2.2 Vectors
23
e r = e x cos θ + e y sin θ
e θ = −e x sin θ + e y cos θ
(2.2)
in terms of the Cartesian base vectors e x and e y . These equations give
∂e r
∂r
=
∂e θ
∂r
= 0
∂e r
∂θ
= −e x sin θ + e y cos θ = e θ
∂e θ
∂θ
= −e x cos θ − e y sin θ = −e r .
(2.3)
To understand the meaning of these results, consider the geometry in Fig. 2.2a.
For a given θ , the vectors e r and e θ maintain the same magnitude (unity) and
direction as they move in the r-direction, and therefore differentiation with respect
to r is zero for both. On the other hand, while moving along a circle in the θ -
direction (constant r), both vectors change direction. Since a vector changes when
either its magnitude or direction changes, differentiating e r and e θ with respect to θ
yields nonzero values.
Using these ideas, we next derive ∂e r /∂θ using geometry alone. Consider two
points P 1 and P 2 located on a circle centered at the origin (Fig. 2.2b). If these points
are separated by the small angle θ , then the vector e r at P 1 becomes e r + e r
at P 2 , where e r ∼ = (∂e r /∂θ) )θ . The geometry of the small triangle in Fig. 2.2b
outside the circle gives
|e r | =
∂e r
∂θ
θ
= |e r | θ
→
∂e r
∂θ
= 1,
which shows that the derivative has unit magnitude. In addition, as θ → 0, the
vector e r , and thus ∂e r /∂θ becomes perpendicular to e r , i.e., it points in the
direction of e θ . Hence, we obtain
∂e r
∂θ
= e θ ,
which agrees with the above result. The result for ∂e θ /∂θ can be found similarly.
2.2.2 Dot and Cross Products
An orthogonal triad of unit vectors {e 1 , e 2 , e 3 } satisfies the relations
e i · e j = δ ij
e i × e j = ij k e k ,
(2.4)