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Lecture 23: Curvilinear Coordinates (RHB 8.10) Flipbook PDF
Lecture 23: Curvilinear Coordinates (RHB 8.10) It is often convenient to work with variables other than the Cartesian co
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Lecture 23: Curvilinear Coordinates (RHB 8.10) It is often convenient to work with variables other than the Cartesian coordinates xi ( = x, y, z). For example in Lecture 15 we met spherical polar and cylindrical polar coordinates. These are two important examples of what are called curvilinear coordinates. In this lecture we set up a formalism to deal with these rather general coordinate systems. Actually we have effectively covered much of the materical in lectures 18 and 19 where we studied surface and volume integrals. There we considered parameterisation of surfaces and volumes. Here we do the same thing but think about how this realises non-Cartesian Coordinate systems. Suppose we change from the Cartesian coordinates (x1 , x2 , x3 ) to the curvilinear coordinates, which we denote ui , each of which are functions of the xi :u1 = u1 (x1 , x2 , x3 ) u2 = u2 (x1 , x2 , x3 ) u3 = u3 (x1 , x2 , x3 ) The ui should be single-valued, except possibly at certain points, so the reverse transformation, xi = xi (u1 , u2 , u3 ) can be made. A point may be referred to by its Cartesian coordinates xi , or by its curvilinear coordinates ui . For example, in 2-D, we might have:-
Now consider coordinate surfaces defined by keeping one coordinate constant. • The Cartesian coordinate surfaces ‘xi = constant’ are planes, with constant unit normal vectors ei (or e1 , e2 and e3 ), intersecting at right angles. • The surfaces ‘ui = constant’ do not, in general, have constant unit normal vectors, nor in general do they intersect at right angles. 89
Example: Spherical polar co-ordinates r =
q
x2
+
y2
+
z2
;
θ = cos
−1
(
z √ 2 x + y2 + z2
)
;
−1
φ = tan
y x
.
The surfaces of constant r, θ, and φ are:r = constant =⇒ spheres centred at the origin unit normal er θ = constant =⇒ cones of semi-angle θ and axis along the z-axis unit normal eθ φ = constant =⇒ planes passing through the z-axis unit normal eφ These surfaces are not all planes, but they do intersect at right angles. If the coordinate surfaces intersect at right angles (i.e. the unit normals intersect at right angles), as in the example of spherical polars, the curvilinear coordinates are said to be orthogonal.
23. 1. Orthogonal Curvilinear Coordinates Unit Vectors and Scale Factors Suppose the point P has position r = r(u1 , u2 , u3 ). If we change u1 by a small amount, du1 , then r moves to position (r + dr ), where dr =
∂r du1 ≡ h1 e1 du1 ∂u1
where we have defined the unit vector e1 and the scale factor h1 by h1
∂r = ∂u1
and e1 =
1 ∂r . h1 ∂u1
• The vector e1 is a unit vector in the direction of increasing u1 . • The scale factor h1 gives the magnitude of dr when we make the change u1 → u1 +du1 . Thus for an infinitessimal change of u1 |dr| = h1 du1 90
Similarly, we can define hi and ei for i = 2 and 3. • The unit vectors ei are not constant vectors. In general they are nonCartesian basis vectors, they depend on the position vector r, i.e. their directions change as the ui are varied. • If ei · ej = δij , then the ui are orthogonal curvilinear coordinates. For Cartesian coordinates, the scale factors are unity and the unit vectors ei reduce to the Cartesian basis vectors we have used throughout the course: r = x e1 + y e2 + z e3
so that
h1 e1 =
∂r = e1 , etc. ∂x
Example: spherical polars: u1 = r, u2 = θ and u3 = φ in that order: r = r sin θ cos φ i + r sin θ sin φ j + r cos θ k, where to avoid confusion in this section we use i, j, k for the Cartesian basis vectors. (By this stage there should be no confusion with the suffices i, j, k). Therefore ∂r = sin θ cos φ i + sin θ sin φ j + cos θ k =⇒ hr = 1 ∂r ∂r = r cos θ cos φ i + r cos θ sin φ j − r sin θ k =⇒ hθ = r ∂θ ∂r = −r sin θ sin φ i + r sin θ cos φ j =⇒ hφ = r sin θ ∂φ Thus er = sin θ cos φ i + sin θ sin φ j + cos θ k = r / r eθ = cos θ cos φ i + cos θ sin φ j − sin θ k eφ = − sin φ i + cos φ j These unit vectors are normal to the level surfaces described above (spheres, cones and planes) and are clearly orthogonal: er · eθ = er · eφ = eθ · eφ = 0 and form a RH orthonormal basis: er × eθ = eφ , eθ × eφ = er , eφ × er = eθ . Example: Cylindrical polars: u1 = ρ, u2 = φ and u3 = z in that order: r = ρ cos φ i + ρ sin φ j + z k . Thus and Therefore
∂r = cos φ i + sin φ j ∂ρ
;
∂r = −ρ sin φ i + ρ cos φ j ∂φ
hρ = 1 ; eρ = cos φ i + sin φ j
;
hφ = ρ ;
∂r = k ∂z
hz = 1
eφ = − sin φ i + cos φ j 91
;
;
ez = k
These unit vectors are normal to the level surfaces described by cylinders about the z-axis (ρ = constant), planes through the z-axis (φ = constant), planes perpendicular to the z axis (z = constant) and are clearly orthogonal. Remark: An example of a curvilinear coordinate system which is not orthogonal is provided by the system of elliptical cylindrical coordinates (see tutuorial 9.4). (a 6= b)
r = a ρ cos θ i + b ρ sin θ j + z k In the following we shall only consider orthogonal systems
Arc Length The arc length ds is the length of the infinitesimal vector dr :(ds)2 = dr · dr . In Cartesian coordinates
(ds)2 = (dx)2 + (dy)2 + (dz)2 .
In curvilinear coordinates, if we change all three coordinates ui by infinitesimal amounts dui , then we have dr =
∂r ∂r ∂r du1 + du2 + du3 ∂u1 ∂u2 ∂u3
= h1 du1 e1 + h2 du2 e2 + h3 du3 e3 For the case of orthogonal curvilinears, because the basis vectors are orthonormal we have (ds)2 = h21 du21 + h22 du22 + h23 du23 For spherical polars, we showed that hr = 1,
hθ = r,
and
hφ = r sin θ
therefore (ds)2 = (dr)2 + r2 (dθ)2 + r2 sin2 θ (dφ)2
92
Lecture 23 continued:
23. 2. Elements of Area and Volume Basically we just repeat using scale factors what we did in lectures 18 and 19. Vector Area If u1 → u1 + du1 , then r → r + dr 1 , where dr 1 = h1 e1 du1 , and if u2 → u2 + du2 , then r → r + dr 2 , where dr 2 = h2 e2 du2 . (Curvature greatly exaggerated in figure!)
On the surface of constant u1 the vector area bounded by dr2 and dr 3 is given by dS 1 = (dr 2 ) × (dr3 ) = (h2 du2 e2 ) × (h3 du3 e3 ) = h2 h3 du2 du3 e1 , since e2 × e3 = e1 for orthogonal systems. Thus dS1 is a vector pointing in the direction of the normal to the surfaces ‘u1 =constant’, its magnitude being the area of the small parallelogram with edges dr 2 and dr 3 . Similarly, one can define dS 2 and dS 3 . For the case of spherical polars, if we vary θ and φ, keeping r fixed, then dS r =
hθ dθ eθ × hφ dφ eφ
= hθ hφ dθ dφ er = r2 sin θ dθ dφ er .
Similarly for dS θ and dS φ . Volume The volume contained in the parallelepiped with edges dr1 , dr2 and dr3 , is dV
= dr1 · dr2 × dr3 = (h1 du1 e1 ) · (h2 du2 e2 ) × (h3 du3 e3 ) = h1 h2 h3 du1 du2 du3
because e1 · e2 × e3 = 1. For spherical polars, we have dV = hr hθ hφ dr dθ dφ = r2 sin θ dr dθ dφ 93
Components of a Vector Field in Curvilinear Coordinates A vector field A(r) can be expressed in terms of curvilinear components Ai , defined as:– A(r) =
X
Ai (u1 , u2 , u3 ) ei
i
where ei is the ith basis vector for the curvilinear coordinate system. For orthogonal curvilinear coordinates, the component Ai is obtained by taking the scalar product of A with the ith (curvilinear) basis vector ei Ai = ei · A(r) NB Ai must be expressed in terms of ui (not x, y, z) when working in the ui basis. Example If A = i in Cartesian coordinates, then in spherical polars,Ar = A·er = sin θ cos φ, etc. A(r, θ, φ) = sin θ cos φ er + cos θ cos φ eθ − sin φ eφ .
Example If A = r then in cylindrical polars Aρ = A · eρ = ρ cos2 φ + ρ sin2 φ = ρ etc, and A(ρ, φ, z) = r = ρ eρ + z ez 23. 3. Grad, Div, Curl, and the Laplacian in Orthogonal Curvilinears We defined the vector operators grad, div, curl firstly in Cartesian coordinates, then most generally through integral definitions without regard to a coordinate system. Here we complete the picture by providing the definitions in any orthogonal curvilinear coordinate system. Gradient In section (2) we defined the gradient in terms of the change in a scalar field f when we let r → r + dr δf = ∇ f (r) · dr (1) Now consider what happens when we write f in terms of orthogonal curvilinear coordinates f = f (u1 , u2 , u3 ). As before, we denote the curvilinear basis vectors by e1 , e2 , and e3 . Let u1 → u1 + du1 , u2 → u2 + du2 , and u3 → u3 + du3 . By Taylor’s theorem, we have δf =
∂f ∂f ∂f du1 + du2 + du3 ∂u1 ∂u2 ∂u3
We have already shown that dr = h1 du1 e1 + h2 du2 e2 + h3 du3 e3 . 94
Using the orthogonality of the basis vectors, ei · ej = δij , we can write δf =
!
∂f ∂f ∂f e1 + e2 + e · (e1 du1 + e2 du2 + e3 du3 ) ∂u1 ∂u2 ∂u3 3 !
=
1 ∂f 1 ∂f 1 ∂f e1 + e2 + e · (h1 e1 du1 + h2 e2 du2 + h3 e3 du3 ) h1 ∂u1 h2 ∂u2 h3 ∂u3 3
=
1 ∂f 1 ∂f 1 ∂f e1 + e2 + e · dr h1 ∂u1 h2 ∂u2 h3 ∂u3 3
!
Comparing this result with equation (1) above, we obtain the following expression for ∇ f in orthogonal curvilinears 1 ∂f 1 ∂f 1 ∂f e1 + e2 + e h1 ∂u1 h2 ∂u2 h3 ∂u3 3
∇f =
3 X 1
=
∂f e hi ∂ui i
i=1
For spherical polars, we obtain ∂f 1 ∂f 1 ∂f + eθ + eφ . ∇ f (r, θ, φ) = er ∂r r ∂θ r sin θ ∂φ For cylindrical polars, we obtain ∂f 1 ∂f ∂f + eφ + ez . ∇ f (ρ, φ, z) = eρ ∂ρ ρ ∂φ ∂z Divergence In orthogonal curvilinear coordinates 1 divA = h1 h2 h3
(
)
∂ ∂ ∂ (A1 h2 h3 ) + (A2 h3 h1 ) + (A3 h1 h2 ) ∂u1 ∂u2 ∂u3
This expression can be obtained by using the integral definition of divA, or alternatively using vector operator identities (see BK 4.13, RHB 8.10). For Cartesian coordinates, we have hi = 1, and we regain the usual expression for ∇ · A in Cartesians. For spherical polars we have 1 divA(r, θ, φ) = 2 r sin θ
(
∂ 2 ∂ ∂ r sin θAr + r sin θAθ + rAφ ∂r ∂θ ∂φ
1 ∂ 2 1 r Ar + = 2 r ∂r r sin θ 95
(
∂ ∂ sin θAθ + A ∂θ ∂φ φ
)
.
)
where Ar , Aθ , and Aφ are the components of the vector field A in the basis (er , eθ , eφ ). For cylindrical polars we have 1 divA(ρ, φ, z) = ρ
(
)
∂ ∂ ∂ ρAρ + Aφ + (ρAz ) ∂ρ ∂φ ∂z
1 ∂ 1 ∂ ∂ ρAρ + Aφ + Az . ρ ∂ρ ρ ∂φ ∂z where Aρ , Aφ , and Az are the components of the vector field A in the basis (eρ , eφ , ez ). Curl In orthogonal curvilinear co-ordinates, curl is most conveniently written as
=
h1 e1 1 ∂ ∇×A = h1 h2 h3 ∂u1 h1 A1
h2 e2 h3 e3 ∂ ∂ ∂u2 ∂u3 h2 A2 h3 A3
e.g. the first component is given by 1 e1 · ∇ × A = h2 h3
(
)
∂ ∂ (A3 h3 ) − (A2 h2 ) ∂u2 ∂u3
and the components of ∇ × A in the e2 and e3 directions may be obtained by cyclic permutations of the suffices. The above formula can be demonstrated by using the line integral definition of curl, as used to proof Stokes’ theorem, (see tutorial) or by vector operator identities (BK. 4.13 or RHB 8.10). For spherical polars we have 1 ∇×A = 2 r sin θ
er r eθ r sin θ eφ ∂ ∂ ∂ ∂r ∂θ ∂φ Ar rAθ r sin θ Aφ
Laplacian of a Scalar Field The Laplacian operator acting on a scalar field is defined by ∇2 f = ∇ · (∇ f ), giving:– 1 ∇f = h1 h2 h3 2
(
∂ ∂u1
h2 h3 ∂f h1 ∂u1
!
∂ + ∂u2
h3 h1 ∂f h2 ∂u2
!
∂ + ∂u3
h1 h2 ∂f h3 ∂u3
!)
1 ∂f sin θ ∂φ
!)
In spherical polars, we have 1 ∇ f (r, θ, φ) = 2 r sin θ 2
(
∂ ∂f r2 sin θ ∂r ∂r
1 ∂ ∂f = 2 r2 r ∂r ∂r
!
!
∂ ∂f + sin θ ∂θ ∂θ (
!
∂ + ∂φ
1 ∂ ∂f + 2 2 sin θ sin θ ∂θ ∂θ r sin θ 96
!
∂2f + ∂φ2
)
.