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Chapter 2 Solutions

Problem 2.1
Let v = a × b, or in indicial notation,
vi e ^j × bk e
^i = aj e ^k = εijk aj bk e
^i

Using indicial notation, show that,

(a) v · v = a2 b2 sin2 θ ,
(b) a × b · a = 0 ,
(c) a × b · b = 0 .

Solution
(a) For the given vector, we have
v · v = εijk aj bk e
^i · εpqs aq bs e
^p = εijk aj bk εpqs aq bs δip = εijk aj bk εiqs aq bs
= (δjq δks − δjs δkq ) aj bk aq bs = aj aj bk bk − aj bk ak bj
= (a · a) (b · b) − (a · b) (a · b) = a2 b2 − (ab cos θ)2
= a2 b2 1 − cos2 θ = a2 b2 sin2 θ

(b) Again, we find

a × b · a = v · a = (εijk aj bk e
^i ) · aq e
^q = εijk aj bk aq δiq = εijk aj bk ai = 0

This is zero by symmetry in i and j.

(c) This is
a × b · b = v · b = (εijk aj bk e
^i ) · bq e
^q = εijk aj bk bq δiq = εijk aj bk bi = 0

Again, this is zero by symmetry in k and and i.

Problem 2.2
With respect to the triad of base vectors u1 , u2 , and u3 (not necessarily unit vectors), the
triad u1 ,u2 , and u3 is said to be a reciprocal basis if ui · uj = δij (i, j = 1, 2, 3). Show that
to satisfy these conditions,
u2 × u3 u3 × u1 u1 × u2
u1 = ; u2 = ; u3 =
[u1 , u2 , u3 ] [u1 , u2 , u3 ] [u1 , u2 , u3 ]
and determine the reciprocal basis for the specific base vectors
u1 ^2 ,
e1 + e
= 2^
u2 ^3 ,
e2 − e
= 2^
u3 ^1 + e
= e ^2 + e^3 .

1
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Answer
1
u1 = 5
(3^ ^2 − 2^
e1 − e e3 )
1
u2 = 5 e1 + 2^
(−^ ^3 )
e2 − e
1
u3 = 5 e1 + 2^
(−^ e2 + 4^
e3 )

Solution
For the bases, we have
u2 × u3 u3 × u1 u1 × u2
u1 ·u1 = u1 · = 1; u2 ·u2 = u2 · = 1; u3 ·u3 = u3 · =1
[u1 , u2 , u3 ] [u1 , u2 , u3 ] [u1 , u2 , u3 ]
since the triple scalar product is insensitive to the order of the operations. Now
u2 × u3
u2 · u1 = u2 · =0
[u1 , u2 , u3 ]

since u2 ·u2 ×u3 = 0 from Pb 2.1. Similarly, u3 ·u1 = u1 ·u2 = u3 ·u2 = u1 ·u3 = u2 ·u3 = 0.
For the given vectors, we have

2 1 0
[u1 , u2 , u3 ] = 0 2 −1 =5
1 1 1

and
^1
e ^2
e ^3
e
1
u2 × u3 = 0 2 −1 ^2 − 2^
e1 − e
= 3^ e3 ; u1 = ^2 − 2^
e1 − e
(3^ e3 )
1 1 1 5

^1
e ^2
e ^3
e
1
u3 × u1 = 1 1 1 e1 + 2^
= −^ ^3 ;
e2 − e u2 = e1 + 2^
(−^ ^3 )
e2 − e
2 1 0 5

^1
e ^2
e ^3
e
1
u1 × u2 = 2 1 0 e1 + 2^
= −^ e2 + 4^
e3 ; u3 = e1 + 2^
(−^ e2 + 4^
e3 )
0 2 −1 5

Problem 2.3
^i , and let b = bi e
Let the position vector of an arbitrary point P (x1 x2 x3 ) be x = xi e ^i be
a constant vector. Show that (x − b) · x = 0 is the vector equation of a spherical surface
having its center at x = 21 b with a radius of 21 b.

Solution
For

(x − b) · x = (xi e ^i ) · xj e
^i − bi e ^j = (xi xj − bi xj ) δij = xi xi − bi xi =
= x21 + x22 + x23 − b1 x1 − b2 x2 − b3 x3 = 0

Now
 2  2  2
1 1 1 1 2
1
x1 − b1 + x2 − b2 + x3 − b3 = b1 + b22 + b23 = b2
2 2 2 4 4
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This is the equation of a sphere with the desired properties.

Problem 2.4
Using the notations A(ij) = 21 (Aij + Aji ) and A[ij] = 21 (Aij − Aji ) show that

(a) the tensor A having components Aij can always be decomposed into a sum of
its symmetric A(ij) and skew-symmetric A[ij] parts, respectively, by the decom-
position,
Aij = A(ij) + A[ij] ,
(b) the trace of A is expressed in terms of A(ij) by

Aii = A(ii) ,

(c) for arbitrary tensors A and B,

Aij Bij = A(ij) B(ij) + A[ij] B[ij] .

Solution
(a) The components can be written as
   
Aij + Aji Aij − Aji
Aij = + = A(ij) + A[ij]
2 2

(b) The trace of A is  

Aii + Aii
A(ii) = = Aii
2
(c) For two arbitrary tensors, we have



Aij Bij = A(ij) + A[ij] B(ij) + B[ij] = A(ij) B(ij) + A[ij] B(ij) + A(ij) B[ij] + A[ij] B[ij]
= A(ij) B(ij) + A[ij] B[ij]

since the product of a symmetric and skew-symmetric tensor is zero

  
Aij + Aji Bij − Bji 1
A(ij) B[ij] = = (Aij Bij + Aji Bij − Aij Bji − Aji Bji )
2 2 4
1
= (Aij Bij + Aji Bij − Aji Bij − Aij Bij ) = 0
4
We have changed the dummy indices on the last two terms.

Problem 2.5
Expand the following expressions involving Kronecker deltas, and simplify where possible.

(a) δij δij , (b) δij δjk δki , (c) δij δjk , (d) δij Aik

Answer
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(a) 3, (b) 3, (c) δik , (d) Ajk

Solution
(a) Contracting on i or j, we have

δij δij = δjj = δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3

(b) Contracting on k and then j gives

δij δjk δki = δij δji = δii = 3

(c) Contracting on j yields

δij δjk = δik
(d) Contracting on i gives
δij Aik = Ajk
Note: It may be helpful for beginning students to write out all terms.

Problem 2.6
If ai = εijk bj ck and bi = εijk gj hk , substitute bj into the expression for ai to show that

a i = g k ck h i − h k ck g i ,

or in symbolic notation, a = (c · g)h − (c · h)g.

Solution
We begin by changing the dummy indices for bj = εjmn gm hn and

ai = εijk bj ck = εijk εjmn gm hn ck = − (εjik εjmn gm hn ck ) = − (δim δkn − δin δkm ) gm hn ck

= −gi hk ck + gk hi ck = gk ck hi − hk ck gi

where we have used the anti-symmetry of εijk = −εjik and the ε−δ identity. Symbolically
a = (c · g)h − (c · h)g

Problem 2.7
By summing on the repeated subscripts determine the simplest form of

(a) ε3jk aj ak , (b) εijk δkj , (c) ε1jk a2 Tkj , (d) ε1jk δ3j vk .

Answer

(a) 0, (b) 0, (c) a2 (T32 − T23 ), (d) −v2

Solution
(a) Summing gives

ε3jk aj ak = ε31k a1 ak + ε32k a2 ak = ε312 a1 a2 + ε321 a2 a1 = a1 a2 − a2 a1 = 0

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(b)
εijk δkj = εij1 δ1j + εij2 δ2j + εij3 δ3j
= εi21 δ12 + εi31 δ13 + εi12 δ21 + εi32 δ23 + εi13 δ31 + εi23 δ32 = 0
(c)
ε1jk a2 Tkj = ε12k a2 Tk2 + ε13k a2 Tk3
= ε123 a2 T32 + ε132 a2 T23 = a2 T32 − a2 T23 = a2 (T32 − T23 )
(d)
ε1jk δ3j vk = ε12k δ32 vk + ε13k δ33 vk = 0 + ε132 δ33 v2 = −v2

Problem 2.8
Consider the tensor Bik = εijk vj .
(a) Show that Bik is skew-symmetric.
(b) Let Bij be skew-symmetric, and consider the vector defined by vi = εijk Bjk
(often called the dual vector of the tensor B). Show that Bmq = 12 εmqi vi .

Solution
(a) For a tensor to be skew-symmetric, one has Aij = −Aji . For the given tensor
Bik = εijk vj = −εkji vj = −Bki
(b) For the dual vector of the tensor B, we have
εmqi vi = εmqi εijk Bjk = (δmj δqk − δmk δqj ) Bjk = Bmq − Bqm = [Bmq − (−Bmq )]
= 2Bmq
since B is skew-symmetric.

Problem 2.9
Use indicial notation to show that
Ami εmjk + Amj εimk + Amk εijm = Amm εijk
where A is any tensor and εijk is the permutation symbol.

Solution
Multiply both sides by εijk and simplify

Amm εijk εijk = 6Amm = Ami εmjk εijk + Amj εimk εijk + Amk εijm εijk

= Ami 2δmi + Amj 2δmj + Amk 2δmk = 6Amm

Problem 2.10
If Aij = δij Bkk + 3Bij , determine Bkk and using that solve for Bij in terms of Aij and its
first invariant, Aii .

Answer
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Bkk = 61 Akk ; Bij = 13 Aij − 1

18 δij Akk

Solution
Taking the trace of Aij gives
Aii = δii Bkk + 3Bii = 3Bkk + 3Bii = 6Bkk
since i and k are dummy indices. This gives
1
Bkk = Akk
6
Substituting for Bkk and solving for Bij gives
1 1 1
3Bij = Aij − δij Akk or Bij = Aij − δij Akk
6 3 18

Problem 2.11
Show that the value of the quadratic form Tij xi xj is unchanged if Tij is replaced by its
symmetric part, 21 (Tij + Tji ).

Solution
The quadratic form becomes
1 1 1
Tij xi xj = (Tij + Tji )xi xj = (Tij xi xj + Tji xi xj ) = (Tij xi xj + Tij xj xi ) = Tij xi xj
2 2 2
since i and j are dummy indices and multiplication commutes.

Problem 2.12
With the aid of Eq 2.7, show that any skew symmetric tensor W may be written in terms
of an axial vector ωi given by
1
ωi = − εijk wjk
2
where wjk are the components of W.

Solution
Multiply by εimn
εimn ωi = − 21 εimn εijk wjk
= − 12 (δmj δnk − δmk δnj ) wjk
= − 12 (wmn − wnm ) = wnm ,
or,
εmni ωi = wnm

Problem 2.13
Show by direct expansion (or otherwise) that the box product λ = εijk ai bj ck is equal to
the determinant
a1 a2 a3
b1 b2 b3 .
c1 c2 c3
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Thus, by substituting A1i for ai , A2j for bj and A3k for ck , derive Eq 2.42 in the form
det A = εijk A1i A2j A3k where Aij are the elements of A.

Solution
Direct expansion gives

λ = εijk ai bj ck = ε1jk a1 bj ck + ε2jk a2 bj ck + ε3jk a3 bj ck

= ε12k a1 b2 ck + ε13k a1 b3 ck + ε21k a2 b1 ck + ε23k a2 b3 ck + ε31k a3 b1 ck + ε32k a3 b2 ck
= ε123 a1 b2 c3 + ε132 a1 b3 c2 + ε213 a2 b1 c3 + ε231 a2 b3 c1 + ε312 a3 b1 c2 + ε321 a3 b2 c1
= a1 b2 c3 − a1 b3 c2 − a2 b1 c3 + a2 b3 c1 + a3 b1 c2 − a3 b2 c1

and
a1 a2 a3
b1 b2 b3 = a1 b2 c3 + a2 b3 c1 + a3 b1 c2 − a1 b3 c2 − a2 b1 c3 − a3 b2 c1 = λ
c1 c2 c3

Using the suggested substitutions for ai , bi , ci , we have A3

λ = εijk A1i A2j A3k = ε1jk A11 A2j A3k + ε2jk A12 A2j A3k + ε3jk A13 A2j A3k

= ε12k A11 A22 A3k + ε13k A11 A23 A3k + ε21k A12 A21 A3k + ε23k A12 A23 A3k
+ε31k A13 A21 A3k + ε32k A13 A22 A3k

= ε123 A11 A22 A33 + ε132 A11 A23 A32 + ε213 A12 A21 A33 + ε231 A12 A23 A31
+ε312 A13 A21 A32 + ε321 A13 A22 A31

= A11 A22 A33 − A11 A23 A32 − A12 A21 A33 + A12 A23 A31 + A13 A21 A32 − A13 A22 A31

and
A11 A12 A13
A21 A22 A23 = A11 A22 A33 − A11 A23 A32 + A12 A23 A31 − A12 A21 A33
A31 A32 A33
+A13 A21 A32 − A13 A22 A31 = λ

Problem 2.14
Starting with Eq 2.42 of the text in the form

det A = εijk Ai1 Aj2 Ak3

show that by an arbitrary number of interchanges of columns of Aij we obtain

εqmn det A = εijk Aiq Ajm Akn

which is Eq 2.43. Further, multiply this equation by the appropriate permutation symbol
to derive the formula
6 det A = εqmn εijk Aiq Ajm Akn .

Solution
Each row or column change introduces a minus sign. After an arbitrary number of row
and column changes, we have

εqmn det A = εijk Aiq Ajm Akn

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Multiplying by εqmn gives

εqmn εqmn det A = (δmm δnn − δmn δnm ) det A = (3 · 3 − δnn ) det A
= (9 − 3) det A = εqmn εijk Aiq Ajm Akn
from the ε − δ identity.

Problem 2.15
Let the determinant of the tensor Aij be given by
A11 A12 A13
det A = A21 A22 A23 .
A31 A32 A33
Since the interchange of any two rows or any two columns causes a sign change in the value
of the determinant, show that after an arbitrary number of row and column interchanges
Amq Amr Ams
Anq Anr Ans = εmnp εqrs det A .
Apq Apr Aps
Now let Aij = δij in the above determinant which results in det A = 1 and, upon expansion,
yields
εmnp εqrs = δmq (δnr δps − δns δpr ) − δmr (δnq δps − δns δpq ) + δms (δnq δpr − δnr δpq ) .
Thus, by setting p = q, establish Eq 2.7 in the form
εmnq εqrs = δmr δns − δms δnr .

Solution
Letting Aij = δij in the determinant gives
δmq δmr δms
δnq δnr δns
δpq δpr δps

= δmq (δnr δps − δns δpr ) − δmr (δnq δps − δns δpq ) + δms (δnq δpr − δnr δpq )
and
εmnp εqrs = δmq (δnr δps − δns δpr ) − δmr (δnq δps − δns δpq ) + δms (δnq δpr − δnr δpq )
since
δ11 δ12 δ13 1 0 0
δ21 δ22 δ23 = 0 1 0 = 1 = ε123 ε123 det A
δ31 δ32 δ33 0 0 1
Setting p = q gives
δmp δmr δms
δnp δnr δns
δpp δpr δps

= δmp (δnr δps − δns δpr ) − δmr (δnp δps − δns δpp ) + δms (δnp δpr − δnr δpp )
= δnr δms − δns δmr − δmr (δns − 3δns ) + δms (δnr − 3δnr )
= δnr δms − δns δmr + 2δmr δns − 2δms δnr = δmr δns − δms δnr
= εpmn εprs
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Problem 2.16
Show that the square matrices
 
1 0 0  
5 2
[Bij ] =  0 −1 0  and [Cij ] =
−12 −5
0 0 1
are both square roots of the identity matrix.

Solution
The product of the matrix with itself should be the identity matrix for it to be a square
root. Thus     
1 0 0 1 0 0 1 0 0
 0 −1 0   0 −1 0  =  0 1 0 
0 0 1 0 0 1 0 0 1
and       
5 2 5 2 25 − 24 10 − 10 1 0
= =
−12 −5 −12 −5 −60 + 60 −24 + 25 0 1

Problem 2.17
Using the square matrices below, demonstrate
(a) that the transpose of the square of a matrix is equal to the square of its transpose
(Eq 2.36 with n = 2),
(b) that (AB)T = BT AT as was proven in Example 2.33
   
3 0 1 1 3 1
[Aij ] =  0 2 4  , [Bij ] =  2 2 5  .
5 1 2 4 0 3

Solution
(a) For the matrix A, we have
    
3 0 1 3 0 1 14 1 5
2
[Aij ] =  0 2 4  0 2 4  =  20 8 16 
5 1 2 5 1 2 25 4 13
and     
3 0 5 3 0 5 14 20 25
T 2
Aij
 
= 0 2 1  0 2 1 = 1 8 4 
1 4 2 1 4 2 5 16 13

T
2
This shows that A2 = AT . Similarly for B, we have
    
1 3 1 1 3 1 11 9 19
[Bij ]2 =  2 2 5   2 2 5  =  26 10 27 
4 0 3 4 0 3 16 12 13
and     
1 2 4 1 2 4 11 26 16
2
BTij =  3
 
2 0  3 2 0 = 9 10 12 
1 5 3 1 5 3 19 27 13
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(b) For (AB)T = BT AT , we have

    
3 0 1 1 3 1 7 9 6
[Aij ] [Bij ] =  0 2 4  2 2 5  =  20 4 22 
5 1 2 4 0 3 15 17 16

and     
1 2 4 3 0 5 7 20 15
BTij ATij =  3
   
2 0  0 2 1 = 9 4 17 
1 5 3 1 4 2 6 22 16

The result is demonstrated.

Problem 2.18
Let A be any orthogonal matrix, i.e., AAT = AA−1 = I, where I is the identity matrix.
Thus, by using the results in Examples 2.9 and 2.10, show that det A = ±1.

Solution
From Example 2.9
det AAT = det A det AT

and from Example 2.10,

det A = det AT

Then
det AAT = det A det AT = det A det A = (det A)2 = det I = 1

and
(det A) = ±1

Problem 2.19
A tensor is called isotropic if its components have the same set of values in every Cartesian
coordinate system at a point. Assume that T is an isotropic tensor of rank two with
0 0 0
components tij relative to axes Ox1 x2 x3 . Let axes Ox1 x2 x3 be obtained with respect
√ to
Ox1 x2 x3 by a righthand rotation of 120◦ about the axis along n ^ = (^e+e ^+e ^) / 3. Show
by the transformation between these axes that t11 = t22 = t33 , as well as other relationships.
00 00 00
Further, let axes Ox1 x2 x3 be obtained with respect to Ox1 x2 x3 by a right-hand rotation
◦
of 90 about x3 . Thus, show by the additional considerations of this transformation that if
T is any isotropic tensor of second order, it can be written as λI where λ is a scalar and I
is the identity tensor.

Solution √
For a 120◦ rotation about the axis n
^ = (^ ^3 ) / 3, the transformation matrix is
^2 + e
e1 + e
 
0 1 0
[aij ] =  0 0 1 
1 0 0