Transcript CEE451Lecture 1 - Ven Te Chow Hydrosystems Lab

CEE 451G ENVIRONMENTAL FLUID MECHANICS
LECTURE 1: SCALARS, VECTORS AND TENSORS
A scalar has magnitude but no direction.
An example is pressure p.
The coordinates x, y and z of Cartesian space are scalars.
A vector has both magnitude and direction
Let ˆi, ˆj, kˆ denote unit vectors in the x, y and z direction. The hat
denotes a magnitude of unity

The position vector x (the arrow denotes a vector that is not a unit
vector) is given as

x  x ˆi  yˆj  zkˆ
z

x
kˆ
ˆi
x
ˆj
y
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LECTURE 1: SCALARS, VECTORS AND TENSORS

The velocity vector u is given as

 dx dx ˆ dy ˆ dz ˆ
u

i
j
k
dt dt
dt
dt

The acceleration vector a is given as


 du du ˆ dv ˆ dw ˆ d2x d2x ˆ d2y ˆ d2z ˆ
a

i
j
k  2  2 i  2 j 2 k
dt dt
dt
dt
dt
dt
dt
dt
The units that we will use in class are length L, time T, mass M and
temperature °. The units of a parameter are denoted in brackets. Thus

[x ]  L

[u]  LT 1

[a ]  ?
LT 2

Newton’s second law is a vectorial statement: where F denotes the
force vector and m denotes the mass (which is a scalar)


F  ma
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LECTURE 1: SCALARS, VECTORS AND TENSORS
The components of the force vector can be written as follows:

F  Fx ˆi  Fy ˆj  Fzkˆ
The dimensions of the force vector are the dimension of mass times
the dimension acceleration

[F]  [Fx ]  MLT 2
Pressure p, which is a scalar, has dimensions of force per unit area.
The dimensions of pressure are thus
[p]  MLT 2 /(L2 )  ML1T2
The acceleration of gravity g is a scalar with the dimensions of (of
course) acceleration:
[g]  LT2
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LECTURE 1: SCALARS, VECTORS AND TENSORS
A scalar can be a function of a vector, a vector of a scalar, etc. For
example, in fluid flows pressure and velocity are both functions of
position and time:

  
p  p(x, t) , u  u(x, t)
A scalar is a zero-order tensor. A vector is a first-order tensor. A
matrix is a second order tensor. For example, consider the stress
tensor .
  xx

   yx

 zx
xy
 yy
 zy
xz 

yz 
zz 
The stress tensor has 9 components. What do they mean? Use the
following mnemonic device: first face, second stress
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LECTURE 1: SCALARS, VECTORS AND TENSORS
Consider the volume element below.
z
y
x
Each of the six faces has a direction.
For example, this face
and this face
are normal to the y direction
A force acting on any face can act in the x, y and z directions.
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LECTURE 1: SCALARS, VECTORS AND TENSORS
Consider the face below.
z
yy yz
yx
y
x
The face is in the direction y.
The force per unit face area acting in the x direction on that face is the
stress yx (first face, second stress).
The forces per unit face area acting in the y and z directions on that
face are the stresses yy and yz.
Here yy is a normal stress (acts normal, or perpendicular to the face)
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and yx and yz are shear stresses (act parallel to the face)
LECTURE 1: SCALARS, VECTORS AND TENSORS
Some conventions are in order
z
yy yz
yx
yx
yz
yy
y
x
Normal stresses are defined to be positive outward, so the orientation
is reversed on the face located y from the origin
Shear stresses similarly reverse sign on the opposite face face are the
stresses yy and yz.
Thus a positive normal stress puts a body in tension, and a negative
normal stress puts the body in compression. Shear stresses always put
the body in shear.`
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LECTURE 1: SCALARS, VECTORS AND TENSORS
Another way to write a vector is in Cartesian form:

x  x ˆi  yˆj  zkˆ  (x, y, z)
The coordinates x, y and z can also be written as x1, x2, x3. Thus the
vector can be written as

x  (x1, x 2, x3 )
or as

x  (xi ) , i  1..3
or in index notation, simply as

x  xi
where i is understood to be a dummy variable running from 1 to 3.
Thus xi, xj and xp all refer to the same vector (x1, x2 and x3) , as the
index (subscript) always runs from 1 to 3.
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LECTURE 1: SCALARS, VECTORS AND TENSORS

Scalar multiplication: let  be a scalar and A = Ai

Then
A  Ai  (Ai,A2,A3 )
be a vector.
is a vector.
Dot or scalar product of two vectors results in a scalar:
 
A  B  A1B1  A2B2  A3B3  scalar
In index notation, the dot product takes the form
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3
3
 
A  B   AiBi  AkBk  ArBr 
i1
k 1
r 1
Einstein summation convention: if the same index occurs twice, always
sum over that index. So we abbreviate to
 
A  B  AiBi  AkBk  ArBr
There is no free index in the above expressions. Instead the indices are
paired (e.g. two i’s), implying summation. The result of the dot product
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is thus a scalar.
LECTURE 1: SCALARS, VECTORS AND TENSORS
Magnitude of a vector:
2  
A  A  A  Ai Ai
A tensor can be constructed by multiplying two vectors (not scalar
product):
 A1B1

AiB j  ( AiB j ) ,i  1..3, j  1..3   A1B2
A B
 1 3
A 2B1
A 2B2
A 2B3
A 3B1 

A 3B3 
A 3B3 
Two free indices (i, j) means the result is a second-order tensor
Now consider the expression
Ai A jBj
This is a first-order tensor, or vector because there is only one free
index, i (the j’s are paired, implying summation).
Ai A jBj  (A1B1  A2B2  A3B2 )(A1, A2, A3 )
That is, scalar times vector = vector.
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LECTURE 1: SCALARS, VECTORS AND TENSORS
Kronecker delta ij
 1 if
ij  
0 if
 1 0 0

i j 
 0 1 0
i j 

0
0
1


Since there are two free indices, the result is a second-order tensor, or
matrix. The Kronecker delta corresponds to the identity matrix.
Third-order Levi-Civita tensor.
 1 if i, j,k cycle clockwise: 1,2,3, 2,3,1 or 3,1,2

ijl   1 if i, j,k cycle counterclockwise: 1,3,2, 3,2,2 or 2,1,3

otherwise
0

Vectorial cross product:
 
AxB  ijk A jBk
One free index, so the result must be a vector.
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LECTURE 1: SCALARS, VECTORS AND TENSORS

Vectorial cross product: Let C be given as
  
C  AxB
Then
ˆj
ˆi
ˆj
 ˆi
kˆ 
kˆ



C  det A1 A 2 A3   A1 A 2 A3 


B
B
B
B1 B2 B3
2
3
 1
ˆj
ˆj   ˆi
ˆj
 ˆi
kˆ  ˆi
kˆ  ˆi


 

 A1 A 2 A3  A1 A 2    A1 A 2 A3  A1


 

B
B
B
B
B
B
B
B
2
3  1
2
2
3  B1
 1
 1
ˆj 

A2  

B2 
A2B3  A3B2 ˆi  A3B1  A1B3 ˆj  A1B2  A2B1kˆ
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LECTURE 1: SCALARS, VECTORS AND TENSORS
Vectorial cross product in tensor notation:
Ci  ijk A jBk
Thus for example
=1
= -1
=0
C1  1jk A jBk  123 A2B3  132 A3B2  111A1B1 
 A2B3  A3B2
a lot of other terms that
all = 0
i.e. the same result as the other slide. The same results are also
obtained for C2 and C3.

The nabla vector operator  :

 ˆ 

  ˆi
j
 kˆ
x1
x 2
x 3
or in index notation

x i
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LECTURE 1: SCALARS, VECTORS AND TENSORS
The gradient converts a scalar to a vector. For example, where p is
pressure,

p ˆ p ˆ p ˆ
grad(p)  p 
i
j
k
x1
x 2
x 3
or in index notation
grad(p) 
p
x i
The single free index i (free in that it is not paired with another i) in the
above expression means that grad(p) is a vector.

The divergence converts a vector into a scalar. For example, where u
is the velocity vector,

u1 u2 u3 ui uk
div(u) 




x1 x 2 x 3 xi xk
Note that there is no free index (two i’s or two k’s), so the result is a
scalar.
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LECTURE 1: SCALARS, VECTORS AND TENSORS

The curl converts a vector to a vector. For example, where u is the
velocity vector,
ˆi
 


curl(u)  xu 
x1
u1
ˆj

x 2
u2
kˆ


x 3
u3
 u3 u2  ˆ  u1 u3  ˆ  u2 u1  ˆ

 i  
 j  
k



 x 2 x 3   x 3 x1   x1 x 2 
or in index notation,

uk
curl(u)  ijk
x j
One free index i (the j’s and the k’s are paired) means that the result is a
vector
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LECTURE 1: SCALARS, VECTORS AND TENSORS
A useful manipulation in tensor notation can be used to change an index
in an expression:
ijuj  ui
This manipulation works because the Kronecker delta ij = 0 except when
i = j, in which case it equals 1.
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