7. Curvilinear Coordinate Systems Flashcards
Curvilinear
Definition
- 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
Curviliear Unit Vector
Definition
-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
Curvilinear Scale Factor
Definition
-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
When can a set of unit vectors be used to express a vector field?
- 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
Orthogonal Curviliear Coordinates
-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
Curviliear Coordinates
Left vs Right Handed Basis
- 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
Vector Algebra for Right-Handed Orthogonal Coordinate Systems
-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
Vector Differentiation for Right-Handed Orthogonal Coordinate Systems
- 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
The Jacobian for Volume Integrals for Right-Handed Coordinates
|J| = h1 h2 h3 dV = h1 h2 h3 dudvdw
The Jacobian for Surface Integrals for Right-Handed Coordinates
-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
Finding Basis Vectors
Method
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)
General Equation for an Ellipse
x²/A² + y²/B² = 1
with A and B constant
Can we use a stretched version of cylindrical polars to model constant elliptical surface?
-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
What coordinate system is used for constant elliptical surfaces?
-elliptic cylindrical coordinates, {u, φ, z}
x = acosh(u)cosφ , y = asinh(u)sinφ, z = z
Elliptic Cylindrical Coordinates
Surfaces of constant u
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)
Elliptic Cylindrical Coordinates
Surfaces of constant φ
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
General Expression for Grad
Equation
grad f = 1/hu ∂f/∂u ^eu + 1/hv ∂f/∂v ^ev + 1/hw ∂f/∂w ^ew
General Expression for Grad
Derivation
-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
General Expression for Divergence
div |F =
1/(huhvhw) * [ ∂/∂u(Fuhvhw) + ∂/∂v(Fvhuhw) + ∂/∂w(Fwhuhv) ]
Laplacian of a Scalar in Curvilinear Coordinates
-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) ]
Application of the Laplacian in Curvilinear Coordinates
Newton’s Law of Gravity Equation
-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
Application of the Laplacian in Curvilinear Coordinates
Poisson’s Equation for Gravitational Potential
-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)
Application for the Laplacian in Curvilinear Coordinates
Variation of ρ with Position
- 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
General Form of Curl in Curvilinear Coordinates
|∇ 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
Other Derivatives
-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
