Introduction to vector and tensor analysis

2.7.1 Cylindrical coordinates
Cylindrical coordinates, x ′ = (ρ, φ, z), are defined in terms of normal cartesian
coordinates x = (x1, x2, x3) = (x, y, z) by the coordinate transformation
x = ρ cos(φ)
y = ρ sin(φ),
z = z,
The domain of variation is 0 ≤ ρ < ∞, 0 ≤ φ < 2π.
The position vector may be written as
Local basis
r(x ′ ) = ρ cos(φ)e1 + ρ sin(φ)e2 + ze3
(2.62)
If we take the partial derivatives with respect to ρ, φ, z, c.f. Eq. (2.33), and
normalize we obtain
e ′ 1 (x′ ) = eρ(x ′ ) = ∂ρr = cos(φ)e1 + sin(φ)e2
e ′ 2 (x′ ) = eφ(x ′ ) = 1
ρ ∂φr = − sin(φ)e1 + cos(φ)e2
e ′ 3 = e3
These three unit vectors, like the Cartesian unit vectors ei, form an orthonormal
basis at each point in space. An arbitrary vector field may therefore be resolved
in this basis
where the vector components
a(x ′ ) = ar(x ′ )er(x ′ ) + aφ(x ′ )eφ(x ′ ) + az(x ′ )ez
ar = a · er, aφ = a · eφ, Vz = a · ez
are the projections of a on the local basis vectors.
Resolution of gradient
The derivatives after cylindrical coordinates are found by differentiation through
the Cartesian coordinates (chain rule)
∂ρ = ∂x
∂ρ ∂x + ∂y
∂ρ ∂y = cos(φ)∂x + sin(φ)∂y
∂φ = ∂x
∂φ ∂x + ∂y
∂φ ∂y = −ρ sin(φ)∂x + ρ cos(φ)∂y
From these relations we can calculate the projections of the gradient operator
∇ =
i ei∂i on the cylindrical basis and we obtain
∇ρ = eρ · ∇ = ∂ρ
∇φ = eφ · ∇ = 1
ρ ∂φ
∇z = ez · ∇ = ∂z
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