Mathematical Methods for Physics and Engineering - Matematica.NET

VECTOR CALCULUS◮Prove the expression for ∇ · a in orthogonal curvilinear coordinates.Let us consider the sub-expression ∇ · (a 1 ê 1 ). Now ê 1 = ê 2 × ê 3 = h 2 ∇u 2 × h 3 ∇u 3 . Therefore∇ · (a 1 ê 1 )=∇ · (a 1 h 2 h 3 ∇u 2 ×∇u 3 ),= ∇(a 1 h 2 h 3 ) · (∇u 2 ×∇u 3 )+a 1 h 2 h 3 ∇ · (∇u 2 ×∇u 3 ).However, ∇ · (∇u 2 ×∇u 3 ) = 0, from (10.43), so we obtain( )ê2∇ · (a 1 ê 1 )=∇(a 1 h 2 h 3 ) · × ê3ê 1= ∇(a 1 h 2 h 3 ) · ;h 2 h 3 h 2 h 3letting Φ = a 1 h 2 h 3 in (10.60) and substituting into the above equation, we find∇ · (a 1 ê 1 )= 1 ∂(a 1 h 2 h 3 ).h 1 h 2 h 3 ∂u 1Repeating the analysis for ∇ · (a 2 ê 2 )and∇ · (a 3 ê 3 ), and adding the results we obtain (10.61),as required. ◭LaplacianIn the expression for the divergence (10.61), leta = ∇Φ = 1 ∂Φê 1 + 1 ∂Φê 2 + 1 ∂Φê 3 ,h 1 ∂u 1 h 2 ∂u 2 h 3 ∂u 3where we have used (10.60). We then obtain∇ 2 Φ= 1 [ ( )∂ h2 h 3 ∂Φ+ ∂ ( )h3 h 1 ∂Φ+ ∂ ( )]h1 h 2 ∂Φ,h 1 h 2 h 3 ∂u 1 h 1 ∂u 1 ∂u 2 h 2 ∂u 2 ∂u 3 h 3 ∂u 3which is the expression for the Laplacian in orthogonal curvilinear coordinates.CurlThe curl of a vector field a = a 1 ê 1 + a 2 ê 2 + a 3 ê 3coordinates is given byin orthogonal curvilinear∣ ∣ ∣∣∣∣∣∣∣∣ h 1 ê 1 h 2 ê 2 h 3 ê 3 ∣∣∣∣∣∣∣∣∇×a = 1 ∂ ∂ ∂. (10.62)h 1 h 2 h 3 ∂u 1 ∂u 2 ∂u 3h 1 a 1 h 2 a 2 h 3 a 3◮Prove the expression for ∇×a in orthogonal curvilinear coordinates.Let us consider the sub-expression ∇×(a 1 ê 1 ). Since ê 1 = h 1 ∇u 1 we have∇×(a 1 ê 1 )=∇×(a 1 h 1 ∇u 1 ),= ∇(a 1 h 1 ) ×∇u 1 + a 1 h 1 ∇×∇u 1 .But ∇×∇u 1 =0,soweobtain∇×(a 1 ê 1 )=∇(a 1 h 1 ) × ê1h 1.368