S and d r is a vector element tangent to the perimeter while the other symbols

# S and d r is a vector element tangent to the

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S , and d r is a vector element tangent to the perimeter while the other symbols are as defined before. Similarly, Stokes theorem for a differentiable rank-2 tensor field A in tensor notation for the first index is given by: ˆ S ijk j A kl n i = ˆ C A il dx i (237)
5.6 Examples of Using Tensor Techniques to Prove Identities 134 while Stokes theorem for differentiable tensor fields of higher rank A in tensor notation for the index k is given by: ˆ S ijk j A lm...k...n n i = ˆ C A lm...k...n dx k (238) 5.6 Examples of Using Tensor Techniques to Prove Identities In this section we provide some examples for using tensor techniques to prove vector and tensor identities. These examples, which are based on the identities given in § 5.4, demonstrate the elegance, efficiency and clarity of the methods and notation of tensor calculus. • ∇ · r = n : ∇ · r = i x i (Eq. 188) = δ ii (Eq. 138) = n (Eq. 138) • ∇ × r = 0 : [ ∇ × r ] i = ijk j x k (Eq. 192) = ijk δ kj (Eq. 137) = ijj (Eq. 134) = 0 (Eq. 123) Since i is a free index, the identity is proved for all components. • ∇ ( a · r ) = a : [ ( a · r )] i = i ( a j x j ) (Eqs. 186 & 172) = a j i x j + x j i a j (product rule)
5.6 Examples of Using Tensor Techniques to Prove Identities 135 = a j i x j ( a j is constant) = a j δ ji (Eq. 137) = a i (Eq. 134) = [ a ] i (definition of index) Since i is a free index, the identity is proved for all components. • ∇ · ( f ) = 2 f : ∇ · ( f ) = i [ f ] i (Eq. 188) = i ( i f ) (Eq. 186) = i i f (rules of differentiation) = ii f (definition of 2nd derivative) = 2 f (Eq. 194) • ∇ · ( ∇ × A ) = 0 : ∇ · ( ∇ × A ) = i [ ∇ × A ] i (Eq. 188) = i ( ijk j A k ) (Eq. 192) = ijk i j A k ( not acting on ) = ijk j i A k (continuity condition) = - jik j i A k (Eq. 142) = - ijk i j A k (relabeling dummy indices i and j ) = 0 (since ijk i j A k = - ijk i j A k ) This can also be concluded from line three by arguing that: since by the continuity condition i and j can change their order with no change in the value of the term while
5.6 Examples of Using Tensor Techniques to Prove Identities 136 a corresponding change of the order of i and j in ijk results in a sign change, we see that each term in the sum has its own negative and hence the terms add up to zero (see Eq. 145). • ∇ × ( f ) = 0 : [ ∇ × ( f )] i = ijk j [ f ] k (Eq. 192) = ijk j ( k f ) (Eq. 186) = ijk j k f (rules of differentiation) = ijk k j f (continuity condition) = - ikj k j f (Eq. 142) = - ijk j k f (relabeling dummy indices j and k ) = 0 (since ijk j k f = - ijk j k f ) This can also be concluded from line three by a similar argument to the one given in the previous point. Because [ ∇ × ( f )] i is an arbitrary component, then each component is zero. • ∇ ( fh ) = f h + h f : [ ( fh )] i = i ( fh ) (Eq. 186) = f∂ i h + h∂ i f (product rule) = [ f h ] i + [ h f ] i (Eq. 186) = [ f h + h f ] i (Eq. 66) Because i is a free index, the identity is proved for all components. • ∇ · ( f A ) = f ∇ · A + A · ∇ f : ∇ · ( f A ) = i [ f A ] i (Eq. 188)
5.6 Examples of Using Tensor Techniques to Prove Identities 137 = i ( fA i ) (definition of index) = f∂ i A i + A i i f (product rule) = f ∇ · A + A · ∇ f (Eqs. 188 & 196) • ∇ × ( f A ) = f ∇ × A + f × A : [ ∇ × ( f A )] i = ijk j [ f A ] k (Eq. 192) = ijk j ( fA k ) (definition of index) = f ijk j A k + ijk ( j f ) A k (product rule & commutativity) = f ijk j A k + ijk [ f ] j A k (Eq. 186) = [ f ∇ × A ] i + [ f × A ] i (Eqs. 192 & 173) = [ f ∇ × A + f × A ] i (Eq. 66)

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• Summer '20
• Rajendra Paramanik
• Tensor, Coordinate system, Polar coordinate system, Coordinate systems

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