Vector Calculus & General Coordinate Systems

Tensor Analysis

Dyadics and Tensors

We have discussed the general notion that a scalar represents a
zero-order tensor and a vector is a first-order tensor. Now
introduce the notion of a second-order tensor and a dyadic.
Begin with an example (Reddy & Rasmussen),

Example: Angular momentum of a rigid body.

H0 ω dH0 = r × (ρdτ )v

= angular momentum

dτ about O for
differential mass
v(r) = ω × r
element dm = ρ dτ.
r

O

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Tensor Analysis

H0 = ∫∫∫R (r × v)ρdτ = angular momentum vector.

Now write HO in a form to show the explicit dependence on
the angular velocity vector ω,

H0 = ∫∫∫R r × (ω × r)ρdτ . (65)

Recall the triple vector product,
a × (b × c) = b(a ⋅c) − c(a ⋅b).
Then eq. (65) becomes,

H0 = ∫∫∫R ⎣⎡r2ω − (r ⋅ω)r⎦⎤ ρdτ . (66)

226

Tensor Analysis

For rigid-body rotation, ω is not a function of position, i.e., ω ≠
ω(r), so we should be able to factor it. To do this, we
introduce the dyad rr.

A dyad is two vectors, side by side, that act as one entity. A
dyadic is a sum of dyads. The unit dyadic is

I (67)
I = e1e1 + e2e2 + e3e3.

Like the identity matrix of linear algebra, the unit dyadic
cIhanges a vector into itself, e.g.,
I ⋅ ω = e1(e1 ⋅ ω) + e2 (e2 ⋅ ω) + e3 (e3 ⋅ ω)

= e1ω1 + e2ω 2 + e3ω 3 = ω.

227

Tensor Analysis

TIhe unit tensor is symmetric, thus, (68)
I = e1e1 + e2e2 + e3e3. (69)
aInd I
I ⋅ω = ω ⋅ I = ω.

Returning to the angular momentum example, with the aid of

the unit dyadic, we can factor ω in eq. (65),

I (70)

H0 = ∫∫∫R ⎣⎡r2I − rr⎤⎦ ⋅ωρdτ .

WIith this formI , we define the moment of inertia tensor,
I 0 = ∫∫∫R ⎡⎣r2I − rr⎤⎦ ρdτ . (71)

and I
H0 = I 0 ⋅ ω. (72)

228

Tensor Analysis I

Note, similar to the square matrix in linear algebra, I 0 is an

operator that transforms a vector (column or row matrix) into

another vector (or into itself if the square matrix is the identity

matrix).

We finish this example by developing a relation for rotational
kinetic energy E of the total mass m,

E = 1 mv2 = 1 ∫∫∫R v ⋅ vρ dτ
2 2

= 1 ∫∫∫R (ω ×r) ⋅ vρ dτ = 1 ∫∫∫R ω ⋅(r × v) ρ dτ
2 2
I
= 1 ω ⋅ ∫∫∫R (r × v)ρd
τ = 1 ω ⋅ ⋅ω
2 2 I0

H0

= rotational kinetic energy in terms

of the moment of inertia tensor.

229

Tensor Analysis

Note the use of the generic term “tensor” generally refers to a
2nd-order tensor or dyad. Before discussing detailed rules of
the tensor algebra and calculus, one more illustrative example
of how tensors are used.
Example: In a continuous medium, the stress at a given
location depends on the force f at that location and the
orientation of the plane area ∆S on which the force acts (stress
= force/area). The stress vector at position r is a function of
two vectors, f (nˆ )and nˆ .

230

Tensor Analysis f (nˆ )

fn ft
∆S
nˆ

The stress tensor contains the information necessary to
determine the state of stress at a point. Consider a tetrahedron
shaped element in the continuum,

231

Tensor Analysis −eˆ1
π(−eˆ1) = −σ1
π(nˆ ) q3
q2
nˆ
−eˆ 2 232

π(−eˆ2 ) = −σ2

q1

−eˆ3
π(−eˆ3 ) = −σ3
From Newton’s second law,

∑ f = fsurface + fbody = ma,

Tensor Analysis S = Snˆ S′ = S′nˆ ′
S′ = S ⋅nˆ ′
nˆ
= Snˆ ⋅nˆ ′
S = projnˆ′S

nˆ ′ 233

S′

Then,
∆S1 = (nˆ ⋅eˆ1)∆S,
∆S2 = (nˆ ⋅eˆ2 )∆S,
∆S3 = (nˆ ⋅eˆ3 )∆S.

Tensor Analysis

Now employ the divergence theorem,

∆h q3 ∫∫∫R divr dτ = w∫∫S r ⋅nˆ dS

r nˆ Shrinking R to an infinitesimal
element,
∆τ q2
∆τ ≅ w∫∫S r ⋅n dS = ∆h∆S .
3
div r

q1

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Tensor Analysis

Newton’s second law then becomes,

∆S[π − σ1(nˆ ⋅ eˆ1) − σ2 (nˆ ⋅ eˆ2 ) − σ3 (nˆ ⋅ eˆ3 )] + ρ∆h ∆S f = ρ∆h ∆S a.

3 3

Solving for the stress vector π,

π = σ1(nˆ ⋅ eˆ1) + σ2 (nˆ ⋅ eˆ2 ) + σ3 (nˆ ⋅ eˆ3 )] + ρ∆h (a − f ).

3

Continuing to shrink R,

lim ⇒ π = nˆ ⋅ (eˆ1σ1 + eˆ 2σ 2 + eˆ 3σ 3 ).

∆h→0

Using tensor notation we write this as,

π(nˆ ) = nˆ ⋅σI . (72)

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Tensor Analysis

Let’s make a careful distinction,

σσI i==σσijieˆjeˆieˆj j stress vector,
stress tensor.

The stress vectors are the stresses on the area perpendicular to
the ith-coordinate, in the eˆ j direction.

The stress vector π is on the area perpendicular to the arbitrary
unit normal vector nˆ . The constitutive relation for the stress

tensor depends upon the properties of the medium.

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Tensor Analysis

Dyadic Rules

As defined earlier, a dyad is a second-order tensor represented
by two vectors adjacent to one another, e.g., ab. A dyadic is

cIomposed of a linear combination of dyads, e.g.,
Φ = a1b1 + a2b2 +" + anbn.
(73)I
Define the dot (scalar) product operation between a dyadic Φ
and a vector v,
I
Φ ⋅Iv = a1(b1 ⋅ v) + a2 (b2 ⋅ v) +" + an (bn ⋅ v), (74)
v ⋅ Φ = (v ⋅ a1)b1 + (v ⋅ a2 )b2 +" + (v ⋅ an )bn.

Note the dot product is generally not commutative. (Can you
think of a situation where it is?)

237

Tensor Analysis

DIefine the transpose (conjugate), (75)
ΦT = b1a1 + b2a2 +" + bnan.

TheIn,
v ⋅ Φ = v ⋅ (a1b1 + a2b2 +" + anbn )

= (Ib1a1 + b2a2 +" + bnan ) ⋅ v (76)
= ΦT ⋅ v.

Note the correspondence of these rules and results with some

of the rules of matrix and vector multiplication in linear

algebra.

Using index notation, let’s explore this further. Each of the
dyads in the dyadic is composed of two vectors, i.e.,

238

Tensor Analysis (77)
(78)
I
Φ = aibi 239

= aije jbikek

= aijbike jek = φ jke jek .

In 3-space,
I

Φ = φ11e1e1 + φ12e1e2 + φ13e1e3
+φ 21e2e1 + φ 22e2e2 + φ 23e2e3
+φ 31e3e1 + φ 32e3e2 + φ 33e3e3

or, using matrix linear algebra notation,

Tensor Analysis

e2( )I ⎡φ 11 φ 12 φ 13 ⎤ ⎛ e1 ⎞ (79)
e3 ⎢⎢φ 21 φ 22 φ 23 ⎥ ⎜ e2 ⎟
Φ = e1 φ 32 φ 33 ⎥ ⎜⎜⎝ e3 ⎟⎠⎟
⎢⎣φ 31 ⎥⎦

= eTj [φ ij ]ei .

Again, there is considerable overlap between the current
analysis and the formal linear algebra. There are differences,
however. For instance, we do not distinguish between row and
coluImn Ivectors when writing
v ⋅Φ = ΦT ⋅ v.
UsinIg theI linear algebra notation, this would be written as
vTΦ = ΦT v.

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Tensor Analysis

IIn general, we can write

Φ = φ jke jek (contravariant scalar components)
= φ jke jek (covariant scalar components)
=φkje jek (mixed scalar components)

Example: Computing stress at a point.

Given the Cartesian system (x1,x2,x3) and the stress tensor

⎡200 400 300 ⎤

[σ ij ] = ⎢⎢400 0 0 ⎥ psi
⎥
Find: T⎢⎣h3e00stress0vec−to1r0π0K ⎦⎥at a point on a plane passing through

the point and parallel to the plane

x1 + 2x2 + 2x3 − 6 = 0.

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Tensor Analysis

Find the normal and tangential components.

Solution:

First we find the normal to the plane,

nˆ = ∇P , P(x1, x2 , x3 ) = x1 + 2x2 + 2x3 − 6
| ∇P |

∇P = ⎛ ˆi1 ∂ + ˆi2 ∂ + ˆi3 ∂ ⎞ + 2 x2 + 2 x3 − 6)
⎜ ∂x1 ∂x2 ∂x3 ⎟ (x1
⎝ ⎠

→ nˆ = ˆi1 + 2ˆi2 + 2ˆi3
3

Now the stress vector,
πK (nˆ ) = nˆ ⋅σI

242

Tensor Analysis

⎛ π1 ⎞ ⎡200 400 300 ⎤ ⎛ 1/ 3 ⎞ ⎛1600 / 3⎞
⎜ ⎟ ⎢⎢400 ⎥ ⎜ 33⎠⎟⎟⎟ ⎜ ⎟
⎜⎜⎝ π 2 ⎟⎟⎠ = ⎣⎢300 0 0 ⎥ ⎝⎜⎜ 2 / = ⎝⎜⎜ 400 / 3 ⎠⎟⎟ psi.
π 3 0 2 / 100 / 3
−100⎦⎥
Note that we actually computed πK = σI ⋅nˆ . We can do this

because the stress tensor is symmetric.

πK (nˆ ) = 1 (1600ˆi1 + 400ˆi2 + 100ˆi3 ) psi ⇐
3

For the normal component,

σ n=πK (nˆ ) ⋅nˆ = 2600 / 9 psi, ⇐

and the tangential component is

σt = | πK | −σ 2 = 100 1781 psi. ⇐
n 81

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Tensor Analysis

Symmetric and Antisymmetric Dyadics

We just showed that the stress tensor is symmetric. Here we
show that the properties of symmetry and antisymmetry (skew-
symmetry) are analogous to those of matrices in linear algebra,
II
Φ = ΦT → symmetric.

φ ijeie j = φ ije jei = φ jieie j → φ ij = φ ji .

II
Φ = −ΦT → antisymmetric.

φ ijeie j = −φ ije jei = −φ jieie j → φ ij = −φ ji .

For the antisymmetric case,

φ ii = −φ ii (no summation) → φ ii = 0.

Therefore, only the j ≠ i components are independent, i.e.,

φ12 = −φ 21, φ13 = −φ 31, φ 23 = −φ 32.

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Tensor Analysis

Thus, there are only three independent components for an
antisymmetric dyadic.

Any dyadic can be separated into a symmetric and

antisymmetric part,

I = 1 I + I ]+ 1 I − I
Φ [Φ ΦT [Φ ΦT ],
2
2


symmetric antisymmetric

or

φ ij = 1 (φ ij + φ ji ) + 1 (φ ij − φ ji ).

22

245

Tensor Analysis

Example:

⎡5 4 3⎤ ⎡5 2 1⎤

[φ ij ] = ⎢⎢2 2 2⎥⎥ , [φ ji ] = ⎢⎢4 2 3⎥⎥ ,

⎢⎣1 3 5⎥⎦ ⎢⎣3 2 5⎥⎦

⎡10 6 4 ⎤ ⎡ 0 2 2 ⎤
1 1
[φ ij ] = 2 ⎢ 6 4 5 ⎥ + 2 ⎢⎢−2 0 −1⎥⎥ .
⎢ ⎥
⎢⎣ 4 5 10⎥⎦ ⎢⎣−2 1 0 ⎥⎦

246

Tensor Analysis

Transformation laws for second-order tensors

The same rules of invariance apply to second-order tensors and
dyadics that apply to vectors (first-order tensors).

Again we use the transformation between two curvilinear

systems, the barred and unbarred systems,

es = ∂q j ej, es = ∂q j ej,
∂q s ∂q s

e s = ∂q s e j , es = ∂qs e j .
∂q j ∂q j

The folIlowing are possible representations of the generic
tensorΦ in the two coordinate systems,

247

Tensor Analysis

I I

Φ = φ ijeie j Φ = φ mn em en
= φijeie j = φmn e m e n
= φ ijeie j = φnm em e n
= φijeie j = φmn e m en

Transforming the basis, invariance requires that,

I = φ ijeie j = φ ij ∂q m ∂q n em en.
Φ ∂qi ∂q j

It then follows that we can write
I

Φ = φ mn em en.

248

Tensor Analysis

So, in general,

φ mn = φ ij ∂q m ∂q n contravariant law (81)
covariant law (82)
∂qi ∂q j (83)

φmn = φij ∂qi ∂q j 249
∂q m ∂q n

φnm = φ i ∂q m ∂q j ⎫
j ∂qi ∂q n ⎪⎪
⎬
∂qi ∂q n ⎪ mixed laws
∂q m ∂q j ⎪⎭
φmn = φij

Tensor Analysis

For the Cartesian system, all transformations are the same, and

ˆii = ∂x j ˆij = βij ˆij ,
∂xi

where βij are the direction cosines (symmetric matrix). Then,

comparing to eq. (81), we get

I

Φ = φ mn ˆim ˆin = φ ij βmi βnjˆiiˆi j

and

φ mn = φ ij βmi βnj

or in matrix format

[φ ] = [β ][φ ][β ]T.

250

Tensor Analysis

The Unit Tensor

WI e already introduced the unit tensor (dyadic) in eq. (68) as
I = eiei = e je j

oIr e
I = δ i ei j = δijeie j .
j (85)

Note this is the invariant component form since the δ i and δi j
j

components are either zero (i ≠ j) or one (i = j) in any

coordinate system.

Recall,
ei = (ei ⋅ em )em = gimem ,

ei = (ei ⋅ em )em = g imem.

251

Tensor Analysis

TIhen, contravariant components (86)
I = g ijeie j covariant components (87)

= gijeie j

= δijeie j ⎪⎫ invariant (mixed) components (88)
⎬
= δ i ei e j ⎪⎭
j

Note, a tensor is always invariant when taken as a whole.
What we are referring to here as invariant components is the
fact that in eqs. (88) the scalar components do not change
during a coordinate transformation as they do in eqs. (86) and
(87), since the gij and gij are system dependent.

252