PART I.3 - 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 3

Contents
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
2

L
nˆ
r
θ
In a plane (e.g., 8½×11 paper)
The vector L comes out of front of page.
m
r⊥
p=mv
p
pxi
ˆpyj
Lz =xpy−ypx
Scalarsarequantitiesrepresented by‘magnitude’ only,e.g.,coordinates (x,y,z), mass (m),
speed (v),volume(V),density(ρ),energy (E),temperature (T),surface (S)&inertia (I ).
Vectors, on the other hand, are ‘oriented’ objects that can also depend on other vectors:
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General Path Rules
)( MomentumLinearvp m=
)()(ˆˆsin component-andMomentumAngular zvmrLz ⊥==== nLnprprL θ××××
where m is theconstantmass and v the velocity vector. Examples of vectors:position (r),
displacement (d), velocity (v), acceleration (a), force (F), torque (ττττ) and surface (S).
0jjjin
0iiinj
0nnnji
==
==
==
ˆˆˆˆˆ
ˆˆˆˆˆ
ˆˆˆˆˆ
××××××××
××××××××
××××××××
and
and
and
The position vector r and momentum vector p. The
vector product of the position r and momentum p
produces angular momentum L = |r||p|sinθ n.ˆ
We can now create an orthogonal basis (i.e., perpendicu-
lar to each other) for a vector space in three dimensions
consisting of mutually perpendicular unit vectors , and :jˆ kˆ
where the symbol ×××× (cross product) represents the vector product, n is a unit vector and
wherethemagnitude (orsize)isrepresentedbythe ‘absolute value’symbol |…| (i.e.,|n|=1).
ˆ
ˆ
jik ˆˆˆ =××××ikj ˆˆˆ =××××kji ˆˆˆ =××××
Setting n for the unit normal k (see Figure) we get the algebra:ˆ ˆ
One of the key properties of vectors is their capability of generating a new vector from
the product of two vectors on a plane–thisnew vector will be perpendicular(out of page):
3
iˆ
kˆ jˆ kˆ jˆ kˆ jˆ
iˆ iˆ iˆ
ˆ

y
xO
∆x
∆y
)( xaf ∆+
)(af
a xa ∆+
Q
P
)(xfy =
Secant
Tangent
d x
P
xp
h
x
∆
≈∆e.g.d y
F(s)
O
P
Q
F(s+∆s)
∆F=F(s +∆s)−−−−F(s)
F(s)+∆F=F(s +∆s)
Dividing both sides of the above equation by the scalar ∆s allows us to obtain ∆F/∆s=
PQ/∆s which is a vector along PQ. If ∆F/∆s approaches a ‘limit’ (i.e., mathematically lim)
as ∆s approaches zero, this ‘limit’ is called the derivative of F with respect to s, that is:
Consider a vector F(s) where s is a scalar variable. Here F is a function of s; for each
value of s, there is a corresponding value of F (see Figure). The increment in F(s) when
s changes to s +∆s is given by:
where d /dx is the symbol for the derivative of a function f (in
this case – along x only). The uncertainty principle limits ∆x.
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PQFFF =∆+=∆ )()( sss −−−−
Illustration of the physical meaning of the variation,
∆F, of a vector F(s) by an amount ∆s (i.e., F(s +∆s))
in the direction of the segment PQ.
x
afxaf
x
y
afxafy
xafxaf
∆
−∆+
=
∆
∆
−∆+=∆
∆+=∆+
)()(
)()(
)()(
hence
havewe
Since
x
xfxxf
xd
xfd
x ∆
−∆+
=
→∆
)()(
lim
)(
0
:SlopemalInfinitesi
Geometrically, the dy/dx number represents the infinitesimal
slope at P:
Given f a continuous function defined by y = f (x) we have:
kji
FFFF ˆˆˆ)()(
limlim
00 sd
Fd
sd
Fd
sd
Fd
s
sss
ssd
d zyx
ss
++++++++
−−−−
=





∆
∆+
=





∆
∆
=
→∆→∆
4

∆C=∆A++++∆B where C=A++++B and C++++∆A=A++++∆A++++B++++∆A.
The derivative of the vector sum A++++B (this also applies to the vector difference:
A−−−−B=−(B−−−−A)) is obtained in the following manner:
Now consider the derivative of the scalar product where C=A•B (same as C=B•A).
Here we obtain:
BAC ∆∆=∆ ++++
sd
d
sd
d
sd
d
sssd
d
ss
BA
BA
BACC
++++++++
++++
==





∆
∆∆
=





∆
∆
=
→∆→∆
)(limlim
00
where
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B
AB
ABA •+•=•=





∆
∆
=
→∆ sd
d
sd
d
sd
d
s
C
sd
Cd
s
)(lim
0
and
BABABA ∆•∆+•∆+∆•=∆C
0limlimlim
000
=
∆
∆
•∆=





∆
∆
•∆
→∆→∆→∆ ss sss
B
A
B
A
since
0)]()([limlim
00
=∆+=∆
→∆→∆
sss
ss
AAA −−−−
On dividing both sides of ∆C=∆A++++∆B by ∆s and taking the ‘limit’ as ∆s→0, we obtain:
Derivative of a
vector C=A++++B
Derivative of a
scalar C =A•B
5

The following relations can also be proved (represents the ‘calculus’ of vector analysis):
The physical (or vector) quantities with F, the Newtonian force, being an outcome are:
( )
( )
( )
( ))()()(
)(
)(
)(
sUUs
sd
Ud
sd
d
UU
sd
d
tss
td
sd
sd
d
td
d
a
sd
d
aa
sd
d
sd
d
sd
d
sd
d
sd
d
===
=⋅=
=
=
=
and
where-
scalarconstantais
vectorconstantais
VVV
V
V
AA
A
A
C0
C
B
AB
ABA
++++
××××++++××××××××
RuleChain
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v
v
aa
vvp
Fr
r
vvp && ≡=⇔====≡=⇔=
td
d
m
td
d
m
td
md
td
d
td
d
m
)(
and
where t is time(parameter)andvand a are calledvelocity andacceleration, respectively.
∇≡≡
∂
∂
∂
∂
∂
∂
=
r
delˆˆˆ kji
zyx
++++++++∇∇∇∇
The del (ornabla) operator, ∇∇∇∇ (N.B., being a vector it is bold) is a differential
operator and is of immense importance in physics. It is defined by (in Cartesian units):
6

êt
r
ωωωω
O
φ
∆θ
x
z
y
∆r
In differentiating a vector, the usual rules of the limiting process apply:
tttd
d
tt ∆
∆
=
∆
∆
=
→∆→∆
rrrrr
00
lim
)(
lim
−−−−++++
If the vector r is referenced to a fixed coordinate system, the ∆r is the vector change
relative to the coordinates which is also the total change, and the equation above is
the total derivative of r.
Differentiation of a vector r referenced to rotating
coordinates with Oxyz set up as a reference frame.
If the vector r is referenced to a fixed coordinate system (e.g., a Cartesian one such
as Oxyz in the Figure), the ∆r is the vector change relative to the coordinates which is
also the total change, and dr/dt=lim∆t→0(∆r/∆t) is the total derivative of r.
If the vector r is referenced to a rotating coordinate system
such as the one shown in the Figure, the vector r remaining
stationary relative to the rotating axes will undergo a change:
φθ sinr∆
where êt is a unit vector along the tangent.
rω××××
along the tangent to the dotted circle, and its rate of change
is established by the limit:
t
t
rr
t
eˆ)sinω(sinlim
0
φφ
θ
=





∆
∆
→∆
Since this expression is equal to the cross product of ωωωω
and r, we conclude that due to the rotation of the vector ωωωω of
the coordinates the vector undergoes a rate of change of:
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7

This ωωωω××××r term occurs in addition to the vector change relative to the coordinate
system, so that the total derivative relative to inertial axes (i.e., the reference frame) is:
rω
rr
××××++++
scoordinate
toRelative
Inertial






=





td
d
td
d
This equation applies to any vector quantity and is of fundamental importance to
dynamics where body-fixed axes are often used.
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Problem: a) If v is the velocity of a particle in a rigid body whose angular velocity is ωωωω,
show that:
ωvkjiv 2ˆˆˆ =





∂
∂
∂
∂
∂
∂
= ××××++++++++××××∇∇∇∇
zyx
b) If the velocity and acceleration of a particle in motion are given by:
show that:
∫=
t
tdtt
0
)()0()( avv ++++
2
2
td
d
td
d
td
d rv
a
r
v === and
and:
∫=
t
tdtt
0
)()0()( vrr ++++
8

As an example of sorts, let us consider the hydrodynamic concepts of streamlines,
pathlines and streaklines. Streamlines represent the loci (i.e., a locus – plural: loci – is a
set of points whose location is determined by one or more specified conditions…) that
are tangent to the velocity vectors in the flow field at a given instant of time (i.e., the
condition!) The Figure shows several of these lines with the tangent velocity vectors v.
There is no limit to the number of streamlines one may draw in a given flow field, since
they are lines and have no thickness. If the streamlines are to be tangent to the velocity
vectors, then the differential equation determining these lines is:
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Streamlines at a given instant.
Remembering that the result of the cross product is
expressed by the determinant:
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kji0
kji
kjikji
v
ˆ)(ˆ)(ˆ)(
ˆ0ˆ0ˆ0
ˆˆˆˆˆˆ
det
xdvydvxdvzdvydvzdv
zdydxd
vvv
zdydxd
vvvd
yxzxzy
zyxzyx
−−−=
=≡










=
++++−−−−
++++++++×××× l
r
y
x
z
dllll v
Streamlines
0v =l
r
d××××
where dllll is a small displacement vector along a streamline. Choosing Cartesian
coordinates x, y, z and inserting the velocity and displacement in the above we get:
0kjikji =)ˆˆˆ()ˆˆˆ( zdydxdvvv zyx ++++++++××××++++++++
zyx v
zd
v
yd
v
xd
==
which gives us after equating like terms in i, j and k:
9
ˆ ˆ ˆ

Solutions of the system of equations given by dx/vx =dy/vy =dz/vz requires the
knowledge of the velocity components of the fluid particles as a function of the space
coordinates and time. Integration of the system of differential equations yields the
equation for the streamline family. Since physical significance is not ascribed to a
velocity with two different directions, streamlines do not intersect in a physical problem.
The only exception to this is at a location where the velocity is zero, since zero or null
vectors, 0, has no inherent direction. Points in a fluid flow (this comprises the definition
of a field) where the velocity is zero are referred to a stagnation points or critical points.
A pathline is defined as the locus or trajectory followed by any given fluid particle for a
given time interval. The path of the particle initially at ro is given by:
),( o trrr =
The velocity of the fluid particle is given by the time derivative of the position vector, that
is:
t
t
td
d
∂
∂
==
)],([ orrr
v
where the partial derivative of the preceding equation indicates that differentiation is to
be carried out for a given particle, that is, holding ro constant. The equation for pathlines
are then the solutions obtained by integrating dr/dt = v(r,t) and setting r=ro at t=0.
In fluid mechanics you might also hear of streakline which is identified by the ‘string’
of fluid particles that has passed through a given point in space at a given instant of
time.
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10

Exercise: The components of a velocity field are given by:
Solution: Using dx/vx =dy/vy (vz =0) and substituting for the velocity components, one
obtains:
( )00e 21
3
≥=== −
tvyCvxCv zy
tC
x forand,
Find the equation that represents the family of streamlines when C1 =C2 =C3 =1. Each of
these constants has a dimension of reciprocal time.
tCtC
x
y
xd
yd
y
yd
x
xd
33
ee −−
=⇔=
The preceding expression may be integrated with time t considered constant since
streamlines describe the tangent loci to the velocity vectors at a given instant of time.
Carrying out this integration gives:
t
xCy e
=
in which C is an integration constant. The last equation represents the streamline family
corresponding to the given velocity field. It should be noted that all of the streamlines
pass through the origin at the same instant of time.
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From what has been stated about the intersection of streamlines at a given instant of
time, one might suspect that the origin is a stagnation point. Examination of the given
velocity components indicates that the velocity is zero at the origin. A specific set of
streamlines may be found by obtaining the integration constant required by the
coordinates of some point.
11

Exercise: For the velocity field of the previous Exercise, what is the pathline for the fluid
particle coincident with the point xo, yo at time t=0?
Solution: To find a pathline equation of the form r=r(ro,t), a description of the velocity
field in the form of v= dr/dt = ∂[r(ro,t)/∂t] is considered in which the coordinates of the
fluid particle are expressed as functions of time. This leads to the following differential
equation for x(t), y(t) and z(t):
0e ====== −
zy
t
x v
td
zd
yv
td
yd
xv
td
xd
and,
in which it is understood that x and y are coordinates of a given particle and as such are
functions of time. Integrating these equations yields:
)(ee e1
o
0o
txxxtd
x
xd tt
t
x
x
===
−
−−
∫∫ or
and:
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)(eo
0o
tyyytd
y
yd t
ty
y
=== ∫∫ or
The desired pathline may be found from the parametric equations for x(t) and y(t) by
elimination of the parameter time by using the first equation and rearranging it so that
ln(x/xo)=1−exp(−t) and using exp(−t)=1/[1−ln(x/xo)] in the second equation. The result is:
)(ln1 o
o
xx
y
y
−
=
12

Exercise: Obtain a streamline through the point xo, yo, and the pathline of the fluid
particle coincident with the point xo, yo at time t=0, if the velocity field is given by:
Solution: For the streamline:
yC
yd
xC
xd
v
yd
v
xd
C
yx 21 e
== −
and
where C is a dimensionless constant, and constants C1 and C2 each have a value of 1
and dimension of reciprocal time.
and for the given point xo, yo the solution is:
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t
ty
y
t
t
C
x
x
yytd
y
yd
xxtd
x
xd C
eee o
0
e
o
0 oo
==== ∫∫∫∫
−
−
orandor
For the pathline:
C
x
x
yy
e
o
o 







=
kjiv ˆ)0(ˆ)(ˆ)e( 21 ++++++++ yCxC C−
=
Eliminating t between the two parametric equations results in:
C
x
x
yy
e
o
o 







=
which is the same as the equation for the streamline (i.e., same locus for both v fields).
13

Exercise: For the velocity field given in the first Exercise, find the equation of the
streakline passing through the point given by r=ai+bj.
Solution: The parametric equations for the pathline in the second Exercise are x=
xoexp[1−exp(−t)] and y=yoexp(t). Recalling that xo and yo are material coordinates of the
fluid particle at t=0, then as they are assigned different pairs of values at t=0, the pairs
identify different fluid particles. Solving these x and y equations for material coordinates:
The fluid particle at the position r=ai+bj at time τ is given by:
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and these are the parametric forms of the vector equation ro =ro[(a,b),τ ]. To find the
equation of the streakline, it is necessary to substitute the values of xo and yo into the
pathline equations. The desired result is:
This solution is of the form r=r{ro[(a,b),τ ],t}. Plotting the locus of the streakline for a
given t requires the calculation of x and y, which are the space coordinates for values of
τ ranging from zero to the value of t. Upon simultaneous solution of the above equations
for x and y, the following equation for the streakline is obtained:
τ−






−=





e1ln
y
b
a
x
ˆ ˆ
t
yyxx
t
−+−
==
−
ee o
e1
o and
ˆ ˆ
( )tbyax ≤≤== −+− −
τττ
0ee o
e1
o forand
ττ
−+−−
==
−−
eeee e1e1 t
byax
t
and
14

dr
r
••••
O P
Field f (e.g.,
scalar field )
Γ
∇∇∇∇f
θ
dr
rd
fd
The vector ∇∇∇∇f is called the gradient of f (r) (and is often written grad f ). Differentiating
both sides of the above equation with respect to r, we obtain the directional derivative:
Hence the gradient of f is a vector whose component in any direction, dr, is the deriva-
tive of f with respect to r. Note that when ∇∇∇∇f is parallel to dr, d f has its maximum value.
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( )rr
rrr
∇∇∇∇∇∇∇∇∇∇∇∇ ≡•=
∂
∂
+
∂
∂
+
∂
∂
= dfzd
z
f
yd
y
f
xd
x
f
fd
)()()(
Illustration of the physical meaning of the
gradient,∇∇∇∇f , of a field f (r) and ∇∇∇∇f•dr/dr is just the
scalar components of ∇∇∇∇f in the direction of dr.
Let f (x,y,z) be a single-valued scalar function with continuous first derivatives in a
certain region of space. If r is the position vector of an object located at P(x,y,z), we
have r=xi++++yj++++zk and dr=dxi++++dyj++++dzk (see Figure). The total derivative of f (r) is:ˆ ˆ ˆ ˆ ˆ ˆ
rd
d
f
rd
fd rr
•= ∇∇∇∇
)(
The scalar operator ∇∇∇∇•∇∇∇∇≡∇2 ≡ is also useful in physics;
it is called the Laplacian operator. In equation form, it is
(e.g., in Cartesian coordinates):
The multiplication operation involving the ∇∇∇∇ operator is
extremely useful in physics. The two vector products
denoted by ∇∇∇∇•F≡divF and ∇∇∇∇××××F≡curlF are called the
divergence of F and the curl of F, respectively.
2
2
2
2
2
2
2
zyx ∂
∂
+
∂
∂
+
∂
∂
=•≡∇≡ ∇∇∇∇∇∇∇∇
since i•i=j•j=k•k=1 and i•j=i•k= j•k =0.ˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆˆ ˆ ˆ
Vector Calculus
Streamlines
15

whereas the curl of the same vector F is (ibid):
The divergence of a vector F is given by (e.g., for Cartesian coordinates):
If at some point P:
Fdiv
r
≡
∂
∂
+
∂
∂
+
∂
∂
=•
z
F
y
F
x
F zyx
F∇∇∇∇
Fˆˆˆ
ˆˆˆ
ˆˆˆ
r
curlkji
kji
kji
F
y
≡







∂
∂
−
∂
∂






∂
∂
−
∂
∂








∂
∂
−
∂
∂
=
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=
∂
∂
∂
∂
∂
∂
=
y
F
x
F
z
F
x
F
z
F
y
F
FF
yx
FF
zx
FF
zy
FFF
zyx
xxzyz
yxzxzy
zyx
++++−−−−
++++−−−−××××∇∇∇∇
The physical interpretation of the curl of a vector is connected with the rotation (or
circulation) of a vector field such as V.
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0
0
0
0
alirrotation
solenoidal
sink
source
asclassifiedisthen
betosaidisthen
atahasthen
atahasthen
VV
V
V
V
V
=





=
<
>
•
××××∇∇∇∇
∇∇∇∇ P
P
16

To find a possible interpretation of the curl of a vector, let us consider a body rotation
with uniform angular speed ω about an axis z. Let us define the vector angular velocity
ωωωω to be the vector of length ω=|ωωωω| extending along z in the direction in which a right-
handed screw would advance if subject to the same rotation as the body. Finally, let r be
the vector drawn from any point P(r) on the axis z to an arbitrary point P in the body (see
Figure – from which we see that the radius at which P rotates is |r|⋅|sinϕ|.) The linear
speed of P is thus v=|v|=ω|r|⋅|sinϕ|=|ωωωω|⋅|r|⋅|sinϕ|=|ωωωω××××r|. The vector velocity v is
directed perpendicular to the plane of ωωωω and r, so that ωωωω, r, and v for a right-handed
system. Hence, the cross product ωωωω××××r gives not only the magnitude of v but the
direction as well.
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A physical interpretation of the curl of a vector.
If we now take the point O as the origin of coordinates, we
can write r=xi ++++ yj ++++zk and ωωωω=ω1i ++++ ω2j++++ ω3k. Hence, the
equation v = ωωωω××××r can be written at length in the form v =(ω2z
−ω3y)i −−−− (ω1z−ω3x)j ++++(ω1y−ω2x)k. If we take the curl (i.e.,
∇×∇×∇×∇×…) of v, we therefore get:
ˆ ˆ ˆ ˆ ˆ ˆ
Expanding this, remembering that ωωωω is a constant vector, we
find∇×∇×∇×∇×v=2ω1 i ++++2ω2 j ++++2ω3 k =2ωωωω which implies that ωωωω= ½∇×∇×∇×∇×v.
The angular velocity of a uniformly rotating body is thus
equal to one-half the curl of the linear velocity of any point of
the body. The name curl in this context is now apparent.
ˆ ˆ ˆ
ˆ ˆ ˆ
ωωωω
r
ϕ
v
O
z
P
|r|⋅|sinϕ|
ω
xyxzyz
zyx
213132 ωω)ωω(ωω
ˆˆˆ
−−−−
∂∂∂∂∂∂=
kji
v××××∇∇∇∇
17

As a general example (and refresher for those new to this vector stuff) of how to use
vectors in calculations, consider two vectors, A=i++++2j++++3k and B=3i++++2j++++k and find:ˆ ˆ ˆ ˆ ˆ ˆ
74.31412374.314321 222322
==++===++= BA and
Magnitude of A and B:
kjikjikjikjiBA ˆ4ˆ4ˆ4ˆ)13(ˆ)22(ˆ)31()ˆˆ2ˆ3()ˆ3ˆ2ˆ( ++++++++++++++++++++++++++++++++++++++++ =+++==
Addition of A and B (or of B and A):
kikjikjikjiBA ˆ2ˆ2ˆ)13(ˆ)22(ˆ)31()ˆˆ2ˆ3()ˆ3ˆ2ˆ( ++++++++++++++++++++−−−−++++++++−−−− −=−−−==
Subtraction of B from A (but not A from B – in that case B−−−−A=−(A−−−−B)):
10343)13()22()31()ˆˆ2ˆ3()ˆ3ˆ2ˆ( =++=⋅+⋅+⋅=•=• kjikjiBA ++++++++++++++++
Scalar product of A with B (or of B with A):
kjikji
kji
BA ˆ4ˆ8ˆ4ˆ)]32()21[(ˆ)]33()11[(ˆ)]23()12[(
123
321
ˆˆˆ
−−−−++++++++−−−−×××× −=⋅−⋅⋅−⋅⋅−⋅==
Vector product of A with B (but not B with A – in that case B××××A=−A××××B):
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kjikji
kji
A
A
A ˆ80.0ˆ53.0ˆ27.0ˆ
14
3ˆ
14
2ˆ
14
1
14
ˆ3ˆ2ˆ
ˆ ++++++++++++++++
++++++++
====
A unit vector A in the direction of A:ˆ
18

y
x
zjˆ
iˆ
kˆ
ττττ
F
r
θ
mN⋅=+=−⋅−⋅=
−
== iikji
kji
Fr ˆ5ˆ)41(ˆ0ˆ0ˆ)]22()11[(
120
210
ˆˆˆ
++++−−−−××××
The angle θ between A and B:
)4.44(714.0
14
10
cos10cos14cos1414cos o
or==⇒====• θθθθBABA
A unit vector n perpendicular to both A and B :
since r =x i++++yj++++zk and length |∇∇∇∇r2|=2r and direction ê(∇∇∇∇r2)=êr (i.e., the direction of r).
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ˆ
kji
kjikjikji
BA
BA
n ˆ
6
1ˆ
6
2ˆ
6
1
616
ˆ4ˆ8ˆ4
96
ˆ4ˆ8ˆ4
)4(8)4(
ˆ4ˆ8ˆ4
ˆ
222
−−−−++++
−−−−++++−−−−++++−−−−++++
××××
××××
−=
⋅
−
=
−
=
−++−
−
==
Another example, this time physical, would be to calculate the moment (i.e., torque ττττ)
of the force F=(−−−−2j+k)N about the origin if the force acts at a point [0,1,2]m.
Here we have r=(j++++2k)m so the moment in this case is:
ˆ ˆ
ˆ ˆ
The gradient of a scalar function φ(r)=φ(r)= r2 =x2 +y2 +z2 (called a 3-sphere):
ˆ ˆ ˆ
)90()0(cos0cos5cos55cos
022)1)(2()2)(1(
1 o
or−
=⇒====•
=+−=+−=•
θθθθFrFr
Fr
Since the scalar product of the two vectors r and
F is 0 this means the two vectors are
perpendicular (i.e. 90°) to each other!
rkjikji 2ˆ2ˆ2ˆ2ˆ)(ˆ)(ˆ)(
)( 2
222
=++=⇒
∂
∂
+
∂
∂
+
∂
∂
= zyxr
z
r
y
r
x
r
r ∇∇∇∇∇∇∇∇φ
19

P(−1,−1,0)
nˆ
x
y
z
As an example,letusevaluatethecurlofafield φ orientedinthex-direction: ∇∇∇∇××××iφ(x,y,z).
In this case, we have (e.g., use the formula for ∇∇∇∇××××F and substitute F for iφ=φφφφx):
φ
φφφφ
φ
φ 





∂
∂
∂
∂
=
∂
∂
∂
∂
=





∂
∂
−





∂
∂
−−=
∂
∂
∂
∂
∂
∂
= kjkjkji
kji
i ˆˆˆˆˆ0ˆ0ˆ)00(
00
ˆˆˆ
ˆ
zzzzzzzyx
−−−−−−−−++++−−−−××××∇∇∇∇
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To plot this surface (or any other curve) visit
www.wolframalpha.com and type x^2+y^2-z=1.
You should obtain this ‘infinite paraboloïd’.
kji
kji
kji
ˆˆ2ˆ2
ˆ)(ˆ)(ˆ)(
ˆˆˆ
222222
−−−−++++
++++++++
++++++++∇∇∇∇
yx
z
zyx
y
zyx
x
zyx
zyx
=
∂
−+∂
∂
−+∂
∂
−+∂
=
∂
∂
∂
∂
∂
∂
=
φφφ
φ
Assume we have a field φ(x,y,z)=1 =x2 + y2 − z; therefore:
As another example, let us find a unit vector that is normal to the surface x2 + y2 − z=1
at the coordinate P(−1,−1,0).
Now, all that remains is to evaluate ∇∇∇∇φ at P(−1,−1,0):
kjikji ˆˆ2ˆ2ˆˆ)1(2ˆ)1(2),,( )0,1,1(
−−−−−−−−++++∇∇∇∇ −=−−−=−−P
zyxφ
then find the unit normal vector n to φ(−1,−1,0):ˆ
kji
kjikji
n ˆ
3
1ˆ
3
2ˆ
3
2
3
ˆˆ2ˆ2
9
ˆˆ2ˆ2
ˆ −−−−−−−−
−−−−−−−−−−−−−−−−
∇∇∇∇
∇∇∇∇
−=
−
=
−
==
φ
φ
ˆ
ˆ
20

zd
z
J
J z
z
∂
∂
+→→→→→→→→zJ
Vd
qd
=ρ
y
x
z
dx
dy
A
B
C
D
E
F
G
H
An infinitesimal quantity of charge dq
A unit volume dV= dxdydz
dV J
dz
vq
Fluid density
z-component flux
As yet another example, physical this time, let us establish a physical meaning for the
divergence of a vector J (i.e., ∇∇∇∇••••J) by use of an illustration from hydrodynamics.
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Consider the flow indicated in the Figure (a flow occurring from the left face to the right
face and reduced to Fz) and let J(r)=ρ(r)v(r), that is, J represents the mass of fluid (i.e.,
the mass flux density) flowing through a unit area normal to side ABCD per unit time.
The density of the fluid is denoted by ρ, and v is its velocity, both functions of position, r.
The z-component of J through the area ABCD (i.e., using the product of the differentials
dx and dy to indicate a surface element) indicated in the Figure per unit time is given by:
ydxdJz
The flow through the area EFGH per unit time may be represented by the following
(Taylor series) expansion:
ydxdzd
z
J
zJydxddzzJ z
zz 





+
∂
∂
+=+ K)()(
where we neglect higher-order terms in this expansion (i.e., the ‘+…’ terms).
N.B., The ABCDEFGH box should
actually be within the limits of the Jz
‘stream’. For clarity, it is magnified.
21

The net increase in the mass of the fluid inside the volume element dV=dxdydz per
unit time due to the flow through the two opposite faces (of the cube in the Figure) is:
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Vd
z
J
zdydxd
z
J
ydxdJydxdJydxdzd
z
J
JydxdJ zz
zz
z
zz
∂
∂
−=
∂
∂
−−=





∂
∂
+−
Similarly, the net increase in the mass of fluid per unit time due to the flow through
BFDH and AECG is:
Vd
y
J
zdydxd
y
J
zdxdJzdxdJzdxdyd
y
J
JzdxdJ
yy
yy
y
yy
∂
∂
−=
∂
∂
−−=







∂
∂
+−
and that through CDGH and ABEF is:
Vd
x
J
zdydxd
x
J
zdydJzdydJzdydxd
x
J
JzdydJ xx
xx
x
xx
∂
∂
−=
∂
∂
−−=





∂
∂
+−
In conclusion, the total increase in the mass of the fluid (as calculated above) per unit
volume (i.e., dV) per unit time due to the excess of inward flow over the outward flow is:
Jkjikji •−=•





∂
∂
∂
∂
∂
∂
−=







∂
∂
+
∂
∂
+
∂
∂
−=








∂
∂
−
∂
∂
−
∂
∂
−
∇∇∇∇++++++++++++++++ )ˆˆˆ(ˆˆˆ
zyx
zyx
zyx
JJJ
zyxz
J
y
J
x
J
Vd
Vd
z
J
y
J
x
J
which is just the rate of increase of the density (i.e., dρ/dt) of the fluid inside of the
volume element dV.
22

The last expression is equivalent to the physical principle:
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t∂
∂
=•−
)(
)(
r
rJ
ρ
∇∇∇∇
The above equation is called the continuity equation, it describes the transport of a
conserved quantity – in this case the density ρ (i.e., mass per unit volume). The
continuity equation is also a stronger, more local form of a conservation law (i.e., nothing
is created and/or nothing is lost in the flow of matter in a ‘closed’ physical system).
t∂
∂
−=•+•=•
ρ
ρρρ vvv ∇∇∇∇∇∇∇∇∇∇∇∇ )(
where we used the identity ∇∇∇∇•(φA)=A•∇∇∇∇φ +φ∇∇∇∇•A.
Substituting J=ρv, we get:
)(00 solenoidor =•=•=
∂
∂
vJ ∇∇∇∇∇∇∇∇
t
ρ
and so it is also required that ∇∇∇∇ρ =0. In this case, the excess or outward flow over
inward flow is zero. In other words, there is no noticeable compression or expansion of
the fluid.
For an incompressible fluid (i.e., a fluid in which there is no change in density of the
fluid particle as its motion is followed) we have:
23

kji ˆˆˆ
zyx vvv ++=v
A frequent need for differentiation in fluid mechanics arises where the ‘change’ in
properties of a fluid particle is desired. While following the particle, such operations are
referred to as differentiation following the motion of the particle, and the derivatives
obtained are referred to as the material derivatives.
A typical fluid particle with two different positions
(for the time interval ∆t).
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To form this derivative, consider a scalar quantity, such as density, expressed as a
function of three space coordinates x, y, z and time t:
]),(),(),([ ttztytxρρ =
Each of these space coordinates is a function of time since the particle is in motion.
During the time interval ∆t, the particle moves from its initial position x, y, z at t to point P
at x+∆x, y+∆y, z+∆z and the density changes by ∆ρ which is approximately given by:
t
t
z
z
y
y
x
x
∆
∂
∂
+∆
∂
∂
+∆
∂
∂
+∆
∂
∂
≅∆
ρρρρ
ρ
The change in each of the space coordinates may be
expressed in terms of the respective velocity components vx,
vy, and vz as shown in the Figure; and these expressions
may be substituted into the expression for ∆ρ to give:
t
t
tv
z
tv
y
tv
x
zyx ∆
∂
∂
+∆
∂
∂
+∆
∂
∂
+∆
∂
∂
≅∆
ρρρρ
ρ
Dividing the receding expression by ∆t and limiting the result
as ∆t approaches zero yields:
t
v
z
v
y
v
xt
zyx
t ∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
=





∆
∆
→∆
ρρρρρ
0
lim
24
x
z
y
(x, y, z, t)
v
(x + ∆x, y + ∆y, z + ∆z, t + ∆t)
∆x = vx ∆t
∆y = vy ∆t
∆z = vz ∆t
t + ∆t
t P

The limit on the left-hand side of the last equation is noted by Dρ/Dt; it is called the
material derivative of the quantity ρ since the particle (material) identity was held cons-
tant during the differentiation process, the material derivative may now be written as:
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{
LocalConvective
t
v
z
v
y
v
xtD
D
zyx
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
=
ρρρρρ
4444 34444 21
The terms labeled ‘convective’ are so named since each of these represent a change
associated with the particle being ‘conveyed’ or changing position. The last term is
referred to as the ‘local’ change since it represents the change in ρ due to any
‘unsteadiness’ at the point x, y, z. In this respect, this equation above is remarkable in
that it relates a change of a quantity associated with a given particle (material identity
held constant) to a change of the same quantity with locality (position in space) held
constant – the connection between the two being the convective change. This statement
may be emphasized by differentiation of the given expression for ρ while the fluid
particle is held constant. This operation may be carried out by recognizing that:
]),,([),( o ttt rrr ρρρ ==
which simply means that the value of the density at a given position in space at a
given time, ρ(r,t), is the value of the density of the fluid particle at this position at the
given time, ρ[r(ro,t),t], a statement that applies to any quantity (as well as density)
that can be attributed to the fluid through the continuum idealization.
25

Let us look at another scenario…
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constantheldidentity
particleFluidparticlefluid
Flowing
,or
ttd
d
tD
D
∂
∂
==
ρρρ
Using the last functional expression (i.e., ρ=ρ(r,t)=ρ[r(ro,t),t]):
]),,(),,(),,([]),,([ oooo ttztytxtt rrrrr ρρ =
Recognizing that the independent variables are ro and t, and employing the rules of
partial differentiation, the following is obtained:
tt
z
zt
y
yt
x
xt ∂
∂
+





∂
∂
∂
∂
+





∂
∂
∂
∂
+





∂
∂
∂
∂
=
∂
∂ ρρρρρ
oooo rrrr
Now the derivatives (∂x/∂t)ro
, (∂y/∂t)ro
, and (∂z/∂t)ro
represent the respective velocity
components of the particle ro, so that:
t
v
z
v
y
v
xtD
D
zyx
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
=
ρρρρρ
which is that same material derivative as obtained earlier.
The differentiation may be indicated alternatively as:
This second method of obtaining Dρ/Dt certainly gives meaning to the name
material derivative.
26

In the analysis of fluid motion, the forces acting on a fluid particle are related to the
acceleration through the use of Newton’s law of motion; and this requires an expression
of acceleration (or differentiation of the velocity vector) following the motion of the fluid
particle. It is then obvious that the desired derivative would be Dv/Dt. The acceleration,
following the fluid particle, is:
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t
v
z
v
y
v
xtD
D
zyx
∂
∂
∂
∂
∂
∂
∂
∂
==
vvvvv
a ++++++++++++
Exercise: Following the fluid particle, calculate the y component of acceleration for a
particle whose velocity vector is given by v=(3z−x2)i++++yt2j++++xz2k in ft/sec at the point x=
1 ft, y=1 ft, z=9 ft, and t=2 sec.
ˆ ˆ ˆ
Solution: The y component of acceleration following the fluid is:
tyty
tyzxttyxz
t
ty
z
ty
zx
y
ty
ty
x
ty
xz
t
v
z
v
v
y
v
v
x
v
v
tD
vD
tzyxa
yy
z
y
y
y
x
y
y
2
2)0()()0)(3(
)()()()(
)3(
),,,(
4
42222
22
2
2
2
2
2
+=
+++−=
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−=
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
==
so that:
ft/secsecftftft 20)2)(1(2)2)(1()2,9,1,1( 4
=+=ya
27

Exercise: A two-dimensional velocity field is given by:
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MRT
jiv ˆ)3(ˆ)2( 22
xxyyx +−= ++++
Solution: a) Since the velocity vector v(x,y) above is not a function of time, the field is
steady. There is no need to form ∂v/∂t; it is obviously zero.
]ˆ3ˆ[)3(]ˆ)23(ˆ4)[2(
]ˆ)3(ˆ)2[(
)3(
]ˆ)3(ˆ)2[(
)2(
22
22
2
22
2
jiji
00
jiji
vvvvv
xxxyxyxyx
y
xxyyx
xxy
x
xxyyx
yx
tz
v
y
v
x
v
tD
D
zyx
+−+++−=
+
∂
++−∂
+
∂
++−∂
−=
∂
∂
∂
∂
∂
∂
∂
∂
=
++++
++++++++
++++++++++++
a) Is this field steady? b) Obtain an expression for the material derivative of v.
b) With the definition of the material derivative, applied to the velocity vector above, we
have:
ji
v ˆ)31527(ˆ)78( 22323
yyxyxxxyxx
tD
D
−+−−−= ++++
and carrying out the algebra we finally get:
28
:78 23
xyxx −− :31527 223
yyxyxx −+−
y
xx
y

∆V
M
fB
FS∆S
∆Sn
V
Drop of water
held together
because of
surface tension.
A fluid is defined as any substance deforming continuously when subjected to a shear
stress regardless of how small the shear stress may be. This means that fluids will ‘flow’
when subjected to a shear stress; and, conversely flowing fluids will generally exhibit the
presence of shear stresses. A detail discussion of hydrodynamics is coming soon…
To really understand the motion of fluids we really need to understand what forces are
involved in producing and maintaining the motion. This is a consequence of applying
Newton’s laws of motion since these laws contain force explicitly. As a result, ‘stress’ will
be introduced to see how this effects the forces that are transmitted through a fluid
continuum.
Control volume V with surface ∆S and body
forces (i.e., fB, FS, &c.) acting on typical
elements. fB is the body force (expressed in
‘per unit mass’ – i.e., fB =FB/mB).
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The best place to start is with the surface and body forces.
Consider a fluid instantaneously occupying the control volume
V shown in the Figure. A force FS is exerted by the
surroundings on ∆S, which is a typical small element of area
on the control surface; M is a moment (i.e., couple) exerted by
the surroundings on ∆S. The force FS may be replaced by two
components, one normal to ∆S, Fn, and one tangential element
to ∆S, Ft. A normal stress is defined by the expression:
σ=
∆→∆ S
F
S 0
lim
M
Ft
Fn
FS∆S
and a shear stress by:
τ=
∆→∆ S
F
S 0
lim
Note the nn and nt subscripts. M does not appear in the defini-
tion of stress since ∆S approaches zero in the limiting process.
Fluid Mechanics
29
We characterize a behavior by considering the average,
or macroscopic, value of a quantity of interest, where the
average is evaluated over a small volume containing a
large number of molecules. The spacing between
molecules is typically very small – 10−6 mm for gases
and 10−7 mm for liquids. The number of molecules per
cubic millimeter is on the order of 1018 for gases and 1021
for liquids!
nn
nt
n
t

Now, let us consider the force due to gravity acting on an element of volume V.
Obviously we need to interpret ρ as the density of the fluid which the mass per unit
volume. Mathematically:
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Infinitesimally:
V
m
=ρ
∫∫∫∫ ==
Vm
BB dVmd ρgfF
where the integration is carried over some small element of volume V. Forces of this
nature are termed body forces.
Since here, fB, the body force per unit mass at a point, is assumed to be the Earth’s
gravity field vector, g, so that the body force is defined as:
mdVdVdmd
ρ
ρ
1
=⇔=
says that an infinitesimal variation in mass generates an infinitesimal change in volume
– an vice versa – provided the density of the material (or fluid) remains constant during
the change (which can be due to temperature and/or pressure, e.g., if related to the ideal
gas law, p=ρRT with ρ =n/V, moles of substance per volume, and R is the gas constant).
Note that the gravity field g=−9.81 m/s2k is valid up to a certain height/altitude
beyond which it is no longer considered a constant. In general the density ρ is
always a constant of the material.
ˆ
30

Hdz
z
g
zdAdVmd ρρ ==
0
Exercise: The column of gas shown has a density distribution given by ρ=ρoexp(−Cz) in
which ρo is the density of the gas at a position given by z=0, and C is a constant with the
dimensions of reciprocal length. The cross-sectional area A of the column (perpendicular
to the z-axis) is constant with respect to z. Neglecting variations of the acceleration due
to gravity (g) with height, calculate the weight of a column of height H of this gas.
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Solution: The weight of the column of gas is the body force due to gravity. To calculate
this body force acting on the entire column, the force acting on a small typical element is
integrated over the mass of gas in the column. The force acting on the element shown,
which has a small height dz, is given by dFB=−gkdm=−gk A[ρoexp(−Cz)]dz in which A is
the constant cross-sectional area. Integration over the mass of the gas may be effected
by allowing z to range from zero to H. This yields:
kkFF ˆ)e1(eˆ o
0
o
HC
H
zC
m
BB
C
Ag
zdAgd −−
−−=−== ∫∫
ρ
ρ
The negative sign introduced into the problem as −gk indicates the force is acting in
the negative direction, that of gravity. Density variations frequently occur in the
Earth’s atmosphere. There are due to partial absorption of energy transmitted
through the atmosphere and the fact that the air closer to the Earth is compressed
by the weight of the air more remote from the Earth’s surface.
ˆ
ˆ ˆ
31

Consider a small element of fluid in a flow field with positive coordinate directions
indicated by the Figure where a face is considered positive or negative according to the
direction of its outward-drawn normal with respect to the coordinate direction. A stress is
positive if it produces a force that acts in the positive coordinate direction on a negative
face, or if it produces a force that acts in a negative coordinate direction on a negative
face. By way of illustration, consider the two stresses σxx and σxx +(∂σxx /∂x)∆x in the
Figure. The one indicated by an arrow in the positive x direction acts on a face that has
an outward-drawn normal in the positive x-direction. This is a positive stress. The other
stress, designated by σxx, is indicated by an arrow in the negative x-direction, and the
direction of the outward-drawn normal to the face over which it acts is in the negative x-
direction. This stress is also positive by definition.
Element in a fluid flow field with several
stresses (stress σ and shear τ ) indicated.
x
y
z z
z
zz
zz ∆
∂
∂
+
σ
σ
xxσ
yxτ
zxτ
x
x
xx
xx ∆
∂
∂
+
σ
σ
y
y
yy
yy ∆
∂
∂
+
σ
σ
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Now, if the direction of the force produced by a stress is posi-
tive (or negative) and the stress acts on a negative (or positive)
face, then the stress is defined as a negative stress. According
to these definitions, negative normal stresses are compressive,
and positive normal stresses are tensile. The state of stress at a
given point in a continuum may be specified by a normal stress
and two shear stresses acting in given coordinate directions on
each of three mutually perpendicular planes intersecting at a
given point. This means that nine components are required to
describe the most general state of stress. These nine quantities
are referred to as the components of the stress tensor σij where
the indices i and j take on values ranging over all coordinates
(i.e., we have σσσσxx,σxy, σxz, σyx, σσσσyy, σyz, σzx, σzy, and σσσσzz).
32

The element shown in the Figure is subject to surface forces due to stresses shown. It
is also subjected to a body force field (e.g., gravity) designated by fB. The direction
cosines of the outward-drawn normal to the oblique face are l, m, and n with respect to
the x, y, and z axes, and the area of the oblique face is A. The problem consists of
expressing σ, τ, and the direction cosines of τ (designated l, m, and n, respectively) in
terms of the given quantities l, m, and n, and the components of the stress tensor.
Summing forces in the x, y, and z directions, by using mean values of stresses with the
appropriate areas and applying Newton’s law of motion to the element, gives:
tD
vDzyx
nA
yxzxzy
nA
zyx
tD
vDzyx
mA
yxzxzy
mA
zyx
tD
vDzyx
lA
yxzxzy
lA
zyx
z
zzzyzxzB
y
yzyyyxyB
x
xzxyxxxB





 ∆∆∆
=+




 ∆∆
−




 ∆∆
−




 ∆∆
−+




 ∆∆∆





 ∆∆∆
=+




 ∆∆
−




 ∆∆
−




 ∆∆
−+




 ∆∆∆





 ∆∆∆
=+




 ∆∆
−




 ∆∆
−




 ∆∆
−+




 ∆∆∆
62226
)(
62226
)(
62226
)(
ρτσττσρ
ρττστσρ
ρτττσσρ
f
f
f
Element of fluid used for obtaining the resultant shear stress τ and normal stress σ on a plane of arbitrary orientation.
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yyσ
x
z
y
∆z∆x
xxσzzσ
yzτ
xzτ
yxτ
zxτ
xyτ
zyτx
y
z
τ
σ
fB
∆y
Rotation of
135° around
the y-axis
O
O
33

You will notice that we have included the material derivative (i.e., D/Dt) in which Dvx/Dt,
Dvy /Dt, and Dvz /Dt are the acceleration components of the center of mass of the
element of fluid.
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yxnAzxmAzylA ∆∆=∆∆=∆∆=
2
1
2
1
2
1
and,
The area of the oblique face, A, may be expressed as a function of its projection on the
coordinate planes as follows:
Substituting these relations into the first of our last set of equations gives:
tD
vDxlA
lAnAmAlAlA
xlA x
xzxyxxxB 




 ∆
=+−−−+




 ∆
33
)( ρτττσσρf
If this equation is divided by A, and then limited by letting ∆x→0, ∆y→0, and ∆z→0 in
such a way that the orientation of the oblique face is preserved, then the mean values of
the stresses in an area take on values of stress at the point O (i.e., here the origin); and:
0=+−−− lnmll xzxyxx τττσσ
By a similar reasoning, we get for the last two of our last set of equations:
0=+−−− mnmlm yzyyyx ττστσ
and:
0=+−−− nnmln zzzyzx τσττσ
These terms containing the body force and accelerations go to zero in the limit.
34

This results in three equation in five unknowns σ, τ, l, m, and n. The remaining two
equations come from geometric considerations, and these are:
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1222
=++ nml
and:
0=++ nnmmll
The first equation is the Pythagorean theorem centered around a unit radius and the last
equation results from the orthogonality of the orientations described by two sets of
direction cosines, l, m, n and l, m, n. All of the equations on the previous slide will then
yield the desired resultant normal and shear stresses and the direction of the resultant
shear stress for a given state of stress at a point on a plane of arbitrary orientation.
An interesting and useful result accrues if the fluid is considered in which shear
stresses are absent. With the shear stresses, τij, zero in our previous boxed equations:
000 =−=−=− zzyyxx σσσσσσ and,
or:
zzyyxx σσσσ ===
This last equation states that the normal stress is isotropic (i.e., independent of the
direction at a or from a point) in a fluid void of shear. One situation in which there are
no shear stresses in a fluid is the case of a static fluid where the pressure is taken as
the negative of the normal stress (i.e., p=−σ!) – moving as a rigid body! This would
follow from the definition of a fluid, for if there were a shear stress present, the fluid
would begin to deform continuously and would no longer remain static! Amazing!
35

Exercise: What is the pressure distribution in the tank shown if it is filled with a liquid
(e.g., stagnant water) of uniform density ρ?
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Assume that the fluid is in static equilibrium and that the only body force is gravity as
shown, and also that the surface of the fluid is open to the atmosphere.
Solution: A selection of the coordinate axes is made as shown in the Figure above.
Since the fluid is in static equilibrium, vx =vy =vz =0, for all times and at all points in the
fluid. The shear stresses are known to be zero since there is no motion of the fluid. The
absence of shear stress ensures isotropy of the normal stress so that σxx =σyy =σzz =−p.
The body force has only a z component in the negative direction, so (fB)z =−g. Our
equations specialize to ∂p/∂x=0, ∂p/∂y=0 and ∂p/∂z+ρg=0. Since p is not a function of
x or y, it must be a function of z only (i.e., no need for partial derivatives) and:
Integrating this, assuming g is independent of z, and recalling that ρ was given as
uniform, one obtains:
0=+ g
zd
pd
ρ
zgpp ρ−= o
where po is the value of the pressure at the surface (i.e.,z=0). This is the required
pressure distribution, which allows calculation of the pressure at any point in the fluid.
x y
z
g
O
36

x
y
z
zy
x
x
xx
xx ∆∆⋅






 ∆
∂
∂
+
2
σ
σ
∆y
∆x
∆z
∆x
zy
x
x
xx
xx ∆∆⋅






 ∆
∂
∂
−
2
σ
σ
zx
y
y
xy
xy ∆∆⋅






 ∆
∂
∂
+
2
τ
τ
yx
z
z
xz
xz ∆∆⋅






 ∆
∂
∂
+
2
τ
τ
zx
y
y
xy
xy ∆∆⋅






 ∆
∂
∂
−
2
τ
τ
yx
z
z
xz
xz ∆∆⋅






 ∆
∂
∂
−
2
τ
τ
(fB)x ⋅ ρ ∆x∆y∆z
In relating the forces acting on a fluid element to the motion of a fluid, a small element
is considered with surface and body forces acting on it. Application of Newton’s law of
motion relates these forces to the acceleration of the fluid element. The Figure shows an
element with stress components acting on the respective faces, and these stresses, as
well as their respective partial derivatives, refer to the center of the fluid element. The
specific body-force component, density, and the acceleration component are also
referred to the center of the element. The stresses in the Figure represent the
approximate values to be used in conjunction with the areas of the corresponding faces,
and the coordinate system shown is an inertial frame of reference.
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Fluid element selected for derivation of
equations of motion with surface forces due to
chosen positive stresses and body-force
components. Only forces in x-direction shown.
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Summing the approximate forces in the x-direction and
equating this to the product of the mass of the element and its
corresponding acceleration yields:
tD
vD
zyxzyx
yx
z
z
yx
z
z
zx
y
y
zx
y
y
zy
x
x
zy
x
x
x
xB
xz
xz
xz
xz
xy
xy
xy
xy
xx
xx
xx
xx
∆∆∆=∆∆∆⋅+
∆∆⋅






 ∆
∂
∂
−−∆∆⋅






 ∆
∂
∂
++
∆∆⋅






 ∆
∂
∂
−−∆∆⋅






 ∆
∂
∂
++
∆∆⋅






 ∆
∂
∂
−−∆∆⋅






 ∆
∂
∂
+
ρρ
τ
τ
τ
τ
τ
τ
τ
τ
σ
σ
σ
σ
)(
22
22
22
f
37

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tD
vD
zyx
x
xB
xzxyxx
ρρ
ττσ
=+
∂
∂
+
∂
∂
+
∂
∂
)(f
The equation above may be simplified by carrying out the algebraic operations
indicated and dividing through by the volume of the element, ∆x∆y∆z. Upon limiting ∆x,
∆y, and ∆z to zero, the equation becomes exact and reduces to:
for the x-direction. Similarly, we get after summing the approximate forces in the y-
direction:
tD
vD
zyx
y
yB
yzyyyx
ρρ
τστ
=+
∂
∂
+
∂
∂
+
∂
∂
)(f
tD
vD
zyx
z
zB
zzzyzx
ρρ
σττ
=+
∂
∂
+
∂
∂
+
∂
∂
)(f
and, finally, in the z-direction:
which are the equations of motion for the fluid particle.
38

zd
C
g
p
pd
−=
Exercise: What is the pressure distribution in a static isothermal atmosphere, if the
pressure, temperature, and density at sea level (z=0) are po, To, and ρo, respectively?
Assume the local acceleration of gravity to be independent of z; also, consider the
atmosphere to be a perfect gas.
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Integrating between limits z=0 (where p=po) and z in general results in:
o
eo
TRzg
pp −
=
From the equation of state for a perfect gas, C=po/ρo =RTo, where R is the specific gas
constant. The pressure distribution may be written as:
C
zg
p
p
−=







o
ln
If this Exercise would have required an answer where z is ‘very large’ then g would
have had to be expressed as a function of z before integration was attempted.
Hint: You can use p/ρ =po/ρo and p/ρ = RT for a perfect gas (c.f., Grade 12 Chemistry).
Solution: Our last boxed equations again reduce to dp/dz+ρg=0 (refer to the last
Exercise). Integration of this ordinary differential equation requires knowing the pressure
as a function of ρ, since ρ was not given as a constant. For a perfect gas at constant
temperature, it is known that p/ρ =po/ρo=C, a constant, or ρ =p/C. Substituting for ρ in
the differential equation, one obtains:
39

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When the fluid is flowing but the shear stresses are absent, we refer to the fluid field
as an ideal fluid. In this special case, our last boxed equations reduce to:
tD
D
p B
v
f ρρ =+− ∇∇∇∇
tD
vD
ztD
vD
ytD
vD
x
z
zB
zzy
yB
yyx
xB
xx
ρρ
σ
ρρ
σ
ρρ
σ
=+
∂
∂
=+
∂
∂
=+
∂
∂
)()()( fff and,
and since the absence of shear stress ensures isotropy of the normal stress compo-
nents we have the pressure being the negative of the strain applied to the fluid element:
tD
vD
z
p
tD
vD
y
p
tD
vD
x
p z
zB
y
yB
x
xB ρρρρρρ =+
∂
∂
−=+
∂
∂
−=+
∂
∂
− )()()( fff and,
These three scalar equations can be written as a single vector expression called Euler’s
equation:
If the body force is conservative (i.e., work done in moving a particle is independent of
the path) then it can be defined by the gradient of a scalar body-force potential (desig-
nate this potential by −VB), thus fB = −∇∇∇∇VB. Our equation above may now be written as:
tD
D
Vp B
v
ρρ =−− ∇∇∇∇∇∇∇∇
This helps us reduce the above equations to:
pzzyyxx −=== σσσ
40

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vdvd
t
vd
tD
D
=•
∂
∂
=• l
r
l
r vv
This last equation may be investigated under several restrictions. Consider first its
integration along a streamline when the velocity field is steady (i.e., ∂v/∂t=0). Let dllll be
a small displacement along a streamline and let us form the scalar product of the last
equation with dllll to obtain:
l
r
l
r
l
r
d
tD
D
dVdp B •=•−•−
v
ρρ∇∇∇∇∇∇∇∇
As ∇∇∇∇p represents the maximum directional change in p, ∇∇∇∇p•dllll=dp (i.e., the change in p
along l, the streamline). Then the term (Dv/Dt)•dllll may be simplified by noting that
Dv/Dt=v(∂v/∂t) because v is steady (Exercise). Since the velocity vector in a steady field
can only change along the streamline, we get:
in which v is the magnitude of the velocity (i.e., v=|v|). The equation of motion may now
be written as:
0
2
2
1
2
1
2
2
12
=+−+
−
∫ ρ
pd
VV
vv
BB
This is one form of (Daniel) Bernoulli’s (1700-1782) equation.
Transferring all terms to one side of the equation and integrating between point 1 and 2
on the streamline, one obtains:
vdvVd
pd
B =−−
ρ
41

2017
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0)(
2
2
1
12
2
1
2
2
=+−+
−
∫ ρ
pd
zzg
vv
The body force commonly considered in elementary fluid mechanics is gravity.
If the negative z axis is arbitrarily selected in the direction of g, the local acceleration due
to gravity, then ΦB becomes gz (N.B., −∇∇∇∇(gz)= −gk, which is the body force per unit mass
for g on the negative z direction). This leads to the specialization:
ˆ
If in addition the density is assumed uniform and steady, then it may be removed from
the integrand, and further specialization gives us:
0)(
2
12
12
2
1
2
2
=
−
+−+
−
ρ
pp
zzg
vv
When ρ is not constant, one must know how ρ varies with p. A relation of the form ρ =
ρ(p) is referred to as a barotropic relation.
The two equations above were obtained by integration along a streamline (any
streamline) in a steady flow field. Upon rearrangement, the last equation may be written
as:
in which C is a constant along a given streamline; however, C can vary for different
streamlines. The sum of the three terms in the above equation is called the
Bernoulli constant. (N.B., no shear stresses – works only for an ideal fluid).
C
p
zg
vp
zg
vp
zg
v
=++++=++
ρρρ 222
2
2
2
2
21
1
2
1
or
42

2017
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tD
vD
zyx
tD
vD
zyx
tD
vD
zyx
z
zB
zzzyzx
y
yB
yzyyyx
x
xB
xzxyxx
ρρ
σττ
ρρ
τστ
ρρ
ττσ
=+
∂
∂
+
∂
∂
+
∂
∂
=+
∂
∂
+
∂
∂
+
∂
∂
=+
∂
∂
+
∂
∂
+
∂
∂
)(
)(
)(
f
f
f
Now the fun stuff! We will establish a relation between stress and rate of strain… That
is, attention is now directed to the problem of finding the stress distribution in a moving
fluid with shear stresses present.
We list the equation we obtained previously (again by using Cartesian coordinates):
0)( =
∂
∂
+•∇
t
ρ
ρv
and the continuity equation:
gives us at least the density, ρ. For our purposes, this continuity equation is written as:
0
)()()(
=
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
z
v
y
v
x
v
t
zyx ρρρρ
An inventory of the dependent variables (i.e., the unknowns – the green boxes above)
indicates that in general there are nine unknown stresses, three unknown velocity
components, the density, and the body force field of three scalar components.
43

2017
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Thus there are a total of 16 unknown scalar quantities, and the above four equations
are obviously inadequate for the determination of these quantities. The problem can be
alleviated in two ways: 1) a reduction in the number of unknowns by virtue of the given
information in a problem and/or; 2) an introduction of more equations from basic laws
and experimental evidence that relate the dependent variables.
Here is the strategy we will adopt since this will be quite the development… The first
task will show that there are only six distinct components required to define the state of
stress! Next, constitutive relations will be introduced to relate stress to the space
derivatives of velocity components. These derivatives are referred to as rates of strain,
and their relation to stresses provides a set of constitutive relations.
Studies in solid mechanics (yes, we will digress a bit and study that too!) entail relating
stress to strain. The constitutive law involved there is referred to as Hooke’s law, which
simply states that stress is proportional to strain for the deformation of an elastic
solid. As an equation it could be written as σ =−Ce where C is a fourth-order tensor and
without such as relation, it would not be possible to solve problems in solid mechanics.
To summarize: Physical phenomena associated with the behavior of material
systems require some ‘cause and effect’ laws or relations (e.g., such as Newton’s law
of motion, F=ma, that relates the force, F, observed to the acceleration, a, which in this
case in particular would be that of fluid elements) to describe the desired behavior.
Fluid flow fields are no exception to this respect. It will be found that ‘constitutive
relations’ provide six equations between stresses and strain rates.
44

As hinted at just earlier, it was Hooke who enunciated (around the end of the 17-th
century) the law of elastic deformation (i.e., Ut tensio sic vis – ‘elongation follows
force’). We can express this law in the following way:
F
k
1
=δ
where δ is the relative displacement (or resulting elongation so that only along the x-axis
δ =x), F is the applied force, and k the rigidity. The rigidity is a constant for a geometry
and given material as long as the condition that the material be elastic in nature.
Hooke’s law has been made general in its applicability and ability to correlate the
deformation and strain of a material. So, for an isotropic material we have the following
linear relations:
σ
E
e
1
=
2017
MRT
and:
τγ
G
1
=
where E is call the Young modulus (a.k.a., tensile modulus or elastic modulus) is a
measure of the stiffness of an elastic material and G is the modulus of rigidity (a.k.a.,
shear modulus) is concerned with the deformation of a solid and also describes the
material’s response to shear stress. A fluid is a material with zero shear modulus!
45

2017
MRT
yxxxyyzzzz IIIM ωω)(ω −+= &
Now back to fluid mechanics. Let us consider a fluid element of volume ∆x∆y∆z in a
flow field in which shear and normal stress are present. If moments are taken about an
axis parallel to the z-axis and passing through the center of mass of the element, one
may formalize the sum of these as*:
in which Ixx, Iyy, and Izz are the principal mass moments of inertia at the center of mass of
the element; ωx and ωy are the angular velocity components about the x- and y-axis,
respectively; ωz is the angular acceleration component about the z-axis; and Mz is the
moment of all the forces about the axis parallel to the z-axis and passing through the
center of mass of the element. In terms of the dimensions of the element and the
approximate values of surface forces acting on a finite-sized element, the preceding
equation may be written as:
⋅⋅⋅⋅
yxz
xy
xy
xy
xy
yx
yx
yx
yx
zy
zyx
zx
zyx
yx
zyx
y
zx
y
y
y
zx
y
y
x
zy
x
x
x
zy
x
x
ωω
12
)()(
12
)()(
ω
12
)()(
2222
2222
222222







 ∆+∆
∆∆∆−
∆+∆
∆∆∆+
∆+∆
∆∆∆=
∆
∆∆






 ∆
∂
∂
−+
∆
∆∆






 ∆
∂
∂
+−
∆
∆∆






 ∆
∂
∂
−+
∆
∆∆






 ∆
∂
∂
+
ρρρ
τ
τ
τ
τ
τ
τ
τ
τ
&
46
* You can review the Space Vehicle Dynamics chapter for more on rotations and moment of inertia where we have labeled this
equation as one of Euler’s equations along the 3-axis: M3 = Cω3 + (B – A)ω1ω2 where Ixx = A, Iyy = B, and Izz = C.

2017
MRT
If this expression is simplified and put on a per unit volume basis by dividing through
by ∆x∆y∆z, one obtains:
12
)()(
ωωω
12
)()( 2222
yxyx
yxzxyyx
∆+∆
+
∆+∆
=− ρρττ &
Taking the limit of both sides of this equation as ∆x, ∆y, and ∆z approach zero makes the
expression exact and yields the following result:
xyyx ττ =
and similarly for the other axes:
yzzy ττ =
and:
zxxz ττ =
It is now apparent that only six of the nine components of the stress tensor are distinct.
We have thus established the symmetry of the stress tensor.
In matrix form we would then have for the symmetric stress tensor σij in Cartesian
coordinates










=










=
zzzyzx
zyyyyx
zxyxxx
zzyzxz
zyyyxy
zxyxxx
ji
σττ
τστ
ττσ
σττ
τστ
ττσ
σ
where the components on either side of the diagonal are equal.
Six distinct components:
σxx, τxy, τxz, σyy, τyz & σzz
47

2017
MRT
There are certain invariants inherent in the state of stress in a continuum. The term
invariant here is employed in the following sense: A quantity is said to be invariant at
a point in the field if this quantity is the same when evaluated with respect to any
and all Cartesian coordinate systems having their origin at the point. In brief,
invariance means ‘invariance to rotation of axes’ at any point in the field.
lmnmlnlnlmlmnml yzzyxzzxxyyxzzyyxx ττττττσσσσ ++++++++= 222
It is not difficult to show that the sum of three normal stresses at a point is invariant
with respect to the rotation of axes. If the forces associated with the stresses (e.g., as
used in the previous Figure with the stresses imposed on a pyramid-looking element)
are summed in a direction normal to the oblique face, the result is, after limiting the
equation as ∆x→0, ∆y→0, and ∆z→0:
in which l, m, and n are the direction cosines of the normal to the oblique face and σ is
the resultant normal stress on this oblique face. As a result of the symmetriesτxy =τyx,τyz
=τzy, and τzx =τxz, one may write the above equation as:
nmlnlmnml zyzxyxzzyyxx τττσσσσ 222222
+++++=
48

2017
MRT
This equation may now be applied to three different planes passing through a given
point in the fluid flow field. Consider an x, y, z coordinate system and a bared x, y, z
coordinate system (i.e., assume that it is rotated arbitrarily about an origin that is home
to both coordinate systems) in which the x, y, z are referred to the x, y, z as follows:
Let l1, m1, n1 = respective direction cosines of x-axis;
l2, m2, n2 = respective direction cosines of y-axis;
l3, m3, n3 = respective direction cosines of z-axis.
Then it follows that the respective direction cosine of the x, y, and z axes referred to the
x, y, and z frame of reference are l1, l2, l3; m1, m2, m3; and n1, n2, n3, respectively.
Now if σ is formulated as σxx from our previous equation, then l=l1, m=m1, n=n1, and:
zyzxyxzzyyxxxx nmnlmlnml τττσσσσ 111111
2
1
2
1
2
1 222 +++++=
Formulating σ as σyy (i.e., l=l2, m=m2, n=n2) gives:
zyzxyxzzyyxxyy nmnlmlnml τττσσσσ 222222
2
2
2
2
2
2 222 +++++=
Finally, considering σ as σzz yields:
zyzxyxzzyyxxzz nmnlmlnml τττσσσσ 333333
2
3
2
3
2
3 222 +++++=
These last three equations are now added to obtain:
zyzxyx
zzyyxxzzyyxx
nmnmnmnlnlnlmlmlml
nnnmmmlll
τττ
σσσσσσ
)(2)(2)(2
)()()(
332211332211332211
2
3
2
2
2
1
2
3
2
2
2
1
2
3
2
2
2
1
+++++++++
++++++++=++
49

2017
MRT
Since l1, l2, and l3 are the direction cosines of a given line, the coefficient of σxx is unity
in the last equation. This is also true for the coefficients of σyy and σzz. The coefficients of
the terms containing shear stresses are zero as was noted in the ll+mm+nn=0 equation
previously obtained. This establishes that:
zzyyxxzzyyxx σσσσσσ ++=++
This equation demonstrates the invariance of the sum of the normal stresses at a
point with respect to rotation of axes.
One-third of this invariant is referred to as the mean bulk stress and is noted by:
)(
3
1
zzyyxx σσσσ ++=
The significance of σ is its relation to pressure and rate of unit volume strain.
50

2017
MRT
The rate of strain at a point in a fluid may be studied by examining the relative
velocity of two closely neighboring fluid particles (e.g., identified by the position vectors
ro – a function of the material coordinates – and ro ++++dro with relative displacement ∆r at
time t) in the flow field at a given instant of time. The x component of the relative
displacement of the two particles ∆r, considered also as a function of the material
coordinates, is:
o
o
o
o
o
o
zd
z
x
yd
y
x
xd
x
x
x
∂
∂
+
∂
∂
+
∂
∂
=∆
neglecting higher-order terms in dxo, dyo, and dzo. The material derivative of ∆x is:
oo
o
o
o
o
o
o rr








∂
∂
+
∂
∂
+
∂
∂
∂
∂
=





∂
∆∂
zd
z
x
yd
y
x
xd
x
x
tt
x
and the order of differentiation may be interchanged on the right-hand side since t and ro
are independent variables and the indicated derivatives have been assumed to be
continuous. This interchange yields:
o
o
o
o
o
o oooo
zd
t
x
z
yd
t
x
y
xd
t
x
xt
x
rrrr






∂
∂
∂
∂
+





∂
∂
∂
∂
+





∂
∂
∂
∂
=





∂
∆∂
The left-hand side of this last equation is identified as dvx, the rate at which ∆r is
changing with respect to time in the x-direction.
51

2017
MRT
The equation:
t
tt
td
d
∂
∂
==
]),,([ orrr
v
permits the substitution of vx for [∂x/∂t]ro
; (∂xo/∂x)dx, (∂yo/∂y)dy, and (∂zo/∂z)dz may also
be substituted for dxo, dyo, and dzo, respectively. Carrying out these substitutions gives:
zd
z
z
z
v
yd
y
y
y
v
xd
x
x
x
v
vd xxx
x
∂
∂
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
= o
o
o
o
o
o
which is equivalent to:
zd
z
v
yd
y
v
xd
x
v
vd xxx
x
∂
∂
+
∂
∂
+
∂
∂
=
By similar formulation, one can obtain:
and:
zd
z
v
yd
y
v
xd
x
v
vd
yyy
y
∂
∂
+
∂
∂
+
∂
∂
=
zd
z
v
yd
y
v
xd
x
v
vd zzz
z
∂
∂
+
∂
∂
+
∂
∂
=
52
We can write these three equations as dvi (where i=1,2,3 for x, y, z and same for k):
∑=
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
=
3
1
3
3
2
2
1
1 k
k
k
iiii
i xd
x
v
xd
x
v
xd
x
v
xd
x
v
vd

2017
MRT
The last three expressions may be written as a single matrix equation of the form (if
you want you can review the Matrix Operations chapter now to familiarize yourself with
matrix concepts) using matrix multiplication:


















∂
∂
+
∂
∂
+
∂
∂
∂
∂
+
∂
∂
+
∂
∂
∂
∂
+
∂
∂
+
∂
∂
=
















⋅


















∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=
















zd
z
v
yd
y
v
xd
x
v
zd
z
v
yd
y
v
xd
x
v
zd
z
v
yd
y
v
xd
x
v
zd
yd
xd
z
v
y
v
x
v
z
v
y
v
x
v
z
v
y
v
x
v
vd
vd
vd
zzz
yyy
xxx
zzz
yyy
xxx
z
y
x
which indicates that the relative velocity of two closely neighboring fluid particles is
related to the relative spatial displacements at a given instant of time through an array
of nine quantities which comprise the components of the velocity gradient tensor.
53
where i=1,2,3 (for x, y, z) is the row index and j =1,2,3 is the column index, so that for
our dvi =Σk(∂vi /∂xk)dxk:
][][][][][][][][][ 332211
3
1
jijiji
k
jkkiji babababac ⋅+⋅+⋅=⋅= ∑=
or:
zd
z
v
yd
y
v
xd
x
v
xd
x
v
xd
x
v
xd
x
v
xd
x
v
vd iiiiii
k
k
k
i
i ⋅
∂
∂
+⋅
∂
∂
+⋅
∂
∂
=⋅
∂
∂
+⋅
∂
∂
+⋅
∂
∂
=⋅





∂
∂
= ∑=
3
3
2
2
1
1
3
1
][][

2017
MRT
To facilitate the interpretation of the above matrix, it may be arranged using the
following identity:
in the form of the sum of a symmetric, [sij]=[sji], and antisymmetric matrix, [aij]=−[aji]:
])[]([
2
1
])[]([
2
1
][][][ ijjiijjijijiji ccccasc −++=+=












































∂
∂
−
∂
∂
−





∂
∂
−
∂
∂
−








∂
∂
−
∂
∂








∂
∂
−
∂
∂
−






∂
∂
−
∂
∂








∂
∂
−
∂
∂
+




































∂
∂








∂
∂
+
∂
∂






∂
∂
+
∂
∂








∂
∂
+
∂
∂
∂
∂








∂
∂
+
∂
∂






∂
∂
+
∂
∂








∂
∂
+
∂
∂
∂
∂
=
















zd
yd
xd
y
v
z
v
x
v
z
v
y
v
z
v
x
v
y
v
x
v
z
v
x
v
y
v
zd
yd
xd
z
v
y
v
z
v
x
v
z
v
y
v
z
v
y
v
x
v
y
v
x
v
z
v
x
v
y
v
x
v
vd
vd
vd
zyzx
zyyx
zxyx
zzyzx
zyyyx
zxyxx
z
y
x
0
2
1
2
1
2
1
0
2
1
2
1
2
1
0
2
1
2
1
2
1
2
1
2
1
2
1
54

2017
MRT
The reason for placing our first matrix in the above form is for the identification of the
rigid body rotation associated with the relative velocity components. This is readily seen
by carrying out the matrix multiplication in the last term of the above [cij] matrix (i.e., the
antisymmetric, [aij], matrix), which results in the following column matrix:
Now, this is a mess unless we put things in terms of the components, ωk (k=1,2,3) of the
rotational rate vector ωωωω=ωxi++++ωyj++++ωzk (this vector was deduced earlier as ωωωω=½∇∇∇∇××××v –
see Vector Calculus chapter):
55




























∂
∂
−
∂
∂
−





∂
∂
−
∂
∂
−








∂
∂
−
∂
∂
+







∂
∂
−
∂
∂
−






∂
∂
−
∂
∂
+







∂
∂
−
∂
∂
=
yd
y
v
z
v
xd
x
v
z
v
zd
y
v
z
v
xd
x
v
y
v
zd
x
v
z
v
yd
x
v
y
v
a
zyzx
zyyx
zxyx
ji
2
1
2
1
2
1
2
1
2
1
2
1
][
kji
kji
v ˆ
2
1ˆ
2
1ˆ
2
1
ˆˆˆ
2
1
2
1








∂
∂
−
∂
∂






∂
∂
−
∂
∂








∂
∂
−
∂
∂
=
∂
∂
∂
∂
∂
∂
==
y
v
x
v
z
v
x
v
z
v
y
v
vvv
zyx
xyxzyz
zyx
++++−−−−××××∇∇∇∇ωωωω
ˆ ˆ ˆ

2017
MRT
This expansion in terms of components gives us the one-for-one correspondence:
so that the messy array we started with may be noted instead by:










+−
−
+−
=
ydxd
zdxd
zdyd
a
xy
xz
yz
ji
ωω
ωω
ωω
][
The three elements of this matrix are clearly the components of ωωωω××××dr representing the
part of the relative velocity due to rigid body rotation; hence these elements do not
contribute to the rate of strain of the fluid particle.
56








∂
∂
−
∂
∂
−=







∂
∂
−
∂
∂
=






∂
∂
−
∂
∂
=





∂
∂
−
∂
∂
−=








∂
∂
−
∂
∂
−=







∂
∂
−
∂
∂
=
x
v
y
v
y
v
x
v
x
v
z
v
z
v
x
v
y
v
z
v
z
v
y
v
yxxy
z
zxxz
y
zyyz
x
2
1
2
1
ω
2
1
2
1
ω
2
1
2
1
ω
We will thus ignore the antisymmetric [aij] matrix from now on since its
components do not contribute to modifying the motion of a fluid in any way
whatsoever.

2017
MRT
However, the first factor in the middle of the [cij] matrix above (i.e., the symmetric, [sij],
matrix) represents the nine components of the rate-of-strain tensor; if the generic
element of this matrix is noted by the symbol e with a double subscript, then:
















=
















=




















∂
∂








∂
∂
+
∂
∂






∂
∂
+
∂
∂








∂
∂
+
∂
∂
∂
∂








∂
∂
+
∂
∂






∂
∂
+
∂
∂








∂
∂
+
∂
∂
∂
∂
=
zzyzxz
zyyyxy
zxyxxx
zzyzxz
zyyyxy
zxyxxx
zyzxz
zyyxy
zxyxx
ji
e
e
e
eee
eee
eee
z
v
z
v
y
v
z
v
x
v
y
v
z
v
y
v
y
v
x
v
x
v
z
v
x
v
y
v
x
v
s
γγ
γγ
γγ
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
][
where the γ s represent the rates of shear strains. Displayed differently, we have:
zxxz
xz
zxxz
yzzy
zy
yzzyxyyx
yx
xyyx
z
zz
y
yy
x
xx
z
v
x
v
ee
y
v
z
v
ee
x
v
y
v
ee
z
v
e
y
v
e
x
v
e
γγ
γγγγ
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
==





∂
∂
+
∂
∂
==
==







∂
∂
+
∂
∂
====







∂
∂
+
∂
∂
==
∂
∂
=
∂
∂
=
∂
∂
=
and
,
,,,
It is noteworthy to specify that the dimensions of each of the components of the rate
of strain is reciprocal time.
57

O
y
x
O
B
A
B
A
tvx∆
ty
y
vx
∆∆
∂
∂
tvy∆
∆y
∆x
α
β
tx
x
vy
∆∆
∂
∂Initial
Final
General motion indicating the Initial () and
Final (−−−− −−−−) locations of fluid element indicated
by solid and dashed outlines, respectively.
2017
MRT
The longitudinal strain in the x-direction is the change of the
length of OA in this direction and is approximately given by:
tx
x
v
xtx
x
v
x xx
∆∆
∂
∂
=∆−





∆∆
∂
∂
+∆
The Figure shows that γxy and exx represent shear and longitudinal rates of strain,
respectively. The shear strain rate is formulated by noting the angles AOB in initially 90°,
and that a time ∆t later, this angle has decreased by α +β. These angle may be
approximated by:
txvxtxxvtyvytyyv yyxx ∆∂∂=∆∆∆∂∂≅∆∂∂=∆∆∆∂∂≅ )(])[()(])[( βα and
The rate of decrease of the angle AOB is given by:
ttxvtyv yx ∆∆∂∂+∆∂∂ ])()[(
which, on limiting with ∆t→0:
yx
yxyx
t x
v
y
v
t
t
x
v
t
y
v
γ=
∂
∂
+
∂
∂
=
∆







∆
∂
∂
+∆
∂
∂
→∆
1
lim
0
The unit rate of strain may be obtained by dividing by ∆t and the
initial length of OA, and by limiting:
xx
x
x
t
e
x
v
xt
tx
x
v
=
∂
∂
=
∆∆
∆∆
∂
∂
→∆ 0
lim
Interpretations may be attributed to γyz, γzx, eyy, and ezz similar
to those for γxy and exx.
58

Our previously obtained equations for σxx, σyy, and σzz provide us with a standard set
of transformation equations expressing the normal stress referred to a set of axes that
have been rotated, and these stresses are given as functions of the stress component
referred to the initial orientation of the axes. The transformation of rates of strain to be
developed next may be used in obtaining relations between stress and these rates.
Rotation of axes showing point fixed in either
set of coordinates by position vector r.
2017
MRT
These may be obtained by projection of the position vector r
on the coordinate axes. This same procedure may be employed
by projecting the velocity vector v of the fluid on the coordinate
axes to obtain:
zyxz
zyxy
zyxx
vnvmvlv
vnvmvlv
vnvmvlv
333
222
111
++=
++=
++=
The Figure shows the two coordinate systems, one (i.e., x, y, z) rotated with respect to
the other (i.e., x, y, z) . We have said earlier that l1, m1, and n1 are the respective direction
cosines of the x-axis, while l2, m2, n2 and l3, m3, n3 are the direction cosines of the y-and z-
axes, respectively. The transformation for the coordinates are given by:
znymxlzznymxlyznymxlx 333222111 ++=++=++= and,
and:
znynxnzzmymxmyzlylxlx 321321321 ++=++=++= and,
y
x
y
z
z
x
(x,y,z) or (x,y,z)
r
v
O
and similarly for vx, vy, and vz. This should be an expected result
since the velocity components are time derivatives of the
coordinates of the fluid particle at the point at any given instant
of time.
59

2017
MRT
To form the rates of strain in the bared coordinate system, the chain rule of calculus is
used as follows:
x
z
z
v
x
y
y
v
x
x
x
v
x
v xxxx
∂
∂
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
From our equations for x(l1,m1,n1), y(l2,m2,n2), z(l3,m3,n3), and x(l1,l2,l3), y(m1,m2,m3),
z(n1,n2,n3) above:
111 n
x
z
m
x
y
l
x
x
=
∂
∂
=
∂
∂
=
∂
∂
and,
and hence:
z
v
n
y
v
m
x
v
l
x
v xxxx
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
111
Partial differentiation of our equation vx =l1vx +m1vy +n1vz yields:
Substituting these three derivatives into our equation above for ∂vx /∂x, utilizing our
values for exx, eyy, ezz, γxy, γyz, and γzx, and simplifying, gives:
zyzxyxzzyyxx
x
xx nmnlmlenemel
x
v
e γγγ 111111
2
1
2
1
2
1 +++++=
∂
∂
=
z
v
n
z
v
m
z
v
l
z
v
y
v
n
y
v
m
y
v
l
y
v
x
v
n
x
v
m
x
v
l
x
v
zyxx
zyxxzyxx
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
111
111111
and
,
60

2017
MRT
In a similar manner:
y
v
x
v
z
v
x
v
z
v
y
v
z
v
y
v zzyyxxzy
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
and,,,,,,
may be generated, with the formulation of the remaining five rates of strain
transformations. These are:
zyzxyxzzyyxxzy
zyzxyxzzyyxxzx
zyzxyxzzyyxxyx
zyzxyxzzyyxxzz
zyzxyxzzyyxxyy
nmnmnlnlmlmlennemmell
nmnlmlenemele
nmnlmlenemele
γγγγ
γγγγ
γγγγ
γγγ
γγγ
)()()(222
)()()(222
)()()(222
233223322332323232
133113311331313131
122112211221212121
333333
2
3
2
3
2
3
222222
2
2
2
2
2
2
++++++++=
++++++++=
++++++++=
+++++=
+++++=
Before relating the six components specifying the state of stress to the rates of strain,
it will be convenient to examine a simple relation between one stress component and a
single rate of strain.
61

The earliest relation between stress and rate of strain may be attributed to Newton who
advanced the following hypothesis: The resistance arising from the want of lubricity in
the parts of a fluid, is, other things being equal, proportional to the velocity with which
the parts of the fluid are separated from one another. Applied to the flow between the
two plates in the Figure (x-y plane) this hypothesis may be formulated for the shear τ as:
Velocity profile for fluid motion between two
parallel plates.
2017
MRT
The viscosity of a fluid is a consequence of the forces among
molecules and the transfer of momentum of the molecules
associated with these forces. Our equation above provides a
simple relation between these effects through a coefficient of
viscosity. The transfer of momentum at a molecular level is
brought out here to emphasize that the relation between stress
and the momentum on the fluid is among molecules. This
regime of flow is referred to as a well-ordered flow or laminar
flow (in contrast to turbulent flow in which transfers of mass,
momentum, and energy are manifest at a microscopic level). A
well-ordered flow is one free of macroscopic velocity
fluctuations.
Stationary plate
vp, velocity of the moving plate
d y
x
v =v( y)
( )0≥≥
∂
∂
== yd
y
v
A
F x
yx µτ
in which F is the force that must be applied to the moving plate of area A and µ (the
constant of proportionality which end up being the slope of the  line) is the coefficient
of viscosity. Both plates are very large compared to the distance of separation d (which
tends to minimize the edge effects). Note that the fluid adjacent to each of the plates
adheres to the plates (this is called the ‘no slip’ condition).
62

∆z
∆r
z
r
zrrr
r
zr
zr ∆∆+⋅







∆
∂
∂
+ )(π2
τ
τ
zrzr ∆⋅ π2τ
r
r
rz
z
p
p
zz
∆




 ∆
+⋅





∆
∂
∂
+
−
2
π2
σ
zz
r
r
rp
σ−
∆




 ∆
+⋅
2
π2
Consider the steady and laminar flow of an incompressible Newtonian fluid in a pipe of
radius R in the absence of body forces. The flow is assumed to have a velocity
component in the direction of the axis of the pipe, which is shown as the z-axis in the
Figure. Let us consider that the velocity component to be a function of the r, θ, and z
coordinates. We will now formulate Newton’s law of motion for a typical element which
requires that the sum of the forces in an axial direction be zero so that:
2017
MRT
Simplifying the preceding expression and limiting it as ∆r and
∆z each approach zero gives:
z
O
Pipe radius, R
r
0)(π2π2
2
π2
2
π2 =∆∆+⋅







∆
∂
∂
++∆⋅−∆




 ∆
+⋅





∆
∂
∂
+−∆




 ∆
+⋅ zrrr
r
zrr
r
rz
z
p
pr
r
rp
zr
zrzr
τ
ττ
Pipe in the absence of body forces and one of
its typical elements. Note that p =−σzz here.
0=
∂
∂
++
∂
∂
−
rrz
p zrzr ττ
This equation would apply for ‘any fluid flowing’ in a pipe under
the given conditions. If the fluid flowing is a Newtonian fluid, then
τrz =µ(∂vz /∂r), and the equation of motion reduces to:
02
2
=





+
∂
∂
−=
∂
∂
+
∂
∂
+
∂
∂
−
rd
vd
r
rd
d
rz
p
r
v
r
v
rz
p zzz µ
µ
µ
To demonstrate that p is a function of z only will require the
condition of a small sector-shaped element rather than a
complete annular cylinder. Subsequent summing of the forces in
the radial direction, equating this sum to zero, and recognizing
that axial symmetry requires no change of p in the θ direction,
yields ∂p/∂r = 0 and it is concluded that p is a function of z only.
63

T
rω
R1
R2
v = ωR2
L
2017
MRT
Exercise: The two cylinders shown below are coaxial; they are long enough so that end
effects may be ignored. The space between them is filled with a fluid of viscosity µ. The
inside cylinder of radius R1 is stationary while a constant torque T is applied to the
outside cylinder for radius R2. Obtain an expression for the steady-state angular speed of
the outside cylinder under the action of fluid shear and the applied torque.
rd
vd
Lr
T
rL
rT
µτ === 2
π2π2
Solution: The fluid adjacent to the outer surface of the inside cylinder has zero velocity,
while that adjacent to the inner surface of the outside cylinder has the velocity of a point
on this surface, which is ωR2. This is the no-slip condition referred to previously.
* The correct expression for the shear is τrθ = µ(dv/dr – V/r); however, V/r is generally small compared to dv/dr.
An applicable form*of Newton’s statement concerning lubricity is τ =F/A=µ(dv/dr).
This may be applied to a typical area through which the torque is transmitted by shear
action. This area is taken to be A=2πrL. Since T=r××××F (and T=|T|=|r||F|sinθ where θ is
90° so T=rF), the force F across this area is T/r, so that:
64

2017
MRT
Separation of variables yields:
∫∫ =
22
1
ω
02
π2
RR
R
vd
r
rd
L
T
µ
This velocity profile in the fluid is indicated in the previous Figure.
The torque T has been taken outside the integral, since it is independent of r when the
outside cylinder has attained a steady-state angular speed ω.
Integration, substitution of limits, and solution for the angular speed gives the requires
result:














−=
rRRL
T 111
π2
ω
12µ
If in addition to this result, the velocity profile is desired, the differential equation may
be integrated with variable upper limits, that is:
∫∫ =
vr
R
vd
r
rd
L
T
02
1π2
µ
and:






−=
rRL
T
rv
11
π2
)(
1µ
65

2017
MRT
The constitutive equations, which relate the stress components to the strain rates,
are based on the following assumptions:
1) Each of the stress components is linearly proportional to the strain rate;
2) The fluid is isotropic and hence there are no preferred directions;
3) In the absence of strain rates, the normal stresses must reduce to the pressure, and
the shear stresses must vanish.
67666564636261
57565554535251
47464544434241
37363534333231
27262524232221
17161514131211
CCCCeCeCeC
CCCCeCeCeC
CCCCeCeCeC
CCCCeCeCeC
CCCCeCeCeC
CCCCeCeCeC
zyzxyxzzyyxxzy
zyzxyxzzyyxxzx
zyzxyxzzyyxxyx
zyzxyxzzyyxxzz
zyzxyxzzyyxxyy
zyzxyxzzyyxxxx
++++++=
++++++=
++++++=
++++++=
++++++=
++++++=
γγγτ
γγγτ
γγγτ
γγγσ
γγγσ
γγγσ
Assumption 1 posits that the most general linear relation between each of the six
distinct stress components (i.e., σxx, σyy, σzz, τxy, τxz, and τyz) and the six rates of strain
(i.e., exx, eyy, ezz, γxy, γxz, and γyz) is the set of six equations:
in which the 42 Cs are constants that must be determined from the assumptions and
experimental evidence.





1st set 





66

2017
MRT
According to Assumption 2, there are no preferred directions (i.e., space or fluid
continuum isotropy) so that the constants in the previous set of equations must be the
same for rotation of the coordinate axes. This may be formalized as:
67666564636261
57565554535251
47464544434241
37363534333231
27262524232221
17161514131211
CCCCeCeCeC
CCCCeCeCeC
CCCCeCeCeC
CCCCeCeCeC
CCCCeCeCeC
CCCCeCeCeC
zyzxyxzzyyxxzy
zyzxyxzzyyxxzx
zyzxyxzzyyxxyx
zyzxyxzzyyxxzz
zyzxyxzzyyxxyy
zyzxyxzzyyxxxx
++++++=
++++++=
++++++=
++++++=
++++++=
++++++=
γγγτ
γγγτ
γγγτ
γγγσ
γγγσ
γγγσ
zyzxyxzzyyxxzz
zyzxyxzzyyxxyy
zyzxyxzzyyxxxx
nmnlmlnml
nmnlmlnml
nmnlmlnml
τττσσσσ
τττσσσσ
τττσσσσ
333333
2
3
2
3
2
3
222222
2
2
2
2
2
2
111111
2
1
2
1
2
1
222
222
222
+++++=
+++++=
+++++=
Recall also the equations we obtained when we investigated the invariance under
rotations of the stress tensor:





2nd set 





We will now proceed and evaluate σxx in quite an amount of detail.
67

2017
MRT
Substitution of our 1st set of equations for σxx, σyy, σzz, τxy, τxz, and τyz into the first of
these last equations gives:
)(2
)(2
)(2
)(
)(
)(
6766656463626111
5756555453525111
4746454443424111
37363534333231
2
1
27262524232221
2
1
17161514131211
2
1
CCCCeCeCeCnm
CCCCeCeCeCnl
CCCCeCeCeCml
CCCCeCeCeCn
CCCCeCeCeCm
CCCCeCeCeCl
zyzxyxzzyyxx
zyzxyxzzyyxx
zyzxyxzzyyxx
zyzxyxzzyyxx
zyzxyxzzyyxx
zyzxyxzzyyxxxx
+++++++
+++++++
+++++++
+++++++
+++++++
++++++=
γγγ
γγγ
γγγ
γγγ
γγγ
γγγσ
The coefficients of the rate of strain in the previous equation may be grouped as follows:
67115711471137
2
127
2
117
2
1
66115611461136
2
126
2
116
2
1
65115511451135
2
125
2
115
2
1
64115411441134
2
124
2
114
2
1
63115311431133
2
123
2
113
2
1
62115211421132
2
122
2
112
2
1
61115111411131
2
121
2
111
2
1
222
)222(
)222(
)222(
)222(
)222(
)222(
CnmCnlCmlCnCmCl
CnmCnlCmlCnCmCl
CnmCnlCmlCnCmCl
CnmCnlCmlCnCmCl
eCnmCnlCmlCnCmCl
eCnmCnlCmlCnCmCl
eCnmCnlCmlCnCmCl
zy
zx
yx
zz
yy
xxxx
++++++
++++++
++++++
++++++
++++++
++++++
+++++=
γ
γ
γ
σ
68

2017
MRT
A second expression for σxx in terms of the rates of strain referred to the x, y, z
coordinate system may be obtained by substitution of exx, eyy, ezz, γxy, γxz, and γyz which
were developed early and are given below for convenience by the set of equations:
zyzxyxzzyyxxxz
zyzxyxzzyyxxzy
zyzxyxzzyyxxyx
zyzxyxzzyyxxzz
zyzxyxzzyyxxyy
zyzxyxzzyyxxxx
nmnlmlenemele
nmnlmlenemele
nmnlmlenemele
γγγγ
γγγγ
γγγγ
γγγ
γγγ
γγγ
)()()(222
)()()(222
)()()(222
133113311331313131
233223322332323232
122112211221212121
333333
2
3
2
3
2
3
222222
2
2
2
2
2
2
111111
2
1
2
1
2
1
++++++++=
++++++++=
++++++++=
+++++=
+++++=
+++++=
into our 2nd set of equations.
69

2017
MRT
Making this substitution:
17
23322332233232323216
13311331133131313115
12211221122121212114
333333
2
3
2
3
2
313
222222
2
2
2
2
2
212
111111
2
1
2
1
2
111
])()()(222[
])()()(222[
])()()(222[
)(
)(
)(
C
nmnmnlnlmlmlennemmellC
nmnlmlenemelC
nmnlmlenemelC
nmnlmlenemelC
zyzxyxzzyyxx
zyzxyxzzyyxx
zyzxyxzzyyxx
zyzxyxzzyyxx
zyzxyxzzyyxx
zyzxyxzzyyxxxx
+
+++++++++
+++++++++
+++++++++
++++++
++++++
+++++=
γγγ
γγγ
γγγ
γγγ
γγγ
γγγσ
17
233216133115122114331322121111
233216133115122114331322121111
233216133115122114331322121111
321631152114
2
313
2
212
2
111
321631152114
2
313
2
212
2
111
321631152114
2
313
2
212
2
111
)]()()([
)]()()([
)]()()([
)222(
)222(
)222(
C
nmnmCnmnmCnmnmCnmCnmCnmC
nlnlCnlnlCnlnlCnlCnlCnlC
mlmlCmlmlCmlmlCmlCmlCmlC
emnCnnCnnCnCnCnC
emmCmmCmmCmCmCmC
ellCllCllClClClC
zy
zx
yx
zz
yy
xxxx
+
+++++++++
+++++++++
+++++++++
++++++
++++++
+++++=
γ
γ
γ
σ
and then grouping the coefficients of the strain rates gives:
70

2017
MRT
These last two boxed equations must be identical for all geometric admissible values
of the nine direction cosines; therefore the coefficients of corresponding rates of strain in
each equation may be equated to one another. The corresponding terms containing only
the Cs may be equated also. Doing this provides the following relations among the 42
constants:
1623321513311412211333221166115611461136
2
126
2
116
2
1
1623321513311412211333221165115511451135
2
125
2
115
2
1
1623321513311412211333221164115411441134
2
124
2
114
2
1
16321531142113
2
3
2
2
2
163115311431133
2
123
2
113
2
1
16321531142113
2
3
2
2
2
162115211421132
2
122
2
1
2
1
16321531142113
2
3
2
2
2
161115111411131
2
121
2
1
2
1
)()()(222
)()()(222
)()()(222
222222
222222
222222
CnmnmCnmnmCnmnmCnmnmnmCnmCnlCmlCnCmCl
CnlnlCnlnlCnlnlCnlnlnlCnmCnlCmlCnCmCl
CmlmlCmlmlCmlmlCmlmlmlCnmCnlCmlCnCmCl
CmnCnnCnnCnnnCnmCnlCmlCnCmCl
CmmCmmCmmCmmmCnmCnlCmlCnCml
CllCllCllClllCnmCnlCmlCnCml
++++++++=+++++
++++++++=+++++
++++++++=+++++
+++++=+++++
+++++=+++++
+++++=+++++
1211
1211
1211
1211
121112
121111
CC
CC
CC
CC
CCC
CCC
1
1
2
3
2
2
2
1
2
3
2
2
2
1
2
3
2
2
2
1
2
3
2
3
2
3
2
2
2
2
2
2
2
1
2
1
2
1
=++=++=++
=++=++=++
nnnmmmlll
nmlnmlnml
One may solve for the constants in terms of C11, C12, and C17 by assigning values to the
direction cosines compatible with the following constraints:
and finally:
1717 CC =+++++ 67115711471137
2
127
2
1
2
1 222 CnmCnlCmlCnCml
0
0
332211332211332211
313131323232212121
=++=++=++
=++=++=++
lnlnlnnmnmnmmlmlml
nnmmllnnmmllnnmmll
and:
71

2017
MRT
The results of this procedure are:
32233113213727
12116655443322 )(
2
1
CCCCCCC
CCCCCCC
=======
−=====
1217
11
CC
C
and
,,
and all other C constants (i.e., 42−15=27 of them) are zero. By substituting these results
in our 1st set of equations for σxx, σyy, σzz, τxy, τxz, and τyz, the constitutive equations
may be written as:
zyzy
zxzx
yxyx
zzyyxxzz
zzyyxxyy
zzyyxxxx
CC
CC
CC
CeCeCeC
CeCeCeC
CeCeCeC
γτ
γτ
γτ
σ
σ
σ
)(
2
1
)(
2
1
)(
2
1
1211
1211
1211
37111212
27121112
17121211
−=
−=
−=
+++=
+++=
+++=
72

2017
MRT
Assumption 3 (e.g., the well-ordered or laminar flow specialization discussed) requires
the normal stress to reduce to the pressure in absence of deformation, and:
pCCC −=== 372717
so that only two constants in the last set of equations are yet to be determined.
A relation between these two may be obtained by reference to the simple one-
dimensional flow between two parallel plates. For that scenario we had τxy =µ(dvx /dy)
and the constitutive equations require:
y
v
CCCC x
yxyx
∂
∂
−=−= )(
2
1
)(
2
1
12111211 γτ
since vy =0 for that kind of laminar flow. This indicates that:
1212 CC +==− µµ 2)(
2
1
1111 CC or
Substituting this value of C11 (C12 remains an unknown) just obtained and the values for
C17, C27, and C37 above into our last boxed equations gives:
zyzyzxzxyxyx
zzyyxxxxzz
zzyyxxxxyy
zzyyxxxxxx
peeeCe
peeeCe
peeeCe
γµτγµτγµτ
µσ
µσ
µσ
===
−+++=
−+++=
−+++=
and,
,
,
,
)(2
)(2
)(2
12
12
12
73

2017
MRT
In summary, the most general relation between each of the six distinct stress
components and the six rates of strain results in six equations including 42 constants
according to Assumption 1. From Assumption 2, the number of constants is reduced to
five, and three are related to the pressure p through Assumption 3. Finally, application of
Newton’s viscosity hypothesis, τ =µ(dv/dy), to the motion of a fluid through the parallel
plate during laminar flow yields a relationship among the two remaining constants and
the coefficient of viscosity µ. The assumed linear relations may now be expressed as:
zyzy
zxzx
yxyx
zzyyxxxxzz
zzyyxxxxyy
zzyyxxxxxx
peeeCe
peeeCe
peeeCe
γµτ
γµτ
γµτ
µσ
µσ
µσ
=
=
=
−+++=
−+++=
−+++=
)(2
)(2
)(2
12
12
12
Since I’ve had a few, one thing I hope for you looking at this slide and having patiently
read through and figured out the details on your own is that you get a feel for the impact
of 1) stress tensors; 2) invariance under rotations; and 3) the ‘shear’ beauty of
mathematics as a whole in providing such elegant derivations that (combined with some
intuition from guys like Newton) help us actually figure out how things work (or flow)
on this small 3 dimensional world subjected to a time dimension that is immutable…
74

2017
MRT
Addition of the first three equations of the above set of equation provides:
peeeCeee zzyyxxzzyyxxzzyyxx 3)(3)(2 12 −+++++=++ µσσσ
a one-third of which is the mean bulk stress:
pC −•





+= v∇∇∇∇12
3
2
µσ
Since the divergence of the velocity (i.e., ∇∇∇∇•v) is the rate of change of volume per unit
volume, the quantity (2/3)µ +C12 is called the volume viscosity.
When the volume viscosity is zero, our last set of equations reduce to:








∂
∂
+
∂
∂
=





∂
∂
+
∂
∂
=







∂
∂
+
∂
∂
=
∂
∂
+•−−=
∂
∂
+•−−=
∂
∂
+•−−=
z
v
y
v
z
v
x
v
y
v
x
v
z
v
p
y
v
p
x
v
p
yz
zy
xz
zx
xy
yx
z
zz
y
yy
x
xx
µτµτµτ
µµσ
µµσ
µµσ
and,
2
3
2
2
3
2
2
3
2
v
v
v
∇∇∇∇
∇∇∇∇
∇∇∇∇
These are the constitutive equations usually credited to Stokes (1819-1903).
75

2017
MRT
Now we constitute the equations of motion with viscosity. This is done by substituting
the values provided by the constitutive equations of Stokes’ just developed into the
equations of motion for a fluid particle. This will be done in detail for the first of these:
tD
vD
zyx
x
xB
xzxyxx
ρρ
ττσ
=+
∂
∂
+
∂
∂
+
∂
∂
)(f
where we substitute for σxx,τxy, and τxz to obtain:
xB
xzxyxx
z
v
x
v
zy
v
x
v
yx
v
p
xtD
vD
)(2
3
2
fv ρµµµµρ +














∂
∂
+
∂
∂
∂
∂
+
















∂
∂
+
∂
∂
∂
∂
+





∂
∂
+•−−
∂
∂
= ∇∇∇∇
and after an easy reduction, similar substitutions can be made for the two other
equations of motion for a fluid particle so that the whole package is given by:
These three equations are referred to as the Navier-Stokes equations of motion.
zB
zxzyzz
yB
zyyxyy
xB
xzxyxx
x
v
z
v
xy
v
z
v
yz
v
zz
p
tD
vD
y
v
z
v
zx
v
y
v
xy
v
yy
p
tD
vD
z
v
x
v
zy
v
x
v
yx
v
xx
p
tD
vD
)(2
3
2
)(2
3
2
)(2
3
2
fv
fv
fv
ρµµµµρ
ρµµµµρ
ρµµµµρ
+














∂
∂
+
∂
∂
∂
∂
+
















∂
∂
+
∂
∂
∂
∂
+





∂
∂
−•
∂
∂
−
∂
∂
−=
+
















∂
∂
+
∂
∂
∂
∂
+
















∂
∂
+
∂
∂
∂
∂
+







∂
∂
−•
∂
∂
−
∂
∂
−=
+














∂
∂
+
∂
∂
∂
∂
+
















∂
∂
+
∂
∂
∂
∂
+





∂
∂
−•
∂
∂
−
∂
∂
−=
∇∇∇∇
∇∇∇∇
∇∇∇∇
76

2017
MRT
A considerable simplification occurs in problems where viscosity can be considered
constant. In this case, the Navier-Stokes equations of motion reduce to the following:
zBz
z
yBy
y
xBx
x
v
zz
p
tD
vD
v
yy
p
tD
vD
v
xx
p
tD
vD
)(
3
1
)(
3
1
)(
3
1
f
v
f
v
f
v
ρµµρ
ρµµρ
ρµµρ
+•+
∂
•∂
+
∂
∂
−=
+•+
∂
•∂
+
∂
∂
−=
+•+
∂
•∂
+
∂
∂
−=
∇∇∇∇∇∇∇∇
∇∇∇∇
∇∇∇∇∇∇∇∇
∇∇∇∇
∇∇∇∇∇∇∇∇
∇∇∇∇
If the fluid is incompressible (i.e., ∇∇∇∇•v=0 vs incomprehensible!) then each of the
second terms on the right-hand side becomes zero, and a further reduction is possible. It
should be observed that the above equations may be reduced to the single vector
equation (N.B., the terms ∇∇∇∇(∇∇∇∇•v)−−−−∇∇∇∇××××(∇∇∇∇××××v) are sometimes replaced by ∇∇∇∇•∇∇∇∇v≡∇2v in
some textbooks on fluid mechanics or hydrodynamics):
Bp
tD
D
fvvv
v
ρµµρ ++++××××∇∇∇∇××××∇∇∇∇−−−−∇∇∇∇∇∇∇∇++++∇∇∇∇∇∇∇∇++++∇∇∇∇ )]()([)(
3
1
••−=
or (Exercise):
Bp
tD
D
fvv
v
ρµµρ ++++××××∇∇∇∇××××∇∇∇∇−−−−∇∇∇∇∇∇∇∇++++∇∇∇∇ )()(
3
4
•−=
77

As an example, let us consider an incompressible Newtonian fluid with constant
viscosity that flows between two parallel plates. One plate is fixed and the other moves
in the positive x-direction with a velocity vp as indicated in the Figure. The motion is
steady and laminar, and none of the variables depend on the z coordinate. There is a
pressure gradient maintained in the x-direction, and there is no y or z component of
velocity at any point in the field. Let us obtain a solution for vx by neglecting body forces
(e.g., gravity) and discuss the influence of several pressure gradients.
An incompressible fluid flows between two
parallel plates.
2017
MRT
With body forces neglected, and ∇∇∇∇•v=0 for the incompres-
sible fluid, the constituted equation of motion reduce to:
Fixed plate
vp, velocity of the moving plate
y
x
O
a
a








∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−=





∂
∂
+
∂
∂
+
∂
∂
+
∂
∂








∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−=







∂
∂
+
∂
∂
+
∂
∂
+
∂
∂








∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−=





∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
z
v
y
v
x
v
z
p
t
v
z
v
v
y
v
v
x
v
v
z
v
y
v
x
v
y
p
t
v
z
v
v
y
v
v
x
v
v
z
v
y
v
x
v
x
p
t
v
z
v
v
y
v
v
x
v
v
zzzzz
z
z
y
z
x
yyyyy
z
y
y
y
x
xxxxx
z
x
y
x
x
µρ
µρ
µρ
and the continuity equation, ∂ρ/∂t+ρ∇∇∇∇•v =∇∇∇∇•v= 0, requires that:
0=
∂
∂
+
∂
∂
+
∂
∂
=•
z
v
y
v
x
v zyx
v∇∇∇∇
Other given conditions require that:
0=
∂
∂
t
vx
since the motion is steady and that all of the derivatives of vy and
vz are zero since vy =0 and vz =0 (i.e., no y or z component of v ).
78

2017
MRT
The continuity equation thus reduces to:
0=
∂
∂
x
vx
Carrying out the operation of ∫(∂vx/∂x)dx yields vx=vx(y,z,t), which means vx=vx(y) only
since vx is not a function of z and time according to the given data. Since ∂vx/∂x=0 and
∂vx/∂t=0, the second derivatives of vx with respect to x or z also vanish:
z
p
y
p
y
v
x
p x
∂
∂
−=
∂
∂
−=
∂
∂
+
∂
∂
−=
0
0
0 2
2
µ
The second and third of these imply that pressure is a function of x only, so that the field
equation may now be written in terms of ordinary derivative as:
xd
pd
yd
vd x
µ
1
2
2
=
79
The three constituted equations of motion may now be written respectively as:
02
2
2
2
=
∂
∂
=
∂
∂
z
v
x
v xx

2017
MRT
A first integration of the previous equation yields:
21
2
21
2
2
1
0
2
1
CaCa
xd
pd
CaCa
xd
pd
vp +−=++=
µµ
and
1
1
Cy
xd
pd
yd
vd x
+=
µ
and then a second integration gives:
21
2
2
1
)( CyCy
xd
pd
yvx ++=
µ
in which C1 and C2 are constants to be specified by the conditions to which the flow is
subjected. For our two parallel plates, these conditions (or constraints) are:
0)()( =−=+ avvav xpx and
using the previously chosen fixed coaxial reference axes as shown in the previous
Figure. These conditions are substituted into the expression for vx(y) to obtain:
Solving for these constraints gives:
2
21
2
1
22
a
xd
pdv
C
a
v
C
pp
µ
−== and
Substituting C1 and C2 into the expression for vx(y), one obtains the following solution:








−−





+= 2
22
1
2
1
2
)(
a
y
xd
pda
a
yv
yv
p
x
µ
80

vp
xx x x
0=
xd
pd
0<
xd
pd
0
2 2
>=
a
v
xd
pd pµ
0
2 2
>>
a
v
xd
pd pµ
vp vp vp
2017
MRT
A discussion of the solution just obtained will provide a better appreciation of the flow
under consideration. First, if the flow were maintained by the motion of one plate with
zero pressure gradient, the velocity profile would be linear because of the effects of
viscosity. For motion of the plate to the right, the established flow between the plates
would be from left to right. Second, if both plates are fixed, that is, vp =0, and the flow
between the plates is maintained by the applied pressure gradient dp/dx, then the
velocity profile is a parabola (or a parabolic wedge since there are no variations of the
dependent variables with the z coordinate). The flow is to the right if dp/dx<0 and it is to
the left if dp/dx>0. Third, if the fluid is flowing under the action of the moving plate and a
pressure gradient, then the velocity profile may take on a number of forms as indicated
by the solution above. A few interesting case are illustrated in the Figures.
Note that it is possible at a cross section to have some flow to the right and the left
when a suitably large positive pressure gradient exists. It is also possible to maintain a
particular value of the pressure gradient, dp/dx=µvp/2a2, such that the velocity profile:








+





+−= 2
2
1
2
1
4
)(
a
y
a
y
a
v
yv
p
x
is perpendicular to the stationary plate at this surface. In this case the constitutive
equation for τyx indicates that the shear stress would be zero at this plate.
81

Problem: With reference to the last Exercise,
a) Obtain an expression for σxx, σyy, σzz, τxy, τyz, and τxz.
b) Show that the normal stress at any point (in any direction in the x, y plane) can be
obtained from:
ml
xd
pd
y
a
v
p
p
2
2 







++−=
µ
σ
c) Obtain an expression for the maximum value σ −σ in the given flow for a positive
pressure gradient. What is the location and the values of l and m for this maximum?
d) What is the fraction of normal stress from the mean bulk stress at a point y=a and a
direction l=m=1/√2 if µ=10−4 lbf -sec/ft2, a=0.01 ft, vp =50 ft/sec, dp/dx=0.694 lbf /in.2-ft
and p=4000 lbf /ft2?
e) Calculate the external force needed to move the plate for the conditions given in (d)
on a per unit length and width basis.
f) If both plates were stationary, determine any location in the fluid flow where the shear
stress would be zero. Would this be true of any x location? What is the direction of the
shear stress at points in the fluid that are adjacent to the wall (at y=a and y=−a) for dp/dx
<0 and dp/dx>0?
g) Obtain an expression for the pressure gradient that:
i. Yields a zero shear stress at the stationary surface;
ii. Yields a zero shear stress at the moving surface;
iii. Will produce zero net mass flow between the two plates.

It is assumed that the following transformations exist:xi = fi(u1,u2,u3)(for i, j=1,2,3), and:
s
rrrr
r dud
u
ud
u
ud
u
ud
u
d
i
i
i
≡
∂
∂
=
∂
∂
∂
∂
∂
∂
= ∑=
3
1
3
3
2
2
1
1
++++++++
The square of the length of this displacement is given by the interval:
∑∑= = ∂
∂
•
∂
∂
=•=
3
1
3
1
2
i j
ji
ji
udud
uu
ddsd
rr
rr
2017
MRT
where gi j =xi ••••xj. Since xi ••••xj =xj ••••xi, gi j =gji (i.e., gij is
symmetric in the interchange of i and j indices). The
quantities gij are called metric coefficients and characterize
the relative nature of the space.
When an object at P is described using coordinates
other than Cartesian, say using coordinates ui=Fi(xj)
we can generalize using curvilinear coordinates.
x2
P
Note that the ∂r/∂ui are tangent to the ui respectively,
since ∂r/∂u1 means u2 and u3 are fixed and r=r(u1) is
constrained to move along the u1 surface. Letting the three
coordinate axes be represented by xi, where xi =∂r/∂ui (see
Figure). Hence the equation above for ds2 =ds•ds becomes:
ê3
x3
x1
∑∑∑ =•=
ji
ji
ji
i j
ji
ji ududgududsd xx2
ê1
ê2
u1
u2
u3
The position vector of a point P is a function of the ui , that is, r=r(u1,u2,u3). An element
of displacement is given by (using the Chain Rule):
),,(),,(),,( 321
3
3321
2
2321
1
1
xxxFuxxxFuxxxFu === and,
83

In an orthogonal curvilinear coordinate system (i.e., the three coordinate surfaces
are everywhere mutually perpendicular) we have xi••••xj =0 (for i≠j). Hence the square of
the element of length becomes:
23
33
22
22
21
11
2
)()()( udgudgudgsd ++=
Note that du2=du3=0 when the element of length ds is along u1. We may therefore write:
3
3
3
333
2
2
2
222
1
1
1
111 udhudgsdudhudgsdudhudgsd ====== and;
where hi =√(gii) are called scale factors.
2017
MRT
We now develop a method for determining gii (i.e., hi) when the xi are known. In
Cartesian coordinates, g11=g22=g33=1 (e.g., we get ds2 =(du1)2+(du2)2 +(du3)2 ) and:
∑∑ ∑∑ ∑∑∑ = = ===








∂
∂
∂
∂
=
















∂
∂








∂
∂
==
3
1
3
1
3
1
3
1
3
1
2
i j
ji
k
j
k
i
k
k j
j
j
k
i
i
i
k
k
kk
udud
u
x
u
x
ud
u
x
ud
u
x
xdxdsd
Comparing this equation with ds2=Σijgij dui du j, we see that:
∑= ∂
∂
∂
∂
=
3
1k
j
k
i
k
ji
u
x
u
x
g
∑=








∂
∂
=
3
1
2
k
i
k
ii
u
x
g
or when i=j:
84

The area and the volume elements are given by (no sum implied):
)(
ˆ
)(
ˆ
)(
ˆ
ˆˆˆ 321
21
3
231
31
2
132
32
1
332211 Fhh
hh
Fhh
hh
Fhh
hh
FFF
eee
eeeF ++++++++++++++++ ==
With this, the divergence of the vector F is given by:








∂
∂
+
∂
∂
+
∂
∂
=
•+•+•=•
)()()(
1
)(
ˆ
)(
ˆ
)(
ˆ
321323121321
321
321
21
3
231
31
2
132
32
1
Fhh
u
Fhh
u
Fhh
uhhh
Fhh
hh
Fhh
hh
Fhh
hh
∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇
eee
F
where we have used the vector identity ∇∇∇∇•(φX)=X•∇∇∇∇φ +φ ∇∇∇∇•X and the fact that
∇∇∇∇•(ê1/h2h3)=∇∇∇∇•(ê2/h1h3)=∇∇∇∇•(ê3/h1h2)=0 (Exercise - Hint: First show that êi = hi∇∇∇∇ui).
2017
MRT
An arbitrary vector F, in terms of general curvilinear components, can be written as:
321
321 udududhhhdududhhd ji
jiji == τσ and
Since ∂f /∂s is the component of ∇∇∇∇f in the ds direction (remember that we said earlier
that ‘the gradient of f is a vector whose component in any direction, dr, is the derivative
of f with respect to r’), the gradient of an arbitrary function f, ∇∇∇∇f, is given by:
3
3
3
2
2
2
1
1
1
3
3
2
2
1
1
3
1
ˆˆˆ
ˆˆˆˆ
u
f
hu
f
hu
f
hs
f
s
f
s
f
s
f
f
ii
i
∂
∂
∂
∂
∂
∂
=
∂
∂
∂
∂
∂
∂
=
∂
∂
= ∑=
eee
eeee ++++++++++++++++∇∇∇∇
where êi are units vectors along dsi and dsi =hi dui =√(gii) dui (still no sum implied).
85

When the Laplacian operator is applied to a scalar function f , it is given by:
















∂
∂
∂
∂
+








∂
∂
∂
∂
+








∂
∂
∂
∂
=








∂
∂
+
∂
∂
+
∂
∂
•=•=∇
3
3
21
32
2
31
21
1
32
1
321
3
3
3
2
2
2
1
1
12
1
ˆˆˆ
u
f
h
hh
uu
f
h
hh
uu
f
h
hh
uhhh
u
f
hu
f
hu
f
h
ff
eee
∇∇∇∇∇∇∇∇∇∇∇∇
In a similar way (Exercise), we find the curl of the vector F:
332211
321
332211
321
ˆˆˆ
1
FhFhFh
uuu
hhh
hhh ∂
∂
∂
∂
∂
∂
=
eee
F××××∇∇∇∇
since F=(ê1/h1)(h1F1)+(ê2/h2)(h2F2)+(ê3/h3)(h3F3) and ∇∇∇∇××××(ê1/h1)=∇∇∇∇××××(ê2/h2)=∇∇∇∇××××(ê3/h3)=0
(Exercise).
In other words, the Laplacian is ‘the divergence of a gradient’ – in this case of a scalar
function f – given in orthogonal curvilinear coordinates (Exercise: Verify in Cartesian).
In case you were wondering, here are a few other gradient related vector identities:
∇×∇∇×∇∇×∇∇×∇φ =0,∇∇∇∇•∇×∇×∇×∇×A=0and∇×∇×∇×∇×∇×∇×∇×∇×A=∇∇∇∇∇∇∇∇•A−−−−∇2Aandpositionones:∇∇∇∇•r=3and∇×∇×∇×∇×r=0.
2017
MRT
86

11cossin
cos)sin(cossin
cossinsincossin
)cos()sinsin()cossin(
111
22
11
2222
22222
222
2222
1
3
2
1
2
2
1
13
1
2
111
===+=
++=
++=






∂
∂
+





∂
∂
+





∂
∂
=






∂
∂
+





∂
∂
+





∂
∂
=








∂
∂
+








∂
∂
+








∂
∂
=








∂
∂
= ∑=
ghg
r
r
r
r
r
r
r
z
r
y
r
x
u
x
u
x
u
x
u
x
g
k
k
andθθ
θϕϕθ
θϕθϕθ
θϕθϕθ
We will now invest some time expanding the details of this method in an example that
will come back frequently – spherical coordinates (r,θ andϕ) and their associated ope-
rators and/or functions – and proceed to calculate the resulting Laplacian operator. Use:
iii
k
i
k
ii gh
u
x
g =








∂
∂
= ∑=
and
3
1
2
θϕθϕθ cossinsincossin rzryrx === and;
which are the transformation relations between x, y & z and r,θ &ϕ.
and defining the coordinates as x1 =x, x2 =y & x3 =z and u1 =r, u2 =θ & u3=ϕ we get:
2017
MRT
First, recall that:
87

rghrrg
r
rrr
rrr
zyx
u
x
u
x
u
x
u
x
g
k
k
===+=
++=
++=






∂
∂
+





∂
∂
+





∂
∂
=






∂
∂
+





∂
∂
+





∂
∂
=








∂
∂
+








∂
∂
+








∂
∂
=








∂
∂
= ∑=
222
2222
22
22222
22222222
222
2222
2
3
2
2
2
2
2
13
1
2
222
)sin(cos
]sin)sin(cos[cos
sinsincoscoscos
andθθ
θϕϕθ
θϕθϕθ
θ
θ
ϕθ
θ
ϕθ
θ
θθθ
θθ
ϕϕθ
ϕθϕθ
θ
ϕ
ϕθ
ϕ
ϕθ
ϕ
ϕϕϕ
sinsin
)]cos(sin[sin
0cossinsinsin
333
22
33
2222
222222
222
2222
3
3
2
3
2
2
3
13
1
2
333
rghrg
r
rr
rrr
zyx
u
x
u
x
u
x
u
x
g
k
k
===
+=
++=






∂
∂
+





∂
∂
+





∂
∂
=






∂
∂
+





∂
∂
+





∂
∂
=








∂
∂
+








∂
∂
+








∂
∂
=








∂
∂
= ∑=
and
2017
MRT
then
and
88

2017
MRT
Secondly, recall that the Laplacian in general curvilinear coordinates was given by:
















∂
∂
∂
∂
+







∂
∂
∂
∂
+







∂
∂
∂
∂
=







∂
∂
+
∂
∂
+
∂
∂
•=•=∇ 3
3
21
32
2
31
21
1
32
1
321
3
3
3
2
2
2
1
1
12 1ˆˆˆ
uh
hh
uuh
hh
uuh
hh
uhhhuhuhuh
eee
∇∇∇∇∇∇∇∇∇∇∇∇
or under the guise of a more compact form:
where:
2
2
22 1
ϕθϕθ ∇+∇=∇
r
rr








∂
∂
+





∂
∂
∂
∂
=∇





∂
∂
∂
∂
=∇ 2
2
2
22
2
2
sin
sin
11
ϕθ
θ
θθ
ϕθ
rr
r
rr
r and
2
2
222
2
2
2
2
2
2
sin
1
sin
sin
11
sin
1
sinsin
sin
1
sin
1sin1
1
sin
sin1
1
ϕθθ
θ
θθ
ϕθϕθ
θ
θ
θ
θ
ϕθϕθ
θ
θ
θ
θ
∂
∂
+





∂
∂
∂
∂
+





∂
∂
∂
∂
=∇














∂
∂
∂
∂
+





∂
∂
∂
∂
+





∂
∂
∂
∂
=














∂
∂⋅
∂
∂
+





∂
∂⋅
∂
∂
+





∂
∂⋅
∂
∂
⋅⋅
=∇
rrr
r
rr
r
r
rr
r
r
r
r
r
rr
rrr
The rest is trivial to calculate. The Laplacian in spherical coordinates is now given by:
89

11sincos
)()sin()cos(
111
22
11
2222222
1
3
2
1
2
2
1
13
1
2
111
===+=






∂
∂
+





∂
∂
+





∂
∂
=





∂
∂
+





∂
∂
+





∂
∂
=







∂
∂
+







∂
∂
+







∂
∂
=







∂
∂
= ∑=
ghg
zzyx
u
x
u
x
u
x
u
x
g
k
k
andϕϕ
ρρ
ϕρ
ρ
ϕρ
ρρρ
Now for cylindrical coordinates (ρ,ϕ andz) and to calculate the resulting Laplacian
operator. Use:
zzyx === and; ϕρϕρ sincos
which are the transformation relations between x, y & z and ρ,ϕ & z.
2017
MRT
So, as before gii =Σk(∂xk/∂ui)2 and hi =√gii and defining the coordinates as x1 =x, x2 =y &
x3 =z and u1 =ρ, u2 =ϕ & u3 =z we get:
ρρ
ϕϕρϕρϕρ
ϕϕ
ϕρ
ϕ
ϕρ
ϕϕϕ
===
+=+=






∂
∂
+





∂
∂
+





∂
∂
=





∂
∂
+





∂
∂
+





∂
∂
=








∂
∂
+








∂
∂
+








∂
∂
=








∂
∂
= ∑=
222
2
22
2222222
2222222
2
3
2
2
2
2
2
13
1
2
222
)cos(sincossin
)()sin()cos(
ghg
zzyx
u
x
u
x
u
x
u
x
g
k
k
and
then
and
11
)()sin()cos(
33333
2222222
3
3
2
3
2
2
3
13
1
2
333
===






∂
∂
+





∂
∂
+





∂
∂
=





∂
∂
+





∂
∂
+





∂
∂
=








∂
∂
+








∂
∂
+








∂
∂
=








∂
∂
= ∑=
ghg
z
z
zzz
z
z
y
z
x
u
x
u
x
u
x
u
x
g
k
k
and
ϕρϕρ
90

2017
MRT
Knowing that the Laplacian in general curvilinear coordinates is given by:
so that the Laplacian in cylindrical coordinates is now given by:














∂
∂⋅
∂
∂
+





∂
∂⋅
∂
∂
+





∂
∂⋅
∂
∂
⋅⋅
=∇
zz 1
111
1
1
11
12 ρ
ϕρϕρ
ρ
ρρ
Then:
















∂
∂
∂
∂
+








∂
∂
∂
∂
+








∂
∂
∂
∂
=∇ 3
3
21
32
2
31
21
1
32
1
321
2 1
uh
hh
uuh
hh
uuh
hh
uhhh














∂
∂
∂
∂
+





∂
∂
∂
∂
+





∂
∂
∂
∂
=∇
zz
z ρ
ϕρϕρ
ρ
ρρ
ϕρ
112
91

Exercise: Generate an expression for the acceleration of a particle in plane polar
coordinates following a fluid particle.
2017
MRT
Solution: Generally, the velocity vector is given by v=vρρρρρ+vϕ ϕϕϕϕ+vz z but since plane polar
coordinates are mentioned we can let go the last term and go with v=vρρρρρ+vϕ ϕϕϕϕ where ρρρρ
and ϕϕϕϕ are unit vectors in the ρ and ϕ directions, respectively, as shown.
t
vvvvvvv
v
t
v
v
tD
D
∂
+∂
+
∂
+∂
+
∂
+∂
=
∂
∂
+
∂
∂
+
∂
∂
==
)ˆˆ()ˆˆ()ˆˆ( ϕϕϕϕρρρρϕϕϕϕρρρρϕϕϕϕρρρρ ϕρϕρϕϕρ
ρ
ϕ
ρ
ϕρρϕρρ
vvvv
a
Now, since ∂ρρρρ/∂ρ=0, ∂ϕϕϕϕ/∂ρ=0, ∂ρρρρ/∂t=0, ∂ϕϕϕϕ/∂t=0, ∂ρρρρ/∂ϕ =ϕϕϕϕ, and ∂ϕϕϕϕ/∂ϕ =−ρρρρ:
Hint: Translate the x, y, z material derivative, D/Dt, to ρ, ϕ, z coordinates.The
acceleration in cylindrical coordinates will then become:
ˆ ˆ ˆ
ˆ ˆ ˆ
ˆ
ϕϕϕϕρρρρ ˆˆ
2








∂
∂
++
∂
∂
+
∂
∂
+








∂
∂
+−
∂
∂
+
∂
∂
=
t
vvvvvv
v
t
vvvvv
v
ϕρϕϕϕϕ
ρ
ρϕρϕρ
ρ
ρϕρρρϕρρ
a
tz
v
v
v
tD
D
z
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
==
vvvvv
a
ϕρρ
ϕ
ρ
z
x
y
ρρρρ ρρρρˆ
ϕϕϕϕˆ
zˆ
ϕ
Substituting v=vρρρρρ+vϕ ϕϕϕϕ:ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
92

Exercise: Consider a steady laminar flow of an incompressible Newtonian fluid in a pipe
in the absence of body forces. The flow is assumed to have a velocity component in the
direction of the axis of the pipe, which is the z-axis as shown. This velocity component is
considered to be a function of ρ,ϕ and z (cylindrical coordinates).
2017
MRT
Solution: The continuity equation (i.e., ∂ρ/∂t +∇∇∇∇•ρv=0 – we are using the ~ here to
highlight the density ρ instead of the cylindrical radial coordinate ρ) demands that the
divergence of the velocity vector vanish (i.e., ∇∇∇∇•v=0) for the flow of an incompressible
fluid. In cylindrical coordinates, this requires:
0
1)(1
=
∂
∂
+
∂
∂
+
∂
∂
z
vvv z
ϕρρ
ρ
ρ
ϕρ
Since there is only a z component of velocity, that is vρ =vϕ=0, only the last partial
derivative is considered, and ∂vz /∂z=0, which upon integration yields vz=vz(ρ,ϕ). But the
dependence of vz on ϕ is ruled out on further assuming symmetry of the fluid flow about
the z-axis, and so vz is established, from conservation of matter and symmetry
considerations, as a function of ρ only:
A solution of the velocity distribution in the fluid as a function of space coordinates, and
the space average of the velocity of the fluid at any cross section in the pipe are desired.
)(ρzz vv =
z
O
Pipe radius, R
ρ
~ ~
~
93

The Navier-Stokes equation in vector form is:
2017
MRT
may be expressed in cylindrical coordinates by use of the expressions for ∇∇∇∇f , ∇∇∇∇•F, and
∇∇∇∇××××F. The Navier-Stokes equation of motion for a compressible fluid with variable
viscosity in cylindrical coordinates is given by (recall that the density here is labelled ρ):
Bp
tD
D
fvv
v
ρµµρ +−•+−= )()(
3
4
××××∇∇∇∇××××∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇
zB
zzzz
B
z
B
z
z
vvv
z
v
z
v
zz
p
tD
vD
vvvvvv
z
vv
z
vvpvv
tD
vD
vvvv
z
v
z
vvv
vpv
tD
vD
)(~111
2
3
2~
)(~1211
22
3
211~
)(~1211
2
3
2~
2
fv
f
v
f
v
ρ
ϕρ
µ
ϕρρ
ρµ
ρρ
µµρ
ρ
ρρϕρρ
µ
ρρϕρ
µ
ρϕρ
µ
ρ
µ
ϕρ
µ
µ
ϕρϕρρ
ρ
ρ
ρϕρρρ
µ
ρ
µ
ρρϕρ
µ
ϕρ
ρ
µµ
ρρρ
ρ
ϕρ
ϕ
ϕϕρϕϕρϕ
ρϕϕρϕ
ρ
ρϕρρϕϕρ
ρϕρ
+
















∂
∂
+
∂
∂
∂
∂
+
















∂
∂
+
∂
∂
∂
∂
+





∂
∂
−•
∂
∂
−
∂
∂
−=
+







−
∂
∂
+
∂
∂
+
















−
∂
∂
+
∂
∂
∂
∂
+
















∂
∂
+
∂
∂
∂
∂
+








−
∂
∂
−•
∂
∂
−
∂
∂
−=







+
+







−
∂
∂
−
∂
∂
+
















∂
∂
+
∂
∂
∂
∂
+
















−
∂
∂
+
∂
∂
∂
∂
+








∂
∂
−•
∂
∂
−
∂
∂
−=








−
∇∇∇∇
∇∇∇∇
∇∇∇∇
~
Dropping out the terms known or assumed to be zero gives:
000 =





∂
∂
∂
∂
+
∂
∂
−=
∂
∂
=
∂
∂
ρ
ρ
ρρ
µ
ϕρ
zv
z
ppp
and,
94

The first two of the preceding equations indicate that the pressure is a function of z
only, so ∂p/∂z=dp/dz; also ∂vz /∂ρ =dvz /dρ =0 since vz =vz(ρ) only. The problem may then
be stated as:
2017
MRT
from axial symmetry (i.e., a maximum). Integrating the first of the above equations
results in:
zd
pd
d
vd
d
d z
µ
ρ
ρ
ρ
ρ
=





with the no-slip condition:
0),,( =zRvz ϕ
and:
0
)],,0([
=
ρ
ϕ
d
zvd z
21
2
1 ln
4
1
)(
1
2
1
CC
zd
pd
vC
zd
pd
d
vd
z
z
++=+= ρρ
µ
ρ
ρ
ρ
µρ
and
The condition from axial symmetry requires that C1 =0, and the no-slip condition yields C2
=−(1/4µ)(dp/dρ)R2. The solution is:
















−= 1
4
1
)( 2
2
2
R
R
zd
pd
vz
ρ
µ
ρ
The profile is shown in the previous Figure as a paraboloid of revolution.
95

ρ
dρ
R dS = 2πρ dρ
z
In the study of fluid mechanics, a space-average velocity is frequently used. This
velocity is designated by vSA and is defined as:
2017
MRT
A
d
v A
SA
ρ
ρ
~
~
∫ •
=
Sv
It is the velocity imagined to prevail uniformly over the surface A for the purpose of
calculating the mass flow rate through the area. If one knows the velocity profile, the
space-average velocity can be obtained by performing the integration indicated by the
definition. For this Exercise, the space-average velocity will be found over an area
transverse to the pipe. The general element of area is shown.
zd
pdR
R
d
Rzd
pdR
R
d
R
R
zd
pd
v
R
A
SA
µρ
ρρ
ρ
µ
ρ
ρ
ρρ
ρ
µ
ρ
ρ
ρ
8π~
1
4
~π2
π~
ˆπ2ˆ1
4
1~
2
2
0 2
22
2
2
2
2
−=








−
=
•
























−
=
∫∫
=
=
kk
This velocity, when multiplied by the cross-sectional area of the pipe and the density
of the fluid, will yield the mass flow rate through the pipe.
From the solution for the velocity profile, the space-average velocity is (on further
assuming that the density ρ is uniform):~
96

Exercise: The inside cylinder shown rotates as a constant angular velocity ω and is
coaxial with the stationary outside cylinder. The space between the two cylinders is
partially filled with a fluid of uniform and steady density ρ and viscosity µ. The velocity of
the Newtonian fluid has only a transverse component vϕ; and this is a function of ρ only.
2017
MRT
Solution: Simplifying the Navier-Stokes equations in cylindrical coordinates by using the
given information (i.e., from the ρ, ϕ and z cylindrical components of the Navier-Stokes
[N-S] equations, respectively), the following equations accrue:
For the orientation of the gravity body force indicated, obtain and expression for the
velocity field vϕ, the pressure p=p(ρ,z), and the equation of the free surface.
0~0
)(1~
2
=+
∂
∂
=





∂
∂
= g
z
p
d
vd
d
dpv
ρ
ρ
ρ
ρρ
µ
ρρ
ρ
ϕϕ
and,
If you haven’t figured out where these come from, I suggest you look over the pre-
vious Exercise and break down the N-S equation components for yourself – 1-by-1!
97
~

R1
R2 vϕ = ωR1
ω
The second equation may be integrated twice to give:
2017
MRT
The velocity profile is given by:
These constants are:
and the constants C1 and C2 are found from the no-slip conditions:
( ) ( )211 0ω RvRRv ==== ρρ ϕϕ forandfor
2
2
2
1
2
2
2
1
22
2
2
1
2
1
1
ωω2
RR
RR
C
RR
R
C
−
−=
−
= and








−
−
=
)(
)(
ω)( 2
2
2
1
2
2
2
1
1
RR
RR
Rv
ρ
ρ
ρϕ
and is shown next.
The result is valid only if ‘end effects’ due to the finite length of the inside cylinder
are neglected. One would not expect the solution to be valid in the region near the
bottom of the inside cylinder.
98
ρ
ρ
ϕ
1
2
21 CCv +=

The pressure distribution may be found from the ρ and z components of the Navier-
Stokes equations by substitution of vϕ into ρvϕ
2/ρ =∂p/∂ρ. This gives, after simplification:
2017
MRT
in which f (z) is an arbitrary function of z that may be determined by starting with the
preceding expression for p (i.e., dp/dz=d f (z)/dz) and substituting it into the expression for
the z component of the Navier-Stokes equations to obtain:
Integrating with respect to ρ (holding z constant), we get:








+−
−
= 3
4
2
2
2
22
2
2
1
4
1
2
)(
ω~
ρρ
ρ
ρ
ρ
RR
RR
R
d
pd
from which:
~
)(
2
ln2
2)(
ω~
2
4
22
2
2
22
2
2
1
4
1
2
zf
R
R
RR
R
p +








+−
−
=
ρ
ρ
ρρ
0~)(
=+ g
zd
zfd
ρ
3
~)( Czgzf +−= ρ
The pressure distribution is given by:
32
4
22
2
2
22
2
2
1
4
1
2
~
2
ln2
2)(
ω~
Czg
R
R
RR
R
p +−








+−
−
= ρ
ρ
ρ
ρρ
The constant C3 requires the specification of the pressure at a given value of ρ and z.
99

If the origin of the coordinate system is selected so that p = pAtm where ρ =R1 and z =0
then:
2017
MRT
The equation of the free surface is given by the locus of points such that p = pAtm, a
constant; and substitution of this condition into the solution gives:
which is indicated on by the solid line on the first Figure.
Atmpzg
R
RR
R
R
RR
R
p +−
















−−





+
−
−
= ρ
ρρ
ρρ
ρ ~11
2
ln2
2)(
ω~
)( 2
1
2
4
212
2
2
1
2
22
2
2
1
4
1
2
















−−





+
−
−
= 2
1
2
4
212
2
2
1
2
22
2
2
1
4
1
2
11
2
ln2
2)(
ω
)(
R
RR
R
R
RRg
R
z
ρρ
ρ
ρ
100

y
x
z
O
dl
d y
d zr
dV
φ
y
x
r
P
ˆ
k
j
i
ˆ
ˆi
kˆ
Constant x plane
Constant
y plane
Constant
z plane
jˆ
ˆ
êz
σx y
êy
σx z
êx
σy z
S
y
x
z
r
O
P
Trajectory
p vˆ
k
i
ki
i
izyx
xx
v
z
F
y
F
x
F
evF ˆ
∂
∂
=•
∂
∂
=•
∂
∂
+
∂
∂
+
∂
∂
=•
σ
σσσσ∇∇∇∇∇∇∇∇∇∇∇∇ and,
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.
kji ˆˆˆ
z
f
y
f
x
f
f
∂
∂
∂
∂
∂
∂
= ++++++++∇∇∇∇
kj
i
kji
F
ˆˆ
ˆ
ˆˆˆ








∂
∂
−
∂
∂






∂
∂
−
∂
∂








∂
∂
−
∂
∂
=
∂
∂
∂
∂
∂
∂
=
y
F
x
F
z
F
x
F
z
F
y
F
FFF
zyx
xyxz
yz
zyx
++++++++
−−−−××××∇∇∇∇
2
2
2
2
2
2
z
f
y
f
x
f
fff
∂
∂
+
∂
∂
+
∂
∂
=•=∇≡ 2
∇∇∇∇∇∇∇∇
kjiFF ˆˆˆ 2222
zyx FFF ∇∇∇=∇≡ ++++++++
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) and
vector field v and a second-order tensor σσσσ (e.g., stress):
The gradient of a scalar function f = f(x,y,z):
2017
MRT
d x
z
2222ˆˆˆ zdydxddzdydxdd ++== ll andkji ++++++++
dl is an infinitesimal differential increment of length:
dV is an infinitesimal differential increment of volume:
The Divergence Theorem is:
The normal surface integral
of σσσσ over a closed boundary:
Green’s Theorem is:
and since ∇∇∇∇φ•n = ∂φ/∂n is
the directional derivative:
ˆ
101

x
x
yP
zˆ
r
ρρρρˆ
ϕϕϕϕˆ
ϕ
θ
Constant z plane
Constant ϕ planeConstant ρ cylinder
z
FFF z
∂
∂
+
∂
∂
+
∂
∂
=•
ϕρρ
ρ
ρ
ϕρ 1)(1
F∇∇∇∇
2222
)( zdddd ++= ϕρρl
Point P at the tip of the distance vector r is determined in Cylindrical Coordinates by the intersection of a
constant z (i.e., a plane), constant ϕ (i.e., a half-plane) and constant ρ (i.e., a cylinder) surfaces.
x
z
y
zρrr ˆˆˆ),,( zz ++++ϕϕϕϕ++++ϕρϕρ ==
dl is an infinitesimal differential
increment of length.
zρ ˆˆ
1
ˆ
z
fff
f
∂
∂
∂
∂
∂
∂
= ++++ϕϕϕϕ++++∇∇∇∇
ϕρρ
z
ρF
ˆ
)(1
ˆˆ
1








∂
∂
−
∂
∂








∂
∂
−
∂
∂








∂
∂
−
∂
∂
=
ϕρ
ρ
ρ
ρϕρ
ρϕ
ρϕ
FF
z
FF
z
FF zz
++++
++++ϕϕϕϕ−−−−××××∇∇∇∇
)2
FFFF ××××(∇(∇(∇(∇××××∇∇∇∇−−−−∇∇∇∇∇∇∇∇ )•(=∇=
dV is an infinitesimal differential
increment of volume.
2017
MRT
2
2
2
2
2
2 11
z
fff
ff
∂
∂
+
∂
∂
+





∂
∂
∂
∂
=•≡∇
ϕρρ
ρ
ρρ
∇∇∇∇∇∇∇∇
Laplacian of a vector function F = F (ρ,ϕ ,z) (identity)
The Laplacian of a scalar function f = f (ρ,ϕ ,z)
The vector product of a vector function F=F(ρ,ϕ,z)
The scalar product of a vector function F= F(ρ,ϕ,z)
Produces another vector
perpendicular to the plane
formed by ∇∇∇∇ and F with unit
vectors given.
ζ
ρ
x
z
y
O
ϕ ζ
ρdϕ
dρ
dllll
dϕ
dz
ρ
•
•
We can make the following geometric objects into
physical realities if we substitute the scalar current I
for f = f (ρ,ϕ ,z) and the current density J for the
vector F= F (ρ,ϕ,z).
The gradient of a scalar function f = f (ρ,ϕ ,z)
•P
102
ˆi
kˆ
jˆ
O

x
z
yP
θθθθˆ
r
rˆ
ϕϕϕϕˆ
ϕ
θ
Constant r sphere
Constant ϕ plane
Constant θ cone
x
z
y
O
ϕ
θ
•
r
rdθ
dr
dθ dllll
rsinθdϕ
dϕ
ϕθθ
θ
θ
ϕθ
∂
∂
+
∂
∂
+
∂
∂
=•
F
r
F
rr
Fr
r
r
sin
1)sin(
sin
1)(1 2
2
F∇∇∇∇
θθ
ϕθθ
ϕθθ
222222
222222
2222
sin
sin
)sin()(
+=ΩΩ+=
++=
++=
dddrdr
drdrdr
drdrdrd
where
l
•
Point P at the tip of the distance vector r is determined in Spherical Coordinates by the intersection of a
constant θ (i.e., a cone), constant r (i.e., a sphere) and constant ϕ (i.e., a half-plane) surfaces.
x
z
y
r
rrr ˆ),,( rr == ϕθ
P
dl is an infinitesimal differential
increment of length.
ϕϕϕϕ++++++++∇∇∇∇ ˆ
sin
1ˆ1
ˆ
ϕθθ ∂
∂
∂
∂
∂
∂
=
f
r
f
rr
f
f θr
ϕϕϕϕ++++
++++−−−−
−−−−××××∇∇∇∇
ˆ
)(1
ˆ
sin
1)(1
ˆ
)sin(
sin
1






∂
∂
−
∂
∂








∂
∂
−
∂
∂








∂
∂
−
∂
∂
=
θ
ϕθ
ϕθ
θ
θ
θ
ϕ
θϕ
r
r
F
r
rF
r
F
r
rF
r
FF
r
θ
rF
)2
FFFF ××××(∇(∇(∇(∇××××∇∇∇∇−−−−∇∇∇∇∇∇∇∇ )•(=∇=
•
ˆi
O
kˆ
dV is an infinitesimal differential
increment of volume.
2017
MRT
2
2
22
2
2
2
2
sin
1
sin
sin
11
ϕθ
θ
θ
θθ
∂
∂
+
+





∂
∂
∂
∂
+





∂
∂
∂
∂
=•=∇
f
r
f
rr
f
r
rr
ff ∇∇∇∇∇∇∇∇
Laplacian of a vector function F=F (r,θ ,ϕ ) (identity)
The Laplacian of a scalar function f = f (r,θ ,ϕ )
The vector product of a vector function F=F(r,θ ,ϕ )
The scalar product of a vector function F= F(r,θ ,ϕ )
The gradient of a scalar function f = f (r,θ ,ϕ )
Produces
another
vector
perpendicular
to the plane
formed by ∇∇∇∇
and F with
unit vectors
given.
We can make the
following geometric
objects into physical
realities if we substitute
the scalar charge density
ρ for f = f (r,θ,ϕ ) and
the current density J for
the vector F = F (r,θ,ϕ ).
Manifolds
jˆ
103

Here we derive the one-dimensional wave equation for transverse waves in a flexible
string of uniform mass per unit length, µ. The string is of length L, and it is fixed at the
ends. It is stretched by a constant tension T which is so great that the gravitational force
(the weight of the string) on the string can be neglected in comparison with the tension.
2017
MRT
Suppose the string is distorted at time t =0, released, and allowed to vibrate in the xy-
plane (see Figure). The main problem is to characterize the vibration of the string. This
motion is represented by u(x,t) for t >0.
2017
MRT
or:
For small oscillations, we may use the small-angle approxim-
ation (i.e., sinθ ≈tanθ ≈θ in radians; hence T1 ≈T2) and obtain:
T1
T2
Ty2
dy
Ty1
Fy1
≠ Fy2dx
x
y
x + dx
y+dy
ds
θ 1
θ 2
2
2
),(),(),(1
t
txu
Tx
txu
x
txu
x xxx ∂
∂
=














∂
∂
−





∂
∂
∆ ∆+
µ
2
2
12
),(
tantan
t
txu
T
s
∂
∂∆
=−
µ
θθ
For equilibrium, we require that 1) for the x-components to have no motion:
22112211 coscos0coscos21
θθθθ TTTTTT xx =⇔=+−=+−
and 2) for the y-components to have motion we use Newton’s second law that the net
force (i.e., −T1 sinθ +T2sinθ) equals mass (i.e.,µ∆s) times acceleration (i.e., ∂2u(x,t)/∂t2 ):
2
2
2211
),(
sinsin21
t
txu
sTTTT yy
∂
∂
⋅∆=+−=+− µθθ
104
Tx1
Tx2

In the limit as ∆x→0, we find the one-dimensional wave equation:
2017
MRT
2
2
22
2
2
2
),(1),(),(
t
txu
vt
txu
Tx
txu
∂
∂
=
∂
∂
=
∂
∂ µ
where v=√(T/µ) which has the dimension of speed. The three-dimensional wave
equation is:
2
2
22
2
2
2
2
2
2 ),(1),(),(),(
),(
t
tu
vz
tu
y
tu
x
tu
tuzyx
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
=∇
rrrr
r
We may attempt to solve the ∂2u/∂t2 =v2 ∂2u/∂x2 differential equation by assuming that
u may be written as a product of a function of t times a function of x. The assumption is
justified if a solution is obtained:
)()(),( tTxXtxu =
or:
2
2
2
2
2
)(
)(
)(
)(
x
xX
tTv
t
tT
xX
∂
∂
=
∂
∂
which may be rewritten as:
2
2
2
2
2
)(
)(
1)(
)(
11
x
xX
xXt
tT
tTv ∂
∂
=
∂
∂
105

The left-hand side of the equation is independent of x, the right-hand side is
independent of t, and since they are equal, each must be independent of both x and t,
and therefore constant. Let the constant be written as −ω2. Thus:
2017
MRT
2
2
2
2
ω
)(
)(
11
−=
∂
∂
t
tT
tTv
or:
tvAtvAtT ωsinωcos)( 21 +=
0)(ω
)( 22
2
2
=+
∂
∂
tTv
t
tT
where v2=T/µ and whose solution is:
where A1 and A2 are undetermined constants. We also have the equation for X(x):
xBxBxX ωsinωcos)( 21 +=
where B1 and B2 are also undetermined constants.
Since the string cannot vibrate at x=0 and x=L, X(x)=X(L)=0. Therefore, B1 =0 and ωL
must be integral multiple of π:
∑
∞
=






=
1
2
π
sin)(
n
n
L
xn
BxX
where n is an integer and there is a solution for every value of n.
106

The time dependence of the solution is determined by the initial conditions – the initial
values of u and ∂u/∂t between x=0 and x=L. If we assume the string is vibrating at
maximum amplitude at t=0, then A2 =0, and we have (N.B., u≠0 and ∂u/∂t=0):
2017
MRT
∑
∞
=








=
1
1
π
cos)(
n
n
L
tnT
AtT
µ
Let A1nB2n =Cn; then the final complete ‘periodic’ solution to the motion may be written:
The Cn depends on the shape in which the string is made to vibrate (e.g., whether it is
plucked in the middle or near the end) and on the amplitude of vibration, and need not
concern us for the present. What is important is that the frequencies (inverse of the
period of the time dependence) are some discrete multiple of √(T/µ)(π/n); only these
frequencies are permitted, as a result of the boundary conditions on a string of finite
length.
∑
∞
=












==
1
π
cos
π
sin)()(),(
n
n
L
tvn
L
xn
CtTxXtxu
107

The equation which characterizes the motion of a vibrating membrane is:
2017
MRT
2
2
22
2
2
2
2 1
t
u
vy
u
x
u
uyx
∂
∂
=
∂
∂
+
∂
∂
=∇
where v =√(T/µ), µ is the mass per unit area of the membrane, T is the tension,u=u(x,y,t)
is the displacement of the membrane at [x,y] and time t and ∇xy
2 is the Laplacian
operator in Cartesian coordinates (N.B., just x and y here). We will develop a solution of
this differential equation for a circular membrane (i.e., a drum).
or:
2
2
2
111
t
u
v
uu
∂
∂
=














∂
∂
∂
∂
+





∂
∂
∂
∂
ϕρϕρ
ρ
ρρ
Here it is convenient to use cylindrical coordinates (i.e., [ρ,ϕ] and here z = 0) where
the Laplacian calculated earlier from general curvilinear coordinates:
2
2
22
2
22
2
111
t
u
v
uuu
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
ϕρρρρ














∂
∂
∂
∂
+





∂
∂
∂
∂
=∇→
∂
∂
+
∂
∂
=∇
ϕρϕρ
ρ
ρρ
ϕρ
112
2
2
2
2
2
yx
yx
In this case, the equation of motion becomes:
108

We now use the method of separation of variables and assume that the solution of the
above partial differential equation has the form:
2017
MRT
)()()(),,( tTtu ϕρϕρ ΦΡ=
On substituting the right-hand side of the above equation into the original partial
differential equation to be solved and then dividing the resulting equation by the right-
hand side of the above equation, we obtain:
where:
2
2
2
22
2
22
2
1111111
α
ϕρρρρ
−==
Φ
Φ
+
Ρ
Ρ
+
Ρ
Ρ td
Td
Tvd
d
d
d
d
d
tBtAtTT
td
Td
ωsinωcos)(0ω2
2
2
+=⇒=+
and ω2 =v2α2. On multiplying (1/v 2)(1/T )d2T/dt2 =−α 2 above by ρ2, we find that:
2
2
2
2
2
2 111
n
d
d
d
d
d
d
=
Φ
Φ
−=
Ρ
Ρ
+
Ρ
Ρ ϕρ
ρ
ρ
ρ
and:
ϕϕϕ
ϕ
nDnCn
d
d
sincos)(02
2
2
+=Φ⇒=Φ+
Φ
109

Multiplying(1/v 2)(1/T )d2T/dt2 =−α2 withρ2Ρ and using −(1/Φ)d2Φ/∂ϕ2 =n2 above,we get:
2017
MRT
0)( 222
2
2
2
=Ρ−+
Ρ
+
Ρ
n
d
d
d
d
ρα
ρ
ρ
ρ
ρ
By means of the change of variable ξ=αρ, this equation above becomes Bessel’s
(1784-1846) equation :
0)( 22
2
2
2
=Ρ−+
Ρ
+
Ρ
n
d
d
d
d
ξ
ξ
ξ
ξ
ξ
with solution called Bessel functions:
)()()( ξξξ nn NFJE +=Ρ
with E and F arbitrary constants and where Jn(ξ) is the Bessel function of the first kind:
for j=0,1,2,…. Consider also n!=n(n−1)L2⋅1 as the factorial and Γ(p)=∫0to∞exp(−t)t p−1dt
(for any p>0) is the gamma function and Nn(ξ)=[Jn(ξ)cos(nπ) − J−n(ξ)]/sin(nπ) are the
Neumann functions which are the Bessel function of the second kind.
∑∑
∞
=
+∞
=
+
+
++Γ
−
=
++Γ
−
=
0
2
0
2
2
)1(!
)2/()1(
)1(!2
)1(
)(
j
njj
j
nj
njj
n
jnjjnj
J
ξξ
ξ
Exercise: Using the power series method, find Jn(ξ) above as a function of the
gamma functions using the fact that (n+1)(n+2)L(n+j)=(n+j)(n+j−1)L(n+2)(n+1)
⋅Γ(n+1)/Γ(n+1)=Γ(n+j+1)/Γ(n+1) and choose a0=1/[2n ⋅Γ(n+1)] for integer n.
110

The solution of the equation of motion of a vibrating membrane is therefore:
2017
MRT
)]ωsin()ωcos([)]sin()cos([)(),,( tBtAnDnCJEtu n +⋅+⋅= ϕϕξϕξ
since we require that u(ρ,ϕ,t) be finite at ξ=αρ =0 (i.e., Nn(ξ=0)→∞).
Here we discuss a few properties of the Bessel functions. The Jn≠0(ξ) functions all
behave like sinξ with decreasing amplitude. The J0(ξ) function equals 1 at ξ=0 and looks
like a damped cosine. As ξ→∞, we have:












+−≅
2
π
2
1
cos
π
2
)( nJn ξ
ξ
ξ
while as ξ→0, we have:
n
n
n
J 





+Γ
≈
2)1(
1
)(
ξ
ξ
where Γ(n+1)=∫0to∞exp(−t)t ndt=n!.
At the edge of the membrane ρ =r (i.e., the radius of the drum), u(ρ,ϕ,t)=0, or Ρ(ξ)=
EJn(ξ)=0 (i.e., a circular node). By use of the properties of the Bessel functions, we
know that there are nodes at αρ=ξk where ξk are values of the ξ for which Jn(ξ) has a
zero. Hence a single term in the solution corresponds to a standing wave whose nodes
are concentric circles. The complete solution is obtained by summing over all such
modes of vibration.
111

y
x
z
O
r
A P
dr
F(r) = F1 + F2Γ
∫ •=
B
P
dV rrFr )()(
∫ •=∆
B
A
dK rrF )(
B
F2
F1
The line integral, IΓ , of a vector F along a curve Γ from A to B is defined as the definite
integral of the scalar component of F in the direction of the tangent to the curve at P(r)
where dr is an element of displacement at P(r):
We can show that the work done on an object by a resultant force F during a displace-
ment from A to B is equal to the change in the kinetic energy (T=½mv2) of the object.
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since d(v• v)/dt =(d v/dt)•v+ v•dv/dt =2v•dv/dt. If the force
F is the gradient of a single-valued scalar function V(r) and
the path of integration is closed, we obtain Work=0.
When an object at P is subjected to a conservative
(E=T+V=constant) force F(r), it is said to possess
potential energy V(r) relative to some fixed point, B.
∫ •=
B
A
drrF )(Work
Tvvmd
m
dmtd
td
d
m AB
t
t
t
t
t
t
B
A
B
A
B
A
∆=−=•=•=•= ∫∫∫ )()(
2
22
2
1vvvvv
v
Work
Zero work means that the final kinetic energy of the object is
equal to the initial kinetic energy. Hence energy is conserved;
such a force, F(r) = −∇∇∇∇V, is said to be a conservative force. The
conservation of total energy (E) is expressed by: E = T +V.
The work is given by ∫F(r)•dr where F=ΣiFi =ma =mdv/dt
(Newton’s second law above) and dr = (dr/dt)dt= vdt. The
expression for the work done above can be put in the form:
Note also that the line integral from A to B is the negative of that from B to A. For
instance, the work done by a variable force F(r) in moving an object from A to B is
defined as (see Figure):
rF∫Γ
Γ •= dI
VTE +=
The Total Energy is:
The Line Integral
112

If the force F acing on a particle moves through a distance dr, the work done is equal
to the scalar product F••••dr. The work done is going from r1 to r2 is then:
∫ •=
2
1
r
r
rF dW
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By substituting for F and changing the variable of integration to time by dr=vdt, the
expression for work becomes:
The scalar quantity ½mv2 is called the kinetic energy of the particle, so the work done on
the particle by the force is equal to the change in kinetic energy of the particle.
2
1
2
2
22
2
2
1
2
1
2
1
2
1
2
1)( 2
1
2
1
2
1
2
1
2
1
2
1
vmvm
vmvdmtd
td
vd
mtd
td
d
mtd
td
d
mdW
v
v
v
v
t
t
t
t
t
t
−=
===
•
=•=•= ∫∫∫∫∫
vv
v
v
rF
r
r
We now define a conservative force system as one in which the work done is a
function only of the position, and independent of the path taken by the force. It follows
then that the work done by a conservative force system around any closed path must be
zero:
0=•= ∫ rF dW
113

We will now define the potential energy V(r1) as the work done by the conservative
force in going from point r1 to some reference point ro:
)( 1
o
1
rrF
r
r
Vd =•∫
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Thus every point in space can be assigned a scalar potential V(r) which will depend on
the reference point.
Consider next the work done by a conservative force in going from r1 to r2. Since the
work done is independent of the path taken, we can go from r1 to r2 as follows:
Thus the work done (in a conservative system) in going from r1 to r2 is the difference in
the scalar potential −[V(r2)−V(r1)], and it is evident that the result is independent of the
reference point. In terms of the differential displacement, the above equation can be
written as:
)()( 21
o
2
o
1
2
o
o
1
2
1
rrrFrFrFrFrF
r
r
r
r
r
r
r
r
r
r
VVddddd −=•−•=•+•=• ∫∫∫∫∫
Vdd −=• rF
which expresses the conservative force in terms of the potential or the potential energy.
This discussion clearly indicates why the reference point for the potential energy is
arbitrary in setting up the differential equations of motion, which are force or moment
equations
114

In a conservative system, the total energy is constant. If we designate the kinetic
energy by the letter T, the equation W=∫F•dr =½mv2
2 −½mv1
2 above can be written as:
)( 1212
2
1
VVTTd −−=−=•∫
r
r
rF
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which illustrates the principle of conservation of energy for the conservative system.
As an example of a conservative force system, we have the gravitational attraction of
the Earth, which is inversely proportional to the square of the distance from the Earth’s
center:
where g and R are the acceleration of gravity at the Earth’s surface and radius of the
Earth, respectively. If we use the Earth’s surface as the reference point, the potential
energy, or potential of a mass m at height h above the Earth’s surface is:
2






−=
r
R
mgF








−≈





+=





+
−=•





−=
−
+∫ R
h
gmhgm
R
h
hgm
hRR
Rgmd
r
R
mhV
R
hR
21
2
2
1
11
)( rg
using the Binomial approximation(1+x)α ≈1 +αx. For the moderate heights h above the
Earth’s surface, h/R is small, and we have for the potential energy the simple equation:
hgmhV ≅)(
115

As another example, let us consider some relativity with the ultimate goal to finding
where Einstein’s famous E=mc2 equation comes about by showing how it is calculated.
2
o
1 





−
=
c
v
m
m
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One of the major developments to come out of the Special Theory of Relativity is that
the mass of a body will vary with its velocity. Einstein showed that all observers will find
classical momentum principles to hold if the mass m of a bodyvarieswithits speed v=|v|
according to:
where mo, the rest mass, is the mass of the body measured when it is at rest with
respect to the observer and c is the speed of light (N.B., the speed of light is a universal
constant wherever you are or how fast you are moving).
The classical expression of Newton’s second law is that the net force on a body is
equal to the rate of change of the body’s momentum, p=mv, with m the relativistic mass.
To include relativistic effects, allowances must be made for the fact that the mass of a
body varies with its velocity.Thus the relativisticgeneralizationof Newton’s second law is:








−
===
2
o
)(1
)(
cv
m
td
d
td
md
td
d vvp
F
116

In relativistic mechanics, as in classical mechanics, the kinetic energy, T, of a body is
equal to the work done by an external force in increasing the speed of the body from
zero to some value v:
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Using Newton’s second law in one linear dimension (say along z):
∫
=
=
•=
vv
v
dT
0
rF
∫∫
∫∫∫∫∫
+=+=
=====
=
=
=
=
zz
zzzzz
v
zzz
v
zzz
v
zz
v
z
v
z
vv
v
z
vv
v
z
mdvdvvmvmdvdvm
vmvd
td
zd
mvdzd
td
mvd
zd
td
pd
zdFT
0
2
0
00000
)()(
)()(
)(
From the expression for the variation of mass velocity we have:
22
o
2222
2
o
)(1
cmvmcm
cv
m
m z
z
=−
−
= or
Taking differentials of both sides of this expression, we obtain:
mdcmdvvdvmmdmvvdvmmdcm zzzzzz
22222
0222 =+⇒=−=
The left-hand sideofthis last equation is exactly the integrand of our integral for T above:
2
o
222
0
2
)()(
o
oo
cmmmcmdcmdcmdvdvvmT
m
m
m
m
m
m
v
zzz
z
−====+= ∫∫∫
117

So, our result is:
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MRT
Basically, the kinetic energy, T, represents the difference between the total energy, E, of
the moving particle and the rest energy, Eo, of the particle when at rest, so that:
If the rest energy is chosen so that Eo=moc2, we obtain Einstein’s famous relation:
2
o
2
cmcmT −=
2
o
2
o cmcmEE −=−
2
cmE =
which shows the equivalence of mass and energy. Thus, even when a body is at rest it
still has an energy content given by Eo=moc2, so that in principle a massive body can be
completely converted into another, more familiar, form of energy. More of this much later.
Exercise: A Uranium 235U92 nucleus (i.e., with T½ =7.13×108 y is an isotope of uranium
making up about 0.72% of natural uranium), when it fissions, releases 200 MeV of ener-
gy (1 MeV=106 eV, 1 eV=1.602×10−19 J). What percentage of the total energy available is
this? If 1u=1.66×10−27 Kg=931.5 MeV/c2, how much 235U92 mass was converted?
Solution: The rest mass of a 235U92 atom, in terms of the unified atomic mass unit (u), is
approximately 235 u. Using the conversion 1u=931.5 MeV, we have: [Total available
energy]=[Rest mass of 235U92]=(235 u)⋅(931.5 MeV/u)=219×103 MeV. This gives us the
[% total energy]=200 MeV/219×103 MeV×100%=0.0913%~0.1%, i.e., a tenth of a
percent of the fissionable material (about 2×10−12 Kg) was converted to pure energy!
118

Since momentum is conserved, but not velocity, it is often useful to express the energy
of a body in terms of its momentum rather than its velocity. To this end, if the expression:
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or:
22
o
22
o
22
)()()( cmcpEcpE +=+=
22
o
222
o )()()( cmcpcmT +=+
Using the results E=mc2, Eo=moc2, and p=|p|=mv, we find the desired relationship
between E and p to be:
2
o
)(1 cv
m
m
−
=
is squared and both sides are multiplied by c4[1−(v/c)2], one obtains:
42
o
22242
cmcvmcm =−
119

The normal surface integral of a vector function F(r) over a closed boundary is
defined as the surface (double) integral of the scalar component of F in the direction of
a normal to the surface S. This is written:
where dS is the area of an element of the boundary surface and n is a unit outward
normal to this element of area. The symbol means double (in general, over any two
coordinates) integration around a closed surface (indicated by the loop around ∫∫S ).
The sign convention for the unit normal n is a follows:
1) For a closed surface (i.e., a surface which encloses a volume), the outward normal
is called positive;
2) For an open surface, the right-hand screw rule is used; the direction of rotation is the
same as that in which the periphery is traversed.
The divergence theorem due to Gauss (1777-1855) is as follows: The normal
surface integral of a vector function F over the boundary of a closed surface dS of
arbitrary shape is equal to the volume (triple) integral of the divergence of F (i.e., ∇∇∇∇••••F)
taken throughout the enclosed volume dV. In equation form, we write:
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∫∫∫∫ •=•≡
SS
S dSdI SFnF ˆ
ˆ
ˆ
∫∫∫∫∫ •=•
VS
dVd FSF ∇∇∇∇
Vector Theorems
120

To prove this theorem, we first expand the right-hand side of the divergence theorem
and obtain:
Although the theorem is valid for an arbitrary shaped closed surface, we choose the
volume in the Figure for convenience (weird – but it is easier than a 3-sphere!) In the
Figure, we have:
∫∫∫∫∫∫ 







∂
∂
+
∂
∂
+
∂
∂
=•
V
zyx
V
zdydxd
z
F
y
F
x
F
dVF∇∇∇∇
4
ˆ Si dzdyd =−y
x
z
dx
dz
dy
−iˆ
−kˆ
jˆ
iˆ
kˆ
x″
y″
(x′,y′,z′)
z″
(x″,y″,z″)
1
ˆ Sk dydxd =−
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zdxdd
zdydd
ydxdd
jS
iS
kS
ˆ)(
ˆ)(
ˆ)(
3
2
1
−=
+=
−=
back
top
left
zdxdd
ydxdd
zdydd
jS
kS
iS
ˆ)(
ˆ)(
ˆ)(
6
5
4
+=
+=
−=
front
right
bottom
(1)
(See Figure)
121

(2)
(4)
Integrating the last term on the right-hand side of Eq. (1) with respect to z from z′ to z″,
we obtain:
∫∫∫∫∫∫∫∫∫ ′−′′=′−′′=
∂
∂
′′
′ 15
),,(),,()],,(),,([
S
z
S
z
S
zz
z
z
z
ydxdzyxFydxdzyxFydxdzyxFzyxFzdydxd
z
F
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∫∫∫∫∫∫∫∫∫ •=•′−•′′=





∂
∂
′′
′
S
z
S
z
S
z
z
z
z
SdFdzyxFdzyxFydxdzd
z
F
nkSknSkn ˆˆˆˆ),,(ˆˆ),,(
15
1155
Note that:
)(ˆˆ
)(ˆˆ
55
11
right
left
ydxdd
ydxdd
+=•
−=•
Skn
Skn
Using Eq. (3) in Eq. (2), we write:
(3)
where:
0)( =∫∫sides
L
since k is perpendicular to dS2, dS3, dS4, and dS6.ˆ
122

(6)
(5)
Similarly, we can show that:
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MRT
∫∫∫∫∫∫∫∫∫ •=•′−•′′=





∂
∂
′′
′
S
x
S
x
S
x
x
x
x
SdFdzyxFdzyxFzdydxd
x
F
niSinSin ˆˆˆˆ),,(ˆˆ),,(
42
4422
and:
∫∫∫∫∫∫∫∫∫ •=•′−•′′=







∂
∂
′′
′
S
y
S
y
S
y
y
y
y
SdFdzyxFdzyxFzdxdyd
y
F
njSjnSjn ˆˆˆˆ),,(ˆˆ),,(
36
3366
Comparing Eqs. (5), (6), and (4), respectively (i.e., with the right-hand side of Eq. (1)),
we obtain:
QED
ˆ)ˆˆˆ(
ˆˆˆˆˆˆ
∫∫
∫∫
∫∫∫∫∫∫∫∫∫∫∫∫
•=
•=
•+•+•=





∂
∂
+
∂
∂
+
∂
∂
=•
′′
′
′′
′
′′
′
S
S
zyx
S
z
S
y
S
x
z
z
y
y
x
x
zyx
V
d
SdFFF
SdFSdFSdFzdydxd
z
F
y
F
x
F
dV
SF
nkji
nknjniF
++++++++
∇∇∇∇
Hence the theorem is proved.
(7)
123

Green’s theorem (1793-1841) is an important corollary of the divergence theorem,
and it has numerous applications in various branches of physics.
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Let ψ and φ be two scalar functions of position with continuous derivatives within a
certain region bounded by a closed surface S. On applying the divergence theorem to
the vector ψ ∇∇∇∇φ in this region, we obtain (i.e., substitute F=ψ ∇∇∇∇φ and dS=ndS in Eq.
(7)):
ˆ
∫∫∫
∫∫∫∫∫∫∫∫
•+=
•+•=•=•
V
VVS
dV
dVdVSd
)(
)()(ˆ
2
φψφψ
φψφψφψφψ
∇∇∇∇∇∇∇∇∇∇∇∇
∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇ n
or:
∫∫∫∫∫ •+=
∂
∂
VS
dVSd
n
)( 2
φψφψ
φ
ψ ∇∇∇∇∇∇∇∇∇∇∇∇
where:
φψφψφψ ∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇ •+•=• )(
and, given the definition of the directional derivative:
n∂
∂
=•
φ
φ nˆ∇∇∇∇
Eq. (8) is known as Green’s theorem of the first form.
(8)
(9)
124

A second form of Green’s theorem is obtained if we consider the following two
equations:
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φψφψφψ ∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇ •+•=• )(
On subtracting Eq. (11) from Eq. (10) and integrating over an arbitrary volume, we
obtain:
∫∫∫∫∫∫ −=−•
VV
dVdV )()( 22
ψφφψψφφψ ∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇
Converting the left-hand side of Eq. (12) to a surface integral by use of the divergence
theorem, we obtain:
Eq. (13) is the Green’s theorem of the second form.
(10)
and:
ψφψφψφ ∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇∇ •+•=• )( (11)
(12)
∫∫∫∫∫ −=





∂
∂
−
∂
∂
VS
dVSd
nn
)( 22
ψφφψ
ψ
φ
φ
ψ ∇∇∇∇∇∇∇∇ (13)
125

A third and equally important vector integration theorem is due to Stokes.
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∫∫∫ •=•
S
dd SFΓF ××××∇∇∇∇
Γ
The curl theorem is as follows: If F and its first derivatives are continuous, the line
integral of F around a closed curve Γ is equal to the normal surface integral of curlF
(i.e., ∇∇∇∇××××F) over an open surface bounded by Γ. In other words, the surface integral of
curlF taken over any open surface S is equal to the line integral of F around the
periphery Γ of the surface. In equation form, we write:
So, we see that the surface integral of the curl of a vector field over an open surface is
equal to the closed line integral of the vector along the contour bonding the surface. In a
nutshell, Stoke’s theorem converts a surface integral of the curl of a vector to a line
integral of the vector, and vice versa.
There are two special cases to consider:
0000∇∇∇∇××××∇∇∇∇∇∇∇∇ =≡ )( ffcurl
1) The curl of a gradient of any scalar field f is identically zero:
2) The divergence of the curl of any vector field F is identically zero:
0)(div =•≡ FFcurl ××××∇∇∇∇∇∇∇∇
126

To prove this theorem, we first expand the right-hand side of the curl theorem above
(i.e., ∫ΓF•dΓΓΓΓ=∫∫S∇∇∇∇ ××××F•dS); it becomes:
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(14)
∫∫∫∫ •=•
S
zyx
S
SdFFFSd )ˆˆˆ(ˆˆ kjinnF ××××∇∇∇∇++++××××∇∇∇∇++++××××∇∇∇∇××××∇∇∇∇
The last integral on the right-hand side of Eq. (14) reduces to:
∫∫∫∫∫∫ 





∂
∂
•−
∂
∂
•=
∂
∂
∂
∂
∂
∂
•=•
S
zz
S
z
S
z
x
F
Sd
y
F
SdSd
F
zyx
SdF )ˆˆ()ˆˆ(
00
ˆˆˆ
ˆ)ˆ(ˆ jnin
kji
nkn ××××∇∇∇∇
Note that the projection of dS onto the yz-plane (see Figure) leads to:
zdydSd +=• in ˆˆ
(15)
(16)
Now we have to find the n•jdS term.ˆ ˆ
y
x
z
Back
dz
dy iˆjˆ
iˆ
kˆ
A
B
Front
ˆn
dS
Γ
dx
jˆ
127

We now let the line segment AB be the intersection of the surface S with a plane that is
parallel to the xy-plane at a distance z from the origin (see Figure - Left). Along the strip
AB, we find that (notice the order of terms in the negative version, c.f., Eq. (19)):
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xd
x
F
yd
y
F
Fdyd
y
F
xd
x
F
Fd zz
z
zz
z
∂
∂
−
∂
∂
=−⇒
∂
∂
−
∂
∂
=
and:
jir ˆˆ ydxdd −−−−=
(17)
The vector dr is tangent to AB at P (see Figure - Right) and perpendicular to n. We may
therefore write:
ˆ
ydxdd jninnr ˆˆˆˆ0ˆ •−•==•
or:
(18)






=•=•
Sd
zdyd
yd
xd
yd
xd
injn ˆˆˆˆ
or better yet, our sought out answer – the projection of dS onto the xz-plane:
zdxdSd +=• jn ˆˆ
A(x1,y1,z)
B(x2,y2,z)
A
jir ˆˆ ydxdd −=
Plane of constant z
A
B
ˆn dS
P
+z
−z
128

On substituting Eqs. (16) and (18), respectively, into Eq. (15), then Eq. (17) we obtain:
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∫∫ ∫∫∫
∫∫∫∫∫∫
−−=−=−=






∂
∂
−
∂
∂
=





∂
∂
−
∂
∂
=•
zdzyxFzyxFFdzdzdFd
zdxd
x
F
yd
y
F
x
F
zdxd
y
F
zdydSdF
zz
zyxB
zyxA
z
S
z
S
zz
S
zz
S
z
)],,(),,([)(
)()()ˆ(ˆ
1122
),,(
),,(
22
11
kn ××××∇∇∇∇
(19)
The sense of the periphery at A is positive (i.e., dz=dΓz) and it is negative at B (i.e.,
dz=−dΓz) (see previous Figure - Right). Hence Eq. (19) becomes:
∫∫∫∫∫ Γ
=+=• zz
A
zz
B
zz
S
z dFdzyxFdzyxFSdF ΓΓ),,(Γ),,()ˆ(ˆ 1122
444 3444 21444 3444 21
)(partfront)(partback
kn ××××∇∇∇∇ (20)
Similarly, we find:
∫∫∫∫∫∫ ΓΓ
=•=• yy
S
yxx
S
x dFSdFdFSdF Γ)ˆ(ˆΓ)ˆ(ˆ jnin ××××∇∇∇∇××××∇∇∇∇ and
Combining the two Eqs. (21) and Eq. (20) with Eq. (14), we obtain:
(21)
QED
)ΓΓΓ()ˆˆˆ(
∫
∫∫∫∫∫
Γ
Γ
•=
++=•=•
ΓF
SkjiSF
d
dFdFdFdFFFd zzyyxx
S
zyx
S
××××∇∇∇∇++++××××∇∇∇∇++++××××∇∇∇∇××××∇∇∇∇
(22)
and the theorem is proved.
129

As an example in the use of the curl theorem, let us evaluate ∫ΓΓΓΓ (zdx+xdy+ydz), where
Γ is the projection as the cylinder y2 +z2 =1 in the plane x+z=2. According to the right-
hand rule, you should orient Γ counter-clockwise as viewed from above (see Figure).
kjiF ˆˆˆ ++++++++××××∇∇∇∇ =
We recall Stoke’s theorem:
∫∫∫ •=•
S
Sdd nFΓF ˆ)(
Γ
××××∇∇∇∇
and identify F(x,y,z)=zi ++++ xj ++++yk (as well dΓΓΓΓ =dxi ++++ dy j ++++ dzk) so this
allows us to evaluate ∇∇∇∇ ×××× F readily:
The given orientation of Γ corresponds to an upward orientation of
the surface S. Thus, if f (x,y,z)= x+z– 2=0 defines the plane, then
with this the gradient of f (x,y,z) is ∇∇∇∇f =i ++++ k and the upper normal n is:
kiki
ki
n ˆ71.0ˆ71.0ˆ
2
1ˆ
2
1
2
ˆˆ
ˆ ++++++++
++++
∇∇∇∇
∇∇∇∇
====
f
f
Hence, from Stoke’s theorem:
∫∫∫∫∫∫∫∫ ==














•=•
SSSS
SddSSdSd 2
2
2ˆ
2
1ˆ
2
1
)ˆˆˆ(ˆ)( kikjinF ++++++++++++××××∇∇∇∇
which will give us the extra √2 factor (e.g., (∂g/∂y)2 =0 while (∂g/∂z)2
=(–1)2 =1) with π/4 the result of the integral over R (i.e., for y2+z2 =1).
π21842222
1
0
2
1
0
1
0Γ
2
=−=








⋅===• ∫∫ ∫∫∫∫∫∫
−
zdzzdyddASdd
z
RS
ΓF
Now, if z=g(x,y) is the equation of a surface S, then the differential
of the surface area is dS =√[1+ (∂g/∂x)2+(∂g/∂y)2]dA. Thus,
considering the y and z plane here (i.e., g(y,z)=x= 2– z) we use the
relation ∫∫S f (x,y,z)dS =∫∫R f (g(y,z),y,z)√[1+ (∂g/∂y)2 +(∂g/∂z)2]dA:2017
MRT
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ
{ {
{ {
)(
:Since
kiΓkjF
ji
ji
F
F
F
kjiF
ˆˆ)0,1,2(ˆˆ)0,1,2(ˆ
ˆˆ
2
ˆˆ
),0,(
),0,(
),0,(ˆ
ˆˆˆ),,(
45.0
5
5
89.0
5
52
08.1
4
23
36.0
4
22
3
2
1
2
1
2
3
2
1
2
3
2
1
2
3
zdxdd
yxzzyx
++++++++
++++
++++
++++++++
−==
===
=
y
x
z
S
dΓΓΓΓ
R
ˆn
dS
dA
∇∇∇∇××××F
Γ
122
=+ zy
2=+ yx
),0,(ˆ
2
1
2
3
F
jF ˆ)0,0,2(ˆ =
)0,1,2(ˆF
iF ˆ)1,0,0(ˆ =
{ {
jiF ˆˆ)1,0,1(ˆ
71.0
2
2
71.0
2
2
++++=
∫ ∫∫∫
−
=
1
0
1
0
2
4
1
zdyddA
z
R
130

We now present the proofs of two recurring integral relations.
∫∫∫∫∫∫∫ ==
SSV
dSdVd Sn φφφ ˆ∇∇∇∇
To prove this, let F=φC, where φ is a constant scalar and C is a constant vector. On
substituting this for F into the divergence theorem (c.f., Eq. (7)), we obtain:
∫∫∫∫∫ •=•
SV
SdVd nCC ˆ)( φφ∇∇∇∇
Note that ∇∇∇∇•(φC)=C•∇∇∇∇φ, since ∇∇∇∇•C=0. By using ∇∇∇∇•(φC)=C•∇∇∇∇φ, we may write Eq.
(23) in the form:
(23)
∫∫∫∫∫ •=•
SV
SdVd )ˆ( nCC φφ∇∇∇∇
or:
0ˆ =





−• ∫∫∫∫∫ SV
SdVd nC φφ∇∇∇∇
or:
since C is an arbitrary constant vector.
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MRT
QEDˆ
∫∫∫∫∫ =
SV
SdVd nφφ∇∇∇∇
The first relation is:
(25)
(24)
131

The second relation is:
∫∫∫∫∫∫∫ −==
SSV
dSdVd SAAnA ××××××××××××∇∇∇∇ ˆ
To prove this, this time let F=A××××B, where A and B is are arbitrary constant vectors,
and on substituting for F into the divergence theorem (c.f., Eq. (7)), we obtain:
∫∫∫∫∫ •=•
SV
SdVd nBABA ˆ)()( ××××××××∇∇∇∇
The integrand on the left-hand side of Eq. (26) may be written as ∇∇∇∇•(A××××B)=B•∇∇∇∇××××A−
A•∇∇∇∇××××B=B•∇∇∇∇××××A, since B is a constant vector. Also note that A××××B•n=B•(n××××A). If we
substitute these former and latter relations into Eq. (26), it becomes:
(26)
∫∫∫∫∫ •=•
SV
SdVd )ˆ( AnBAB ××××××××∇∇∇∇
since B is an arbitrary constant vector.
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MRT
ˆ ˆ
(27)
0ˆ =





−• ∫∫∫∫∫ SV
SdVd AnAB ××××××××∇∇∇∇
QEDˆ
∫∫∫∫∫ =
SV
SdVd AnA ××××××××∇∇∇∇ (28)
or:
or:
132

Let us consider two examples.
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MRT
In the first example, if S is a closed surface which encloses a volume V, prove that:
0n =∫∫S
Sdˆ
In Eq. (25), let φ=1. We therefore obtain:
QEDˆ
∫∫∫∫∫ ==
SV
SdVd n0φ∇∇∇∇
and the result is proved.
In the second example, if S is a closed surface which encloses a volume V, prove that:
0Sr =∫∫S
d××××
In Eq. (28), let A=r and ndS=dS . We therefore obtain:
since ∇∇∇∇××××r=0 and the result is proved.
QED0rSr ==− ∫∫∫∫∫ VS
Vdd ××××∇∇∇∇××××
ˆ
(29)
(30)
133

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
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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
134
Appendix
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).

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135
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

136
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
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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
137
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138
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
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139
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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140
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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141
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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142
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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143
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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

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References
144
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.

And now, let us enjoy some humor…
As quoted by Sidney Coleman:*
Class: Why not use Feynman’s lecture notes?
Murray Gell-Mann: Because Feynman uses a different method than we do.
Class: What is Feynman’s method?
Murray Gell-Mann: You write down the problem. Then you look at it and you think. Then you write down the answer.
*Murray Gell-Mann (1929- ) is an American physicist who received the 1969 Nobel Prize in physics for his work on the theory of
elementary particles. Sidney Richard Coleman (1937-2007) was an American theoretical physicist who studied under Murray Gell-
Mann. His 1976 course is available online from Harvard at: https://www.physics.harvard.edu/events/videos/Phys253 and lecture
notes by Brian Hill, typeset and edited by Yuan-Sen Ting and Bryan Chen Gin are available at: http://arxiv.org/abs/1110.5013.
Happy face math…
−1
2
3
=
=
=
)sup ( =
∂ ( ) =
sin ( ) =
Re( ) = no ‘i’s
Im ( ) =
∇∇∇∇ ×××× ( ) =
∇∇∇∇( ) =
log ( ) =
‘i ’s

PART I.3 - Physical Mathematics

PART I.3 - Physical Mathematics

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PART I.3 - Physical Mathematics