Appendix A. Curvilinear coordinates. A.1 Lamé coefficients. Consider set of equations. ξ i = ξ i (x 1,x 2,x 3 ), i = 1,2,3
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1 Appendix A Curvilinear coordinates A. Lamé coefficients Consider set of equations ξ i = ξ i x,x 2,x 3, i =,2,3 where ξ,ξ 2,ξ 3 independent, single-valued and continuous x,x 2,x 3 : coordinates of point P in Cartesian system with radius-vector x ξ,ξ 2,ξ 3 : coordinates of point P in curvilinear system e,e 2,e 3 : base unit vectors in Cartesian coordinates e ξ,e ξ2,e ξ3 : base unit vectors in curvilinear system we restrict ourselves to orthogonal curvilinear coordinates systems e ξi e ξj = δ ij Consider x = xx i ξ j and take the total differential summation over double appearing indices understood dx = x ξ j dξ j x ξ partial derivative keeping ξ 2,ξ 3 fixed: vector tangential to coordinate line ξ in general x ξ = h e ξ dx = h dξ e ξ +h 2 dξ 2 e ξ2 +h 3 dξ 3 e ξ3 h i Lamé coefficients Those coefficients might depend on the coordinates ξ i h i = h i ξ,ξ 2,ξ 3 57
2 h i dξ i no summation components of length element along e ξi Without restriction to orthogonal systems the length element ds = dx is defined as Coefficients ds 2 = dx dx = dx j dx j = x j ξ l x j ξ m dξ l dξ m = g lm dξ l dξ m g lm = x j ξ l x j ξ m are called the elements of the metric tensor For an orthogonal curvilinear system the base vectors e ξ,e ξ2,e ξ3 are mutually perpendicular and form a right-handed system We get the length element squared the Lamé coefficients ds 2 = h 2 dξ 2 +h 2 2dξ 2 2 +h 2 3dξ 3 2 g = h 2, g 22 = h 2 2, g 33 = h 2 3, g ik = 0 for i k the volume element elementary rectangular box d 3 x dv = h h 2 h 3 dξ dξ 2 dξ 3 58
3 A.2 Some special orthogonal coordinates Cartesian coordinates < x <, < x 2 <, < x 3 < h = h 2 = h 3 = ds 2 = dx 2 +dx 2 2 +dx 2 3 d 3 x = dx dx 2 dx 3 Cylindrical coordinates 0 ρ <, 0 ϕ < 2π, < z < x = ρ cosϕ, x 2 = ρ sinϕ, x 3 = z h ρ = h z =, h ϕ = ρ ds 2 = dρ 2 +ρ 2 dϕ 2 +dz 2 d 3 x = ρdρdϕdz Spherical coordinates 0 r <, 0 θ π, 0 ϕ < 2π x = r sinθ cosϕ, x 2 = r sinθ sinϕ, x 3 = r cosθ h r =, h θ = r, h ϕ = r sinθ ds 2 = dr 2 +r 2 dθ 2 +r 2 sinθ 2 dϕ 2 d 3 x = r 2 dr sinθdθdϕ = r 2 drdcosθdϕ r 2 drdω Parabolic cylindrical coordinates parametrization in Mathematica 0 u <, 0 v <, < z < x = 2 u2 v 2, x 2 = uv, x 3 = z h u = h v = u 2 +v 2, h z = ds 2 = u 2 +v 2 du 2 +dv 2 +dz 2 d 3 x = u 2 +v 2 dudvdz parametrization of Arfken 0 ξ <, 0 η <, < z < x x 2, u η, v ξ Parabolic coordinates parametrization of Arfken, Landau/Lifshitz 0 ξ <, 0 η <, 0 ϕ < 2π x = ξη cosϕ, x 2 = ξη sinϕ, x 3 = ξ η 2 59
4 h ξ = ξ +η, h η = ξ +η, h ϕ = ξη 2 ξ 2 η ds 2 = ξ +η dξ 2 + ξ +η dη 2 +ξηdϕ 2 4 ξ 4 η d 3 x = ξ +ηdξdηdϕ 4 parametrization in Mathematica 0 u <, 0 v <, 0 ϕ < 2π ξ = u 2, η = v 2, x = uv cosϕ, x 2 = uvsinϕ, x 3 = 2 u2 v 2 h u = h v = u 2 +v 2, h ϕ = uv ds 2 = u 2 +v 2 du 2 +dv 2 +uvdϕ 2 d 3 x = uvu 2 +v 2 dudvdϕ 60
5 A.3 Vector operations in orthogonal coordinates Gradient The length element along coordinate line ξ is h dξ form of the nabla operator e ξ gradψ = e ξ ψ = h ψ ξ ψ = h ψ ξ e ξ + h 2 ψ ξ 2 e ξ2 + h 3 ψ ξ 3 e ξ3 = 3 i= e ξi h i ξ i Divergence and Laplacian Use the divergence definition with A = lim V 0 V S A nda A = A ξ,ξ 2,ξ 3 e ξ +A 2 ξ,ξ 2,ξ 2 e ξ +A 3 ξ,ξ 2,ξ 3 e ξ3 and choose as volume V an elementary rectangular box of sides h ξ h 2 ξ 2 h 3 ξ 3 Analogously to the derivation in Cartesian coordinates compare Chapter.2 we calculate the net outward flux in direction of coordinate line ξ divided by the volume: lim V 0 [ A ξ + ξ,ξ 2,ξ 3 h 2 ξ + ξ,ξ 2,ξ 3 ξ 2 h 3 ξ + ξ,ξ 2,ξ 3 ξ 3 ] A ξ,ξ 2,ξ 3 h 2 ξ,ξ 2,ξ 3 ξ 2 h 3 ξ,ξ 2,ξ 3 ξ 3 /h ξ h 2 ξ 2 h 3 ξ 3 h 2 h 3 A h h 2 h 3 ξ we get for the divergence A = h h 2 h 3 { h 2 h 3 A + h 3 h A 2 + } h h 2 A 3 ξ ξ 2 ξ 3 The Laplacian operator follows 2 = 2 = : = h h 2 h 3 { ξ h2 h 3 h ξ + ξ 2 h3 h h 2 ξ 2 + ξ 3 h h 2 h 3 ξ 3 } 6
6 Curl Using a similar derivation as in Cartesian coordinates we get for the curl of vector A A = { h 3 A 3 } h 2 A 2 e ξ + cyclic permutations h 2 h 3 ξ 2 ξ 3 Representation as determinant: A = h h 2 h 3 h e ξ h 2 e ξ2 h 3 e ξ3 ξ ξ 2 ξ 3 h A h 2 A 2 h 3 A 3 62
7 A.4 Some explicit forms of vector operations Cartesian coordinates x,x 2,x 3 with base vectors e,e 2,e 3 ψ = ψ x e + ψ x 2 e 2 + ψ x 3 e 3 A = A + A 2 + A 3 x x 2 x 3 A3 A = A 2 e + x 2 x 3 ψ = 2 ψ + 2 ψ + 2 ψ x 2 x 2 2 x 2 3 A A 3 A2 e 2 + A e 3 x 3 x x x 2 Cylindrical coordinates ρ,ϕ,z with base vectors e ρ,e ϕ,e z ψ = ψ ρ e ρ + ψ ρ ϕ e ϕ + ψ z e z A = ρ ρ ρa ρ+ A ϕ ρ ϕ + A z z A z A = ρ ϕ A ϕ e ρ + z ψ = ρ ψ + 2 ψ ρ ρ ρ ρ 2 ϕ + 2 ψ 2 z 2 Aρ z A z ρ e ϕ + ρ ρ ρa ϕ A ρ e z ϕ Spherical coordinates r,θ,ϕ with base vectors e r,e θ,e ϕ ψ = ψ r e r + ψ r θ e θ + A = r 2 r r2 A r + r sinθ A = + ψ = r 2 r θ sinθa ϕ A θ ϕ r sinθ A r r sinθ ϕ r r 2 ψ + r ψ r sinθ ϕ e ϕ θ sinθa θ+ r sinθ r ra ϕ r 2 sinθ θ e r A ϕ ϕ e θ + r r ra θ A r θ sinθ ψ 2 ψ + θ r 2 sin 2 θ ϕ 2 e ϕ Note r 2 r r 2 ψ r r 2 r 2 rψ 63
8 A.5 Relationbetweenunitbasevectorsandtheirtime derivatives Relation between the local unit base vectors of cylindrical coordinates e ρ t,e ϕ t,e z and their time derivatives and the global constant unit base vectors of Cartesian coordinates e,e 2,e 3 eρ = cosϕe +sinϕe2 e = cosϕeρ sinϕeϕ e ϕ = sinϕe +cosϕe 2 e z = e 3 e 2 = sinϕe ρ +cosϕe ϕ e 3 = e z ė ρ = sinϕ ϕe +cosϕ ϕe 2 = ϕe ϕ ė = 0 ė ϕ = ϕe ρ ė 2 = 0 ė z = 0 ė 3 = 0 Same for the local unit base vectors of spherical coordinates e r t,e θ t,e ϕ t and their time derivatives e r = sinθ cosϕe +sinθ sinϕe 2 +cosθe 3 e θ = cosθ cosϕe +cosθ sinϕe 2 sinθe 3 e ϕ = sinϕe +cosϕe 2 e = sinθ cosϕe r +cosθ cosϕe θ sinϕe ϕ e 2 = sinθ sinϕe r +cosθ sinϕe θ +cosϕe ϕ e 3 = cosθe r sinθe θ ė r = θe θ +sinθ ϕe ϕ ė θ = θe r +cosθ ϕe ϕ ė ϕ = ϕ sinθe r +cosθe θ Relation between the local unit base vectors e ρ t,e ϕ t,e z and e r t,e θ t,e ϕ t e ρ = sinθe r +cosθe θ e ϕ = e ϕ e z = cosθe r sinθe θ e r = sinθe ρ +cosθe z e θ = cosθe ρ sinθe z e ϕ = e ϕ 64
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