Mathematical Physics Vol 1

Chapter 4. Field theory

194

Exercise 146

d ϕ d t

d e ρ d t

d e ϕ d t

= ˙ ϕ e ϕ ,

= − ˙ ϕ e ρ , where ˙ ϕ =

Prove that

.

Solution It was shown in Example 144 on p.192 (see (4.218)) that e ρ = cos ϕ i + sin ϕ j , e ϕ = − sin ϕ i + cos ϕ j , e z = k .

By differentiating, we obtain from here d e ρ d t

= − sin ϕ ˙ ϕ i + cos ϕ ˙ ϕ j =( − sin ϕ i + cos ϕ j ) ˙ ϕ = ˙ ϕ e ϕ , d e ϕ d t = cos ϕ ˙ ϕ i + sin ϕ ˙ ϕ j =( cos ϕ i + sin ϕ j ) ˙ ϕ = − ˙ ϕ e ρ ,

which was to be proved.

Exercise 147 Express the velocity v and acceleration a of a particle in cylindrical coordinates.

Solution The position vector is r = x i + y j + z k , and the velocity and acceleration are, by definition v = d r d t = ˙ x i + ˙ y j + ˙ z k , that is a = d v d t = ¨ x i + ¨ y j + ¨ z k . Let us now express the position vector in cylindrical coordinates r = x i + y j + z k =( ρ cos ϕ )( cos ϕ e ρ − sin ϕ e ϕ ) +( ρ sin ϕ )( sin ϕ e ρ + cos ϕ e ϕ )+ z e z = ρ e ρ + z e z . Differentiating by time we obtain the velocity

d ρ d t

d e ρ d t

d r d t

d z d t

e ρ + ρ

v =

e z

=

+

= ˙ ρ e ρ + ρ ˙ ϕ e ϕ + ˙ z e z .