distributivity δ il δjm A j l B m A l l B i δ ilδ jm B j l Am B l lA i A l l Bi

Distributivity δ il δjm a j l b m a l l b i δ ilδ

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(distributivity) δ il δ jm A j l B m - A l l B i + δ il δ jm B j l A m - B l l A i + ( A l l ) B i + ( B l l ) A i = (Eq. 134) δ il δ jm A j l B m - ( A l l ) B i + δ il δ jm B j l A m - ( B l l ) A i + ( A l l ) B i + ( B l l ) A i = (grouping) δ il δ jm A j l B m + δ il δ jm B j l A m = (cancellation) A m i B m + B m i A m =
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5.6 Examples of Using Tensor Techniques to Prove Identities 141 (Eq. 134) i ( A m B m ) = (product rule) [ ( A · B )] i (Eqs. 186 & 188) Because i is a free index, the identity is proved for all components. • ∇ · ( A × B ) = B · ( ∇ × A ) - A · ( ∇ × B ) : ∇ · ( A × B ) = i [ A × B ] i (Eq. 188) = i ( ijk A j B k ) (Eq. 173) = ijk i ( A j B k ) ( not acting on ) = ijk ( B k i A j + A j i B k ) (product rule) = ijk B k i A j + ijk A j i B k (distributivity) = kij B k i A j - jik A j i B k (Eq. 142) = B k ( kij i A j ) - A j ( jik i B k ) (commutativity & grouping) = B k [ ∇ × A ] k - A j [ ∇ × B ] j (Eq. 192) = B · ( ∇ × A ) - A · ( ∇ × B ) (Eq. 172) • ∇ × ( A × B ) = ( B · ∇ ) A + ( ∇ · B ) A - ( ∇ · A ) B - ( A · ∇ ) B : [ ∇ × ( A × B )] i = ijk j [ A × B ] k (Eq. 192) = ijk j ( klm A l B m ) (Eq. 173) = ijk klm j ( A l B m ) ( not acting on ) = ijk klm ( B m j A l + A l j B m ) (product rule) = ijk lmk ( B m j A l + A l j B m ) (Eq. 142)
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5.6 Examples of Using Tensor Techniques to Prove Identities 142 = ( δ il δ jm - δ im δ jl ) ( B m j A l + A l j B m ) (Eq. 151) = δ il δ jm B m j A l + δ il δ jm A l j B m - δ im δ jl B m j A l - δ im δ jl A l j B m (distributivity) = B m m A i + A i m B m - B i j A j - A j j B i (Eq. 134) = ( B m m ) A i + ( m B m ) A i - ( j A j ) B i - ( A j j ) B i (grouping) = [( B · ∇ ) A ] i + [( ∇ · B ) A ] i - [( ∇ · A ) B ] i - [( A · ∇ ) B ] i (Eqs. 196 & 188) = [( B · ∇ ) A + ( ∇ · B ) A - ( ∇ · A ) B - ( A · ∇ ) B ] i (Eq. 66) Because i is a free index, the identity is proved for all components. ( A × B ) · ( C × D ) = A · C A · D B · C B · D : ( A × B ) · ( C × D ) = [ A × B ] i [ C × D ] i (Eq. 172) = ijk A j B k ilm C l D m (Eq. 173) = ijk ilm A j B k C l D m (commutativity) = ( δ jl δ km - δ jm δ kl ) A j B k C l D m (Eqs. 142 & 151) = δ jl δ km A j B k C l D m - δ jm δ kl A j B k C l D m (distributivity) = ( δ jl A j C l ) ( δ km B k D m ) - ( δ jm A j D m ) ( δ kl B k C l ) (commutativity) = ( A l C l ) ( B m D m ) - ( A m D m ) ( B l C l ) (Eq. 134) = ( A · C ) ( B · D ) - ( A · D ) ( B · C ) (Eq. 172) = A · C A · D B · C B · D (definition) ( A × B ) × ( C × D ) = [ D · ( A × B )] C - [ C · ( A × B )] D :
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5.7 Exercises 143 [( A × B ) × ( C × D )] i = ijk [ A × B ] j [ C × D ] k (Eq. 173) = ijk jmn A m B n kpq C p D q (Eq. 173) = ijk kpq jmn A m B n C p D q (commutativity) = ijk pqk jmn A m B n C p D q (Eq. 142) = ( δ ip δ jq - δ iq δ jp ) jmn A m B n C p D q (Eq. 151) = ( δ ip δ jq jmn - δ iq δ jp jmn ) A m B n C p D q (distributivity) = ( δ ip qmn - δ iq pmn ) A m B n C p D q (Eq. 134) = δ ip qmn A m B n C p D q - δ iq pmn A m B n C p D q (distributivity) = qmn A m B n C i D q - pmn A m B n C p D i (Eq. 134) = qmn D q A m B n C i - pmn C p A m B n D i (commutativity) = ( qmn D q A m B n ) C i - ( pmn C p A m B n ) D i (grouping) = [ D · ( A × B )] C i - [ C · ( A × B )] D i (Eq. 174) = [[ D · ( A × B )] C ] i - [[ C · ( A × B )] D ] i (index definition) = [[ D · ( A × B )] C - [ C · ( A × B )] D ] i (Eq. 66) Because i is a free index, the identity is proved for all components. 5.7 Exercises 5.1 Write, in tensor notation, the mathematical expression for the trace, determinant and inverse of an n × n matrix. 5.2 Repeat the previous exercise for the multiplication of a matrix by a vector and the multiplication of two n × n matrices.
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  • Summer '20
  • Rajendra Paramanik
  • Tensor, Coordinate system, Polar coordinate system, Coordinate systems

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