Appendix A Curvilinear coordinates A.1 Lam´ e coefficients Consider set of equations ξi = ξi (x1 , x2 , x3 ) , i = 1, 2, 3 where ξ1 , ξ2 , ξ3 independent, single-valued and continuous (x1 , x2 , x3 ) : coordinates of point P in Cartesian system with radius-vector x (ξ1 , ξ2 , ξ3 ) : coordinates of point P in curvilinear system e1 , e2 , e3 : base unit vectors in Cartesian coordinates eξ1 , eξ2 , eξ3 : base unit vectors in curvilinear system we restrict ourselves to orthogonal curvilinear coordinates systems eξi · eξj = δij Consider x = x(xi (ξj )) and take the total differential (summation over double appearing indices understood) dx = ∂x dξj ∂ξj ∂x partial derivative keeping ξ2 , ξ3 fixed: vector tangential to coordinate line ξ1 ∂ξ1 −→ ∂x = h1 eξ1 ∂ξ1 in general dx = h1 dξ1 eξ1 + h2 dξ2 eξ2 + h3 dξ3 eξ3 hi Lam´e coefficients Those coefficients (might) depend on the coordinates ξi hi = hi (ξ1 , ξ2 , ξ3 ) 157 hi dξi (no summation) components of length element along eξi Without restriction to orthogonal systems the length element ds = |dx| is defined as ∂xj ∂xj dξl dξm = glm dξl dξm ∂ξl ∂ξm ds2 = dx · dx = dxj dxj = Coefficients glm = ∂xj ∂xj ∂ξl ∂ξm are called the elements of the metric tensor For an orthogonal curvilinear system the base vectors eξ1 , eξ2 , eξ3 are mutually perpendicular and form a right-handed system We get the length element squared ds2 = h21 (dξ1 )2 + h22 (dξ2 )2 + h23 (dξ3 )2 the Lam´e coefficients g11 = h21 , g22 = h22 , g33 = h23 , gik = 0 for i 6= k the volume element (elementary rectangular box) d3 x ≡ dV = h1 h2 h3 dξ1 dξ2 dξ3 158 A.2 Some special orthogonal coordinates • Cartesian coordinates (−∞ < x1 < ∞ , −∞ < x2 < ∞ , −∞ < x3 < ∞) h1 = h2 = h3 = 1 ds2 = dx21 + dx22 + dx23 d3 x = dx1 dx2 dx3 • Cylindrical coordinates (0 ≤ ρ < ∞ , 0 ≤ ϕ < 2π , −∞ < z < ∞) x1 = ρ cos ϕ , x2 = ρ sin ϕ , x3 = z hρ = hz = 1 , hϕ = ρ ds2 = dρ2 + ρ2 dϕ2 + dz 2 d3 x = ρ dρ dϕ dz • Spherical coordinates (0 ≤ r < ∞ , 0 ≤ θ ≤ π , 0 ≤ ϕ < 2π) x1 = r sin θ cos ϕ , x2 = r sin θ sin ϕ , x3 = r cos θ hr = 1 , hθ = r , hϕ = r sin θ ds2 = dr2 + r2 dθ2 + r2 sin θ2 dϕ2 d3 x = r2 dr sin θ dθ dϕ = r2 dr d cos θ dϕ ≡ r2 dr dΩ • Parabolic cylindrical coordinates parametrization in Mathematica (0 ≤ u < ∞ , 0 ≤ v < ∞ , −∞ < z < ∞) 1 x1 = (u2 − v 2 ) , x2 = u v , x3 = z 2 √ h u = h v = u2 + v 2 , h z = 1 ds2 = (u2 + v 2 ) (du2 + dv 2 ) + dz 2 d3 x = (u2 + v 2 ) du dv dz parametrization of Arfken (0 ≤ ξ < ∞ , 0 ≤ η < ∞ , −∞ < z < ∞) x1 ↔ x2 , u → η , v → ξ • Parabolic coordinates parametrization of Arfken, Landau/Lifshitz (0 ≤ ξ < ∞ , 0 ≤ η < ∞ , 0 ≤ ϕ < 2π) x1 = p ξ η cos ϕ , x2 = p 159 1 ξ η sin ϕ , x3 = (ξ − η) 2 1 hξ = 2 s 1 ξ+η , hη = ξ 2 s p ξ+η , hϕ = ξ η η 1ξ +η 2 1ξ +η 2 dξ + dη + ξη dϕ2 4 ξ 4 η 1 d3 x = (ξ + η) dξ dη dϕ 4 ds2 = parametrization in Mathematica (0 ≤ u < ∞ , 0 ≤ v < ∞ , 0 ≤ ϕ < 2π) ξ = u2 , η = v 2 , 1 x1 = u v cos ϕ , x2 = u v sin ϕ , x3 = (u2 − v 2 ) 2 √ hu = hv = u2 + v 2 , hϕ = uv ds2 = (u2 + v 2 ) (du2 + dv 2 ) + uv dϕ2 d3 x = uv(u2 + v 2 ) du dv dϕ 160 A.3 Vector operations in orthogonal coordinates Gradient The length element along coordinate line ξ1 is h1 dξ1 −→ ∇ψ = eξ1 · gradψ = eξ1 · ∇ψ = 1 ∂ψ h1 ∂ξ1 1 ∂ψ 1 ∂ψ 1 ∂ψ eξ1 + eξ2 + eξ h1 ∂ξ1 h2 ∂ξ2 h3 ∂ξ3 3 ⇒ form of the nabla operator ∇= 3 X eξ i i=1 1 ∂ hi ∂ξi Divergence and Laplacian Use the divergence definition 1 ∇ · A = lim ∆V →0 ∆V with I S A · n da A = A1 (ξ1 , ξ2 , ξ3 ) eξ1 + A2 (ξ1 , ξ2 , ξ2 ) eξ1 + A3 (ξ1 , ξ2 , ξ3 ) eξ3 and choose as volume ∆V an elementary rectangular box of sides h1 ∆ξ1 h2 ∆ξ2 h3 ∆ξ3 Analogously to the derivation in Cartesian coordinates (compare Chapter 1.2) we calculate the net outward flux in direction of coordinate line ξ1 divided by the volume: lim A1 (ξ1 + ∆ξ1 , ξ2 , ξ3 ) h2 (ξ1 + ∆ξ1 , ξ2 , ξ3 )∆ξ2 h3 (ξ1 + ∆ξ1 , ξ2 , ξ3 )∆ξ3 − ∆V →0 A1 (ξ1 , ξ2 , ξ3 ) h2 (ξ1 , ξ2 , ξ3 )∆ξ2 h3 (ξ1 , ξ2 , ξ3 )∆ξ3 /(h1 ∆ξ1 h2 ∆ξ2 h3 ∆ξ3 ) 1 ∂ (h2 h3 A1 ) → h1 h2 h3 ∂ξ1 ⇒ we get for the divergence 1 ∇·A= h1 h2 h3 ∂ ∂ ∂ (h2 h3 A1 ) + (h3 h1 A2 ) + (h1 h2 A3 ) ∂ξ1 ∂ξ2 ∂ξ3 The Laplacian operator follows (∇2 = ∇2 = ∆): 1 ∆= h1 h2 h3 ∂ ∂ξ1 h2 h3 ∂ h1 ∂ξ1 ∂ + ∂ξ2 161 h3 h1 ∂ h2 ∂ξ2 ∂ + ∂ξ3 h1 h2 ∂ h3 ∂ξ3 Curl Using a similar derivation as in Cartesian coordinates we get for the curl of vector A ∂ ∂ 1 (h3 A3 ) − (h2 A2 ) eξ1 + cyclic permutations ∇×A= h2 h3 ∂ξ2 ∂ξ3 Representation as determinant: h 1 eξ 1 h 2 eξ2 h 3 eξ 3 1 ∂ ∂ ∂ ∇×A= h1 h2 h3 ∂ξ1 ∂ξ2 ∂ξ3 h1 A1 h2 A2 h3 A3 162 A.4 Some explicit forms of vector operations Cartesian coordinates (x1 , x2 , x3 ) with base vectors e1 , e2 , e3 ∂ψ ∂ψ ∂ψ e1 + e2 + e3 ∂x1 ∂x2 ∂x3 ∂A1 ∂A2 ∂A3 ∇·A = + + ∂x1 ∂x2 ∂x 3 ∂A1 ∂A3 ∂A2 ∂A1 ∂A3 ∂A2 ∇×A = e1 + e2 + e3 − − − ∂x2 ∂x3 ∂x3 ∂x1 ∂x1 ∂x2 ∂ 2ψ ∂ 2ψ ∂ 2ψ ∆ψ = + + ∂x21 ∂x22 ∂x23 ∇ψ = Cylindrical coordinates (ρ, ϕ, z) with base vectors eρ , eϕ , ez 1 ∂ψ ∂ψ ∂ψ eρ + eϕ + ez ∂ρ ρ ∂ϕ ∂z 1 ∂ 1 ∂Aϕ ∂Az ∇·A = (ρ Aρ ) + + ρ ∂ρ ρ ∂ϕ ∂z ∂Aρ 1 ∂Aρ ∂Az ∂ 1 ∂Az ∂Aϕ − − (ρ Aϕ ) − eρ + eϕ + ez ∇×A = ρ ∂ϕ ∂z ∂z ∂ρ ρ ∂ρ ∂ϕ 1 ∂ ∂ψ 1 ∂ 2ψ ∂ 2ψ ∆ψ = + 2 ρ + 2 ρ ∂ρ ∂ρ ρ ∂ϕ2 ∂z ∇ψ = Spherical coordinates (r, θ, ϕ) with base vectors er , eθ , eϕ ∇ψ = ∇·A = ∇×A = + ∆ψ = 1 ∂ψ 1 ∂ψ ∂ψ er + eθ + eϕ ∂r r ∂θ r sin θ ∂ϕ 1 ∂ 2 ∂ 1 1 ∂Aϕ (r Ar ) + (sin θ Aθ ) + 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ ∂ ∂Aθ 1 er (sin θ Aϕ ) − r sin θ ∂θ ∂ϕ 1 ∂Ar 1 ∂ 1 ∂ ∂Ar − (r Aϕ ) eθ + (r Aθ ) − eϕ r sin θ ∂ϕ r ∂r r ∂r ∂θ 1 ∂ ∂ 2ψ 1 ∂ ∂ψ 1 2 ∂ψ r + sin θ + r2 ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin2 θ ∂ϕ2 Note 1 ∂ r2 ∂r ∂ψ r ∂r 2 ≡ 163 1 ∂2 (r ψ) r ∂r2 A.5 Relation between unit base vectors and their time derivatives Relation between the local unit base vectors of cylindrical coordinates eρ (t), eϕ (t), ez and their time derivatives and the global constant unit base vectors of Cartesian coordinates e1 , e2 , e3 eρ = cos ϕ e1 + sin ϕ e2 eϕ = − sin ϕ e1 + cos ϕ e2 ez = e3 e1 = cos ϕ eρ − sin ϕ eϕ e2 = sin ϕ eρ + cos ϕ eϕ e3 = ez e˙ ρ = − sin ϕ ϕ˙ e1 + cos ϕ ϕ˙ e2 = ϕ˙ eϕ e˙ ϕ = −ϕ˙ eρ e˙ z = 0 e˙ 1 = 0 e˙ 2 = 0 e˙ 3 = 0 Same for the local unit base vectors of spherical coordinates er (t), eθ (t), eϕ (t) and their time derivatives er = sin θ cos ϕ e1 + sin θ sin ϕ e2 + cos θ e3 eθ = cos θ cos ϕ e1 + cos θ sin ϕ e2 − sin θ e3 eϕ = − sin ϕ e1 + cos ϕ e2 e1 = sin θ cos ϕ er + cos θ cos ϕ eθ − sin ϕ eϕ e2 = sin θ sin ϕ er + cos θ sin ϕ eθ + cos ϕ eϕ e3 = cos θ er − sin θ eθ e˙ r = θ˙ eθ + sin θ ϕ˙ eϕ e˙ θ = −θ˙ er + cos θ ϕ˙ eϕ e˙ ϕ = −ϕ˙ (sin θ er + cos θ eθ ) Relation between the local unit base vectors eρ (t), eϕ (t), ez and er (t), eθ (t), eϕ (t) eρ = sin θ er + cos θ eθ e ϕ = eϕ ez = cos θ er − sin θ eθ er = sin θ eρ + cos θ ez eθ = cos θ eρ − sin θ ez eϕ = eϕ 164
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