Indicial Notations
Dr. Rakesh K Kapania
Aerospace and Ocean Engineering Department
Virginia Polytechnic Institute and State University, Blacksburg, VA
AOE 5024, Vehicle Structures
Summer, 2016
c
2016
Rakesh K. Kapania, Mitchell Professor, Aerospace and Ocean Engineering, Virginia
Polytechnic Institute and State University, Blacksburg, VA, 24061-0203.
Definitions
Scalar: A Quantity that has no direction, only magnitude;
Examples: Energy, Temperature
Vector: A quantity that has both magnitude and direction;
Examples: Force, Velocity, Displacement
~ in the Cartesian Co-ordinate system (X , Y , Z )
Consider the Vector V
and ~i, ~j, and ~k are unit vectors along x, y , and z directions, respectively.
In Indicial Notation the co-ordinate system is represented as (X1 , X2 , X3 )
and ~e1 , ~e2 and ~e3 as the unit vectors.
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Rakesh
K. Kapania
AOE 5024, Vehicle Structures
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Definitions (cont’d)
The co-ordinate system in Indicial Notation,
Indicial Notation: ~ei , Xi
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Rakesh
K. Kapania
AOE 5024, Vehicle Structures
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Definitions (cont’d)
~ can be represented as
The vector V
~ = Vx~i + Vy~j + Vz~k
V
= V1 e~1 + V2 e~2 + V3 e~3
i =3
=
∑ Vi ~ei
i =1
= Vi ~ei Einstein 0 s Summation Notation
= Vi 0~ei 0 New Co − ordinate System
i is called a Free Index, takes values 1, 2, and 3.Vi Represents the
Vector V.
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Rakesh
K. Kapania
AOE 5024, Vehicle Structures
4
Definitions (cont’d)
In Einstein’s summation notation, an index repeated twice indicates a
sum.
e.g. ai bi = a1 b1 + a2 b2 + a3 b3
Two Free Indices


A11 A12 A13
Aij =  A21 A22 A23 
A31 A32 A33
The 3 by 3 Matrix A is represented
simply as Aij
Here i and j are two free indices
and take the values 1, 2, and 3.
Note: Aii = A11 + A22 + A33 indicates a sum. Reduces to a sum over
the three terms on the diagonal.
c
Rakesh
K. Kapania
AOE 5024, Vehicle Structures
5
Definitions (cont’d)
Kronecker Delta
δij is called Kronecker Delta and is given as:
δij = 1 if i = j
= 0 if i 6= j
Alternatively

1 0 0
δ= 0 1 0 
0 0 1

Note: δii represents δ11 + δ22 + δ33 , a summation over the index i,
hence δii = 3
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Rakesh
K. Kapania
AOE 5024, Vehicle Structures
6
Definitions (cont’d)
Alternating Tensor
eijk is the so-called Alternating Tensor. It has 27 components as all
the three free indices i, j, and k can take any value 1, 2, and 3.
eijk = 1 if i, j, and k are in Cyclic order
= −1 if i, j, and k are in Countercyclic order
= 0 Otherwise
e123 = e231 = e312 = 1
e321 = e213 = e132 = −1
e112 = e223 = e231 = e331 = e221 . . . = 0
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Rakesh
K. Kapania
AOE 5024, Vehicle Structures
7
Example 1
Example: Consider a fiber of length ds and components dx1 , dx2 , and
dx3 in x1 , x2 and x3 respectively.
ds 2 = dx12 + dx22 + dx32
Using Einstein’s summation notation:
ds 2 = dxi dxi
Here the repeated index implies a summation over the three values of
i. At times, it will be convenient to use the following form of the above
equation.
ds 2 = δij dxi dxj
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Rakesh
K. Kapania
AOE 5024, Vehicle Structures
8
Example 1 (cont’d)
Note that the RHS in the above equation really represents a sum of 9
terms as the values of the two free indices i and j varies independently
from 1 to 3. Out of 9 terms, only three are nonzero, terms for which
i = j.
Prove: Aij δjk = Aik
Hint: The index j is being repeated and δij = 1 only if i = j. How many
equations does the aforementioned equation represents ? (Answer: 9)
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Rakesh
K. Kapania
AOE 5024, Vehicle Structures
9
Example 2
Alternating Tensor Example: Determinant of a 3 by 3 matrix
Consider a matrix A that can be written as:




A1 A2 A3
a11 a12 a13
A =  B1 B2 B3  or  a21 a22 a23 
C1 C2 C3
a31 a32 a33
|A| =
−
=
=
a11 a22 a33 − a12 a21 a33 + a13 a21 a32
a11 a32 a23 + a12 a31 a23 − a13 a31 a22
eijk ai1 aj2 ak3
eijk Ai Bj Ck
This equation represents a sum of 27 terms. Only six of these are
nonzero. These are the ones containing:
e123 = e231 = e312 = 1 and
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Rakesh
K. Kapania
e321 = e213 = e132 = −1
AOE 5024, Vehicle Structures
10
Cross Product of Two vectors
The equation for the determinant of a matrix can be used to represent
the cross-product of two vectors (assuming a right handed coordinate
system)
~u = u1~e1 + u2~e2 + u3~e3 and ~v = v1~e1 + v2~e2 + v3~e3 .
~e1 ~e2 ~e3 ~u x ~v = u1 u2 u3 v1 v2 v3 = (u2 v3 − u3 v2 )~e1 − (u1 v3 − u3 v1 )~e2 + (u1 v2 − u2 v1 )~e3
= eijk~ei uj vk
c
Rakesh
K. Kapania
AOE 5024, Vehicle Structures
11
Dot Product of Two vectors
The Dot Product of the vectors ~u and ~v is:
~u · ~v = u1 v1 + u2 v2 + u3 v3 = ui vi
c
Rakesh
K. Kapania
AOE 5024, Vehicle Structures
12
Gradient of a Scalar
~ represents the vector:
Let Φ be a scalar and ∇
~ = ∂ ~i + ∂ ~j + ∂ ~k
∇
∂x
∂y
∂z
~ Φ represents the Gradient of the scalar Φ and is given as:
∇
~ Φ = ∂Φ~i + ∂Φ~j + ∂Φ~k
∇
∂x
∂y
∂z
c
Rakesh
K. Kapania
AOE 5024, Vehicle Structures
13
Gradient of a Scalar (cont’d)
~ Φ becomes:
In indicial notation, ∇
~ Φ = ∂Φ ~e1 + ∂Φ ~e2 + ∂Φ ~e3 = Φ,i ~ei
∇
∂x1
∂x2
∂x3
In Indicial Notation, a comma represents a partial derivative, Thus Φ,i
represents the partial derivative of Φ with respect to xi and the ith
component of teh vector ∇φ
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Rakesh
K. Kapania
AOE 5024, Vehicle Structures
14
Gradient of a Vector
~ = Vx~i + Vy~j + Vz~k.
Consider a vector V
~,
~ (div V
~ ) is its dot product with the vector ∇
The Divergence of V
i.e.
~ ·V
~ =∇
~ = ∂Vx + ∂Vy + ∂Vz
div V
∂x
∂y
∂z
In Indicial Notation,
~ = ∂Vi = Vi ,i
div V
∂xi
Note: the divergence of a vector is a scalar and the gradient of a scalar
is a vector that represents the direction for the maximum change for
that scalar.
c
Rakesh
K. Kapania
AOE 5024, Vehicle Structures
15
Curl of a Vector
~ = Vx~i + Vy~j + Vz~k is defined as the cross
The Curl of a vector V
~ with V
~.
product of ∇
~i
~k ~e1 ~e2 ~e3 ~j
∂
∂
∂
∂
∂
∂ ~ ×V
~ =
∇
=
∂x ∂y ∂z ∂x ∂x ∂x 2
3
1
Vx Vy Vz V1 V2 V3 {z
}
{z
} |
|
Classic Notation
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Rakesh
K. Kapania
Indicial Notation
AOE 5024, Vehicle Structures
16
Curl of a Vector (cont’d)
The determinant on the right hand side can be expressed using the
Alternating Tensor eijk .
~ ×V
~ = eijk~ei ∂Vk
∇
∂xj
Using Vk ,j to represent the partial derivative of Vk with respect to xj .
~ ×V
~ = eijk~ei Vk ,j
∇
We reiterate the fact that the right hand side represents a sum of 27
terms, only six of which are nonzero.
c
Rakesh
K. Kapania
AOE 5024, Vehicle Structures
17
Some Other Identities
δii = 3
δij δij = 3
eijk eijk = 6
eijk Aj Ak = 0
(3 Equations)
δij δjk = δik
δij eijk = 0
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Rakesh
K. Kapania
(3 Equations)
AOE 5024, Vehicle Structures
18
Misc. Example 1
Given vector A = 2x1 iˆ + 5x22 jˆ + 3x32 k̂. Determine the values of the
following.
• Ai,i
I For Ai,i
• Ai,i2
• Ai,i3
∂A1 ∂A2 ∂A3
∂Ai
=
+
+
∂xi
∂x1
∂x2
∂x3
Ai,i = 2 + 10x2 + 6x3
Ai,i =
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Rakesh
K. Kapania
AOE 5024, Vehicle Structures
19
Misc. Example 1 (cont’d)
I
For Ai,i2
Ai,i2 =
∂2 Ai
∂ ∂A1 ∂ ∂A2 ∂ ∂A3 =
+
+
∂xi ∂x2
∂x1 ∂x2
∂x2 ∂x2
∂x3 ∂x2
Ai,i2 = 0 +
I
∂
(10x2 ) + 0 = 10
∂x2
For Ai,i3
Ai,i3 =
∂2 Ai
∂ ∂A1 ∂ ∂A2 ∂ ∂A3 =
+
+
∂xi ∂x3
∂x1 ∂x3
∂x2 ∂x3
∂x3 ∂x3
Ai,i3 = 0 + 0 +
c
Rakesh
K. Kapania
∂
(6x3 ) = 6
∂x3
AOE 5024, Vehicle Structures
20
Misc. Example 2
Represent the following in Indicial Notations
• ∇4 φ
I
• (A × B ).C
For ∇4 φ
∇4 φ = ∇2 (∇2 φ) = ∇2 (φ,ii ) = (φ,ii )jj = φ,iijj
I
For (A × B ).C
(A × B ).C = (eijk Aj Bk ).~ei .Cp e~p
(A × B ).C = eijk Aj Bk Cp δip
(A × B ).C = eijk Aj Bk Ci
c
Rakesh
K. Kapania
AOE 5024, Vehicle Structures
21