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The Products of Tensors
CHAPTER
4
The Products of Tensors 4.1 The Product of Two Tensors Consider, as a vehicle for explanation, two second order tensors Spq and T . Each tensor has nine components so that it is possible to form 81 products. Letting these products be identified by the letter Q and associated subscripts, vw
(4.1)
Qpqvw — Spq Tvw
On changing axes the S,T components transform according to the now familiar law and the right-hand side becomes In this expression the unrepeated suffices (XipX] Spq)(X XnwTvw). ij,m,n survive the summation process so this is Q'^ the bracketed components being respectively S' and T' . q
mv
mn9
tj
Thence
Qijmn
i.e.
Qijmn
=
=
mn
^ipXjqAmvXnwSpqTvw, ^ipXjqXmvXnwQpqvw*
(4.2)
It is seen that the 81 componented quantity β is a fourth order tensor. The extension of the above argument to tensors of order other than two is obvious. Multiplication of anrathorder tensor's components by an nih. order tensor's components gives the components of an (m + /i)th order tensor. An example is the multiplication of the components of two vectors (first order tensors) to give the components of a second order tensor, given in Section 2.5(a). 50
51
T H E P R O D U C T S OF T E N S O R S
4.2
The Product of more than Two Tensors
Systematic application of the product of tensors two at a time shows that the product of any number of tensors is a tensor, where "product" means the formation of all possible products of components, e.g. S T Rij = QpqvwRij — L ij, where L is a sixth order tensor. pq vw
pqvW
4.3 Products Involving the Kronecker Delta. Contraction
The reader is invited to consider the meaning of the expression obtained by multiplying the right-hand side of eqn. (4.1) by ôq . Noticing the repetition of the suffices q and w, which implies that these are now dummy suffices over which summation is effected one concludes that there are nine such expressions, one for each pairing of values of ρ and v, and that each expression contains nine terms. The unrepeated suffices ρ and ν are those which survive the summation process so that the expression quoted is the (p,v)th component of some entity, say K . it is readily proved that Κ is a second With Kp = SpqT ôq order tensor. Summation over w is effected by adopting the pattern "change w into q and drop ô ", so that W
v v
V
vw
Wy
qw
Kpv
(4.3)
— SpqTvq.
On changing axes, the right-hand side becomes
(λαρλ^Ξ'αυ)
x
Applying X\>qXdq — St>d this reduces to X pX SbdSabTcd and on using "change b into d and drop
0
pv
α
0
cv
αο
αο
Q vq
PQ V
P
QW
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CARTESIAN TENSORS IN ENGINEERING
SCIENCE
Kronecker Delta. This use of the Kronecker Delta to provide repetition of one of the suffices is called contraction and its effect is to reduce the order of the resulting tensor quantity by two. S a yRepeated , SpqTvwOqwdpvi contraction is of course possible. If one starts with, this expression reduces to SpqTyqôpv and thence to SpqTpq. There is now no free suffix left and so the quantity SpqTpq is a zeroth order tensor, i.e. a scalar. It is therefore concluded that from the components of an mth order and an nth order tensor one may form tensors of the (m + «)th, (m + η — 2)th, (m + η — 4)th, etc. orders. In particular it is possible, by contracting often enough to arrive at either a scalar or a vector quantity, according as (m + n) is even or odd. 4.4 Examples of Contraction (a) If A and Β are vectors, then the nine quantities A B are the components of a second order tensor. Consider however the expression A B ô . With summation over ρ and q, this means the following: v
p
q
q pq
(i) Take in turn each element of A B P
ΑιΒι A2B1 A3B1
viz. each element of
A1B2 A±Bs A2B2 A2B2 A3B2 A3Bz
(ii) Multiply each of the A B p
Q9
q
by its associated element of 3
PQi
0 1 0
viz. by its associated element of (iii) Add the results together. The result is clearly (A1B1 + A B 2
2
+
^3^3).
In suffix notation, the above is done very rapidly by using the
53
THE P R O D U C T S OF T E N S O R S
pattern "write ρ in place of q and drop the § \ One may then write directly A B ô = A B and the right-hand side is immediately (ΑιΒι + A2B2 + A3B3). The significant fact about A B is that it is a tensor quantity, and since there is no free suifix surviving the summation process it is a zeroth order tensor, i.e. a scalar. This scalar, which is thus an invariant quantity under change of axes is called the scalar product of the two vectors. p q
p
P
q pq
P
P
P
(b) The above is a particular case of forming T S , where T is a second order tensor. The quantity T S is clearly T on using the pattern "write ρ in place of q and drop the ô \ and so T is a scalar invariant of the tensor T. Now T ( = Tn + T22 + Γ33) is the sum of the principal diagonal elements of the Γ matrix. This invariant quantity is called the "trace" or "spur" of the tensor. The reader is asked at this point to refer back to questions 3 and 4 at the end of Chapter 2 and be satisfied that the sum of the diagonal elements of the two matrices shown are the same; this is an example of the trace remaining constant under rotation of axes. pq
pq
pq
pq
pq
pp
pq
pp
pp
(c) If one takes the components of a second order tensor and those of afirstorder tensor (vector), one can form the components of a third order tensor or, by contracting, afirstorder tensor. In the latter case, if the tensor be called Γ and the vector B, we have, say, T B ô . This is the pth component of a vector, say A , and (4.4) pq
r
qr
p
In eqn. (4.4) the tensor Τ may be thought of as an operator which, when applied to a given vector B, produces a new vector A. The relationship of eqn. (4.4) may, of course, be displayed in a familiar way using matrix notation, viz. Βι' B2 Bz
(4.4a)
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CARTESIAN TENSORS IN E N G I N E E R I N G
SCIENCE
or, more shortly, {A } = [Τ] {B}. 4.5
The Invariants of a Second Order Tensor
The discussion of contraction given above has introduced one of the invariants of a second order tensor. Other invariants are now sought. As a preliminary it is readily proved that if S and Τ are both second order tensors then so is (S + T) ; this proof is left as an exercise to the reader. The particular case in which T= —kôpq enables invariant quantities of the tensor S to be found. (S
+ T)
is then S
- kôpq,
i.e
Sn — k S12
Sis
S21
S22 — k S 2 3
S31
S32
S33 —
k
On changing axes (S + T) becomes (S' + T) but Τ is still — kôpq (see the work in the previous chapter) because Τ is an isotropic tensor having the same components referred to all sets of axes. Thus with respect to the new axes the tensor's components are Su — k S 1 2 S21
S31
S13
S 22~& S 2 3
S32
S33 —
k
One now uses the law of transformation of a tensor (eqn. (2.20c)) together with the known fact that the determinants of [λ] and of [λ] are both equal to unity (eqn. (2.12) et seq.). Thence using the multiplication theorem for determinants (see Section 2.4) it is concluded that the determinants of the two arrays shown above are equal, viz. τ
Su —
S 1 2
S13
S21
S22 — k
S23
S31
S32
S33 — k
Su— A: S 1 2 S13 S21 S 2 2— k S 2 3 S31 S32 S33 — k
(4.5)
Each side of eqn. (4.5) is a cubic in k. Since this equation must
55
THE P R O D U C T S OF T E N S O R S
be satisfied for every value of k it follows that the coefficients of each power of k must be the same. These conditions give 2
From the coefficient of k
Su + S22 + S 3 3
^11
^33
(4.6a)
From the coefficient of k Su S12 S21S22
_j s s3 22 2
S32 S33
From the constant
L Sn S13 S31S33
S n S12 S2I S 2 2
Su S12 S13 S21 S22 S 2 3 S31 S32
S33
+
S 22 S 2 3 S32 S33
+
S u S13 S31
S33
(4.6b)
S u S12 S13 S 2 1 S 22 S 2 3 S31
s
3 2
s
(4.6c)
3 3
i.e. the three left-hand sides are invariants of the tensor S. It is instructive to write these invariants in suffix form, and in so doing, the third of them calls for the definition of a new quantity. Thefirstinvariant is clearly Ji = S,
(4.7)
The reader will find it instructive to verify that the second invariant is 1 (4.8) J2 — y[Sj>pSqq SpqSpq] For the third invariant, the determinant may be written out in full as S n S 22 S 3 3
S11S23S32 +
S12S23S31 — S12S21S33 +
S13S21S32
S13S22S31.
The pattern of these terms is as follows : (i) In a given term the leading subscripts are all different and are 1, 2, 3 in that order. (ii) The second subscripts are all different.
56
CARTESIAN TENSORS IN E N G I N E E R I N G
SCIENCE
(iii) Where the second subscripts are in the cyclic order 1231231..., the term is positive. (iv) Where the second subscripts are in the anticyclic order 321321..., the term is negative. It is concluded that the value of the determinant may be expressed as SpqrSipS^qSsr (carrying out the implied summation over p,q and r) provided that e is defined as follows : pqr
1, if p,q,r are all different and in the cyclic order 12312... = — 1, if p,q,r are all different and in the anticyclic order 32132... = 0, if any two of p,q r are the same.
£pqr =
9
This function is commonly called the Levi-Civita density. In some books the set of e are called permutation symbols. Hence the third invariant of the tensor S is pqr
J% —
4.6
The Levi-Civita Density as a Third Order Tensor
The value of the expression XipX2 l Ep q 3r
whence
(4.9)
£pqrSlpS2qS$r
£123 —
All
^12
Λ-13
Xzi
Λ-22
^23
^31
Λ-32 Λ-33
qr
= +1 =
is £123
XlpX^qXzr^pqr*
Similarly, by taking all possible combinations w,v,h> for the leading suffices of the three lambdas, it is readily verified that in all cases (4.10) Noting that the Levi-Civita density is defined the same with respect to all coordinate axes, so that e w and e w are the same thing, it follows that eqn. (4.10) is the law of transformation UV
UV
57
T H E P R O D U C T S OF T E N S O R S
appropriate to a third order tensor. 4.7
The Products of Vector Components
(a) The scalar product of two vectors A and Β has already been defined as A B . This scalar quantity often written A . Β and called the "dot product", is ABcosO, where A and Β are magnitudes of A and B, and θ is the angle between them. (b) The vector product arises from the equation C = ε ΑΒ. The term on the right is made up of components of third, first and first order tensors. With two contractions (i.e. summation over ρ and q) the result is a vector C of which C is one component. The vector C is called the vector product, or cross product of A with B. In a familiar notation C = Α χ Β. In Suffix notation P
P
r
ραν
ρ
ς
r
C — Cflf —
(4.11)
SpqrApBql',
where i is a unit vector in the x direction. It will be noted that the order in which A and Β are written down is important because if, in eqn. (4.11) the roles of A and Β are reversed, the associated e changes sign owing to the fact that the subscripts ρ and q change places. r
r
pqr
Hence
Α χ Β = - Β χ Α.
(4.12)
Important particular examples of eqn. (4.11) are the cross products of unit vectors ii,i2 and 13. Using eqn. (4.11), letting A = ii so that A \ = 1, A2 = 0, Az = 0, and Β = i so that Bi = 0, B2 = 1, Bz = 0 one finds ii X 12 = 13. There are five more such equations. The complete set is 2
ii X i 2 =
13, 12 X 13 =
ii, 13 X ii =
12 x ii = - 1 3 , 13 X Î2 = - h ,
12 ^|
h X 13 = - 1 2
J
It will also be appreciated that the cross product of a vector with itself is zero, i.e. A χ A = 0 because this cross product Ε
58
CARTESIAN
TENSORS
I N E N G I N E E R I N G
SCIENCE
is e qrApAq\ the terms of which vanish in identical pairs (see a similar example below). In particular P
r
ii χ h = 0, h X i = 0, i Χ i = 0. 2
3
3
(4.14)
The cross product C of two vectors A and Β is perpendicular to both A and B, i.e. is perpendicular to the plane defined by A and B. To see that this is so it is only necessary to examine the scalar product of C with A and of C with B. The scalar product C . A is C
r
A
r
= ερατΑρΒςΑτ.
(4.15)
The right-hand side of eqn. (4.15) gives zero identically, since the terms vanish in identical pairs. For example, with q = 2 there are two terms, namely £123^1^2^3 and £321^3^2^1; these give B2(AiA — Λ3Λ1), i.e. zero. Similarly, for all the other contributions. Since C . A is zero it is concluded that C is perpendicular to A. Similarly, one may show that C is perpendicular to B. Examining the form of eqn. (4.11), and recalling that the definition of the Levi-Civita density was suggested by the need to express a determinant in suffix notation, it is clear that 3
H
A X B = Αι Bi
12
13 A2 A3
B2
(4.16)
B3
Alternatively, the reader will readily verify that A x B = (ΑΒ$\ηθ)η, where η is the unit vector perpendicular to the plane of A and Β and in the direction such that rotation of vector A into vector Β through an angle smaller than 180° would cause a righthand screw to advance along n. Figure 26 shows some examples of cross products. It has been remarked above that A χ Β is perpendicular to A and B. Thus, if A and Β are themselves mutually perpendicular, then A,B and (Α χ Β) constitute a right-handed triad of perpendicular axes. This fact provides a means of solving problems
THE
P R O D U C T S
59
OF TENSORS
\Rotation
Rotation
"BxA
(b)
(a)
Rotation Rotation
(d)
(c) FIG. 2 6 .
such as question 2 at the end of Chapter 2. In that question, two /2 2 mutually perpendicular vectors were given, viz. I + and
I
1
[—7^1 \ \ 0
+
2
\
A/
/
(0)i2 — —7713land 5
+
1
ν
the third vector of the triad was
sought. Taking the cross product of these two, either by expanding the determinant eqn. (4.16) or (what amounts to the same thing) making repeated use of eqns. (4.13) and (4.14), one finds
/ · Γνΐ 4
1 1
-
5
+
^
ΐ
Λ ~ ^ τ 2
2
This enables the matrix of direction cosines to be completed.
60
CARTESIAN TENSORS IN ENGINEERING SCIENCE
4.8 The Vector Operator V The vector operator V , pronounced "del" or (more rarely) d
d
e
dxi
dxz
cxz
"nabla" is defined as V = — ii + -—12 + In suffix notation
V=
àxp
13. (4.17)
ip.
There are three main applications of the vector operator. (i) If / is a scalar function of (xi xz xa), i.e. if / = f(xi,X2 X3), so that to every point in (xi X2 X3) space there corresponds a value of 9
9
df
'•
t h e n
v
/
i
s
W
df 1
+
W
9
9
9
2
df +
W * 3
In suffix notation V/ = -—\ .
(4.18)
v
OXp
V/is called the gradient of/, often written grad/, and is a vector, the operator V can thus be regarded as operating on a scalar / and producing a vector V/. In order to prove that the triple / df df df\ 9 9 \dx' dx dx~)
^
S a
v e c t or
^ '
s
n e c e s s a r
y
t o
show that these
components transform, under change of axes, in accordance with the law given in eqn. (1.8). Using the chain rule of partial differentiation,
61
T H E P R O D U C T S OF T E N S O R S
df
df
Similarly for —— and — . All three equations may be expressed (4.19b)
dx'p dXq
dXq
This last equation is important and it is well worth while to acquire a clear geometrical "feel" for the chain rule. Figure 27 f here
f + Sf here
FIG. 2 7 .
shows the development of this expression for the case q = 1. At two points x i , X 2 , * 3 and x\+δχ\,χ ,χζ, the values of the scalar function are /and f+Sf. Because the change takes place owing to a change in JCI, whilst x and x$ stay constant, the limiting value of 2
2
8f
çT— as Sxi is made indefinitely small is written
df
Now, as shown
in Fig. 27, let the length Sxi be considered to be the body diagonal of a small box whose sides are in the directions of the coordinate axes x[,x and x . Then the lengths of the sides of the box are AB = AnÄxi"^ 2
3
BC = λ2ιδχχ CD=
^ (4.20)
λζιδχι J
For example, the triangle ABD is right-angled at Β and the cosine of angle BAD is λη.
62
C A R T E S I A N TENSORS IN E N G I N E E R I N G S C I E N C E
The rate at which/ changes as one moves along the line AB is df
—, ; consequently, the increase of / between A and Β is df
TTAB
Ξ
ΟΧγ
df -—,-ληδχι.
CXX
Similarly, the increases in/ from Β to C and C to D are respectively df ^-τλ2ιδχι ox 2
A
df
and -τ-τλζιδχι. The total change in / a s one moves from ox
df
3
df
df
df
to D is -— δχχ and so - — 8 χ ι = —,- /Ιιιδχι + to
df -^τλζιδχι, dx3
- — r
dxl
to
-
df
df
i.e. on cancelling to, — =-τ-τλιι -
to
to
+
^
dx2
df
^
+
df
—/I21 + a dx2
In suffix notation eqn. (4.21a) is written - = -^-7—
Τ-7Ά31.
^*3
(4.21a)
dxv
0
to fop to
^/
A2ito
·
df
The corresponding equations for -— and -~— are obvious. All df
three equations may be expressed as -— = CXq
df dx
(4.21b)
— - 7
uXp
CXq
Equations (4.19b) and (4.21b) are identical because λ
dx
ν α
= - -· CXq
But eqn. (4.21b) expresses a transformation which is in accordance with the definition of a vector given in Chapter I. Hence the triple
(OLK df df \dxidx2
1)
df\ dxsj
is aisvector. directionThe of this vector at conany point in the space quicklyThe established. equation / = stant represents a surface in the space; at a given point on the surface choose axes so that the tangent plane is the plane of axes x and X3, then xi is normal to the surface. Because the x 2
2
T H E P R O D U C T S OF
63
TENSORS
axis is tangent to the surface / = constant, it follows that — for
df
2
df
is zero and similarly for -—. Hence grad / = -—iiandisperpenCX%
GX\
dicular to the surface / = constant. This direction is the "line of steepest ascent" of the scalar function /. It is instructive to think of the contours drawn on a map of a mountainous district, the line of steepest descent down a mountainside cuts through the contours at right angles. (ii) If A is a vector the dot product of V with A, viz. V . A, is dA\
dA2
dAs
dAp
-λ— dxi + -7Γ~ dx + ^—5 dxs i.e. in suffix notation V . A =——. dxp The scalar V . A is called the divergence of the vector A and is written divA. It arises quite naturally, for example, when considering theflowof an incompressible fluid; let V be the velocity vector at a point 2
of such a fluid, then considering flow into and out of a cuboid whose sides are δχι, δχ2 and δχζ (Fig. 28) one may equate
X1X2X3
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CARTESIAN
TENSORS
I N E N G I N E E R I N G
SCIENCE
total inflow and total outflow, whence / VlÔX2ÔXz
+
V2ÔX3ÔX\
I dV2 \ \V2 + ^δχ2\δχ3δχι dVi g i v i n g
+
\
( Vi + ~fa-àxi
=
VzÔXlÔX2
dVi
\δχ2δχ3
+
ι dV3 \ + I Vz + ~^δχζ\δχιδχ2
dV2
dVs
Έί + Έί +
°· "·
=
divV
=
α
(iii) If A is a vector, the cross product of V with A, viz. V χ A is id A3 \dx2
dA2\ dx3 /
(dAi
dAz\ 2 dxi Γ
\dX3
In suffix notation V χ A = ε.pqr
ldA2
\ dx±
ΘΑ dXp
'
dAi\ 3 dx2 Γ "
(4.22)
Γ
A very useful form is also VXA: dxi Αχ
12
13
d
d
(4.23)
dx2 dxz A2 A3
The vector V χ A is usually called "curl A" or, more rarely in recent years, "rot A". The latter name emphasises the link between
FIG. 2 9 .
65
T H E P R O D U C T S OF T E N S O R S
this quantity and the phenomenon of rotation. If a rigid body is imagined to rotate at uniform angular velocity ω around the x$ axis (Fig. 29) the velocity V at X1X2X3 is — (0x2*1 + coxih and curl V is, using the notation of eqn. (4.23), | ii 12 13
e
d
e
dxi
dx2
dxz
— ωχ2
coxi
I
0
i.e. curl V = 2ωΐ3, which is twice the angular velocity vector of the motion. A useful mental picture is provided by imagining a leaffloatingin a straight line down a stream; if the leaf preserves always the same attitude, i.e. does not spin around a vertical axis, its motion belongs to that class called irrotational, a class of motions of great importance influidmechanics and characterised by curl V = 0. If the leaf spins as it floats down stream, the motion is rotational and the curl of V has magnitude twice the angular velocity of spin. The suffix notation provides a convenient way of establishing relationships between vector quantities. Two worked examples follow; others are given in the problems at the end of the chapter. Example (a). The scalar product of A with Β χ C is written A . Β χ C. Prove that A . B x C = Α χ Β . C A . Β Α
χ
X C Β.
C
=
AfSpqrBpCq
=
si]kAiBfk
=
EpqrAfBpCq)
^ }
^'
But r p,q is the same cyclic order as p,qs, so thefirstof these expressions is e pqA BpCq which is identical with e^A^^Cjc, merely using different dummy suffices. Example (b). The vector product of A with Β χ C is written Α χ (Β χ C) and is called the triple vector product. Write this in suffix form and show that this vector is of the form μΒ + vC, where μ and ν are scalar s. The rth component of Β χ C is 8pq BpC . 9
r
r
r
q
V
66
CARTESIAN TENSORS IN ENGINEERING SCIENCE
Thence
Α χ (Β X C) = e rtA (e B C )U. S
s
pqr
p
(4.25)
q
To show that this vector is of the form μΒ + vC, suppose without loss of generality that the χι axis is chosen to be the direction of vector B, so that Β = Bih, and that the plane of Β and C is chosen to be the plane of x± and x so that C = CJi + C2I2; then in eqn. (4.25) for a non-zero contribution ρ must be unity and q must be 1 or 2. But if ρ = q = 1, the e is zero and therefore a non-zero contribution is only forthcoming if ρ = 1 and q = 2, whence r must be 3. The right-hand side then takes the form ZsztAseimBiCzH, i.e. e ztA BiC \ . For e zt to be non-zero, t must be either 1 or 2. Hence the available unit vectors must lie in the plane of Β and C which proves the proposition. 2
pqr
s
s
2 t
s
4.9 The Eigenvectors of a Second Order Tensor Τ It has already been noticed (above, Section 4.4(c)) that an equation A = T B , which in matrix notation is written {A} = [Τ] {B} can be regarded as the tensor Τ operating on a vector Β and giving rise to a new vector A. There is one particular case of great interest in the study of vibrations and also in the stability of structures; this is the case in which the vector A which results is in the same direction as the vector B, so that A can be written A = kB, where k is a scalar constant. Any vector Β which has this property is called an eigenvector of the tensor Γ, and the associated value of k is called an eigenvalue. Some properties of the eigenvectors are developed below. The equation which must be satisfied is k {B} = [T]{B}> p
i.e.
pq
q
kB! = TnBi + T12B2 + TizB3 kB2 = Τ2ιΒι + 722#2 + T2ZBs kBs = T31B1 + T32B2 + TzsBd.
"] ± J
(4.26)
The necessary and sufficient condition that there shall exist values of i?i,Z?2,#3, not all of which are zero, satisfying these three
THE
P R O D U C T S
OF
equations is that the determinant Tn-
67
TENSORS
7"i3
7^12
721
Γ22 — k
781
7^32
Γ23
733 -
shall be zero. This is a cubic equation in k. A sufficient, though not necessary, condition for the roots of this equation to be real is that the Τ matrix shall be symmetric (see the problems at the end of this chapter). On limiting consideration to symmetric tensors it is very advantageous to take advantage of the fact, already demonstrated, that there exist three mutually perpendicular axes, called principal axes, with respect to which the Τ matrix is diagonal. Taking the principal axes as coordinate axes ( x i , X 2 , * 3 ) the equation for eigenvalues becomes Ti-k 0 0 = 0 0 0 T -k 0 n-k 0 2
the three roots of which are ki = Γι, k = T , ks = Ts, i.e. the eigenvalues are the values of the principal components of the tensor. To find the eigenvectors Β for each of these values of k, let it be supposedfirstthat the values are all different. The vector corresponding to the eigenvalue ki must then satisfy eqns. (4.26) which now appear as 2
kiBi = kiBi
kiB2 =
2
~)
k2B2
±
(4.26a)
ksBs. J
kiB3 =
Hence B = Bs = 0 and Bi is arbitrary, whence 2
Β =
Gh.
(4.27)
Similarly, the other eigenvectors are C212, C313, where the C's are arbitrary constants. One may conclude that when the principal axes are uniquely defined these axes are the axes of the three distinct eigenvectors.
68
CARTESIAN
TENSORS
IN
E N G I N E E R I N G
SCIENCE
If two values of k are the same, say k\ = £2, with kz different, tofinda vector corresponding to the eigenvalue k\ one substitutes into eqn. (4.26) and finds Ί
kiBi = kiBi kiBi = kiB3 =
(4.26b)
Hence B == 0 and B\ and B are arbitrary. It is concluded that any vector in the plane of the two equal principal components of the tensor is an eigenvector. The eigenvector in the third direction has a uniquely defined direction. For the case where k\ = k = kz (the case of the isotropic tensor), it is left as an exercise to the reader to show that any vector is an eigenvector. 3
2
2
Summary of Chapter 4
1. The product of two tensors is a tensor, where product means multiplication of the components of the one tensor by components of the other. 2. The product of anrathand «th order tensor is an (ra + n)th order tensor, i.e. Qabc- . >pqr. . . = Sabc-. . T . . . p q r
3. Repetition of a suffix in the product of two tensors is effected by multiplying by the appropriate Kronecker Delta, e.g.
OapSabc* · *TPqr
=
Sabc* · »Taqr> · ·
4. The operation of item 3 is called contraction. It reduces the order of the resulting tensor by two, i.e. Sabc- . >T qr. . . = Übe- . .ffr. . ., where U is of the order two less than the Q of item 2 a
5. By contracting enough times any product of tensors can be made to yield a scalar or a vector. 6. If T are the components of a second order tensor, then = T . This is equal to T + T22 + T33 and is the PQ
TpqôPq
p p
n
T H E P R O D U C T S OF
69
TENSORS
first invariant of the tensor. 7. By contracting once, a second order tensor can operate on a vector and generate another vector, viz. A — T B , i.e. {A} = p
pq
q
mi*}. 8. The Levi-Civita density e is defined as £pqr = + 1 , if PMS are all different and in cyclic order 1231231... = — 1, if p,qs are all different and in anticyclic order 3213213. .. = 0, if any two of p,q,r are the same. pqr
9. e
pqr
is a third order tensor.
10. The invariants of a second order tensor are Ji = Τη + T22 + Γ33 h
=
Τη Tn
Ta
+
T22
^3 = I Tn
Tl2
T2I
T22
Γ22
723
732
733
+
7Ίι
7l3
731
733
T12 T2Z
T$l Tz2 733 I These are conveniently remembered as the sums of determinants of size l x l , 2 x 2 and 3 x 3 respectively, involving the elements of the principal diagonal.
11. In suffix notation the invariants are J1
=z
Tpp
J2 = ^TppTqq Jz =
— TpqTpq^j
£pqrTipT2qTzr.
12. The scalar product of two vectors A and Β is A B . P
13. The vector product of two vectors A and Β is
P
e AB\. pqr
p
q r
14. If A,B are perpendicular then Α,Β,Α χ Β form a righthanded triad of mutually perpendicular axes.
70
CARTESIAN TENSORS IN ENGINEERING SCIENCE
15. If a tensor operates on a vector and generates a parallel vector, viz. [T]{B} = k{B} with k as scalar, then {B} is an eigenvector and k the corresponding eigenvalue. 16. If the principal components T\,Ti,T of a symmetric tensor are all different, the eigenvectors are in the directions of the principal axes and there are three distinct eigenvalues. 3
17. If two of the 7 \ , 7 2 , 7 3 are equal, any vector in the plane of the two equal ones is an eigenvector. If all three 7 i , 7 2 , 7 3 are equal, any vector is an eigenvector.
Examples on Chapter 4 1. Find the scalar product of (4ii + 2Î2 + 13) and ( —ii + 3Î2 + 7i ). 8
Ans. 9. 2. Find the vector product of the two vectors given in question 1. Ans. llii - 29i + 14i . 2
3
3. Three mutually perpendicular vectors are sought. The first is to be (4ii + 2Î2 + 13). The second is to be (—ii + 3Î2 + ah), where a is at present unknown. The third vector has no component specified. Find the three vectors. Ans. For thefirsttwo vectors to be mutually perpendicular, their scalar product must be zero, hence — 4 + 6 + a = 0 and a = —2. Having now two vectors a suitable third vector is the cross product of the two, i.e. (4ii + 2Î2 + 13) X ( — h + 3i2 — 213) which yields — 7ii + 7Î2 + 14Î3. 4. Write down the array of coefficients of the vectors of question 3. Is this an array of direction cosines?
T H E P R O D U C T S OF
Ans.
4 -1 -7
2 1 3 -2 7 14
TENSORS
71
No.
5. What must be done to make the array of numbers in question 4 into direction cosines and what is this process called? Ans. Divide the rows by \/2\, \/\4 and ly/6 respectively. Normalisation. 6. The determinant of a stress tensor matrix is zero. What conclusion can be drawn from this? (Consider the invariants.) Ans. At least one of the principal stresses is zero, i.e. a state of plane stress prevails. 7. Prove that Sp Spst mutation identity). qr
=
ôqsôrt
—
Sqtôrs
(this is called the per-
8. Using the identity of question 7 prove that A Χ (Β X C) = (A . C)B - (A . B)C. 9. A symmetric tensor has components [ 5 2 5 2 2 2 5 2 5 Use the invariants to find the principal components. Ans. 10*9, 1-1, 0 very closely.
4
The Products of Tensors 4.1 The Product of Two Tensors Consider, as a vehicle for explanation, two second order tensors Spq and T . Each tensor has nine components so that it is possible to form 81 products. Letting these products be identified by the letter Q and associated subscripts, vw
(4.1)
Qpqvw — Spq Tvw
On changing axes the S,T components transform according to the now familiar law and the right-hand side becomes In this expression the unrepeated suffices (XipX] Spq)(X XnwTvw). ij,m,n survive the summation process so this is Q'^ the bracketed components being respectively S' and T' . q
mv
mn9
tj
Thence
Qijmn
i.e.
Qijmn
=
=
mn
^ipXjqAmvXnwSpqTvw, ^ipXjqXmvXnwQpqvw*
(4.2)
It is seen that the 81 componented quantity β is a fourth order tensor. The extension of the above argument to tensors of order other than two is obvious. Multiplication of anrathorder tensor's components by an nih. order tensor's components gives the components of an (m + /i)th order tensor. An example is the multiplication of the components of two vectors (first order tensors) to give the components of a second order tensor, given in Section 2.5(a). 50
51
T H E P R O D U C T S OF T E N S O R S
4.2
The Product of more than Two Tensors
Systematic application of the product of tensors two at a time shows that the product of any number of tensors is a tensor, where "product" means the formation of all possible products of components, e.g. S T Rij = QpqvwRij — L ij, where L is a sixth order tensor. pq vw
pqvW
4.3 Products Involving the Kronecker Delta. Contraction
The reader is invited to consider the meaning of the expression obtained by multiplying the right-hand side of eqn. (4.1) by ôq . Noticing the repetition of the suffices q and w, which implies that these are now dummy suffices over which summation is effected one concludes that there are nine such expressions, one for each pairing of values of ρ and v, and that each expression contains nine terms. The unrepeated suffices ρ and ν are those which survive the summation process so that the expression quoted is the (p,v)th component of some entity, say K . it is readily proved that Κ is a second With Kp = SpqT ôq order tensor. Summation over w is effected by adopting the pattern "change w into q and drop ô ", so that W
v v
V
vw
Wy
qw
Kpv
(4.3)
— SpqTvq.
On changing axes, the right-hand side becomes
(λαρλ^Ξ'αυ)
x
Applying X\>qXdq — St>d this reduces to X pX SbdSabTcd and on using "change b into d and drop
0
pv
α
0
cv
αο
αο
Q vq
PQ V
P
QW
52
CARTESIAN TENSORS IN ENGINEERING
SCIENCE
Kronecker Delta. This use of the Kronecker Delta to provide repetition of one of the suffices is called contraction and its effect is to reduce the order of the resulting tensor quantity by two. S a yRepeated , SpqTvwOqwdpvi contraction is of course possible. If one starts with, this expression reduces to SpqTyqôpv and thence to SpqTpq. There is now no free suffix left and so the quantity SpqTpq is a zeroth order tensor, i.e. a scalar. It is therefore concluded that from the components of an mth order and an nth order tensor one may form tensors of the (m + «)th, (m + η — 2)th, (m + η — 4)th, etc. orders. In particular it is possible, by contracting often enough to arrive at either a scalar or a vector quantity, according as (m + n) is even or odd. 4.4 Examples of Contraction (a) If A and Β are vectors, then the nine quantities A B are the components of a second order tensor. Consider however the expression A B ô . With summation over ρ and q, this means the following: v
p
q
q pq
(i) Take in turn each element of A B P
ΑιΒι A2B1 A3B1
viz. each element of
A1B2 A±Bs A2B2 A2B2 A3B2 A3Bz
(ii) Multiply each of the A B p
Q9
q
by its associated element of 3
PQi
0 1 0
viz. by its associated element of (iii) Add the results together. The result is clearly (A1B1 + A B 2
2
+
^3^3).
In suffix notation, the above is done very rapidly by using the
53
THE P R O D U C T S OF T E N S O R S
pattern "write ρ in place of q and drop the § \ One may then write directly A B ô = A B and the right-hand side is immediately (ΑιΒι + A2B2 + A3B3). The significant fact about A B is that it is a tensor quantity, and since there is no free suifix surviving the summation process it is a zeroth order tensor, i.e. a scalar. This scalar, which is thus an invariant quantity under change of axes is called the scalar product of the two vectors. p q
p
P
q pq
P
P
P
(b) The above is a particular case of forming T S , where T is a second order tensor. The quantity T S is clearly T on using the pattern "write ρ in place of q and drop the ô \ and so T is a scalar invariant of the tensor T. Now T ( = Tn + T22 + Γ33) is the sum of the principal diagonal elements of the Γ matrix. This invariant quantity is called the "trace" or "spur" of the tensor. The reader is asked at this point to refer back to questions 3 and 4 at the end of Chapter 2 and be satisfied that the sum of the diagonal elements of the two matrices shown are the same; this is an example of the trace remaining constant under rotation of axes. pq
pq
pq
pq
pq
pp
pq
pp
pp
(c) If one takes the components of a second order tensor and those of afirstorder tensor (vector), one can form the components of a third order tensor or, by contracting, afirstorder tensor. In the latter case, if the tensor be called Γ and the vector B, we have, say, T B ô . This is the pth component of a vector, say A , and (4.4) pq
r
qr
p
In eqn. (4.4) the tensor Τ may be thought of as an operator which, when applied to a given vector B, produces a new vector A. The relationship of eqn. (4.4) may, of course, be displayed in a familiar way using matrix notation, viz. Βι' B2 Bz
(4.4a)
54
CARTESIAN TENSORS IN E N G I N E E R I N G
SCIENCE
or, more shortly, {A } = [Τ] {B}. 4.5
The Invariants of a Second Order Tensor
The discussion of contraction given above has introduced one of the invariants of a second order tensor. Other invariants are now sought. As a preliminary it is readily proved that if S and Τ are both second order tensors then so is (S + T) ; this proof is left as an exercise to the reader. The particular case in which T= —kôpq enables invariant quantities of the tensor S to be found. (S
+ T)
is then S
- kôpq,
i.e
Sn — k S12
Sis
S21
S22 — k S 2 3
S31
S32
S33 —
k
On changing axes (S + T) becomes (S' + T) but Τ is still — kôpq (see the work in the previous chapter) because Τ is an isotropic tensor having the same components referred to all sets of axes. Thus with respect to the new axes the tensor's components are Su — k S 1 2 S21
S31
S13
S 22~& S 2 3
S32
S33 —
k
One now uses the law of transformation of a tensor (eqn. (2.20c)) together with the known fact that the determinants of [λ] and of [λ] are both equal to unity (eqn. (2.12) et seq.). Thence using the multiplication theorem for determinants (see Section 2.4) it is concluded that the determinants of the two arrays shown above are equal, viz. τ
Su —
S 1 2
S13
S21
S22 — k
S23
S31
S32
S33 — k
Su— A: S 1 2 S13 S21 S 2 2— k S 2 3 S31 S32 S33 — k
(4.5)
Each side of eqn. (4.5) is a cubic in k. Since this equation must
55
THE P R O D U C T S OF T E N S O R S
be satisfied for every value of k it follows that the coefficients of each power of k must be the same. These conditions give 2
From the coefficient of k
Su + S22 + S 3 3
^11
^33
(4.6a)
From the coefficient of k Su S12 S21S22
_j s s3 22 2
S32 S33
From the constant
L Sn S13 S31S33
S n S12 S2I S 2 2
Su S12 S13 S21 S22 S 2 3 S31 S32
S33
+
S 22 S 2 3 S32 S33
+
S u S13 S31
S33
(4.6b)
S u S12 S13 S 2 1 S 22 S 2 3 S31
s
3 2
s
(4.6c)
3 3
i.e. the three left-hand sides are invariants of the tensor S. It is instructive to write these invariants in suffix form, and in so doing, the third of them calls for the definition of a new quantity. Thefirstinvariant is clearly Ji = S,
(4.7)
The reader will find it instructive to verify that the second invariant is 1 (4.8) J2 — y[Sj>pSqq SpqSpq] For the third invariant, the determinant may be written out in full as S n S 22 S 3 3
S11S23S32 +
S12S23S31 — S12S21S33 +
S13S21S32
S13S22S31.
The pattern of these terms is as follows : (i) In a given term the leading subscripts are all different and are 1, 2, 3 in that order. (ii) The second subscripts are all different.
56
CARTESIAN TENSORS IN E N G I N E E R I N G
SCIENCE
(iii) Where the second subscripts are in the cyclic order 1231231..., the term is positive. (iv) Where the second subscripts are in the anticyclic order 321321..., the term is negative. It is concluded that the value of the determinant may be expressed as SpqrSipS^qSsr (carrying out the implied summation over p,q and r) provided that e is defined as follows : pqr
1, if p,q,r are all different and in the cyclic order 12312... = — 1, if p,q,r are all different and in the anticyclic order 32132... = 0, if any two of p,q r are the same.
£pqr =
9
This function is commonly called the Levi-Civita density. In some books the set of e are called permutation symbols. Hence the third invariant of the tensor S is pqr
J% —
4.6
The Levi-Civita Density as a Third Order Tensor
The value of the expression XipX2 l Ep q 3r
whence
(4.9)
£pqrSlpS2qS$r
£123 —
All
^12
Λ-13
Xzi
Λ-22
^23
^31
Λ-32 Λ-33
qr
= +1 =
is £123
XlpX^qXzr^pqr*
Similarly, by taking all possible combinations w,v,h> for the leading suffices of the three lambdas, it is readily verified that in all cases (4.10) Noting that the Levi-Civita density is defined the same with respect to all coordinate axes, so that e w and e w are the same thing, it follows that eqn. (4.10) is the law of transformation UV
UV
57
T H E P R O D U C T S OF T E N S O R S
appropriate to a third order tensor. 4.7
The Products of Vector Components
(a) The scalar product of two vectors A and Β has already been defined as A B . This scalar quantity often written A . Β and called the "dot product", is ABcosO, where A and Β are magnitudes of A and B, and θ is the angle between them. (b) The vector product arises from the equation C = ε ΑΒ. The term on the right is made up of components of third, first and first order tensors. With two contractions (i.e. summation over ρ and q) the result is a vector C of which C is one component. The vector C is called the vector product, or cross product of A with B. In a familiar notation C = Α χ Β. In Suffix notation P
P
r
ραν
ρ
ς
r
C — Cflf —
(4.11)
SpqrApBql',
where i is a unit vector in the x direction. It will be noted that the order in which A and Β are written down is important because if, in eqn. (4.11) the roles of A and Β are reversed, the associated e changes sign owing to the fact that the subscripts ρ and q change places. r
r
pqr
Hence
Α χ Β = - Β χ Α.
(4.12)
Important particular examples of eqn. (4.11) are the cross products of unit vectors ii,i2 and 13. Using eqn. (4.11), letting A = ii so that A \ = 1, A2 = 0, Az = 0, and Β = i so that Bi = 0, B2 = 1, Bz = 0 one finds ii X 12 = 13. There are five more such equations. The complete set is 2
ii X i 2 =
13, 12 X 13 =
ii, 13 X ii =
12 x ii = - 1 3 , 13 X Î2 = - h ,
12 ^|
h X 13 = - 1 2
J
It will also be appreciated that the cross product of a vector with itself is zero, i.e. A χ A = 0 because this cross product Ε
58
CARTESIAN
TENSORS
I N E N G I N E E R I N G
SCIENCE
is e qrApAq\ the terms of which vanish in identical pairs (see a similar example below). In particular P
r
ii χ h = 0, h X i = 0, i Χ i = 0. 2
3
3
(4.14)
The cross product C of two vectors A and Β is perpendicular to both A and B, i.e. is perpendicular to the plane defined by A and B. To see that this is so it is only necessary to examine the scalar product of C with A and of C with B. The scalar product C . A is C
r
A
r
= ερατΑρΒςΑτ.
(4.15)
The right-hand side of eqn. (4.15) gives zero identically, since the terms vanish in identical pairs. For example, with q = 2 there are two terms, namely £123^1^2^3 and £321^3^2^1; these give B2(AiA — Λ3Λ1), i.e. zero. Similarly, for all the other contributions. Since C . A is zero it is concluded that C is perpendicular to A. Similarly, one may show that C is perpendicular to B. Examining the form of eqn. (4.11), and recalling that the definition of the Levi-Civita density was suggested by the need to express a determinant in suffix notation, it is clear that 3
H
A X B = Αι Bi
12
13 A2 A3
B2
(4.16)
B3
Alternatively, the reader will readily verify that A x B = (ΑΒ$\ηθ)η, where η is the unit vector perpendicular to the plane of A and Β and in the direction such that rotation of vector A into vector Β through an angle smaller than 180° would cause a righthand screw to advance along n. Figure 26 shows some examples of cross products. It has been remarked above that A χ Β is perpendicular to A and B. Thus, if A and Β are themselves mutually perpendicular, then A,B and (Α χ Β) constitute a right-handed triad of perpendicular axes. This fact provides a means of solving problems
THE
P R O D U C T S
59
OF TENSORS
\Rotation
Rotation
"BxA
(b)
(a)
Rotation Rotation
(d)
(c) FIG. 2 6 .
such as question 2 at the end of Chapter 2. In that question, two /2 2 mutually perpendicular vectors were given, viz. I + and
I
1
[—7^1 \ \ 0
+
2
\
A/
/
(0)i2 — —7713land 5
+
1
ν
the third vector of the triad was
sought. Taking the cross product of these two, either by expanding the determinant eqn. (4.16) or (what amounts to the same thing) making repeated use of eqns. (4.13) and (4.14), one finds
/ · Γνΐ 4
1 1
-
5
+
^
ΐ
Λ ~ ^ τ 2
2
This enables the matrix of direction cosines to be completed.
60
CARTESIAN TENSORS IN ENGINEERING SCIENCE
4.8 The Vector Operator V The vector operator V , pronounced "del" or (more rarely) d
d
e
dxi
dxz
cxz
"nabla" is defined as V = — ii + -—12 + In suffix notation
V=
àxp
13. (4.17)
ip.
There are three main applications of the vector operator. (i) If / is a scalar function of (xi xz xa), i.e. if / = f(xi,X2 X3), so that to every point in (xi X2 X3) space there corresponds a value of 9
9
df
'•
t h e n
v
/
i
s
W
df 1
+
W
9
9
9
2
df +
W * 3
In suffix notation V/ = -—\ .
(4.18)
v
OXp
V/is called the gradient of/, often written grad/, and is a vector, the operator V can thus be regarded as operating on a scalar / and producing a vector V/. In order to prove that the triple / df df df\ 9 9 \dx' dx dx~)
^
S a
v e c t or
^ '
s
n e c e s s a r
y
t o
show that these
components transform, under change of axes, in accordance with the law given in eqn. (1.8). Using the chain rule of partial differentiation,
61
T H E P R O D U C T S OF T E N S O R S
df
df
Similarly for —— and — . All three equations may be expressed (4.19b)
dx'p dXq
dXq
This last equation is important and it is well worth while to acquire a clear geometrical "feel" for the chain rule. Figure 27 f here
f + Sf here
FIG. 2 7 .
shows the development of this expression for the case q = 1. At two points x i , X 2 , * 3 and x\+δχ\,χ ,χζ, the values of the scalar function are /and f+Sf. Because the change takes place owing to a change in JCI, whilst x and x$ stay constant, the limiting value of 2
2
8f
çT— as Sxi is made indefinitely small is written
df
Now, as shown
in Fig. 27, let the length Sxi be considered to be the body diagonal of a small box whose sides are in the directions of the coordinate axes x[,x and x . Then the lengths of the sides of the box are AB = AnÄxi"^ 2
3
BC = λ2ιδχχ CD=
^ (4.20)
λζιδχι J
For example, the triangle ABD is right-angled at Β and the cosine of angle BAD is λη.
62
C A R T E S I A N TENSORS IN E N G I N E E R I N G S C I E N C E
The rate at which/ changes as one moves along the line AB is df
—, ; consequently, the increase of / between A and Β is df
TTAB
Ξ
ΟΧγ
df -—,-ληδχι.
CXX
Similarly, the increases in/ from Β to C and C to D are respectively df ^-τλ2ιδχι ox 2
A
df
and -τ-τλζιδχι. The total change in / a s one moves from ox
df
3
df
df
df
to D is -— δχχ and so - — 8 χ ι = —,- /Ιιιδχι + to
df -^τλζιδχι, dx3
- — r
dxl
to
-
df
df
i.e. on cancelling to, — =-τ-τλιι -
to
to
+
^
dx2
df
^
+
df
—/I21 + a dx2
In suffix notation eqn. (4.21a) is written - = -^-7—
Τ-7Ά31.
^*3
(4.21a)
dxv
0
to fop to
^/
A2ito
·
df
The corresponding equations for -— and -~— are obvious. All df
three equations may be expressed as -— = CXq
df dx
(4.21b)
— - 7
uXp
CXq
Equations (4.19b) and (4.21b) are identical because λ
dx
ν α
= - -· CXq
But eqn. (4.21b) expresses a transformation which is in accordance with the definition of a vector given in Chapter I. Hence the triple
(OLK df df \dxidx2
1)
df\ dxsj
is aisvector. directionThe of this vector at conany point in the space quicklyThe established. equation / = stant represents a surface in the space; at a given point on the surface choose axes so that the tangent plane is the plane of axes x and X3, then xi is normal to the surface. Because the x 2
2
T H E P R O D U C T S OF
63
TENSORS
axis is tangent to the surface / = constant, it follows that — for
df
2
df
is zero and similarly for -—. Hence grad / = -—iiandisperpenCX%
GX\
dicular to the surface / = constant. This direction is the "line of steepest ascent" of the scalar function /. It is instructive to think of the contours drawn on a map of a mountainous district, the line of steepest descent down a mountainside cuts through the contours at right angles. (ii) If A is a vector the dot product of V with A, viz. V . A, is dA\
dA2
dAs
dAp
-λ— dxi + -7Γ~ dx + ^—5 dxs i.e. in suffix notation V . A =——. dxp The scalar V . A is called the divergence of the vector A and is written divA. It arises quite naturally, for example, when considering theflowof an incompressible fluid; let V be the velocity vector at a point 2
of such a fluid, then considering flow into and out of a cuboid whose sides are δχι, δχ2 and δχζ (Fig. 28) one may equate
X1X2X3
64
CARTESIAN
TENSORS
I N E N G I N E E R I N G
SCIENCE
total inflow and total outflow, whence / VlÔX2ÔXz
+
V2ÔX3ÔX\
I dV2 \ \V2 + ^δχ2\δχ3δχι dVi g i v i n g
+
\
( Vi + ~fa-àxi
=
VzÔXlÔX2
dVi
\δχ2δχ3
+
ι dV3 \ + I Vz + ~^δχζ\δχιδχ2
dV2
dVs
Έί + Έί +
°· "·
=
divV
=
α
(iii) If A is a vector, the cross product of V with A, viz. V χ A is id A3 \dx2
dA2\ dx3 /
(dAi
dAz\ 2 dxi Γ
\dX3
In suffix notation V χ A = ε.pqr
ldA2
\ dx±
ΘΑ dXp
'
dAi\ 3 dx2 Γ "
(4.22)
Γ
A very useful form is also VXA: dxi Αχ
12
13
d
d
(4.23)
dx2 dxz A2 A3
The vector V χ A is usually called "curl A" or, more rarely in recent years, "rot A". The latter name emphasises the link between
FIG. 2 9 .
65
T H E P R O D U C T S OF T E N S O R S
this quantity and the phenomenon of rotation. If a rigid body is imagined to rotate at uniform angular velocity ω around the x$ axis (Fig. 29) the velocity V at X1X2X3 is — (0x2*1 + coxih and curl V is, using the notation of eqn. (4.23), | ii 12 13
e
d
e
dxi
dx2
dxz
— ωχ2
coxi
I
0
i.e. curl V = 2ωΐ3, which is twice the angular velocity vector of the motion. A useful mental picture is provided by imagining a leaffloatingin a straight line down a stream; if the leaf preserves always the same attitude, i.e. does not spin around a vertical axis, its motion belongs to that class called irrotational, a class of motions of great importance influidmechanics and characterised by curl V = 0. If the leaf spins as it floats down stream, the motion is rotational and the curl of V has magnitude twice the angular velocity of spin. The suffix notation provides a convenient way of establishing relationships between vector quantities. Two worked examples follow; others are given in the problems at the end of the chapter. Example (a). The scalar product of A with Β χ C is written A . Β χ C. Prove that A . B x C = Α χ Β . C A . Β Α
χ
X C Β.
C
=
AfSpqrBpCq
=
si]kAiBfk
=
EpqrAfBpCq)
^ }
^'
But r p,q is the same cyclic order as p,qs, so thefirstof these expressions is e pqA BpCq which is identical with e^A^^Cjc, merely using different dummy suffices. Example (b). The vector product of A with Β χ C is written Α χ (Β χ C) and is called the triple vector product. Write this in suffix form and show that this vector is of the form μΒ + vC, where μ and ν are scalar s. The rth component of Β χ C is 8pq BpC . 9
r
r
r
q
V
66
CARTESIAN TENSORS IN ENGINEERING SCIENCE
Thence
Α χ (Β X C) = e rtA (e B C )U. S
s
pqr
p
(4.25)
q
To show that this vector is of the form μΒ + vC, suppose without loss of generality that the χι axis is chosen to be the direction of vector B, so that Β = Bih, and that the plane of Β and C is chosen to be the plane of x± and x so that C = CJi + C2I2; then in eqn. (4.25) for a non-zero contribution ρ must be unity and q must be 1 or 2. But if ρ = q = 1, the e is zero and therefore a non-zero contribution is only forthcoming if ρ = 1 and q = 2, whence r must be 3. The right-hand side then takes the form ZsztAseimBiCzH, i.e. e ztA BiC \ . For e zt to be non-zero, t must be either 1 or 2. Hence the available unit vectors must lie in the plane of Β and C which proves the proposition. 2
pqr
s
s
2 t
s
4.9 The Eigenvectors of a Second Order Tensor Τ It has already been noticed (above, Section 4.4(c)) that an equation A = T B , which in matrix notation is written {A} = [Τ] {B} can be regarded as the tensor Τ operating on a vector Β and giving rise to a new vector A. There is one particular case of great interest in the study of vibrations and also in the stability of structures; this is the case in which the vector A which results is in the same direction as the vector B, so that A can be written A = kB, where k is a scalar constant. Any vector Β which has this property is called an eigenvector of the tensor Γ, and the associated value of k is called an eigenvalue. Some properties of the eigenvectors are developed below. The equation which must be satisfied is k {B} = [T]{B}> p
i.e.
pq
q
kB! = TnBi + T12B2 + TizB3 kB2 = Τ2ιΒι + 722#2 + T2ZBs kBs = T31B1 + T32B2 + TzsBd.
"] ± J
(4.26)
The necessary and sufficient condition that there shall exist values of i?i,Z?2,#3, not all of which are zero, satisfying these three
THE
P R O D U C T S
OF
equations is that the determinant Tn-
67
TENSORS
7"i3
7^12
721
Γ22 — k
781
7^32
Γ23
733 -
shall be zero. This is a cubic equation in k. A sufficient, though not necessary, condition for the roots of this equation to be real is that the Τ matrix shall be symmetric (see the problems at the end of this chapter). On limiting consideration to symmetric tensors it is very advantageous to take advantage of the fact, already demonstrated, that there exist three mutually perpendicular axes, called principal axes, with respect to which the Τ matrix is diagonal. Taking the principal axes as coordinate axes ( x i , X 2 , * 3 ) the equation for eigenvalues becomes Ti-k 0 0 = 0 0 0 T -k 0 n-k 0 2
the three roots of which are ki = Γι, k = T , ks = Ts, i.e. the eigenvalues are the values of the principal components of the tensor. To find the eigenvectors Β for each of these values of k, let it be supposedfirstthat the values are all different. The vector corresponding to the eigenvalue ki must then satisfy eqns. (4.26) which now appear as 2
kiBi = kiBi
kiB2 =
2
~)
k2B2
±
(4.26a)
ksBs. J
kiB3 =
Hence B = Bs = 0 and Bi is arbitrary, whence 2
Β =
Gh.
(4.27)
Similarly, the other eigenvectors are C212, C313, where the C's are arbitrary constants. One may conclude that when the principal axes are uniquely defined these axes are the axes of the three distinct eigenvectors.
68
CARTESIAN
TENSORS
IN
E N G I N E E R I N G
SCIENCE
If two values of k are the same, say k\ = £2, with kz different, tofinda vector corresponding to the eigenvalue k\ one substitutes into eqn. (4.26) and finds Ί
kiBi = kiBi kiBi = kiB3 =
(4.26b)
Hence B == 0 and B\ and B are arbitrary. It is concluded that any vector in the plane of the two equal principal components of the tensor is an eigenvector. The eigenvector in the third direction has a uniquely defined direction. For the case where k\ = k = kz (the case of the isotropic tensor), it is left as an exercise to the reader to show that any vector is an eigenvector. 3
2
2
Summary of Chapter 4
1. The product of two tensors is a tensor, where product means multiplication of the components of the one tensor by components of the other. 2. The product of anrathand «th order tensor is an (ra + n)th order tensor, i.e. Qabc- . >pqr. . . = Sabc-. . T . . . p q r
3. Repetition of a suffix in the product of two tensors is effected by multiplying by the appropriate Kronecker Delta, e.g.
OapSabc* · *TPqr
=
Sabc* · »Taqr> · ·
4. The operation of item 3 is called contraction. It reduces the order of the resulting tensor by two, i.e. Sabc- . >T qr. . . = Übe- . .ffr. . ., where U is of the order two less than the Q of item 2 a
5. By contracting enough times any product of tensors can be made to yield a scalar or a vector. 6. If T are the components of a second order tensor, then = T . This is equal to T + T22 + T33 and is the PQ
TpqôPq
p p
n
T H E P R O D U C T S OF
69
TENSORS
first invariant of the tensor. 7. By contracting once, a second order tensor can operate on a vector and generate another vector, viz. A — T B , i.e. {A} = p
pq
q
mi*}. 8. The Levi-Civita density e is defined as £pqr = + 1 , if PMS are all different and in cyclic order 1231231... = — 1, if p,qs are all different and in anticyclic order 3213213. .. = 0, if any two of p,q,r are the same. pqr
9. e
pqr
is a third order tensor.
10. The invariants of a second order tensor are Ji = Τη + T22 + Γ33 h
=
Τη Tn
Ta
+
T22
^3 = I Tn
Tl2
T2I
T22
Γ22
723
732
733
+
7Ίι
7l3
731
733
T12 T2Z
T$l Tz2 733 I These are conveniently remembered as the sums of determinants of size l x l , 2 x 2 and 3 x 3 respectively, involving the elements of the principal diagonal.
11. In suffix notation the invariants are J1
=z
Tpp
J2 = ^TppTqq Jz =
— TpqTpq^j
£pqrTipT2qTzr.
12. The scalar product of two vectors A and Β is A B . P
13. The vector product of two vectors A and Β is
P
e AB\. pqr
p
q r
14. If A,B are perpendicular then Α,Β,Α χ Β form a righthanded triad of mutually perpendicular axes.
70
CARTESIAN TENSORS IN ENGINEERING SCIENCE
15. If a tensor operates on a vector and generates a parallel vector, viz. [T]{B} = k{B} with k as scalar, then {B} is an eigenvector and k the corresponding eigenvalue. 16. If the principal components T\,Ti,T of a symmetric tensor are all different, the eigenvectors are in the directions of the principal axes and there are three distinct eigenvalues. 3
17. If two of the 7 \ , 7 2 , 7 3 are equal, any vector in the plane of the two equal ones is an eigenvector. If all three 7 i , 7 2 , 7 3 are equal, any vector is an eigenvector.
Examples on Chapter 4 1. Find the scalar product of (4ii + 2Î2 + 13) and ( —ii + 3Î2 + 7i ). 8
Ans. 9. 2. Find the vector product of the two vectors given in question 1. Ans. llii - 29i + 14i . 2
3
3. Three mutually perpendicular vectors are sought. The first is to be (4ii + 2Î2 + 13). The second is to be (—ii + 3Î2 + ah), where a is at present unknown. The third vector has no component specified. Find the three vectors. Ans. For thefirsttwo vectors to be mutually perpendicular, their scalar product must be zero, hence — 4 + 6 + a = 0 and a = —2. Having now two vectors a suitable third vector is the cross product of the two, i.e. (4ii + 2Î2 + 13) X ( — h + 3i2 — 213) which yields — 7ii + 7Î2 + 14Î3. 4. Write down the array of coefficients of the vectors of question 3. Is this an array of direction cosines?
T H E P R O D U C T S OF
Ans.
4 -1 -7
2 1 3 -2 7 14
TENSORS
71
No.
5. What must be done to make the array of numbers in question 4 into direction cosines and what is this process called? Ans. Divide the rows by \/2\, \/\4 and ly/6 respectively. Normalisation. 6. The determinant of a stress tensor matrix is zero. What conclusion can be drawn from this? (Consider the invariants.) Ans. At least one of the principal stresses is zero, i.e. a state of plane stress prevails. 7. Prove that Sp Spst mutation identity). qr
=
ôqsôrt
—
Sqtôrs
(this is called the per-
8. Using the identity of question 7 prove that A Χ (Β X C) = (A . C)B - (A . B)C. 9. A symmetric tensor has components [ 5 2 5 2 2 2 5 2 5 Use the invariants to find the principal components. Ans. 10*9, 1-1, 0 very closely.