Cartesian coordinates (x, y, z) are simple but not always convenient. For problems with cylindrical or spherical symmetry, using coordinates that match the symmetry (like cylindrical or spherical coordinates) simplifies the math significantly. These are examples of orthogonal curvilinear coordinates.
A point P in space can be defined by the intersection of three surfaces:
u₁(x,y,z) = c₁, u₂(x,y,z) = c₂, u₃(x,y,z) = c₃
The values (u₁, u₂, u₃) are the curvilinear coordinates of P.
The system is orthogonal if these three coordinate surfaces are mutually perpendicular at every point of intersection. This means the tangent vectors to the coordinate lines are orthogonal.
The coordinates u₁, u₂, u₃ may not have units of length (e.g., angles). The scale factors (or "metric coefficients") `h₁, h₂, h₃` convert an infinitesimal change in a coordinate `duᵢ` into an actual physical displacement `dsᵢ`.
Let r = x(u₁,u₂,u₃)î + y(u₁,u₂,u₃)ĵ + z(u₁,u₂,u₃)k̂ be the position vector.
The scale factors are defined as the magnitudes of the tangent vectors:
h₁ = |∂r/∂u₁|, h₂ = |∂r/∂u₂|, h₃ = |∂r/∂u₃|
An infinitesimal displacement `dsᵢ` along the uᵢ-axis is `dsᵢ = hᵢ duᵢ`.
The total displacement `dr` is `dr = h₁du₁ ê₁ + h₂du₂ ê₂ + h₃du₃ ê₃`, where `ê₁, ê₂, ê₃` are the orthogonal unit vectors in the direction of increasing u₁, u₂, u₃.
The square of the arc length `ds` is:
ds² = |dr|² = (h₁du₁)² + (h₂du₂)² + (h₃du₃)²
This involves the set of equations that relate (x,y,z) to (u₁,u₂,u₃) and vice versa.
We need the functions `x(u₁,u₂,u₃)`, `y(u₁,u₂,u₃)`, `z(u₁,u₂,u₃)` and their inverses `u₁(x,y,z)`, `u₂(x,y,z)`, `u₃(x,y,z)`.
Example: Cylindrical Coordinates (ρ, φ, z)
Example: Spherical Coordinates (r, θ, φ)
Once we have the scale factors `h₁, h₂, h₃`, we can define all the differential elements needed for integration.
An infinitesimal displacement vector `dr` is given by:
dr = h₁du₁ ê₁ + h₂du₂ ê₂ + h₃du₃ ê₃
The magnitude of the displacement is the arc length `ds`. From the Pythagorean theorem in 3D:
ds² = |dr|² = (h₁du₁)² + (h₂du₂)² + (h₃du₃)²
An area element `dA` is a small patch on a coordinate surface. It's a vector `dA = n̂ dA`, where `n̂` is the normal unit vector.
The infinitesimal volume element `dV` is a small box formed by the three displacements.
dV = (ds₁) (ds₂) (ds₃) = (h₁du₁) (h₂du₂) (h₃du₃) = h₁h₂h₃ du₁du₂du₃
The term J = h₁h₂h₃ is called the Jacobian of the transformation.
Using the scale factors, we can write general expressions for the vector operators. Let `f` be a scalar field and `V = V₁ê₁ + V₂ê₂ + V₃ê₃` be a vector field.
∇f = (1/h₁)(∂f/∂u₁)ê₁ + (1/h₂)(∂f/∂u₂)ê₂ + (1/h₃)(∂f/∂u₃)ê₃
∇ · V = (1 / h₁h₂h₃) [ ∂/∂u₁(h₂h₃V₁) + ∂/∂u₂(h₁h₃V₂) + ∂/∂u₃(h₁h₂V₃) ]
This is best remembered as a determinant:
            
        |  h₁ê₁    h₂ê₂    h₃ê₃   |
∇ × V = (1 / h₁h₂h₃) * | ∂/∂u₁   ∂/∂u₂   ∂/∂u₃  |
        |  h₁V₁    h₂V₂    h₃V₃   |
            
        
        The Laplacian is the divergence of the gradient (∇²f = ∇ · (∇f)).
∇²f = (1 / h₁h₂h₃) [ ∂/∂u₁( (h₂h₃/h₁)(∂f/∂u₁) ) + ∂/∂u₂( (h₁h₃/h₂)(∂f/∂u₂) ) + ∂/∂u₃( (h₁h₂/h₃)(∂f/∂u₃) ) ]
These formulas are the "master templates". You just need to plug in the scale factors for any specific coordinate system.
The coordinates are `(u₁, u₂, u₃) = (ρ, φ, z)`.
Transformations: `x = ρ cos(φ)`, `y = ρ sin(φ)`, `z = z`
From `r = ρcos(φ)î + ρsin(φ)ĵ + zk̂`:
So, (h₁, h₂, h₃) = (1, ρ, 1).
Jacobian: J = h₁h₂h₃ = (1)(ρ)(1) = ρ
Plug (h₁, h₂, h₃) = (1, ρ, 1) into the general formulas:
The coordinates are `(u₁, u₂, u₃) = (r, θ, φ)`.
Transformations: `x = r sin(θ) cos(φ)`, `y = r sin(θ) sin(φ)`, `z = r cos(θ)`
From `r = rsinθcosφ î + rsinθsinφ ĵ + rcosθ k̂`:
So, (h₁, h₂, h₃) = (1, r, r sin(θ)).
Jacobian: J = h₁h₂h₃ = (1)(r)(r sin(θ)) = r² sin(θ)
Plug (h₁, h₂, h₃) = (1, r, r sin(θ)) into the general formulas:
∇²f = (1/r²)∂/∂r(r² ∂f/∂r) + (1/(r²sinθ))∂/∂θ(sinθ ∂f/∂θ) + (1/(r²sin²θ))(∂²f/∂φ²)
You should be able to derive this from the general formula or, in many cases, memorize it directly.