Exercise 1.1
edit
p.15
Problem : If
p
i
{\displaystyle p_{i}}
and
q
i
{\displaystyle q_{i}}
are vectors, then
p
1
/
q
1
{\displaystyle p_{1}/q_{1}}
,
p
2
/
q
2
{\displaystyle p_{2}/q_{2}}
,
p
3
/
q
3
{\displaystyle p_{3}/q_{3}}
are three numbers whose values are defined for any set of axes. Are they the components of a vector?
Solution : If
p
i
/
q
i
{\displaystyle p_{i}/q_{i}}
are components of a vector they should transform like the components of a vector. See Table 2 p. 13
p
i
′
=
a
i
j
p
j
{\displaystyle p_{i}'=a_{ij}p_{j}}
A transformation with
a
i
j
{\displaystyle a_{ij}}
p
i
′
=
a
i
j
p
j
{\displaystyle p_{i}'=a_{ij}p_{j}}
q
i
′
=
a
i
j
q
j
{\displaystyle q_{i}'=a_{ij}q_{j}}
p
i
′
q
i
′
=
a
i
j
p
j
a
i
j
q
j
=
p
j
q
j
≠
a
i
j
p
j
q
j
{\displaystyle {\frac {p_{i}'}{q_{i}'}}={\frac {a_{ij}p_{j}}{a_{ij}q_{j}}}={\frac {p_{j}}{q_{j}}}\neq a_{ij}{\frac {p_{j}}{q_{j}}}}
shows that
p
i
/
q
i
{\displaystyle p_{i}/q_{i}}
are not the components of a vector.
Answer : No, because they do not transform like vector components.
Exercise 1.2
edit
p.22
Exercise 1.3
edit
p. 31
Problem:
Electrical conductivity tensor with axes
x
1
{\displaystyle x_{1}}
,
x
2
{\displaystyle x_{2}}
,
x
3
{\displaystyle x_{3}}
[
σ
i
j
]
=
[
25
0
0
0
7
−
3
3
0
−
3
3
13
]
⋅
10
7
o
h
m
−
1
m
−
1
{\displaystyle {\begin{bmatrix}\sigma _{ij}\end{bmatrix}}={\begin{bmatrix}25&0&0\\0&7&-3{\sqrt {3}}\\0&-3{\sqrt {3}}&13\end{bmatrix}}\cdot 10^{7}\mathrm {ohm} ^{-1}\mathrm {m} ^{-1}}
is transformed to a new set of axes
x
1
{\displaystyle x_{1}}
,
x
2
{\displaystyle x_{2}}
,
x
3
{\displaystyle x_{3}}
with the following angles
x
1
′
O
x
1
=
0
∘
{\displaystyle x_{1}'Ox_{1}=0^{\circ }}
,
x
2
′
O
x
2
=
30
∘
{\displaystyle x_{2}'Ox_{2}=30^{\circ }}
,
x
2
′
O
x
3
=
60
∘
{\displaystyle x_{2}'Ox_{3}=60^{\circ }}
,
x
3
′
O
x
3
=
30
∘
{\displaystyle x_{3}'Ox_{3}=30^{\circ }}
.
Draw up a transformation table similar to (11) on p.9 and check that the sum of the squares of
a
i
j
{\displaystyle a_{ij}}
for each row and column is 1.
Solution : It follows from the given angles that
x
3
′
O
x
2
=
30
∘
+
60
∘
+
30
∘
=
120
∘
{\displaystyle x_{3}'Ox_{2}=30^{\circ }+60^{\circ }+30^{\circ }=120^{\circ }}
.
a
i
j
{\displaystyle a_{ij}}
consists of the direction cosines of the angles:
(
a
i
j
)
=
(
cos
(
0
∘
)
0
0
0
cos
(
30
∘
)
cos
(
60
∘
)
0
cos
(
120
∘
)
cos
(
30
∘
)
)
=
(
1
0
0
0
3
/
2
1
/
2
0
−
1
/
2
3
/
2
)
{\displaystyle {\begin{pmatrix}a_{ij}\end{pmatrix}}={\begin{pmatrix}\cos(0^{\circ })&0&0\\0&\cos(30^{\circ })&\cos(60^{\circ })\\0&\cos(120^{\circ })&\cos(30^{\circ })\end{pmatrix}}={\begin{pmatrix}1&0&0\\0&{\sqrt {3}}/2&1/2\\0&-1/2&{\sqrt {3}}/2\end{pmatrix}}}
Checking the columns and rows
cos
(
0
∘
)
2
=
1
2
=
1
{\displaystyle \cos(0^{\circ })^{2}=1^{2}=1}
cos
(
30
∘
)
2
+
cos
(
60
∘
)
2
=
3
/
4
+
1
/
4
=
1
{\displaystyle \cos(30^{\circ })^{2}+\cos(60^{\circ })^{2}=3/4+1/4=1}
cos
(
30
∘
)
2
+
cos
(
120
∘
)
2
=
1
{\displaystyle \cos(30^{\circ })^{2}+\cos(120^{\circ })^{2}=1}
Problem : Transform
σ
i
j
{\displaystyle \sigma _{ij}}
to
σ
i
j
′
{\displaystyle \sigma _{ij}'}
and interpret the result.
Solution : A second-rank tensor transforms according to (22) on p.11:
σ
i
j
′
=
a
i
k
a
j
l
σ
k
l
{\displaystyle \sigma _{ij}'=a_{ik}a_{jl}\sigma _{kl}}
:
This can be reduced because we have some zero components in the electrical conductivity tensor
σ
i
j
{\displaystyle \sigma _{ij}}
:
σ
i
j
′
{\displaystyle \sigma _{ij}'}
=
a
i
1
a
j
1
σ
11
⏞
25
+
a
i
1
a
j
2
σ
12
⏞
0
+
a
i
1
a
j
3
σ
13
⏞
0
+
{\displaystyle =a_{i1}a_{j1}\overbrace {\sigma _{11}} ^{25}+a_{i1}a_{j2}\overbrace {\sigma _{12}} ^{0}+a_{i1}a_{j3}\overbrace {\sigma _{13}} ^{0}+}
+
a
i
2
a
j
1
σ
21
⏞
0
+
a
i
2
a
j
2
σ
22
⏞
7
+
a
i
2
a
j
3
σ
23
⏞
−
3
3
+
{\displaystyle +a_{i2}a_{j1}\overbrace {\sigma _{21}} ^{0}+a_{i2}a_{j2}\overbrace {\sigma _{22}} ^{7}+a_{i2}a_{j3}\overbrace {\sigma _{23}} ^{-3{\sqrt {3}}}+}
+
a
i
3
a
j
1
σ
31
⏞
0
+
a
i
3
a
j
2
σ
32
⏞
−
3
3
+
a
i
3
a
j
3
σ
33
⏞
13
{\displaystyle +a_{i3}a_{j1}\overbrace {\sigma _{31}} ^{0}+a_{i3}a_{j2}\overbrace {\sigma _{32}} ^{-3{\sqrt {3}}}+a_{i3}a_{j3}\overbrace {\sigma _{33}} ^{13}}
=
a
i
1
a
j
1
25
+
a
i
2
a
j
2
7
−
a
i
2
a
j
3
3
3
−
a
i
3
a
j
2
3
3
+
a
i
3
a
j
3
13
{\displaystyle =a_{i1}a_{j1}25+a_{i2}a_{j2}7-a_{i2}a_{j3}3{\sqrt {3}}-a_{i3}a_{j2}3{\sqrt {3}}+a_{i3}a_{j3}13}
now it is just a matter of calculating the components
σ
11
′
=
a
11
a
11
⏟
1
25
+
…
⏟
0
=
25
{\displaystyle \sigma _{11}'=\underbrace {a_{11}a_{11}} _{1}25+\underbrace {\ldots } _{0}=25}
σ
12
′
=
0
{\displaystyle \sigma _{12}'=0}
σ
13
′
=
0
{\displaystyle \sigma _{13}'=0}
σ
21
′
=
0
{\displaystyle \sigma _{21}'=0}
σ
22
′
=
a
21
a
21
⏟
0
25
+
a
22
a
22
⏟
3
/
4
7
−
a
22
a
23
⏟
3
/
4
3
3
−
a
23
a
22
⏟
3
/
4
3
3
+
a
23
a
23
⏟
1
/
4
13
=
1
4
(
21
−
9
−
9
+
13
)
=
4
{\displaystyle \sigma _{22}'=\underbrace {a_{21}a_{21}} _{0}25+\underbrace {a_{22}a_{22}} _{3/4}7-\underbrace {a_{22}a_{23}} _{{\sqrt {3}}/4}3{\sqrt {3}}-\underbrace {a_{23}a_{22}} _{{\sqrt {3}}/4}3{\sqrt {3}}+\underbrace {a_{23}a_{23}} _{1/4}13={\frac {1}{4}}\left(21-9-9+13\right)=4}
σ
23
′
=
a
21
a
31
⏟
0
25
+
a
22
a
32
⏟
−
3
/
2
7
−
a
22
a
33
⏟
3
/
4
3
3
−
a
23
a
32
⏟
−
1
/
4
3
3
+
a
23
a
33
⏟
3
/
4
13
=
3
4
(
−
7
−
9
+
3
+
13
)
=
0
{\displaystyle \sigma _{23}'=\underbrace {a_{21}a_{31}} _{0}25+\underbrace {a_{22}a_{32}} _{-{\sqrt {3}}/2}7-\underbrace {a_{22}a_{33}} _{3/4}3{\sqrt {3}}-\underbrace {a_{23}a_{32}} _{-1/4}3{\sqrt {3}}+\underbrace {a_{23}a_{33}} _{{\sqrt {3}}/4}13={\frac {\sqrt {3}}{4}}\left(-7-9+3+13\right)=0}
σ
31
′
=
0
{\displaystyle \sigma _{31}'=0}
σ
32
′
=
a
31
a
21
⏟
0
25
+
a
32
a
22
⏟
−
3
/
4
7
−
a
32
a
23
⏟
−
1
/
4
3
3
−
a
33
a
22
⏟
3
/
4
3
3
+
a
33
a
23
⏟
3
/
4
13
=
3
4
(
−
7
+
3
−
9
+
13
)
=
0
{\displaystyle \sigma _{32}'=\underbrace {a_{31}a_{21}} _{0}25+\underbrace {a_{32}a_{22}} _{-{\sqrt {3}}/4}7-\underbrace {a_{32}a_{23}} _{-1/4}3{\sqrt {3}}-\underbrace {a_{33}a_{22}} _{3/4}3{\sqrt {3}}+\underbrace {a_{33}a_{23}} _{{\sqrt {3}}/4}13={\frac {\sqrt {3}}{4}}\left(-7+3-9+13\right)=0}
σ
33
′
=
a
31
a
31
⏟
0
25
+
a
32
a
32
⏟
1
/
4
7
−
a
32
a
33
⏟
−
3
/
4
3
3
−
a
33
a
32
⏟
−
3
/
4
3
3
+
a
33
a
33
⏟
3
/
4
13
=
1
4
(
7
+
18
+
49
)
=
16
{\displaystyle \sigma _{33}'=\underbrace {a_{31}a_{31}} _{0}25+\underbrace {a_{32}a_{32}} _{1/4}7-\underbrace {a_{32}a_{33}} _{-{\sqrt {3}}/4}3{\sqrt {3}}-\underbrace {a_{33}a_{32}} _{-{\sqrt {3}}/4}3{\sqrt {3}}+\underbrace {a_{33}a_{33}} _{3/4}13={\frac {1}{4}}\left(7+18+49\right)=16}
or written in array notation
[
σ
i
j
′
]
=
[
1
0
0
0
4
0
0
0
16
]
⋅
10
7
o
h
m
−
1
m
−
1
{\displaystyle {\begin{bmatrix}\sigma _{ij}'\end{bmatrix}}={\begin{bmatrix}1&0&0\\0&4&0\\0&0&16\end{bmatrix}}\cdot 10^{7}\mathrm {ohm} ^{-1}\mathrm {m} ^{-1}}
Interpretation : The new set of axes
x
i
′
{\displaystyle x_{i}'}
is the principle axes of the tensor.
Problem : Radial vector
O
P
{\displaystyle OP}
with direction cosines in
x
i
{\displaystyle x_{i}}
axes
(
0
,
1
2
,
3
2
)
{\displaystyle {\begin{pmatrix}0,&{\frac {1}{2}},&{\frac {\sqrt {3}}{2}}\end{pmatrix}}}
. Find the electrical conductivity in that direction with an analytical expression.
Solution : Get the vector components in the principle axes
x
i
′
{\displaystyle x_{i}'}
: Using the transformation for polar vector components
l
i
′
=
a
i
j
l
j
{\displaystyle l_{i}'=a_{ij}l_{j}}
leads to
(
0
,
3
2
,
1
2
)
{\displaystyle {\begin{pmatrix}0,&{\frac {\sqrt {3}}{2}},&{\frac {1}{2}}\end{pmatrix}}}
. Using equation (32) on p.25
σ
=
l
i
2
σ
i
{\displaystyle \sigma =l_{i}^{2}\sigma _{i}}
gives us
σ
=
3
4
4
o
h
m
−
1
m
−
1
+
1
4
16
o
h
m
−
1
m
−
1
=
7
o
h
m
−
1
m
−
1
{\displaystyle \sigma ={\frac {3}{4}}4\mathrm {ohm} ^{-1}\mathrm {m} ^{-1}+{\frac {1}{4}}16\mathrm {ohm} ^{-1}\mathrm {m} ^{-1}=7\mathrm {ohm} ^{-1}\mathrm {m} ^{-1}}
.
Problem : An electric field
E
=
1
v
o
l
t
m
−
1
{\displaystyle E=1\;\mathrm {volt} \mathrm {m} ^{-1}}
is applied in direction
O
P
{\displaystyle OP}
. Calculate the components of
E
i
{\displaystyle E_{i}}
and the current density
j
i
{\displaystyle j_{i}}
along the
x
i
{\displaystyle x_{i}}
axes.
Solution : The components of the electric field are
E
i
=
l
i
E
{\displaystyle E_{i}=l_{i}E}
:
E
1
=
0
{\displaystyle E_{1}=0}
E
2
=
1
2
v
o
l
t
/
m
{\displaystyle E_{2}={\frac {1}{2}}\;\mathrm {volt} /\mathrm {m} }
E
3
=
3
2
v
o
l
t
/
m
{\displaystyle E_{3}={\frac {\sqrt {3}}{2}}\;\mathrm {volt} /\mathrm {m} }
The components of the current density are
j
i
=
σ
i
k
E
k
=
σ
i
k
l
k
E
{\displaystyle j_{i}=\sigma _{ik}E_{k}=\sigma _{ik}l_{k}E}
:
j
1
=
0
{\displaystyle j_{1}=0}
j
2
=
(
7
2
−
9
2
)
⋅
10
7
a
m
p
s
/
m
2
=
−
10
7
a
m
p
s
/
m
2
{\displaystyle j_{2}=\left({\frac {7}{2}}-{\frac {9}{2}}\right)\cdot 10^{7}\mathrm {amps} /\mathrm {m} ^{2}=-10^{7}\mathrm {amps} /\mathrm {m} ^{2}}
j
3
=
(
−
3
3
2
+
13
3
2
)
a
m
p
s
/
m
2
=
5
3
⋅
10
7
a
m
p
s
/
m
2
{\displaystyle j_{3}=\left(-{\frac {3{\sqrt {3}}}{2}}+{\frac {13{\sqrt {3}}}{2}}\right)\mathrm {amps} /\mathrm {m} ^{2}=5{\sqrt {3}}\cdot 10^{7}\mathrm {amps} /\mathrm {m} ^{2}}
Problem : determine the magnitude and direction of the current density
j
i
{\displaystyle j_{i}}
.
Solution :
Magnitude:
j
=
j
1
2
+
j
2
2
+
j
3
2
=
76
⋅
10
7
a
m
p
s
m
−
2
{\displaystyle j={\sqrt {j_{1}^{2}+j_{2}^{2}+j_{3}^{2}}}={\sqrt {76}}\cdot 10^{7}\mathrm {amps} \mathrm {m} ^{-2}}
Direction:
j
=
(
0
,
−
1
/
76
,
5
3
/
76
)
{\displaystyle j={\begin{pmatrix}0,&-1/{\sqrt {76}},&5{\sqrt {3/76}}\end{pmatrix}}}
. It lies in the
x
2
{\displaystyle x_{2}}
,
x
3
{\displaystyle x_{3}}
plane with angles
O
x
3
=
cos
−
1
(
5
3
/
76
)
≈
6
∘
35
′
{\displaystyle Ox_{3}=\cos ^{-1}(5{\sqrt {3/76}})\approx 6^{\circ }35'}
and
−
O
x
2
=
cos
−
1
(
1
/
76
)
≈
83
∘
25
′
{\displaystyle -Ox_{2}=\cos ^{-1}(1/{\sqrt {76}})\approx 83^{\circ }25'}
.
Problem : Repeat [6] and [7] but with the
x
i
′
{\displaystyle x_{i}'}
axes.
Solution :
Components:
j
i
=
σ
i
l
i
E
{\displaystyle j_{i}=\sigma _{i}l_{i}E}
j
1
=
0
{\displaystyle j_{1}=0}
j
2
=
4
1
2
10
7
a
m
p
s
/
m
2
=
2
⋅
10
7
a
m
p
s
/
m
2
{\displaystyle j_{2}=4{\frac {1}{2}}10^{7}\mathrm {amps} /\mathrm {m} ^{2}=2\cdot 10^{7}\mathrm {amps} /\mathrm {m} ^{2}}
j
3
=
16
3
2
10
7
a
m
p
s
/
m
2
=
8
3
⋅
10
7
a
m
p
s
/
m
2
{\displaystyle j_{3}=16{\frac {\sqrt {3}}{2}}10^{7}\mathrm {amps} /\mathrm {m} ^{2}=8{\sqrt {3}}\cdot 10^{7}\mathrm {amps} /\mathrm {m} ^{2}}
Magnitude:
j
=
2
2
+
8
2
3
10
7
a
m
p
s
/
m
2
=
14
⋅
10
7
a
m
p
s
/
m
2
{\displaystyle j={\sqrt {2^{2}+8^{2}3}}10^{7}\mathrm {amps} /\mathrm {m} ^{2}=14\cdot 10^{7}\mathrm {amps} /\mathrm {m} ^{2}}
Direction:
(
0
,
2
/
7
,
4
/
7
)
{\displaystyle {\begin{pmatrix}0,&2/7,&4/7\end{pmatrix}}}
with angles
O
x
2
=
73
∘
24
′
{\displaystyle Ox_{2}=73^{\circ }24'}
and
O
x
3
=
55
∘
9
′
{\displaystyle Ox_{3}=55^{\circ }9'}
.