7. Curvilinear Coordinate Systems

Question 1

Curvilinear

Definition

Answer 1

• in performing volume and surface integrals we have often used non-Cartesian coosdinate systems in which the coordinate lines and surfaces are curved
• such coordinate systems are called curvilinear
  e. g. cylindrical and spherical polars

Question 2

Curviliear Unit Vector

Definition

Answer 2

-consider a general curvilinear coordinate system (u,v,w) given locally by:
x = x(u,v,w) , y = y(u,v,w) and z = z(u,v,w)
-consider an infinitesimal displacement:
d|x = (dx, dy, dz)
-using the chain rule for partial derivatives:
d|x = (∂|x/∂u du , ∂|x/∂v dv , ∂|x/∂w dw)
-sub in tangent vectors: tu, tv, tw :
d|x = (tu du , tv dv , tw dw)
-we can define a unit vector in the u direction by
|eu = tu / |tu| , where tu is the tangent vector tu=∂|x/∂u

Question 3

Curvilinear Scale Factor

Definition

Answer 3

-given a unit vector in the u direction in terms of a tangent vector tu :
|eu = tu / |tu|
-the associated scale factor hu is the magnitude of tu:
hu = |tu|
-so the unit vector is
|eu = tu / hu
-and a small change in displacement is given by:
d|x = hu eu du + hv ev dv + hw ew dw

Question 4

When can a set of unit vectors be used to express a vector field?

Answer 4

• provided eu . (ev x ew) ≠ 0, we can express vector fields in terms of the basis vectors eu, ev and ew
• this formula gives the volume of a parallelepiped formed by the three tangent vectors tu, tv and tw
• so long as none of the vectors are coplanar, the volume will not be zero

Question 5

Orthogonal Curviliear Coordinates

Answer 5

-although in principal we can use any system of coordinates for which eu . (ev x ew) ≠ 0
-in practice it is much easier to use systems in which eu, ev and ew are mutually perpendicular
i.e. where:
eu . ev = eu . ew = ev . ew = 0

Question 6

Curviliear Coordinates

Left vs Right Handed Basis

Answer 6

• if we choose an orthogonal coordinate system, the basis will be either left or right handed
• if, eu . (ev x ew) = 1 then the basis is right handed and obeys the right hand rule
• if eu . (ev x ew) = -1 then the basis is left handed and obeys the left hand rule

Question 7

Vector Algebra for Right-Handed Orthogonal Coordinate Systems

Answer 7

-since ^eu, ^ev and ^ew satisfy the same relationships with scalar and vector products as ^e1, ^e2 and ^e3, i.e.
^ei ^ej = 𝛿ij
^ej x ^ek = εijk ^ei
-the rules of vector algebra are identical in the (u,v,w) coordinate system
-so dot products are calculated by summing the products of the ith components of two vectors and cross products can be calculated by finding a determinant if the e1 e2 and e3 unit vectors are replaced with the new basis vectors

Question 8

Vector Differentiation for Right-Handed Orthogonal Coordinate Systems

Answer 8

• vector differentiation does not treat the components in the same way as Cartesian components, since ^eu, ^ev and ^ew vary with position
• thus we need to find general formulae for grad, div curl and ∇²

Question 9

The Jacobian for Volume Integrals for Right-Handed Coordinates

Answer 9

|J| = h1 h2 h3 
dV = h1 h2 h3 dudvdw

Question 10

The Jacobian for Surface Integrals for Right-Handed Coordinates

Answer 10

-consideran element of surface with normal in the positive ^eu direction
-this is a surface of constant u so we can paramaterise it using v and w so that:
d|S = (|tv x |tw) dv dw = hv hw (^ev x ^ew) dvdw
-but, ^eu = ^ev x ^ew , so:
d|S = hv hw ^eu dvdw
and
dS = hv hw dv dw

Question 11

Finding Basis Vectors

Method

Answer 11

1) write the position vector |x = (x,y,z) in terms of the new coordinate system
2 )find tangent vectors for each new variable
e.g. |tr = d|x / dr
3) find the correspondig scale factor h
e.g. hr = | |tr |
4) the basis vector is the unit vector in the direction of the tangent vector
e.g. ^er = |tr / hr
5) check orthogonality by finding dot products between each of the new basis vectors (these should all be zero)
6) check handedness with scalar triple product, this should be either 1 (right handed) or -1 (left handed)

Question 12

General Equation for an Ellipse

Answer 12

x²/A² + y²/B² = 1

with A and B constant

Question 13

Can we use a stretched version of cylindrical polars to model constant elliptical surface?

Answer 13

-a stretched version of polar coordinates:
x = aRcosφ , y = bRsinφ , z = z , so that surfaces of constant R are elliptical
-however this transformation also means that surfaces of constant R are no longer perpendicular to surfaces of constant φ
-hence we cannot derive an orthogonal basis from this transformation

Question 14

What coordinate system is used for constant elliptical surfaces?

Answer 14

-elliptic cylindrical coordinates, {u, φ, z}

x = acosh(u)cosφ , y = asinh(u)sinφ, z = z

Question 15

Elliptic Cylindrical Coordinates

Surfaces of constant u

Answer 15

x = acosh(u)cosφ
y = asinh(u)sinφ
-rearrange
cosφ = x /acosh(u), sinφ = y/asinh(u)
sin²φ + cos²φ = 1 = x²/a²cosh²u + y²/a²sinh²u
-which for CONSTANT u is the equation of an elliptical surface with:
A = a cosh(u) and B = a sinh(u)

Question 16

Elliptic Cylindrical Coordinates

Surfaces of constant φ

Answer 16

x = acosh(u)cosφ
y = asinh(u)sinφ
-rearrange
cosh(u) = x/acosφ , sinh(u) = y/asinφ
-using the identity, cosh²(u) - sinh²(u) = 1
x² / a²cos²φ - y² / a²sin²φ = 1
-therefore surfaces of constant φ are hyperbolic

Question 17

General Expression for Grad

Equation

Answer 17

grad f = 1/hu ∂f/∂u ^eu + 1/hv ∂f/∂v ^ev + 1/hw ∂f/∂w ^ew

Question 18

General Expression for Grad

Derivation

Answer 18

-let f be a scalar field expressed in terms of curvilinear coordinates u, v and w
-recall that the change in f due to an infinitesimal change in displacement d|x is:
df = (grad f) . d|x
-and from the complete derivative definition:
df = ∂f/∂u du + ∂f/∂v dv + ∂f/∂w dw
-in curvilinear coordinates,:
d|x = hu ^eu du + hv ^ev dv + hw ^ew dw
-if we choose a displacement such that dv=dw=0 :
df = ∂f/∂u du = (grad f ) . ^eu hu du
-this gives
(grad f) . ^eu = 1/hu ∂f/∂u
-by definition the L.H.S. is the u component of grad f
-similarly for the other two directions:
grad f = 1/hu ∂f/∂u ^eu + 1/hv ∂f/∂v ^ev + 1/hw ∂f/∂w ^ew

Question 19

General Expression for Divergence

Answer 19

div |F =

1/(huhvhw) * [ ∂/∂u(Fuhvhw) + ∂/∂v(Fvhuhw) + ∂/∂w(Fwhuhv) ]

Question 20

Laplacian of a Scalar in Curvilinear Coordinates

Answer 20

-to find ∇²f we use the identity ∇²f = div(grad f)
-sub in the formulas for grad and div in curvilinear coordinates:
∇²f = 1/huhvhw * [ ∂/∂u (hvhw/hu ∂f/∂u) + ∂/∂v (huhw/hv ∂f∂v + ∂/∂w (huhv/hw ∂f/∂w) ]

Question 21

Application of the Laplacian in Curvilinear Coordinates

Newton’s Law of Gravity Equation

Answer 21

-Newton’s Law of Gravity can be written as:
|∇ . |g(|x) = - 4πG ρ(|x)
-where ρ is mass density, a scalar field
-G is the universal gravitational constant
-|g is the gravitational field

Question 22

Application of the Laplacian in Curvilinear Coordinates

Poisson’s Equation for Gravitational Potential

Answer 22

-gravitational field |g is conserevative so can be written as grad of a potential:
|g = ∇Φ
-so:
|∇ . (|∇Φ) = - 4πG ρ(|x)
using the identity ∇²f = div(grad f)
∇²Φ = - 4πG ρ(|x)

Question 23

Application for the Laplacian in Curvilinear Coordinates

Variation of ρ with Position

Answer 23

• consider a spherical ball of mass M and radius a centred on the origin
• the mass density out side of the ball i.e. for r>a is 0
• the mass density within the ball i.e. fo r≤a is M/(4πr³/3 = 3M/4πr³
• in general we expect Φ to be independent of θ and φ, we expect spherical symmetry as ρ only depends on r

Question 24

General Form of Curl in Curvilinear Coordinates

Answer 24

|∇ x |F = 1/huhvhw * det (3x3)
where det(3x3) is the determinant of a 3x3 matrix:
-first row: hu ^eu , hv ^ev , hw ^ew
-second row: ∂/∂u , ∂/∂v , ∂/∂w
-third row: hu Fu , hv Fv , hw Fw

Question 25

Other Derivatives

Answer 25

-other scalar operators such as (|G.|∇) cannot be applied component by component to |F as they can in the Cartesian basis
-instead these identities need to either be written in terms of coordinate invariant functions
OR transform to Cartesians for part of the calculation and then transform back to curvilinear coordinates afterwards