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
