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AppendixA
FundamentalsofVectorAnalysis
Abstract The purpose of this appendix is to present a consistent but brief
introduction to vector calculus. For the sake of completeness, we shall begin with a
brief review of vector algebra. It should be emphasized that this appendix cannot be
seen as a textbook on vector algebra and analysis. In order to learn the subject in a
systematic way,thereadercanusespecialtextbooks.Atthesametime,wewillcon-
sider here a content which is supposed to be sufficient for applications in Classical
Mechanics,at the level used in this book.
A.1 Vector Algebra
In this paragraph, we use a constant and global basis which consists of orthonor-
ˆ ˆ ˆ
mal vectors i,j,k. A vector is represented geometrically by an oriented segment
(arrow), which is characterized by length (also called absolute value, or modulus,
or magnitude of a vector) and direction. Any vector a can be expressed as a linear
combinationof the basis vectors,
ˆ ˆ ˆ
a=a1i+a2j+a3k. (A.1)
Thelinear operations on vectors include:
(i) The multiplication by a constant k, which is equivalent to the multiplication of
all componentsof a vector by the same constant:
ˆ ˆ ˆ
ka =(ka )i+(ka )j+(ka )k. (A.2)
1 2 3
(ii) The sum of two vectors, a and b, obtained by the addition operation is a vector
with componentsequal to the sum of the componentsof the original vectors,
ˆ ˆ ˆ
a+b=(a1+b1)i+(a2+b2)j+(a3+b3)k. (A.3)
I.L. Shapiro and G. de Berredo-Peixoto, Lecture Notes on Newtonian Mechanics, 227
Undergraduate Lecture Notes in Physics, DOI 10.1007/978-1-4614-7825-6,
©Springer Science+Business Media, LLC 2013
228 A Fundamentals of Vector Analysis
Besidesthelinearoperations,wedefinethefollowingtwotypesofmultiplication.
Thescalar product(dot product) between the two vectors, a and b,isdefinedas
a·b=(a,b)=abcosϕ, (A.4)
whereaandbrepresentabsolutevaluesofthevectorsaandb,givenbya=|a|and
b=|b|,andϕ isthesmallest angle between these vectors.
Themainpropertiesof the scalar product (A.4) are commutativity,
a·b=b·a,
andlinearity,
a·(b +b )=a·b +a·b . (A.5)
1 2 1 1
Asaresult,wecanusetherepresentation(A.1)forbothvectorsaandb,inthisway
wearriveat the formula for the scalar product expressed in components,
a·b=a1b1+a2b2+a3b3. (A.6)
Thedefinitionofvectorproduct(crossproduct)betweenthetwovectorsisalittle
bit more complicated. The vector c is the vector product of the vectors a and b,and
is denoted as
c=a×b= a,b , (A.7)
if the following three conditions are satisfied:
• Thevectorcisorthogonaltotheothertwo vectors,i.e., c⊥a and c⊥b.
• Themodulusofthevectorproductisgivenby
c =|c| =absinϕ,
whereϕ is the smallest angle between the vectors a and b.
• The orientation of the three vectors, a, b,andc, is right handed (or dextrorota-
tory). This expression, as discussed in Chap.2, means the following. Imagine a
rotation of a until its direction matches with the direction of b, by the smaller
angle between them. The vector c belongs to the axis of rotation and the only
question is how to choose the direction of this vector. This direction must be
chosensuchthat, by lookingat the positive direction c, the rotation is performed
clockwise. A simple way to memorize this guidance is to remember about the
motion of a corkscrew. When the corkscrew turns a up to b, it advances in the
direction of c. Another useful rule is the right-hand rule, rather commonplace in
the textbooksof Physics.
Themainpropertiesof the vector productare the antisymmetry,
a×b=b×a (A.8)
A.1 Vector Algebra 229
(or anticommutativity), and the linearity,
a,(b +b ) = a,b + a,b . (A.9)
1 2 1 2
Thevectorproductcanbeexpressedas a determinant,namely,
ˆ ˆ ˆ
i j k a a a a a a
ˆ 2 3 ˆ 1 3 ˆ 1 2
a×b=a1a2 a3 = ib b jb b +kb b . (A.10)
b b b 2 3 1 3 1 2
1 2 3
Someimportantrelations involving vector and scalar products will be addressed in
the form of exercises.
Exercises
1. Using the definition of vector product, check that
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
i ×j = k, k×i = j, j×k=i. (A.11)
Show that the magnitude of the vector product, |a×b|, is equal to the area of the
parallelogram with vectors a and b as edges.
2. We can combinebothtypesofmultiplicationand buildthe so-called mixed prod-
uct, which involves three independent vectors a, b and c,
a1 a2 a3
(a, b, c)= a, b,c =b1 b2 b3. (A.12)
(a) Verify the second equality of (A.10). c1 c2 c3
(b) Showthat the mixed producthas the following cyclic property:
(a, b, c)=(c, a, b)=(b,c,a).
(c) Show that the modulus of the mixed product, |(a, b, c)|, is equal to the volume
of the parallelepiped with the three vectors a, b,andc as edges.
(d) Show that the mixed product (a, b, c) has a positive sign in the case when the
three vectors a, b, c (in this order!) have dextrorotatory orientation.
3. Another, interesting for us, quantity is the double vector product,
Derive the following relation a, b, c. (A.13)
a, b, c =b a,c c a,b . (A.14)
This identity is useful in many cases.
230 A Fundamentals of Vector Analysis
A.2 Scalar and Vector Fields
In the next paragraph we will consider differential operations performed on
the scalar or vector fields. For this reason, here we introduce the notion of a field,
including scalar and vector cases.
Thescalar field is a function f(r) of a point in space. Each point of the space M
is associated with a real number, regardless of how we parameterize the space, i.e.,
regardless of the choice of a system of coordinates. As examples of scalar fields in
physics, we can mention the pressure of air or its temperature at a given point.
In practice, we used coordinatesto parametrize the space, and the scalar field be-
comesafunctionof the coordinates, f(r)=f(x,y,z). In order to have the property
of coordinate-independence, mentioned above, the function f(x,y,z) should obey
the following condition: If we change the coordinates and consider, instead of x, y
′ ′ ′
and z, some other coordinates, say x , y ,andz , the form of the functional depen-
dence f(x,y,z) should be adjusted such that the value of this function in a given
geometric point M would remain the same. This means that the new form of the
function, denoted by f′ is defined by the condition
′ ′ ′ ′
f (x , y , z )=f(x,y,z). (A.15)
Example1. Consider a scalar field defined on the coordinates x, y and z by the
formula
f (x,y,z)=x2+y2.
Find the shape of the field f′ for other coordinates,
′ ′ ′
x =x+1, y =z, z =y. (A.16)
′ ′ ′ ′
Solution. To find f (x , y , z ),wewilluseEq.(A.15).Thefirststepistosolve(A.16)
withrespecttothecoordinatesx,y,x andthenjustreplacethesolutionintheformula
for the field. We find
′ ′ ′
x =x 1, y =z , z =y (A.17)
hence
f ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ 2 ′2
(x , y , z )=f x(x , y , z ), y(x , y , z ), z(x , y , z ) = x 1 +z .
In addition to the coordinates, the scalar field may depend on the time variable,
but since time and spatial coordinates are independent in Classical Mechanics, this
appendixdoesnotconsidertemporaldependenceforscalarfields.Thesameisvalid
for other types of fields.
Thenextexampleofourinterestisthevectorfield.Thedifferencewiththescalar
field is that, in the vector case, each point of the space is associated with a vector,
say, A(r). If we parameterize the points in space by their radius-vectors, we have
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