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 denote

unit

vectors in the x, y and z direction. The

hat

denotes a magnitude of unity The

position vector

vector) is given as  x (the arrow denotes a vector that is not a unit  x  x iˆ  y jˆ  z kˆ x z kˆ iˆ jˆ  x y 1

LECTURE 1: SCALARS, VECTORS AND TENSORS

The

velocity vector

 u   d x dt   u is given as dx dt iˆ  dy dt jˆ  dz kˆ dt The

acceleration

 a   d u dt  du dt iˆ   vector is given as dv dt jˆ  dw dt kˆ  d 2 dt  x 2  d 2 x dt 2 iˆ  d 2 y dt 2 jˆ  d 2 z kˆ dt 2 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  u  [ a ] ] ]    L LT ?

 1 LT  2

Newton’s second law

 is a vectorial statement: where denotes the force vector and m denotes the mass (which is a scalar)  F  m  a 2

LECTURE 1: SCALARS, VECTORS AND TENSORS

The components of the force vector can be written as follows:  F  F x iˆ  F y jˆ  F z kˆ The

dimensions

of the force vector are the dimension of mass times the dimension acceleration [  F ]  [ F x ]  MLT  2 Pressure p, which is a scalar, has dimensions of force per unit area. The dimensions of pressure are thus [ p ]  MLT  2 /( L 2 )  ML  1 T  2 The acceleration of gravity g is a scalar with the dimensions of (of course) acceleration: [ g ]  LT  2 3

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  y x  zx  xy  y y  zy  xz  y z  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 x  yy  yz  yx y 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 and  yx is a

normal stress

and  yz are (acts normal, or perpendicular to the face)

shear stresses

(act parallel to the face) 6

LECTURE 1: SCALARS, VECTORS AND TENSORS

Some conventions are in order z x  yy  yz  yx   yz yx  y yy 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.` 7

LECTURE 1: SCALARS, VECTORS AND TENSORS

Another way to write a vector is in

Cartesian

 x  x iˆ  y jˆ  z kˆ  ( x , y , z ) form: The coordinates x, y and z can also be written as x 1 , x 2 , x 3 . Thus the vector can be written as  x  ( x 1 , x 2 , x 3 ) or as  x  ( x i ) , i  1 ..

3 or in

index notation

, simply as  x  x i where i is understood to be a dummy variable running from 1 to 3.

Thus x i , x j and x p all refer to the same vector (x 1 , x 2 index (subscript) always runs from 1 to 3.

and x 3 ) , as the 8

LECTURE 1: SCALARS, VECTORS AND TENSORS

Scalar multiplication

: let  Then   A   A i  (  A i ,  A 2 ,  A 3  A i ) be a vector. is a vector.

Dot or scalar product

 A   B  A 1 of two vectors results in a scalar: B 1  A 2 B 2  A 3 B 3  scalar In index notation, the dot product takes the form  A   B  i 3   1 A i B i  3  k  1 A k B k  r 3   1 A r B r  Einstein summation convention: if the same index occurs twice,

always sum over that index

. So we abbreviate to  A   B  A i B i  A k B k  A r B r 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 is thus a scalar.

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LECTURE 1: SCALARS, VECTORS AND TENSORS

Magnitude

of a vector:  A 2   A   A  A i A i A

tensor

can be constructed by multiplying two vectors (not scalar product): A i B j  ( A i B j ) , i  1 ..

3 , j  1 ..

3 A 1 B 1    A 1 B A 1 B 3 2 A A A 2 2 2 B B B 1 2 3 A A 3 3 B B A 3 B 3 1 3   Two free indices (i, j) means the result is a

second-order

tensor Now consider the expression A i A j B j This is a

first-order tensor

, or

vector

because there is only one free index, i (the j’s are paired, implying summation).

A i A j B j  ( A 1 B 1  A 2 B 2  A 3 B 2 )( A 1 , A 2 , A 3 ) 10 That is, scalar times vector = vector.

LECTURE 1: SCALARS, VECTORS AND TENSORS

Kronecker delta

 ij  ij  1 0 if if i  i  1 j j    0 0 0 1 0 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.

 ijl      1 1 if if 0 ,i ,j k ,i ,j k cycle clockwise: 1,2,3, 2,3,1 or 3,1,2 cycle counterclockwise: 1,3,2, 3,2,2 or 2,1,3 otherwise Vectorial

cross product

:  A  x B   ijk A j B k One free index, so the result must be a

vector

.

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LECTURE 1: SCALARS, VECTORS AND TENSORS

 C  C   A x  B Then  C      A iˆ B 1 1  det      A iˆ B 1 1 A B jˆ 2 2 A B jˆ 2 2 kˆ A 3 B 3           A iˆ B 1 1 kˆ A 3 B 3       A B iˆ 1 1 B jˆ A 2 2            A iˆ B 1 1 A jˆ B 2 2 A jˆ B 2 2 kˆ A B 3 3  kˆ A 3 B 3           iˆ A 1 B 1 B jˆ A 2 2        A 2 B 3  A 3 B 2 A 3 B 1  A 1 B 3   A 1 B 2  A 2 B 1  kˆ 12

LECTURE 1: SCALARS, VECTORS AND TENSORS

Vectorial cross product in tensor notation: C i   ijk A j B k Thus for example C 1   1 jk A j B k  = 1  123 A 2 B 3  = -1  132 A 3 B 2  = 0  111 A 1 B 1   A 2 B 3  A 3 B 2 a lot of other terms that all = 0 i.e. the same result as the other slide. The same results are also obtained for C 2 and C 3 . The

nabla vector operator

  :    iˆ   x 1  jˆ   x 2  kˆ   x 3 or in index notation   x i 13

LECTURE 1: SCALARS, VECTORS AND TENSORS

The

gradient

converts a scalar to a vector. For example, where p is pressure, grad or in index notation ( p )    p   p  x 1 iˆ   p  x 2 jˆ   p  x 3 kˆ 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

 u is the velocity vector, div (  u )   u 1  x 1   u 2  x 2   u 3  x 3   u i  x i   u k  x k 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

 u velocity vector, curl (  u )    x  u   iˆ  x 1 u 1  jˆ  x 2 u 2 kˆ   x 3 u 3     u 3  x 2   u 2  x 3   iˆ     u 1  x 3   u 3  x 1   jˆ     u 2  x 1   u 1  x 2   kˆ or in index notation, curl (  u )   ijk  u k  x j One free index i (the j’s and the k’s are paired) means that the result is a vector 15

LECTURE 1: SCALARS, VECTORS AND TENSORS

A useful manipulation in tensor notation can be used to change an index in an expression:  ij u j  u i This manipulation works because the Kronecker delta  ij i = j, in which case it equals 1.

= 0 except when 16