Mathematical Physics Vol 1

Chapter 4. Field theory

192

If we first square and then add equations (4.215) and (4.216), we obtain ρ 2 ( cos 2 ϕ + sin 2 ϕ )= x 2 + y 2 , that is, ρ = p x 2 + y 2 . Here we used the basic trigonometric identity cos 2 ϕ + sin 2 ϕ = 1 and the fact that ρ , by definition (distance), is positive. Dividing the left and right side of equation (4.216) by equation (4.215) yields

ρ sin ϕ ρ cos ϕ = tan ϕ , that is, ϕ = arctan y x . From here follow the transformations

y x

=

ρ = p x 2 + y 2 ϕ = arctan

y x

z = z .

Exercise 144 Show that the cylindric coordinate system is orthogonal.

Solution The position vector is

r = x i + y j + z k = ρ cos ϕ i + ρ sin ϕ j + z k . The tangent vectors, which correspond to coordinates ρ , ϕ , z are determined by

∂ r ∂ρ

,

∂ r ∂ϕ

∂ r ∂ z

and

, that is

∂ r ∂ρ ∂ r ∂ϕ

= cos ϕ i + sin ϕ j ,

= − ρ sin ϕ i + ρ cos ϕ j ,

∂ r ∂ z

= k . The corresponding unit vectors are e 1 = e ρ = ∂ r / ∂ρ | ∂ r / ∂ρ | =

cos ϕ i + sin ϕ j q cos 2 ϕ + sin 2 ϕ ρ sin ϕ i + ρ cos ϕ j q ρ 2 cos 2 ϕ + ρ 2 sin 2 ϕ

= cos ϕ i + sin ϕ j ,

∂ r / ∂ϕ | ∂ r / ∂ϕ | ∂ r / ∂ z | ∂ r / ∂ z |

= −

= − sin ϕ i + cos ϕ j ,

e 2 = e ϕ =

(4.218)

e 3 = e z =

= k .

and thus

e 1 · e 2 =( cos ϕ i + sin ϕ j ) · ( − sin ϕ i + cos ϕ j )= 0 , e 1 · e 3 =( cos ϕ i + sin ϕ j ) · k = 0 , e 2 · e 3 =( − sin ϕ i + cos ϕ j ) · k = 0 .