Mathematical Methods for Physics and Engineering - Matematica.NET

VECTOR CALCULUS∇Φ = ∂Φ∂ρ êρ + 1 ∂Φρ ∂φ êφ + ∂Φ∂z êz∇ · a = 1 ∂ρ ∂ρ (ρa ρ)+ 1 ∂a φρ ∂φ + ∂a z∂z∣ ∇×a = 1 ê ρ ρê φ ê z ∣∣∣∣∣∣∣∂ ∂ ∂ρ∂ρ ∂φ ∂z∣ a ρ ρa φ a z∇ 2 Φ = 1 (∂ρ ∂Φ )+ 1 ∂ 2 Φρ ∂ρ ∂ρ ρ 2 ∂φ + ∂2 Φ2 ∂z 2Table 10.2 Vector operators in cylindrical polar coordinates; Φ is a scalarfield and a is a vector field.defined by the vectors dρ ê ρ , ρdφê φ and dz ê z :dV = |dρ ê ρ · (ρdφê φ × dz ê z )| = ρdρdφdz,which again uses the fact that the basis vectors are orthonormal. For a simplecoordinate system such as cylindrical polars the expressions for (ds) 2 and dV areobvious from the geometry.We will now express the vector operators discussed in this chapter in termsof cylindrical polar coordinates. Let us consider a vector field a(ρ, φ, z) andascalar field Φ(ρ, φ, z), where we use Φ for the scalar field to avoid confusion withthe azimuthal angle φ. We must first write the vector field in terms of the basisvectors of the cylindrical polar coordinate system, i.e.a = a ρ ê ρ + a φ ê φ + a z ê z ,where a ρ , a φ and a z are the components of a in the ρ-, φ- andz- directionsrespectively. The expressions for grad, div, curl and ∇ 2 can then be calculatedand are given in table 10.2. Since the derivations of these expressions are rathercomplicated we leave them until our discussion of general curvilinear coordinatesin the next section; the reader could well postpone examination of these formalproofs until some experience of using the expressions has been gained.◮Express the vector field a = yz i − y j + xz 2 k in cylindrical polar coordinates, and hencecalculate its divergence. Show that the same result is obtained by evaluating the divergencein Cartesian coordinates.The basis vectors of the cylindrical polar coordinate system are given in (10.49)–(10.51).Solving these equations simultaneously for i, j and k we obtaini =cosφ ê ρ − sin φ ê φj =sinφ ê ρ +cosφ ê φk = ê z .360