA =)(
For example:
63232
)( AAA == ⋅
6)222(22232
)( AAAAAA ==⋅⋅= ++
27

Reciprocal quantities are expressed in a similar way with the addition of a minus sign
in the exponent, as follows:
It is important to remember that any quantity raised to the zeroth power, e.g., A0, is
equal to 1. Hence:
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4
4
3
3
2
2
1 1111 −−−−
==== A
A
A
A
A
A
A
A
and,,
Exactly the same rules as before are used in combining quantities raised to negative
powers with one another and with some quantities raised to a positive power. Thus
6)71(7123443
2)2(1213)25(25
)(
)(
−−−−⋅−−
−⋅−−−−−
===
====
AAAAAAA
AAAAAAA
and
,,
10)22(22
=== −−
AAAA
This is more easily seen if we write A−2 as 1/A2:
1
1
2
2
2
222
==⋅=−
A
A
A
AAA
28

The square root has that name because the length of each side of a square of area A
is given by √A. Some examples of square roots are as follows:
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22
93339
25.425.6565.625.42
100101010100
164441611111
AAAAA =⋅=
=⋅=
=⋅=
=⋅==⋅=
.
because
andbecause
,because
,because,because
The square root of a number less than 1 is larger than the number itself:
49.07.07.07.049.0
01.01.01.01.001.0
=⋅=
=⋅=
because
andbecause
29

The square root of a quantity may be either positive or negative because (+A)(+A)=
(−A)(−A)=A2. Using the exponents we see that, because:
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we can express square roots by the exponent ½:
AAAA === ⋅ 1)21(2221
)(
21
AA =
Other roots may be expressed similarly. The cube root of a quantity A, written ,
when multiplied by itself twice equals A. That is:
3
A
AAAAA ==⋅⋅ 33333
)(
which may be more conveniently written:
1)31(3331
)( AAA == ⋅
where .313
AA =
30

In general the n-th roots of a quantity, , may be written A1/n, which is a more
convenient form for most purposes. Some examples may be helpful:
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161)41)(41(4141
1)3)(31(313
47)41)(7(741
41)21)(21(2121
2)21)(4(4214
2)4)(21(2144
)(
)(
)(
)(
)()(
)(
AAA
AAA
AAA
AAAA
AAAA
AAAA
==
==
==
===
===
===
−−−
−−−
and
,
,
,
,
n
A
31

An equation that involves power or roots or both is subject to the same basic principles
that govern the manipulation of simpler equations: Whatever is done to one side must be
done to the other. Hence the following rules: An equation remains valid when both
sides are raised to the same power, that is, when each side is multiplied by itself
the same number of times as the other side and An equation remains valid when
the same root is taken on both sides an equal sign.
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Exercise: Newton’s (1642-1727) law of gravitation states that the force F between two
massive bodies whose masses are m1 and m2 a distance r apart is given by the formula:
where G is a universal constant. Solve this formula for r and explain its proportionality.
Solution: We begin by transposing r2 with F to give:
and finally we take the square root √ of both sides:
The distance r is proportional to the square root of the product of the masses, √(m1⋅m2),
and inversely proportional to the square root of the force between them, 1/√F. The
constant of proportionality would be the square root of the universal constant,√G≡G½.
2
21
r
mm
GF =
F
mm
Gr 212
=
F
mmr
F
mm
F
mm
G
F
mm
G
F
mm
Grr
1
21
212121212
⋅⋅∝⇒
⋅
∝⋅=⋅=⋅==
32

Quadratic equations, which have the general form:
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02
=++ cxbxa
are often encountered in algebraic calculations. The term quadratic describes something
that pertains to squares, to the operation of squaring, &c. Many physical phenomena
involve quadratic behavior such as dropping a ball from the roof of a building and calcu-
lating the distance s travelled as a function of time gives: s=½gt2. The time t is quadratic!
a
cabb
x
a
cabb
x
2
4
2
4 22
−−
=
−−
= −+
−−−−++++
and
By looking at these formulas, we see that the nature of the solution depends upon the
value of the quantity b2 −4ac. When b2 =4ac, √(b2 −4ac)=0 and the two solutions are
equal to just x=−b/2a. When b2 >4ac, √(b2 −4ac) is a real number and the solutions x+
and x− are different. When b2 <4ac, √(b2 −4ac) is the square root of a negative number,
and the solutions are different. The square root of a negative number is called an imagi-
nary number because squaring a real number, whether positive or negative, always
gives a positive number: (+2)2 =(−2)2 =+4, hence √(−4) cannot be either +2 or −2.
In such an equation, a, b, and c are constants. A quadratic equation is satisfied by the
values of x given by the formulas:
33

Exercise: Solve the quadratic equation:
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0932 2
=−− xx
Solution: Here a=2, b=−3, and c=−9, so that:
The two solutions are:
81729)9)(2(4)3(4 22
=+=−−−=− acb
2
3
4
6
4
93
)2(2
81)3(
2
4
3
4
12
4
93
)2(2
81)3(
2
4
2
2
−=
−
=
−
=
−−−
=
−−−
=
==
+
=
+−−
=
−+−
=
−
+
a
cabb
x
a
cabb
x and
Exercise: Solve the quadratic equation:
0442
=+− xx
Solution: Here a=1, b=−4, and c=4, so that b2 −4ac = 0 hence x=−b/2a =−(−4)/2 =2.
34

A right triangle is a triangle, two sides of which are perpendicular. Such triangles are
frequently encountered in physics, and it is necessary to know how their sides and
angles are related. The hypotenuse of a right triangle is the side opposite the right angle,
as shown in the Figure. It is always the longest side. The three basic trigonometric
functions, the sine, cosine, and tangent of an angle, are defined as follows:
ehypothenus
sideadjacent
and
ehypothenus
sideopposite
====
c
b
c
a
θθ cossin
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A right triangle – 2 sides of which are perpendicular.
θ
b = Adjacent side
a=Oppositeside
ϕ
o
90
tancossin
222
=+=+
===
ϕθ
θθθ
cba
b
a
c
b
c
a
90°
sideadjacent
sideopposite
====
b
a
cb
ca
θ
θ
θ
cos
sin
tan
From these definitions we see that:
Numerical values of sinθ , cosθ, and tanθ for angles from 0°
to 90°. These figures may be used for angles from 90° to
360° with the help of the Table:
f (θ ) θ 90° + θ 180° + θ 270° + θ
sinθ sinθ cosθ −sinθ −cosθ
cosθ cosθ −sinθ −cosθ sinθ
tanθ tanθ −1/tanθ tanθ −1/tanθ
For example, if we require the value of sin120°, we first note
that 120° =90° +30°. Then, since sin(90° +θ)=cosθ we have
sin120° =sin(90° +30°) =cos30° =0.866.
35
2
1
1

The inverse of a trigonometric function is the angle whose function is given. For
instance, the inverse of sinθ is the angle θ. If sinθ =x, then the angle θ may be
designated as θ =arcsinx or as θ =sin−1 x. It is important to keep in mind that an
expression such as sin−1 x does not mean 1/sin−1 x. The inverse trigonometric functions
are:
xxx
x
iswhoseangle sine=≡=
=
−1
sinarcsin
sin
θ
θ
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zzz
z
iswhoseangle tangent=≡=
=
−1
tanarctan
tan
θ
θ
yyy
y
iswhoseangle cosine=≡=
=
−1
cosarccos
cos
θ
θ
and:
and, finally, the ratio of the above two:
Inverse functions are cotanθ =1/tanθ, secθ =1/sinθ (i.e., secant) and cosecθ =1/cosθ.
36

To solve a given triangle means to find the values of any unknown sides or angles in
terms of the values of the known sides and angles. A triangle has three sides and three
angles, and we must know the values of at least three of these six quantities, including
one of the sides, to solve the triangle for the others. In a right triangle, one of the angles
is always 90°, and so all we need here are the lengths of any two of its sides or the
lengths of one side and the value of one of the other angles to find the remaining sides
and angles.
θ
θθθ
sin
sintantan
a
c
c
a
ba
b
a
=⇒==⇒= and
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Suppose we know the length of the side b and the angle θ in the right triangle of the
previous Figure. From the definitions of sine and tangent we see that:
This gives us the two unknown sides a and c. To find the unknown angle ϕ, we can use
any of these formulas:






=





=





= −−−
a
b
c
a
c
b 111
tancossin ϕϕϕ or,
Alternatively, we can use the fact that the sum of the angles in any triangle is 180°.
Because one of the angles in a right triangle is 90°, the sum of the other two must be
90°! Hence, since θ +ϕ =90° we get ϕ =90°−θ here.
37

Another useful relationship in the right triangle is the Pythagorean theorem, which
states that: The sum of the squares of the sides of such a triangle adjacent to the
right angle is equal to the square of its hypotenuse. For the triangle in the previous
Figure:
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Thus we can always express the length of any of the sides of the right triangle in terms
of the other sides:
Exercise: In the triangle in the previous Figure, a=7cm and b=10cm. Find c, θ, and ϕ.
222
cba =+
222222
bacacbbca +=−=−= or,
Solution: From the Pythagorean theorem:
cmcmcmcm 2.1214910049107 2222
==+=+=+= bac
Since tanθ =a/b:
o
35)7.0(tan
10
7
tantan 111
==





=





= −−−
cm
cm
b
a
θ
The value of the other angle ϕ is given by ϕ =90°−θ =90°−35° =55°.
38

f (x)
x
O
Secant
Tangent
yo
xo
Po
P
∆x= dx
dy
∆x
θ
Now we complicate things… We define the derivative of a function y= f (x) at the point
x which is defined as the slope of the tangent to a function at the point x (see Figure).
The difference quotient is the slope of the secant through the points P(x,y) and
Po(xo,yo):
o
o )()()(
xx
xfxf
x
xf
x
y
−
−
=
∆
∆
=
∆
∆
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Derivative of a function f (x).
The differential quotient f ′(x) is the limit (i.e., identified by the mathematical
terminology lim) value of the difference quotient for P→Po and ∆x→0:
x
xfxxf
x
y
xf
xd
yd
xx ∆
−∆+
=
∆
∆
=′≡
→∆→∆
)()(
limlim)(
00
The derivative of a function at the point Po corresponds to the gradient of its graph at the
point Po, f ′(xo)= tanθ (i.e., the magnitude of the rate of increase of the graph is the slope).
39

Applying ‘derivation’ comes with a set of rules. An effort of memorization is needed…
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Factor rule: A constant factor c remains unchanged when the derivative is taken:
0=
xd
cd
xd
xfd
c
xd
xfcd )()]([
⋅=
⋅
Constant rule: The derivative of a constant c is equal to zero:
Power rule: When carrying out the derivative of a power function, the exponent is
lowered by unity, and the old exponent enters as a factor:
1−
⋅= n
n
xn
xd
xd
Sum rule: The derivative of a sum (or difference) is equal to the sum (or difference) of
the derivatives:
xd
xgd
xd
xfd
xd
xgxfd )()()]()([
±=
±
40

Product rule (N.B., These can be written shorthand by letting u ≡ f (x), v ≡g(x), &c.):
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Quotient rule:
)()(
)(
)(
)(
)(
)(
)()(
)]()()([
)(
)(
)(
)(
)(
)]()([
xhxg
xd
xfd
xh
xd
xgd
xf
xd
xhd
xgxf
xd
xhxgxfd
udvvduvud
xd
xfd
xg
xd
xgd
xf
xd
xgxfd
⋅⋅+⋅⋅+⋅⋅=
⋅⋅
+=⇔⋅+⋅=
⋅
Chain rule:
xd
xgd
xgxd
xgd
xgxd
d
xg
xd
xgd
xf
xd
xfd
xg
xd
xgxfd
xg
xf
xd
d
)(
)(
1)]([
)(
1
)(
)(
)(
)(
)(
)]()([
)(
)(
2
1
2
1
−
==





⋅−⋅
=
⋅
=





−
−
gd
fd
xd
gd
xd
fd
xgd
xgfd
xd
xgd
xd
xgfd
⋅=⋅= or
)(
)]([)()]([
where d f/dg is called the exterior derivative and d g/dx is the interior derivative.
41

Logarithmic derivative: The derivative of the logarithm lny of the function y for y>0:
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xd
yd
yy
xdyd
xd
yd 1ln
==
42
Exercise: Find the derivative of y= f (x)=expx/(1+x)≡ex/(1+x) (not y =exp[x/(1+x)]≡ex/(1+x)).
Solution: Let g(x)=expx and h(x)=1+x (i.e., where the function f (x) is the quotient of
functions g(x) and h(x): f (x)=g(x)/h(x).
1
)(
)(e
)(
)( ==′==′
xd
xhd
xh
xd
xgd
xg x
,
We use the quotient rule, f ′(x)=df (x)/dx=[h(x)g′(x) − g(x)h′(x)]/h2(x), to find the derivative
of the function f (x):
since g′(x)=expx and h′(x)=1 and h2(x)=(1+x)2. Simplifying the algebra will then give us:
22
)1(
)1(ee)1(
)1(
)()e()()1(
)(
x
x
x
xhxgx
xf
xxx
+
−+
=
+
′−′+
=′
2
)1(
e
)(
x
x
xf
x
+
=′

Simply said, ‘integration’ is the reverse (i.e., the inverse function…) of ‘differentiation’.
The antiderivative function is the integral function F(x) of a function f (x). The
derivative F′(x) of the integral function is equal to f (x). The function F(x) is defined over
the same interval as f (x).
Integration of a function f (x), determination of the integral function F(x) of f (x), the
derivative of which is again the original function f (x).
To any integrable function, there exists infinitely many integral functions F(x)+C that
differ only by an additive integration constant C. All integral functions have the same
slope at a fixed value x.
Indefinite integral I: The integration constant C is not fixed:
f (x)
x
O
a b
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Definite integral A of a function f (x).
Definite integral A: The upper and lower boundary of
integration are fixed (see Figure). The definite integral is a
number given by the operation:
The definite integral A corresponds to the area between the
function f (x) and the x-axis between x=a and x =b. If f (x)
becomes negative in the integration interval, then the
definite integral is equal to the difference of the areas above
and below the x-axis.
)()()()( aFbFxFxdxfA
b
a
b
a
−=== ∫
CxFxdxfI +== ∫ )()(
43
∫=
b
a
xdxfA )(

Applying ‘integration’ also comes with a set of rules.
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( )1
1
1
−≠
+
=
+
∫ n
n
x
xdx
n
n
Constant rule: A constant factor C may be pulled out of the integral
Sum rule: The integral over the sum (or difference) of terms is equal to the sum (or
difference) of the integrals over the terms:
Power rule:
Inversion rule: Inversion of the sign of the definite integral under inversion of the
integration boundaries:
∫∫∫ ±=± xdxgxdxfxdxgxf )()()]()([
∫∫ −=
a
b
b
a
xdxfxdxf )()(
∫∫ ⋅=⋅ xdxfCxdxfC )()(
44
By n≠−1 we mean that n cannot take a value that equals −1 since it would then become a
singular (as in ‘singularity’) integral which would lead to an term being divided by zero.

Equality of upper and lower boundary: The integral vanishes if the interval is a=a:
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Interval rule: Definite integrals may be decomposed into integrals over parts of the
interval ab that can be split in two intervals ac and cb:
Partial integration: Also called integration by parts which is the inversion of the
product rule of differentiation:
Substitution rule:
0)( =∫
a
a
xdxf
∫∫∫ +=
b
c
c
a
b
a
xdxfxdxfxdxf )()()(
∫∫ ⋅−⋅=⋅ xdxg
xd
xfd
xgxfxd
xd
xgd
xf )(
)(
)()(
)(
)(
( ))()(
)(
)]([ xgzzdzfxd
xd
xgd
xgf ==⋅ ∫∫
Logarithmic integration:
Cxfxd
xd
xfd
xf
+=⋅∫ )(ln
)(
)(
1
45

F (x)f (x) f ′(x)F (x)f (x)
Now we show the derivatives and integrals of elementary functions as given below
in the Table where the original function f (x), its derivative f ′(x)=d f (x)/dx and the integral
function ∫ f (x)dx=F (x)+C are displayed respectively.
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c 0 xc
2
2
1
x1x
1
1
1 +
+
a
a
x1−a
xaa
x
x1 2
1 x− xln
xsin xcos xcos−
xcos xsin− xsin
xtan x2
cos1 xcosln−
xsinlnx2
sin1−xcot
x
e x
e x
e
aax
lnaax
lnx
a
xxx −lnx1xln
xalog axln1 axxx ln)ln( −
xarcsin 2
11 x− 2
1arcsin xxx −+
xarccos 2
11 x−− 2
1arccos xxx −−
xarctan 2
11 x+ )1(lnarctan 2
2
1
xxx +−
)1(lncotarc 2
2
1
xxx ++2
11 x+−xcotarc
46
f ′(x)

Exercise: Integrate the indefinite integral ∫5expx dx.
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47
Solution: With ∫C f (x)dx=C∫ f (x)dx and ∫expx dx=expx+C (not ∫axdx=ax/lnx+C) we get:
Cxdxd xxx
+== ∫∫ e5e5e5
Problem: Evaluate the definite integral ∫0 to (1/7)ln 214exp(7x)dx.
Solution: Using the change of variable:
xxfu 7)( ==
so that:
xdud 7=
or:
xdud =
7
1
In addition, the range of x-values for the infinitesimal increment dx is x:0→(1/7)ln2 so
that the range of u-values is x:7⋅0→7⋅(1/7)ln2 or x:0→ln2. Substituting this into the
original problem and replacing all forms of x we get:
∫ ∫∫∫ =





==
=
=
2ln
0
2ln
0
7
7
2ln)71(
0
7
2ln)71(
0
7
e2
7
e14e14e14 ud
ud
xdxd uu
ux
duxd
xx
The integral then reduces to 2expu|0 to ln2 =2exp(ln2)–2exp(0)=2⋅2−2⋅1=4−2=2 so that
the answer to this definite integral is 2 where we have used exp(lnx)=x and exp(0)=1.

We will now summarize the essentials of infinite series without detailed proofs.
48








=⇔= ∑=
∞→∞→
k
n
n
k
k
k
wsss
1
limlim
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Infinite sequences and series are important in mathematicalphysics. Let w1,w2,…,wn,…
be a sequence of numbers (real or complex). If we define an infinite series as being:
(N.B., limk – from1to ∞ – is called the limit as k approached infinity) then the series
Σn=1tok wn (N.B., Σn – from 1 to k – is called the summation sign over n) is said to be
convergent. The number s is called the sum (or value) of the series and is given by:
)11( 2
−=⇒−=+= iibias
where:
∑∑
∞
=
∞
=
==
11 n
n
n
n vbua and
for wn =un +ivn, a complex number with both un and vn real (N.B., ibid for s=a+ib above).
Now, if the series:
...321
1
+++=∑
∞
=
wwww
n
n
is convergent,thenΣn=1to∞ wn is said to beabsolute convergent.If Σn=1to∞wn is conver-
gent but Σn=1to∞ |wn| diverges, then Σn=1to∞wn is said to be conditionally convergent.

Now for simple convergence tests.
49





=
>
<
=






 +
∞→
antindeterministestthethen
isseriesthethen
isseriesthethen
1
1
1
lim 1
divergent
convergent
n
n
n a
a
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And now for some important series in mathematical physics.
The sequence:
12
...1 −
++++= n
n zzzs
is called a geometric series (or sequence). On multiplying this sequence by z and
subtracting the resulting sequence zsn from sn, we obtain:
( )1
1
1
1)1( ≠
−
−
=−=− z
z
z
szsz
n
n
n
n or
The corresponding infinite geometric series converges, for |z|<1, to:
( )1
1
1
1
<
−
== ∑
∞
=∞→
z
z
zs
n
n
n
n
The region |z|<1 is called the circle of convergence of the infinite geometric series.
First, the comparison test: If Σnan is a convergent series and un≤an for all n, then Σnun
is a convergent series. If Σnbn is a divergent series and vn ≤bn for all n, then Σnbn is also a
divergent series. Second, the ratio test: Consider the series Σnan . If:

Now for a series of functions: Consider:
50
)(...)()()( 21 zuzuzuzs nn +++=
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and:
( )Nnzszfzuzszf n
n
nn
n
≥<−== ∑
∞
=
∞→
allforif ε)()()()(lim)(
1
where N is independent of z in the region a≤|z|≤b and ε is an arbitrarily small quantity
greater than zero, then the series sn(z) is said to be uniformly convergent in the closed
region a≤|z|≤b.
∑∫∫ ∑∫
∞
=
∞
=
==
11
)()()(
n
b
a
n
b
a
n
n
b
a
zdzuzdzuzdzf
The derivative of f (z):
∑∑
∞
=
∞
=
=
11
)()(
)(
n
n
n
n zu
zd
d
zu
zd
d
zd
zfd
equals
only if un(z), and dun(z)/dz are continuous in the region and Σndun(z)/dz is uniformly
convergent in the region.
If the individual terms, un(z), of a uniformly convergent series are continuous, the
series may be integrated term by term, and the resultant series will always be
convergent. Thus:

A determinant is a square array of quantities called elements an,bn,cn,…,rn (n ≠0) which
may be combined according to the rules given below. In symbolic matrix form, we write:
51
nnnn rcba
rcba
rcba
L
MOMMM
L
L
2222
1111
=∆
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MRT
Here n is called the order of the determinant. The value of the determinant (i.e., the
outcome of calculating ∆) in terms of the elements ai,bj,ck,…,rl is defined as:
∑=∆
n
ji
kjikji rcba
l
ll L
,...,,
...ε
where the Levi-Civita symbol, εijk…l, has the following property:





−
+
=
repeatedisindexanif
ofnpermutatiooddanfor
ofnpermutatioevenanfor
0
),,,,(1
),,,,(1
... lK
lK
l kji
kji
kjiε





=−
=+
=
)(0
)132,213,321(1
)312,231,123(1
Otherwize
if
if
kji
kji
kjiε
For example:

On applying the equation ∆=Σijk…l εijk…l ai bj ck…rl above to the third-order determinant:
52
333
222
111
cba
cba
cba
=∆
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MRT
we obtain, by first carrying the sum over i:
Then, by carrying the sum over j, this reduces to:
∑∑∑
∑∑∑∑∑∑
++=
++===∆
== = =
kj
kjkj
kj
kjkj
kj
kjkj
kj
kjkjkjkjkjkj
kj
kjkj
i j k
kjikji
cbacbacba
cbacbacbacbacba
332211
332211
3
1
3
1
3
1
3
1
)(
εεε
εεεεε
i
ii
∑
∑
∑
+++
+++
++=∆
k
kkkkkk
k
kkkkkk
k
kkkkkk
cbacbacba
cbacbacba
cbacbacba
)(
)(
)(
333323321331
322322221221
311321121111
εεε
εεε
εεε

333
222
111
)( 312231123
213132321
123213132
312231321
cba
cba
cba
cbacbacba
cbacbacba
cbacbacba
cbacbacba
++−
+++=∆
−++
−−=∆
or
−−−−
++++
++++
++++
−−−−−−−−
which finally reduces to (i.e.,a formulafora third-order determinant):
Since ε11k=ε22k=ε33k=0 (i.e., an index is repeated) the equation above becomes:
53
2016
MRT
and finally expanding the rest, by finally carrying the sum over k, we get:
000 233213313223122131132112 ++++++++=∆ ∑∑∑∑∑∑ k
kk
k
kk
k
kk
k
kk
k
kk
k
kk cbacbacbacbacbacba εεεεεε
323332223232123132
313331213231113131
332323232223132123
312321212221112121
331133231132131131
321312221212121112
cbacbacba
cbacbacba
cbacbacba
cbacbacba
cbacbacba
cbacbacba
εεε
εεε
εεε
εεε
εεε
εεε
+++
+++
+++
+++
+++
++=∆
With the aid of the property for εijk, the above equation means that:
323223123313213113332232132
312212112331231131321221121
)0()0()1()0()1()0()0()0()1(
)1()0()0()0()1()0()1()0()0(
cbacbacbacbacbacbacbacbacba
cbacbacbacbacbacbacbacbacba
++−+++++++++
−++++−+++++=∆
a3b2c1
a1b2c3 a3b1c2a2b3c1
a1b3c2 a2b1c3
+a2b3c1
+a1b2c3
+a3b1c2 −a3b2c1
−a1b3c1
−a1b3c1

The result of the previous equation can be written in the form:
54
22
11
3
33
11
2
33
22
1
122131331223321 )()()(
cb
cb
a
cb
cb
a
cb
cb
a
cbcbacbcbacbcba
+−=
−+−−−=∆
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MRT
where:
2332
33
22
cbcb
cb
cb
−=
is called the minor of this determinant.
The procedure of expressing ∆ in the form above may be generalized to obtain the
value of the n-th-order determinant. In the last few sides we saw that the expansion of a
third-order determinant is expressed as a linear combination of the product of an
element and a second-order determinant. Careful examination of the above equations
reveals that the second-order determinant is the determinant obtained by omitting the
elements in the row and column in which the multiplying element appears in the original
determinant. The resulting second-order determinant is called a minor. Thus the minor
of a1 is obtained in the following way:
33
22
1
333
222
111
cb
cb
a
cba
cba
cba
:thereforeisofminorThe
out

In the general n-th-order determinant, the sign (−1)i +j is associated with the minor of
the element in the i-th row and the j-th column. The minor with its sign (−1)i +j is called
the cofactor. For the determinant:
55
nnnn
n
n
aaa
aaa
aaa
AA
L
MOMM
L
L
11
12221
11211
det ==
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MRT
the value in terms of cofactors is given by the Laplace (1749-1827) development:
( )iAaAA
n
j
ji
jidet
1
anyfor∑=
==
Expanding this along the first row gives i=1. For example:
12
12
11
11
2
1
1
1
2221
1211
AaAaAaA
aa
aa
A
j
j
j +==⇔= ∑=
where:
212121
2112
222222
1111
)1()1( aaaAaaaA −=−=−==+=−= ++
and
On substituting this last result into the expression for |A|, we obtain the number:
21122211 aaaaA −=

The following are properties of determinants (i.e., their inherent rules):
56
321
321
321
333
222
111
ccc
bbb
aaa
cba
cba
cba
==∆
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1. The value of a determinant is not changed if corresponding rows and columns are
interchanged (N.B., this is more appropriately called transposing along the diagonal),
e.g.:
2. If a multiple of one column is added (row by row) to another column or if a multiple of
one row is added (column by column) to another row, the value of the determinant is
unchanged,e.g.:
3333
2222
1111
333
222
111
cbbka
cbbka
cbbka
cba
cba
cba
+
+
+
=
3. If each element of a column or row is zero, the value of the determinant is zero,e.g.:
0000
333
111
=
cba
cba

4. If two columns or rows are identical, the value of the determinant is zero,e.g.:
57
2016
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0
111
222
111
=
cba
cba
cba
5. If two columns or rows are proportional, the value of the determinant is zero,e.g.:
0
2
2
2
133
222
111
=
caa
caa
caa
6. If two columns or rows are interchanged, the sign of the determinant is changed,e.g.:
333
222
111
333
222
111
abc
abc
abc
cba
cba
cba
−=
7. If each element of a column or row is multiplied by the same number, the resulting
determinant is multiplied by that same number,e.g.:
333
222
111
333
222
111
abc
abc
abc
k
ckba
ckba
ckba
=

As the story goes, on the evening of October 16, 1843, William R. Hamilton (1805-
1865) was walking with his wife along the Royal Canal in Dublin when the answer leap-
ed to his mind, the fruit of years of reflection. With his penknife he then and there carved
on a stone on Broom Bridge the fundamental formula for quaternion multiplication:
58
zyx iiiiii σσσσσσσσσσσσ −=−=−=−=−=−=−=−=−= 321213132
ˆˆˆ kji and,
kji ˆ
10
01ˆ
0
0ˆ
01
10
321 ii
i
i
i zyx ==





−
===




 −
===





= σσσσσσ and,
1ˆˆˆˆˆˆ 222
−==== kjikji
31221
2222222222
3
2
2
2
1 1ˆˆˆˆˆˆ σσσσσσσσ iiii =−==−−−=++=++ andkjikji











=
≠
=





−+
−
=
)(1
)(0
321
213
ji
ji
i
i
ji
kkk
kkk
k δ
δδδ
δδδ
σ
This also means that they all take on the form (Exercise):
which in today’s notation is provided using the Pauli matrices σ in the directions 1,2,or3:
The Pauli matrices, σk, can be written using:
which means that:
or x, y, and z as shown, and where the imaginary number is i=√(−1), or otherwise:
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101
10
01
det1)(0
0
0
det110
01
10
det 3
2
21 −=−−=
−
=−==−−=
−
=−=−== σσσ and, iii
i
i

bc
a
θ
In a plane
(e.g., a sheet
of 8½×11
paper on a
large table)
d
Vector d
comes
out from
the front
of the
sheet!
rrrr
O
where the symbol ×××× (cross product) represents the vector
product. In the Figure, the variable θ is the angle between
the vectors a and c. Notice how this new vector d provides a
way to reach out, mathematically, into another dimension!
Vectors, on the other hand, are ‘oriented’ objects and the vector ‘addition’ rule is:
Examples of vectors are: position R, displacement d, velocity v, acceleration a, force F
and torque ττττ, all of which are characterized buy the fact that they also provide direction!
a c====++++ b
c××××ad ====
Adding the vectors a and b together produce a
resultant vector c. The product of the vectors a
and c, respectively, produces a new vector d.
rrrr
rrrr
rrrr
rrrr rrrr
rrrr
Scalars are quantities represented by magnitude only. For example: mass m, volume V,
density ρ (i.e., m/V ), energy E and temperature T. The magnitude is represented by the
‘absolute value’ symbol |…| which means |a|=|±a|=a, the magnitude of the vector a.
rrrr rrrr
One of the key properties of vectors is the capability of
generating a new vector from the product of two vectors on a
plane – this new vector will be perpendicular (out of page):
ˆ
×××× ====
ˆ
k
j
i
ˆ
ˆ
ˆ
k
j
i
ˆ
ˆ
ˆ
k
j
i
ˆ
ˆ
î ˆj ˆk ×××× ====ˆj ˆk î ×××× ====ˆk î ˆj
and
×××× ====î î 0 ×××× ====ˆj ˆj 0 ×××× ====ˆk ˆk 0
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rrrr rrrr
With this definition, we can create an orthogonal basis for
a vectors space in three dimensions consisting in mutually
perpendicular unit vectors i, j and k:
ˆ ˆ
T
m
E
59
rrrrrrrr
rrrr
rrrr rrrr rrrr
rrrr

a
c
θ
b
In Space (e.g.,
look around!)
d Vector d comes out from
bottom of the sheet
rrrr
O
e
Unit vector e
comes out
from the
top of the
plane
ˆ
ˆ
c×××× = a sinθad = c
The vector product is given by:
The scalar product is given by:
The vector commutativity addition rule is given by (i.e., the ‘sum’ order is irrelevant!):
c ==== a ++++ b a++++b====
c××××ad ==== c ×××× a==== −
Adding arbitrarily the vectors a and b together
produce a resultant vector c. The product of
the vectors a and c, respectively, produces a
vector d and the angle θ is shown.
rrrr
rrrr
rrrr
rrrr
rrrr
rrrr
while the vector associativity addition rule – introducing a new vector e (e.g., a unit
‘normal to the plane’ vector) – is:
e( b++++a b++++a====)++++ e ++++ ( )
where the scalar product symbol ‘ • ’ (dot product) of a
vector a with a vector b.
rrrr rrrr
The magnitude of the vector product of two vectors can be
constructed by taking the product of the magnitudes of the
vectors times the sine of the angle (<180 degrees) between
them.
b• = ba cosθ b a•=a
where the symbol ‘ ××××’ (cross product) represents the vector
product whereas the magnitude of this new vector is:
ˆ
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ˆˆ
60

The position vector, R, of an object located at a Point
P(R) =P(X,Y,Z) is then given in a ‘Cartesian’ coordinate
system as a function of Cartesian unit vectors: I, J and K:
The angles αααα, ββββ, and γγγγ are the angles that OP
makes with the three coordinate axes. Here we
have X = Rl, Y = Rm, and Z = Rn where l = cosαααα, m =
cosββββ, and n = cosγγγγ (with 1= l2 + m2 + n2). The
quantities l, m, and n are called direction cosines.
Now the fun stuff begins… In the language of line segments (i.e., OA, AB, BP, OP, &c –
see Figure), an object located at P from an origin O is represented by the relation:
BPABOA ++≡
The position of a physical object is completely specified by its position vector (some-
times called radius vector). To start with, the position vector is a vector drawn (pretty
much) from the origin of a (say Cartesian) coordinate system to the object in question.
ˆ
The vectors RX I, RY J and RZ K are three components of R;
they are the vector representations of the projections of R
on the three coordinate axes, respectively. The quantities
RX, RY, and RZ are the magnitudes of the vector
components in the three respective directions (e.g., in
Cartesian: X, Y, and Z). Using the Pythagorean theorem:
ˆ
1
ˆcosˆcosˆcos
=
++=≡
and
KJI
2016
MRT
!
)()()()(
222
22222
ZYX
BPABOA
++=
++==
R
R
RRR ZYX
=
++=
++==
)ˆcosˆcosˆ(cos
ˆˆˆ
KJI
KJI
ˆ ˆ
ˆ ˆ
61
ˆ
K
J
I
ˆ
ˆ
Y
X
Z
Rαααα
ββββ
γγγγ
P(x,y,z)B
A
ˆR
OC
R
R
Rˆ
Rˆ
ROP
OP
αααα ββββ γγγγ
OP R
Rˆ
αααα ββββ γγγγ

r
•••• P
ˆr
ϕ
r • rˆ
ˆr
ϕ ab
a××××b
14243
|r|cosϕ
The unit vector in the direction of r (e.g., a position1meter away) is thus identified as:
The dot product of two vectors a and b with angle ϕ between them is a scalar
quantity defined by the equation:
The cross product of two vectors a and b is a vector
defined by the equation (the product is non-commutative!):
where r is a unit vector in the direction perpendicular to both
a and b. From the Figure it is seen that the magnitude is
equal to the product of the length of one of the two vectors
and the projection of the other on a line perpendicular to the
first vector, which is equal to the area of the parallelogram
formed with a and b as sides.
The result is not dependent on the order of multiplication and, hence, the dot product is
commutative (i.e., independent of the order of summation):
kji
r
r
r
r ˆcosˆcosˆcosˆ γβα ++==≡
r
ϕcosab=•ba
abbaabba •=•⇒=•−• 0
The equation suggests a convenient procedure for determining the component of a
vector r along any chosen direction r, the result being (see Figure):
ϕcosˆ r=•rr
abrba ×××××××× −== ˆ)sin( ϕba
ˆ
Component of r along r by dot product and cross
product of two vectors is normal to the plane of the
two vectors.
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ˆ
ˆ
62

When resolved into rectangular (i.e., Cartesian) components, the dot product is:
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MRT
ijk
kjkiji
0kjkikj0jikiji0
kkjkikkjjjijkijiii
kkjkikkjjjijkijiii
kjikjiba
ˆ)(ˆ)(ˆ)(
)ˆˆ)(()ˆˆ)(()ˆˆ)((
)ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ(
)ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ(
ˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆ
)ˆˆˆ()ˆˆˆ(
yzzyzxxzxyyx
yzzyxzzxxyyx
yzxzzyxyzxyx
zzyzxzzyyyxyzxyxxx
zzyzxzzyyyxyzxyxxx
zyxzyx
babababababa
babababababa
babababababa
bababababababababa
bababababababababa
bbbaaa
−−−=
−−−=
+=
=
=
=
++++++++
××××++++××××++++××××
××××++++××××−−−−××××++++++++××××−−−−××××++++××××++++
××××++++××××++++××××++++××××++++××××++++××××++++××××++++××××++++××××
××××++++××××++++××××++++××××++++××××++++××××++++××××++++××××++++××××
++++++++××××++++++++××××
↑↑↑↑
since i××××j=k (j××××i=−k), j××××k=i (k××××j=−i) & k××××i=j (i××××k=−j) and i××××i=0, j××××j=0 & k××××k=0.
The result of the cross product can be conveniently expressed by the determinant:
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
zzyyxx
zzyzxzzyyyxyzxyxxx
zzyzxzzyyyxyzxyxxx
zyxzyx
bababa
bababababababababa
bababababababababa
bbbaaa
++=
•••••••••=
•••••••••=
•=•
)ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ()ˆˆ(
ˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆˆ
)ˆˆˆ()ˆˆˆ(
kkjkikkjjjijkijiii
kkjkikkjjjijkijiii
kjikjiba
++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++
++++++++++++++++
since i••••j=j••••i=0, j••••k=k••••j=0 & k••••i=i••••k=0 and i••••i=1, j••••j=1 & k••••k=1 while the cross
product becomes:
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
kjikji
kji
ba ˆ)(ˆ)(ˆ)(ˆˆˆ
ˆˆˆ
xyyxxzzxyzzy
yx
yx
zx
zx
zy
zy
zyx
zyx babababababa
bb
aa
bb
aa
bb
aa
bbb
aaa −−−=+== ++++−−−−−−−−××××
↑↑↑↑↑↑↑↑
63

Certain multiple products of vectors are occasionally encountered and we list two of
the most common ones in the following:
cbabcacbabacacbcba )()()()()()( ••=•=•=• −−−−×××××××××××××××××××× and
Exercise: Find a++++b and a−−−−b for a=2i−−−−j++++k and b=i−−−−3j−−−−5k.ˆ ˆ ˆ ˆ ˆ ˆ
Exercise: What is the direction cosines of the vector 2i−−−−2j−−−−4k.ˆ ˆ ˆ
Exercise: Find the unit vector perpendicular to i−−−−2j−−−−3k and i++++2j−−−−k.ˆ ˆ ˆ ˆ ˆ ˆ
Exercise: Find a•b××××c if a=2i, b=3j, and c=4k.ˆ ˆ ˆ
Exercise: Show that a=2i−−−−j++++k, b=i−−−−3j–5k, and c=3i−−−−4j−−−−4k form the sides of a right
triangle [HINT: What equation do you need to show that two vectors are perpendicular?]
ˆ ˆ ˆ ˆ ˆ ˆ
Exercise: If ai−−−−2j++++k is perpendicular to i−−−−2j−−−−3k, find a.ˆ ˆ ˆ ˆ ˆ ˆ
Exercise: a) Calculate the distance from P1(−1, −4,5) to P2(3,−2,2). b) What are the
direction cosines of P1P2? c) What is the equation of the line connecting P1 and P2?
Exercise: Show that a=i++++4j++++3k and b=4i++++2j−−−−4k are perpendicular.ˆ ˆ ˆ ˆ ˆ ˆ
Exercise: Find the angle between 2i and 3i++++4k.ˆ ˆ ˆ
Exercise: Find the component of 8i++++j in the direction of i++++2j−−−−2k.ˆ ˆ ˆ ˆ ˆ
Exercise: Compute (a××××b)××××c and a××××(b××××c) directly for a=2i++++2j, b=3i–j++++k, and c=8i.ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ
Exercise: Show that (p−−−−eA)××××(p−−−−eA)=iehB where p= –ih∇∇∇∇, B=∇∇∇∇ ××××A, i=√(−1), e and
h are constants while ∇∇∇∇=i∂/∂x++++j∂/∂y++++k∂/∂z. Physically, A is just a vector potential.ˆ ˆ ˆ
I definitely suggest that you attempt these exercises as to appreciate these concepts:
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64

Speed,v[mi/h]
Useful unit conversions:
• 1 mile [mi]=1.61 kilometers [km]
• 1 foot [ft]=0.3048 meter [m].
• 1 km=1000 m=103 m.
• 1 m=3.281 ft.
• 1 mi=5280 ft.
For this example, we have:
• 5 mi/h=8 km/h=2.2 m/s.*
• 10 mi/h=16.09 km/h=4.2 m/s.
• 50 mi/h=80.47 km/h=37.8 m/s.
• 100 mi/h=160.93 km/h=71.5 m/s.
.2.2
3600
1
1
10
1
1.6
1
5.0
0.5
3
s
m
s
h
km
m
mi
km
h
mi
h
mi
=

























=
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MRT
a) Determine the average acceleration in mi/h/s for each of the three speeds;
b) Suppose that the acceleration is constant (i.e., uniform Acceleration) from rest until
the shifting from 3rd to 4th speed. What would be the distance traveled?
c) Determine the real distance traveled up to the shifting from the 3rd to the 4th speeds.
* For example, to convert ‘mi/h’ into ‘m/s’:
For this non-uniform acceleration stuff, there is pretty much no better way than to look
over an example such as an accelerating 1966 Alpha Romeo GT and calculating it!
1st-2nd
2nd-3rd
3rd-4th
Time, t [s]
The graph below represents the speed, v, as a
function of time, t, of a 1966 Alpha Romeo GT.
(Reprinted from Road and Track, November 1986.)
65

Time, t [s]
Speed,v[mi/h]
2016
MRT
1st-2nd
2nd-3rd
3rd-4th
a) Using a=∆v/∆t=(vf −vi)/(tf −ti) (see Plot below), we get
• up to 1st → 2nd: 9.2 mi/h/s (~15 km/h/s);
• up to 2nd → 3rd: 4.8 mi/h/s (~8 km/h/s);
• up to 3rd → 4th: 3.2 mi/h/s (~5 km/h/s).
b) Using P1:(0s,0mi/h) and P2:(17s,84 mi/h), we get a=4.94 (mi/h)/s (~8 km/h/s). Then,
by checking units and using the kinematics formula and solving for s=(vf
2 −vo
2)/2a, we
get the distance 2000 ft.
c) 1200 ft.
(4s,37mi/h)
(9.5s,61mi/h)
(17s,84mi/h)
smi/h
ss
mi/hmi/h
/2.9
04
037
21 =
−
−
=→a
smi/h
ss
mi/hmi/h
/8.4
5.45.9
3716
32 =
−
−
=→a
smi/h
ss
mi/hmi/h
/2.3
1071
6148
43 =
−
−
=→a
)( 2
tfv =
∆v
t [s]
v [mi/h]
9.5 s4.5 s
37 mi/h
61 mi/h
if
if
tt
vv
t
v
a
−
−
=
∆
∆
=
The slope of a v(t) plot:∆t
This acceleration is
considered to be constant
from rest until the shifting
from 3rd to 4th speed:
a = (vf – vo)/(tf – to)
= (84 – 0)/(17– 0)
a ≅ 5 mi/h/s







P1
P2
is the acceleration a(t).
66

α
θo
Efficiency in service delivery:
otan
2
tan θα −=
L
h
vo
L
2016
MRT
† Basketball players in the United States of America would
understand g as: The speed of an object in free-fall will
increase by 32 feet per second with each second it falls or 32
ft/s2, which is always directed towards the center of the Earth.
* The time (t-) dependent parabolic trajectory of a basketball is
given by (g/2)t2 + (vo)t + (so − s) = 0 which is of the parametric
form k1t2 + k2t + k3 = 0 where k1=|g|/2, k2 = |vo| and k3 = |so−−−− s| are
parameters of the trajectory. Note that the absolute value
means |x| = |±x| = x.
A player throws a basketball with an initial velocity vo=|vo| along an initial angle θo
towards a hoop situated at a horizontal distance L and at a height h above the throw
stance. Assume no air-resistance,nor jump from the player,nor rotationof the basketball.
Show that the scalar initial velocity required to reach the hoop is given by the relation:
)(tancos2 oo
2o
Lh
Lg
v
−
=
θθ
The symbol g represents the constant acceleration
due to Earth’s gravity (∼10 meters per second, per
second or 10 m/s2 †) and is given by (we will see why
later):
where G is Newton’s constant of universal attraction
(6.673×10−11 m3/kg/s2), M⊕ is the mass of the Earth
(5.9742×1024 kg) and R⊕ is the equatorial radius of the
Earth (6,378,100 m).
2
⊕
⊕
=
R
MG
g
67
A basketball follows a path in xy-t Space-Time –
showing a parabolic trajectory.*
2010
PMO
Allstream

z
F
y
F
x
F zyx
∂
∂
+
∂
∂
+
∂
∂
=•F∇∇∇∇
Point P at the tip of the distance vector r is given in Rectangular (Cartesian) Coordinates by the intersection of
constant x, constant y and constant z planes (i.e., a box in 3D!) – relative to the frame of the basketball.
zyx ˆˆˆ
z
f
y
f
x
f
f
∂
∂
∂
∂
∂
∂
= ++++++++∇∇∇∇
zy
x
zyx
F
ˆˆ
ˆ///
ˆˆˆ








∂
∂
−
∂
∂






∂
∂
−
∂
∂








∂
∂
−
∂
∂
=∂∂∂∂∂∂=
y
F
x
F
x
F
z
F
z
F
y
F
FFF
zyx
xyzx
yz
zyx
++++++++
++++××××∇∇∇∇
2
2
2
2
2
2
z
f
y
f
x
f
ff
∂
∂
+
∂
∂
+
∂
∂
=•=∇2
∇∇∇∇∇∇∇∇
y
x
r
P
ˆ
k
j
i
ˆ
ˆ
k
i
ˆ
Constant y plane
j
Constant x plane
kjirr ˆˆˆ),,( zyxzyx ++++++++==
Constant
z plane
y
x
z
r
O
zyxF ˆˆˆ zyx FFF 2222
∇∇∇=∇ ++++++++
y
x
z
O
dllll
dy
dx
dz
z
O
kji ˆˆˆ dzdydxd ++++++++=l
dl is an infinitesimal differential
increment of length.
dV is an infinitesimal differential
increment of volume.
2222
dzdydxd ++=l
2016
MRT
The Laplacian of a vector function F=F(x,y,z):
The Laplacian of a scalar function f = f (x,y,z):
As a vector product of a vector function F=F(x,y,z):
As a scalar product of a vector function F=F(x,y,z):
The gradient of a scalar function f = f (x,y,z):We can make the
following geometric
objects into physical
realities if we
substitute the scalar
initial velocity vo for f
= f (x,y,z) and gravity
g =−j=−y for the
vector F = F (x,y,z).
so
g
Fans P
ˆ
x
y
ˆ
2
2
1
oo tt gvss ++++−−−− +=
As a function of the distance of the ball from the fans,
the displacement s=s(x,y,z) is given by the quadratic
equation represented as a function of time variable t:
where vo is the initial velocity vector and g is gravity.
Basketball Path
r
s
s−so
Plane of
free throw
vo
yx ˆˆ -
ˆ
ˆ
ˆ ˆ
•
•
ˆx
yˆ
68

x
)(tancos2
),,(
cos2
)(tan
coscos
sin
)sin(
cos
)cos(
oo
2oo
o
22
o
2
o
2
oo
2
1
oo
oo
2
2
1
oo
2
2
1
oo
oo
oo
2
2
1
oo
Lh
Lg
hLv
v
Lg
Lh
v
L
g
v
L
vh
tgtvhy
tatvyy
v
L
ttvLx
tatvxx
yy
xx
−
=
−=








−







=
−==
++=
=⇒==
++=
θθ
θ
θ
θ
θθ
θ
θ
θ
θ
whereas, along the y-axis, we have:
The initial velocity vo is obtained by using the Kinematic Equation of Motion of a projectile:
This is the required initial velocity (speed)
of the basketball.
The acceleration components are given by |ax|=0 and |ay|=g. The kinematics along the x-axis are:
The required velocity vector vo will depend on gravity g andonthe scalar parameters θo,Landh.
ŷ
[m/s]
According to Einstein’s Theory of Gravity: Space
acts on matter, telling it how to move; in turn, matter
reacts back on space, telling it how to curve.
(g =−gy) vo= vox + voy
vox=vocosθo
voy=vosinθo
vo
θo
Vector Addition
hmax
Gravity
o
yxyx
tt
yxtt aa
v
avv
v
vyxsavss ++++++++++++++++++++ =
∆
∆
==
∆
∆
=== ,, andwhere oo
o
o
2
2
1
oo ˆˆ
so
2016
MRT
ˆ
(since ay = −g)
Note: vy = 0000 at hmax (θ = 0) and since there is no air-resistance, vx = vox = cte.
69
ˆ

vo
Path of the projectile
φ
θo
d
a) Show that the projectile travels a distance d up the incline, where:





 −
=
φ
θφθ
2
oo
2
o
cos
cos)sin(2
g
v
d
b) For what value of θo is d a maximum, and what is that maximum value of d?
* Problem from Physics for Scientists and Engineers with Modern Physics, by R. A. Serway and J. W. Jewett, Brooks Cole; 9th edition
(2013) – Problem P4.86. Solution adapted from Doug Davis, © 2001.
Here is another interesting problem that will be a primer for the ballistic stuff we will do
later and setup to highlightthe use of various trigonometric identities to solve problems.
2016
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Let us first define the slope. Using x=d⋅cosφ and y=d⋅sinφ
we get y/x=dsinφ /dcosφ =sinφ /cosφ =tanφ giving us:
(N.B., y here is a function of the slope φ, i.e., y(φ)). Next, we
isolate the projectile information using vox=vo⋅cosθo and voy =
vo⋅sinθo , x=voxt and y=voyt −½gt2. Substituting vox and voy we
find:
φφ tan)( xy =
(N.B., y here is a function of the initial angle θo , i.e., y(θo )).
and:
ooo cosθv
x
v
x
t
x
==
2
ooo
2
1
sin)( tgtvy −= θθ
A projectile* is fired up an incline (with incline angle φ) with an initial speed vo =|vo|
(v-naught), at an angle θo, with respect to the horizontal (θo >φ) as shown in the Figure.
70
x
y
φ
d sinφ
vo cosθo
(vox)
(voy)
vosinθo
d
θo
g=−gj
d cosφ
ˆ

o
22
o
2
o
2
oooo
ooo
cos2
1
tan
cos2
1
cos
sin)(
θ
θ
θθ
θθ
v
x
gx
v
x
g
v
x
vy
−=








−







=
Since y(φ)= x tanφ, we get the equality with our result for y(θo) just above:
o
22
o
2
oo
cos2
1
tantan)()(
θ
θφθφ
v
x
gxxyy −=⇔=
o
22
o
o
cos2
1
tantan
θ
θφ
v
x
g−=
Substituting the time t=x/vocosθo into y(θo)=vosinθot −½gt2 we get:
Dividing throughout with x, we get:
and adding −tanθo to both sides and then multiplying by cos2θo throughout:
o
2
o2
oo
22
o
o cos)tan(tan
2cos2
1
tantan θφθ
θ
θφ −=⇒−=− x
v
g
x
v
g
2016
MRTo
2
o
2
o
cos)tan(tan
2
θφθ −=
g
v
x
Then by multiplying by 2vo
2/g throughout, we thus isolate x:
71

This difference in tangents (i.e., the term tanθo −tanφ) is troublesome and needs to be
reduced in sin and cos functions for θo and φ. From the trigonometric identities we find
that tan(α −β)=(tanαααα −tanββββ)/(1+tanα tanβ), we obtain by using it (i.e., for θθθθo=αααα and φφφφ =ββββ):








+
−
−
=+−=−
φ
φ
θ
θ
φθ
φθ
φθφθφθ
cos
sin
cos
sin
1
)cos(
)sin(
)tantan1)(tan(tantan
o
o
o
o
ooo
Since x=dcosφ:
We then get for d:
o
2
o
2
o
cos)tan(tan
2
cos θφθφ −==
g
v
dx
we find d by dividing through with cosφ:
φ
θ
cos
cos
)(
2 o
22
o
φφφφθθθθ tantan o −=
g
v
d
φ
θ
φθ
φθφθ
φθ
φθ
φ
θ
φθ
φθ
φθ
φθ
cos
cos
coscos
sinsincoscos
)cos(
)sin(2
cos
cos
coscos
sinsin
1
)cos(
)sin(2 o
2
o
oo
o
o
2
oo
2
o
o
o
o
2
o





 +
−
−
=
















+
−
−
=
g
v
g
v
d
but cos(θo −φ)=cosθo cosφ +sinθo sinφ, so:
φ
θφθ
φ
θ
φθ
φθ
φθ
φθ
2
oo
2
oo
2
o
o
o
o
2
o
cos
cos)sin(2
cos
cos
coscos
)cos(
)cos(
)sin(2 −
=




 +
−
−
=
g
v
g
v
d 2016
MRT
72

But upon closer inspection we see that there is a trigonometric identity that can
reduce the sum cosθθθθocosφφφφ ++++sinθθθθosinφφφφ to the cosine of the difference of angles cos(θo −φ)
(i.e., we use the trigonometric identity cos(α −β)=cosαααα cosββββ ++++sinαααα sinββββ). So:
giving us the solution for the distance d a projectile travels up an incline of angle φ:
φ
θ
φθ
φθ
φθ
φθ
cos
cos
coscos
)cos(
)cos(
)sin(2 o
2
o
o
o
o
2
o





 −
−
−
=
g
v
d
Now we reduce this mess such that trigonometric terms start canceling each other:





 −
=
φ
θφθ
2
oo
2
o
cos
cos)sin(2
g
v
d








⋅
−
⋅
−
−
=
φ
θθ
φθ
φθ
φθ
φθ
cos
coscos
coscos
)cos(
)cos(
)sin(2 oo
o
o
o
o
2
o
g
v
d
This is the answer to a). Now for b)… If the slope φ is null (i.e., when φ=0), we get:
)cossin2(
0cos
cos)0sin(2
oo
2
o
2
oo
2
o
θθ
θθ
g
v
g
v
d =
−
=
and since sin(2α)=2sinα cosα, we obtain (as it should be):
g
v
d o
2
o 2sin θ
=
2016
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73

Now that we have an expression for d, we can find the maximum value of d, called the
range, R, by taking the derivative of R=(2vo
2/g)[sin(θ −φ)cosθ/cos2φ] (remembering that φ
is a constant) with respect to the angle θ (not θo explicitly) and setting that equal to zero
and solving for the angle θ and then evaluating R for that value of θ:
θ
θφθ
φθ
θφθ
φ
θφθ
φθθ d
d
g
v
d
d
g
v
g
v
d
d
d
Rd ]cos)[sin(
cos
2]cos)[sin(
cos
2
cos)sin(
cos
2
2
2
o
2
2
o
2
2
o −
⋅=
−
=








−=
Now the derivative of sin(θ −φ)cosθ with respect to θ is:
θφθθφθθφθθφθ
θ
θ
φθθ
θ
φθ
θφθ
θ
sin)sin(cos)cos()sin()sin(cos)cos(
][cos
)sin(cos
)][sin(
]cos)[sin(
−−−=−⋅−+⋅−=
⋅−+⋅
−
=⋅−
d
d
d
d
d
d
where we have used the product rule d(u⋅v)/dx=u⋅dv/dx +v⋅du/dx (e.g., for u=sin(θ −φ)
and v=cosθ). Using the trigonometric identity cos(α +β)=cosα cosβ −sinα sinβ (for α=θ −φ
and β=θ), we get:
)2cos(])cos[(]cos)[sin( φθθφθθφθ
θ
−=+−=−
d
d
which gives us the result:
)2cos(
cos
2
2
2
o
φθ
φθ
−=
g
v
d
Rd 2016
MRT
74

Now, set dR/dθ =0 (this provides THE way to find the maximum):
0)2cos(
cos
2
2
2
o
=−= φθ
φθ g
v
d
Rd
For this to happen (i.e., dR/dθ =0) we need cos(2θ −φ)=0 since vo
2, g and cosφ are non-
zero. We plot the function y=cosx next and look for a value that corresponds to cosx=0:
We see that the curve crosses over at π/2 or 90°. This means that cos(π/2)=0 so 2θ −φ =
π/2 or 2θ −φ =90°. This gives us an equation to solve for θ: θ =90°/2+φ/2=45°+φ/2. You
will notice that this is consistent with the maximum horizontal range occurring for θo=
45° for φ =0… the answer to the first part of b) (i.e., what value of θo is d a maximum).
Now we must evaluate R for θ =45°+φ/2:
φ
φφ
φ
φ
φ
φ
2
2
o
2
2
o
cos
2
45cos
2
45sin
2
cos
2
45cos
2
45sin
2






+





−
=






+





−





+
=
g
v
g
v
R
We must now look for a way to simplify this sin(45°−φ/2)cos(45°+φ/2) monster!
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y = cosx y = sin x y = sin x/cosx = tanx
75
π/2

Again, using the trigonometric identities sin(α −β)=sinα cosβ −cosα sinβ and cos(α +β)=
cosα cosβ −sinα sinβ we find sin(45°−φ/2)=sin45°cosφ/2 −cos45° sinφ/2=(1/√2)(cosφ/2 −
sinφ/2) and cos(45°+φ/2)=cos45°cosφ/2 −sin45°sinφ/2=(1/√2)(cosφ/2−sinφ/2) so that the
product sin(45°−φ/2)cos(45°+φ/2)=½(cosφ/2 −sinφ/2)(cosφ/2−sinφ/2) becomes:






−





+=





+−=





+





−
2
sin
2
cos2
2
sin
2
cos
2
1
2
sin
2
sin
2
cos2
2
cos
2
1
2
45cos
2
45sin 2222 φφφφφφφφφφ
But since cos2α +sin2α =1 and sin(2α)=2sinα cosα (here with α =φ/2 for both), we obtain:
)sin1(
2
1
2
2sin)1(
2
1
2
45cos
2
45sin φ
φφφ
−=











−=





+





−
If we put this back into R=(2vo
2/g)[sin(45°−φ/2)cos(45°+φ/2)], we get:
φ
φ
2
2
o
cos
)sin1(½2 −
=
g
v
R
and crossing out the factor of 2 with its fraction (i.e.,the ½ factor) gives us our final result:







 −
=
φ
φ
2
2
o
cos
sin1
g
v
R
This is the answer to the second part of b) (i.e., What is that maximum value of d?).
When the incline angle φ is zero, we’re back to a horizontal range problem. This
expression reduces to R=vo
2/g as it should.
2016
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76

2016
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77
Appendix
PART II – MODERN PHYSICS
Charge and Current Densities
Electromagnetic Induction
Electromagnetic Potentials
Gauge Invariance
Maxwell’s Equations
Foundations of Special Relativity
Tensors of Rank One
4D Formulation of Electromagnetism
Plane Wave Solutions of the Wave
Equation
Special Relativity and
Electromagnetism
The Special Lorentz Transformations
Relativistic Kinematics
Tensors in General
The Metric Tensor
The Problem of Radiation in
Enclosures
Thermodynamic Considerations
The Wien Displacement Law
The Rayleigh-Jeans Law
Planck’s Resolution of the Problem
Photons and Electrons
Scattering Problems
The Rutherford Cross-Section
Bohr’s Model
Fundamental Properties of Waves
The Hypothesis of de Broglie and Einstein
Appendix: The General Theory of
Relativity
References
We list here, as a reference, the Contents of the remaining parts of this 10-PART Series
which as a whole makes for quite a thorough review of Theoretical Physics (N.B., Since
Superstring Theory is still being developed, this content is accurate up to year 1990-ish).

78
PART III – QUANTUM MECHANICS
Introduction
Symmetries and Probabilities
Angular Momentum
Quantum Behavior
Postulates
Quantum Angular Momentum
Spherical Harmonics
Spin Angular Momentum
Total Angular Momentum
Momentum Coupling
General Propagator
Free Particle Propagator
Wave Packets
Non-Relativistic Particle
Appendix: Why Quantum?
References
2016
MRT

PART IV – QUANTUM FIELDS
Review of Quantum Mechanics
Galilean Invariance
Lorentz Invariance
The Relativity Principle
Poincaré Transformations
The Poincaré Algebra
Lorentz Transformations
Lorentz Invariant Scalar
Klein-Gordon & Dirac
One-Particle States
Wigner’s Little Group
Normalization Factor
Mass Positive-Definite
Boosts & Rotations
Mass Zero
The Klein-Gordon Equation
The Dirac Equation
References
79
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PART V – THE HYDROGEN ATOM
What happens at 10−−−−10 m?
The Hydrogen Atom
Spin-Orbit Coupling
Other Interactions
Magnetic & Electric Fields
Hyperfine Interactions
Multi-Electron Atoms and Molecules
Appendix – Interactions
The Harmonic Oscillator
Electromagnetic Interactions
Quantization of the Radiation Field
Transition Probabilities
Einstein’s Coefficients
Planck’s Law
A Note on Line Broadening
The Photoelectric Effect
Higher Order Electromagnetic Interactions
References
2016
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81
PART VI – GROUP THEORY
Symmetry Groups of Physics
Basic Definitions and Abstract Vectors
Matrices and Matrix Multiplication
Summary of Linear Vector Spaces
Linear Transformations
Similarity Transformations
Dual Vector Spaces
Adjoint Operator and Inner Product
Norm of a Vector and Orthogonality
Projection, Hermiticity and Unitarity
Group Representations
Rotation Group SO(2)
Irreducible Representation of SO(2)
Continuous Translational Group
Conjugate Basis Vectors
Description of the Group SO(3)
Euler Angles α, β & γ
Generators and the Lie Algebra
Irreducible Representation of SO(3)
Particle in a Central Field
Transformation Law for Wave Functions
Transformation Law for Operators
Relationship Between SO(3) and SU(2)
Single Particle State with Spin
Euclidean Groups E2 and E3
Irreducible Representation Method
Unitary Irreducible Representation of E3
Lorentz and Poincaré Groups
Homogeneous Lorentz Transformations
Translations and the Poincaré Group
Generators and the Lie Algebra
Representation of the Poincaré Group
Normalization of Basis States
Wave Functions and Field Operators
Relativistic Wave Equations
General Solution of a Wave Equation
Creation and Annihilation Operators
References
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82
PART VII–QUANTUM ELECTRODYNAMICS
Particles and Fields
Second Quantization
Yukawa Potential
Complex Scalar Field
Noether’s Theorem
Maxwell’s Equations
Classical Radiation Field
Quantization of Radiation Oscillators
Klein-Gordon Scalar Field
Charged Scalar Field
Propagator Theory
Dirac Spinor Field
Quantizing the Spinor Field
Weyl Neutrinos
Relativistic Quantum Mechanics
Quantizing the Maxwell Field
Cross Sections and the Scattering Matrix
Propagator Theory and Rutherford
Scattering
Time Evolution Operator
Feynman’s Rules
The Compton Effect
Pair Annihilation
Møller Scattering
Bhabha Scattering
Bremsstrahlung
Radiative Corrections
Anomalous Magnetic Moment
Infrared Divergence
Lamb Shift
Overview of Renormalization in QED
Brief Review of Regularization in QED
Appendix I: Radiation Gauge
Appendix II: Path Integrals
Appendix III: Dirac Matrices
References
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83
Fermion Masses and Couplings
Why Go Beyond the Standard Model?
Grand Unified Theories
General Consequences of Grand
Unification
Possible Choices of the Grand Unified
Group
Grand Unified SU(5)
Spontaneous Symmetry Breaking in
SU(5)
Fermion Masses Again
Hierarchy Problem
Higgs Scalars and the Hierarchy
Problem
Appendix
References
PART VIII – THE STANDARD MODEL
The Particles
The Forces
The Hadrons
Scattering
Field Equations
Fermions
Particle Propagators
Noether’s Theorem and Global Invariance
Local Gauge Invariance in QED
Yang-Mills Gauge Theories
Quantum Chromodynamics (QCD)
Renormalization
Strong Interactions and Chiral Symmetry
Spontaneous Symmetry Breaking (SSB)
Weak Interactions
The SU(2)⊗U(1) Gauge Theory
SSB in the Electroweak Model
Gauge Boson Masses
Gauge Boson Mixing and Coupling
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PART IX – SUPERSYMMETRY
Motivation
Introduction to Supersymmetry
The SUSY Algebra
Realizations of the SUSY Algebra
The Wess-Zumino Model
Lagrangian with Mass and Interaction
Terms
The Superpotential
Supersymmetric Gauge Theory
Spontaneous Breaking of
Supersymmetry
F-type SUSY Breaking
D-type SUSY Breaking
The Scale of SUSY Breaking
The SUSY Particle Spectrum
Supersymmetric Grand Unification
General Relativity
The Principle of Equivalence
General Coordinates
Local Lorentz Frames
Local Lorentz Transformations
General Coordinate Transformations
Covariant Derivative
The Einstein Lagrangian
The Curvature Tensor
The Inclusion of Matter
The Newtonian Limit
Local Supersymmetry
A Pure SUGRA Lagrangian
Coupling SUGRA to Matter and Gauge
Fields
Higher-dimensional Theories
Compactification
The Kaluza Model of Electromagnetism
Non-Abelian Kaluza-Klein Theories
Kaluza-Klein Models and the Real World
N=1 SUGRA in Eleven Dimensions
References
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PART X – SUPERSTRING THEORY
A History of the Origins of String
Theories
The Classical Bosonic String
The Quantum Bosonic String
The Interacting String
Fermions in String Theories
String Quantum Numbers
Anomalies
The Heterotic String
Compactification and N=1 SUSY
Compactification and Chiral Fermions
Compactification and Symmetry
Breaking
Epilogue: Quantum Gravity
Appendix I: Feynman’s Take on
Gravitation
Appendix II: Review of Supersymmetry
Appendix III: A Brief Review of Groups
and Forms
Appendix IV: The Gamma Function
Appendix V: The Beta Function
References

2016
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References
86
M. Klein, Calculus, 2-nd Edition, Dover, 1977.
New York University
Believe it or not, my father had the first edition in 2 volumes and when I found them and browsed through them I was amazed at the
Motion in One Dimension in a Resisting Medium part which eventually ended up being the case for this treatment of air resistance in
ballistic problems. Since then, I’ve always enjoyed Klein’s way of presenting calculus using physical concepts. The only problem
though is his use of 32 [ft/sec] everywhere! Most of the gravitation discussion related to hollow or filled up spheres is from this book.
H. Benson, University Physics, Revised Edition, 1996.
Vanier College
Amazing course which is somewhat similar to the one I had with Halliday and Resnick back in 1986-88. It is a great page turner with
special topics everywhere. The problems in themselves are worth solving (e.g., kinematics of a basketball shot!) and many have an
equation to prove that is actually displayed! My edition has 44 chapters spread across 942 pages covering everything from vectors,
kinematics, inertia, particle dynamics, work and energy, conservation of energy, momentum, rotations about a fixed axis, gravitation,
solids and fluids, oscillations, waves and sound, temperature and the ideal gas law, thermodynamics and entropy, electrostatics, the
electric field, pretty much everything ‘electric and magnetic’, Maxwell’s equations, light and optics, special relativity , quantum theory
and wave mechanics, atoms and solids. Finally nuclear physics and elementary particles and a (1997) view of Grand Unified Theory.
C. Harper, Introduction to Mathematical Physics, Prentice Hall, 1976.
California State University, Haywood
This is my favorite go-to reference for mathematical physics. Most of the differential equations presentation and solutions, complex
variable and matrix definitions, and most of his examples and problems, &c. served as the primer for this work. Harper’s book is so
concise that you can pretty much read it in about 2 weeks and the presentation is impeccable for this very readable 300 page
mathematical physics volume.
D.G. Zill, W. S. Wright, Advanced Engineering Mathematics, 4-th Edition, Jones & Bartlett, 2011.
Loyola Mary-mount University
If you are going to go into some scientific field that makes you learn engineering physics you should get this book as a reference.
Besides being set in color, it is easily readable and with just enough conceptual ‘meat’ around the physical ‘bone’ to capture your
attention without being to mathematical about it. I mean, when I browsed through it I wished I had used this book for my course!
F. P. Beer, E. R. Johnson, W. E. Clausen, Vector Mechanics for Engineers, Volume II - Dynamics, 8-Edition, McGraw-Hill, 2007.
Lehigh University, University of Connecticut, Ohio State University
Great reference and a constant page turner with amazing color graphics. It is fairly advanced though but requires some study.
T. Allen Jr, R.L. Ditsworth, Fluid Mechanics, McGraw Hill, 1972.
College of Engineering Sciences at Arizona State University
My father’s book. I first skimmed through this book when I was 14 and fell in love with it instantly. The fluid mechanics here is all theirs.
W.J. Thomson, Introduction to Space Dynamics, Dover, 1986.
University of California at Santa Barbara
Great introduction to vectors, kinematics, dynamics and earth & satellite-related applications. Most of the mathematical treatment for
kinematics, rotation transformations and satellite or ballistic dynamics here is based on this book since it is succinct and to the point.

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PART I.1 - Physical Mathematics

PART I.1 - Physical Mathematics

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PART I.1 - Physical Mathematics