34
2 Vector and Tensor Analysis
where Eq. (2.20) 1 has been used. This equation shows that Q T · Q = I, which
indicates that Q T = Q −1 . Hence, we conclude that Q must satisfy the relation
Q T · Q = Q · Q T = I.
(2.35)
Any tensor Q satisfying this equation is called an orthogonal tensor.
In addition, we note that det(Q T · Q) = det Q T det Q = det Q 2 = det I = 1,
which shows that det Q = ±1. If det Q = +1, Q is called a proper orthogonal
tensor or a rotation tensor. (The minus sign corresponds to a reflection.)
These results show that Q rotates the vector a into the vector ¯
a = Q · a. Dotting
both sides of this equation with Q −1 gives a = Q −1 · ¯
a = Q T · ¯
a, showing that Q T
produces the reverse rotation of ¯
a to a.
To better understand what a rotation tensor does, consider the vector a = a i e i ,
where the e i are Cartesian base vectors. The tensor Q transforms a into the rotated
vector
¯
a = Q · a = Q · (a i e i ) = a i Q · e i
= a i ¯
e i ,
(2.36)
where the ¯
e i = Q · e i are unit vectors obtained by rotating the e i . Notably, the
components of a relative to e i are the same as the components of ¯
a relative to ¯
e i .
The reason for this is that the base vectors rotate with a, making the components
appear unchanged relative to the rotated basis. Of course, the components of ¯
a
change relative to the unrotated basis.
Finally, inspecting the equation ¯
e i = Q · e i reveals that we can write
Q = ¯
e i e i
(2.37)
as confirmed by the substitution
Q · e i = (¯ e j e j ) · e i = ¯
e j δ ji = ¯
e i .
With Q T = e i ¯
e i , it is easy to show that these representations satisfy Eq. (2.35). If
there is no rotation, then ¯
e i = e i and Q reduces to the identity tensor I = e i e i .
Example 2.4 Consider rigid-body rotation about the z-axis of a Cartesian coordinate system (x, y, z). (a) Determine the rotation tensor Q if the rotation angle is θ .
(b) Use Q to rotate the vector u = u x e x .
Solution
(a) To find Q, we consider rotation of the unit base vectors (e x , e y , e z ) into the unit
vectors (¯ e x , ¯
e y , ¯
e z ) through the equations
¯
e x = Q · e x ,
¯
e y = Q · e y ,
¯
e z = Q · e z .
(2.38)
2 Vector and Tensor Analysis
where Eq. (2.20) 1 has been used. This equation shows that Q T · Q = I, which
indicates that Q T = Q −1 . Hence, we conclude that Q must satisfy the relation
Q T · Q = Q · Q T = I.
(2.35)
Any tensor Q satisfying this equation is called an orthogonal tensor.
In addition, we note that det(Q T · Q) = det Q T det Q = det Q 2 = det I = 1,
which shows that det Q = ±1. If det Q = +1, Q is called a proper orthogonal
tensor or a rotation tensor. (The minus sign corresponds to a reflection.)
These results show that Q rotates the vector a into the vector ¯
a = Q · a. Dotting
both sides of this equation with Q −1 gives a = Q −1 · ¯
a = Q T · ¯
a, showing that Q T
produces the reverse rotation of ¯
a to a.
To better understand what a rotation tensor does, consider the vector a = a i e i ,
where the e i are Cartesian base vectors. The tensor Q transforms a into the rotated
vector
¯
a = Q · a = Q · (a i e i ) = a i Q · e i
= a i ¯
e i ,
(2.36)
where the ¯
e i = Q · e i are unit vectors obtained by rotating the e i . Notably, the
components of a relative to e i are the same as the components of ¯
a relative to ¯
e i .
The reason for this is that the base vectors rotate with a, making the components
appear unchanged relative to the rotated basis. Of course, the components of ¯
a
change relative to the unrotated basis.
Finally, inspecting the equation ¯
e i = Q · e i reveals that we can write
Q = ¯
e i e i
(2.37)
as confirmed by the substitution
Q · e i = (¯ e j e j ) · e i = ¯
e j δ ji = ¯
e i .
With Q T = e i ¯
e i , it is easy to show that these representations satisfy Eq. (2.35). If
there is no rotation, then ¯
e i = e i and Q reduces to the identity tensor I = e i e i .
Example 2.4 Consider rigid-body rotation about the z-axis of a Cartesian coordinate system (x, y, z). (a) Determine the rotation tensor Q if the rotation angle is θ .
(b) Use Q to rotate the vector u = u x e x .
Solution
(a) To find Q, we consider rotation of the unit base vectors (e x , e y , e z ) into the unit
vectors (¯ e x , ¯
e y , ¯
e z ) through the equations
¯
e x = Q · e x ,
¯
e y = Q · e y ,
¯
e z = Q · e z .
(2.38)