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Mathematical Methods for Physics and Engineering - Matematica.NET

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26.19 COVARIANT DIFFERENTIATIONconstant (this term vanishes in Cartesian coordinates). Using (26.75) we write∂v∂u j = ∂vi∂u j e i + v i Γ k ije k .Since i <strong>and</strong> k are dummy indices in the last term on the right-h<strong>and</strong> side, we mayinterchange them to obtain( )∂v∂u j = ∂vi∂v∂u j e i + v k Γ i ikje i =∂u j + vk Γ i kj e i . (26.86)The reason <strong>for</strong> the interchanging the dummy indices, as shown in (26.86), is thatwe may now factor out e i . The quantity in parentheses is called the covariantderivative, <strong>for</strong> which the st<strong>and</strong>ard notation isv i ; j ≡ ∂vi∂u j +Γi kjv k , (26.87)the semicolon subscript denoting covariant differentiation. A similar short-h<strong>and</strong>notation also exists <strong>for</strong> the partial derivatives, a comma being used <strong>for</strong> theseinstead of a semicolon; <strong>for</strong> example, ∂v i /∂u j is denoted by v i ,j. In Cartesiancoordinates all the Γ i kjare zero, <strong>and</strong> so the covariant derivative reduces to thesimple partial derivative ∂v i /∂u j .Using the short-h<strong>and</strong> semicolon notation, the derivative of a vector may bewritten in the very compact <strong>for</strong>m∂v∂u j = vi ; je i<strong>and</strong>, by the quotient rule (section 26.7), it is clear that the v i ; j are the (mixed)components of a second-order tensor. This may also be verified directly, usingthe trans<strong>for</strong>mation properties of ∂v i /∂u j <strong>and</strong> Γ i kjgiven in (26.84) <strong>and</strong> (26.78)respectively.In general, we may regard the v i ; j as the mixed components of a secondordertensor called the covariant derivative of v <strong>and</strong> denoted by ∇v. In Cartesiancoordinates, the components of this tensor are just ∂v i /∂x j .◮Calculate v i ; i in cylindrical polar coordinates.Contracting (26.87) we obtainNow from (26.83) we havev i ; i = ∂vi∂u i +Γ i kiv k .Γ i 1i =Γ 1 11 +Γ 2 12 +Γ 3 13 =1/ρ,Γ i 2i =Γ 1 21 +Γ 2 22 +Γ 3 23 =0,Γ i 3i =Γ 1 31 +Γ 2 32 +Γ 3 33 =0,969

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