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Formulae involving Ñ Vector Identities with Proofs: Nabla Formulae for Vector Analysis 李国华 (Kok-Wah LEE) @ 08 May 2009 (Version 1.0) No. 4, Jalan Bukit Beruang 5, Taman Bukit Beruang, 75450 Bukit Beruang, Melaka, Malaysia. Email: contact@xpreeli.com; E96LKW@hotmail.com Tel.: +6013-6134998; +606-2312594; +605-4664998 www.xpreeli.com All rights reserved. Vector: Scalar: Nabla: A = A1 i + A2 j + A3 k j = j(x,y,z) Ñ =i ¶x + j ¶y +k ¶z B = B1 i + B2 j + B3 k C = C1 i + C2 j + C3 k y = y(x,y,z) (1) (A x B).C ≡ (B x C).A ≡ (C x A).B (2) A x (B x C) ≡ (A.C)B - (A.B)C (3) Prove Ñ(j + y) = Ñj + Ñy ¶xi + ¶y j + ¶z k( +y)= ¶ j¶xy)i+ ¶( ¶yy) j + ¶( ¶zy)k = ¶x i+ ¶x i+ ¶y j + ¶y j + ¶z k + ¶z k =  ¶x i + ¶y j + ¶z k+ ¶x i + ¶y j + ¶z k = ¶xi + ¶y j + ¶z kj +¶xi + ¶y j + ¶z k \ Ñ(j + y) = Ñj + Ñy (4) Prove Ñ(j y) = j Ñy + y Ñj ¶xi + ¶y j + ¶z k( y)= ¶ jx )i+ ¶ jy ) j + ¶ ¶z )k 1 = j ¶x i +y ¶x i +j ¶x j +y ¶x j +j ¶x k +y ¶x k = j¶x i+ ¶y j + ¶ z k+y¶x i + ¶y j + ¶z k = j¶xi + ¶y j + ¶z k y +y¶xi+ ¶y j + ¶z kj \ Ñ(j y) = j Ñy + y Ñj (5) Prove Ñ.(A + B) = Ñ.A + Ñ.B Ñ.(A+ B) = ¶xi + ¶y j + ¶z k ( 1 + B1 i +(A2 + B2 )j +(A3 + B3 )k] ¶(A + B ) ¶(A + B2 ) ¶(A + B3 ) ¶x ¶y ¶z Ñ.A+Ñ.B = ¶xi + ¶y j + ¶z k(Ai + A j + A k)+¶xi + ¶y j + ¶z k(Bi + B2 j + B3k) ¶ 1 ¶ 2 ¶ 3 ¶ 1 ¶B2 ¶B3 ¶x ¶y ¶z ¶x ¶y ¶z ¶(A + B ) ¶(A + B2 ) ¶(A + B3) ¶x ¶y ¶z LHS = RHS \ Ñ.(A + B) = Ñ.A + Ñ.B (6) Prove Ñx(A + B) = ÑxA + ÑxB Ñx(A+ B) = ¶xi + ¶y j + ¶z kx (A + B i +(A + B2 )j +(A + B3)k] i j k = ¶ ¶ ¶ ¶x ¶y ¶z A + B A2 + B2 A3 + B3 = ¶(A¶y B3 )− ¶(A¶z B2 )i −¶(A¶x B3 )− ¶(A¶z 1 )j +¶(A¶x B2 )− ¶(A¶y 1 )k 2 = ¶A ¶A  ¶A ¶A  ¶A ¶A   ¶B3 ¶B2  ¶B ¶B  ¶B2 ¶B    ¶y ¶z   ¶x ¶z   ¶x ¶y    ¶y ¶z   ¶x ¶z   ¶x ¶y   i j k i j k = ¶ ¶ ¶ + ¶ ¶ ¶ ¶x ¶y ¶z ¶x ¶y ¶z A A2 A B B2 B3 \ Ñx(A + B) = ÑxA + ÑxB (7) Prove Ñ.(jA) = (Ñj).A + j(Ñ.A) Ñ.( A) = ¶xi + ¶y j + ¶z k.jAi +jA j +j 3k) = ¶ ¶x1)+ ¶ ¶y2 )+ ¶ ¶z3 ) = LHS (Ñj).A+j(Ñ.A)=  ¶x i + ¶y j + ¶z k. Ai + A j + A k)+j¶xi + ¶y j + ¶z k . Ai + A j + A k) =  ¶j ¶j ¶j  ¶A ¶A ¶A   1 ¶x 2 ¶y 3 ¶z   ¶x ¶y ¶z  = A ¶x +j ¶x +A ¶y +j ¶ y +A ¶z +j ¶ z  = ¶ jA ) ¶( A2 ) ¶( A ) = RHS ¶x ¶y ¶z LHS = RHS \ Ñ.(jA) = (Ñj).A + j(Ñ.A) (8) Prove Ñx(jA) = (Ñj)xA + j(ÑxA) i Ñx( A)= ¶x jA j k ¶ ¶ ¶y ¶z j 2 j 3 = ¶ ¶y3 )− ¶( z2 )i −¶( x3 )− ¶( z1)j +¶( x2 )− ¶( y1)k =  ¶ 3 ¶j ¶ 2 ¶j   ¶ 3 ¶j ¶ 1 ¶j   ¶ 2 ¶j ¶ 1 ¶j   ¶y 3 ¶y ¶y 2 ¶y   ¶x 3 ¶x ¶z 1 ¶z   ¶x 2 ¶x ¶y 1 ¶y  = A ¶y − A ¶y i−A ¶x − A ¶z j +A ¶x − A ¶y k 3 + j ¶A3 −j ¶A2 i −j ¶A3 −j ¶A1 j +j ¶A2 −j ¶A1 k i j k i j k = ¶j ¶j ¶j + ¶ ¶ ¶ ¶x ¶y ¶z ¶x ¶y ¶z A A2 A 1 A A \ Ñx(jA) = (Ñj)xA + j(ÑxA) (9) Prove Ñ.(AxB) = B.(ÑxA) - A.(ÑxB) Ñ.(AxB) =  ¶ i + ¶ j + ¶ k. A 1 j k A2 A B2 B3 = ¶xi+ ¶y j + ¶z k.(A2B3 − 3B2 i−( 1B3 − 3 1 )j +( 1B2 − A2 1 )k] ¶(A2B3 − A B2 ) ¶(A B3 − A B ) ¶(A B2 − A B ) ¶x ¶y ¶z B.(ÑxA) = (Bi + B2 j + B3k . ¶y − ¶z i −¶ x − ¶ z j +¶ x − ¶y k = ¶A3 ¶A2  ¶A ¶A  ¶A2 ¶A  1 ¶y ¶z  2 ¶x ¶z  3 ¶x ¶y  Similarly, by interchanging the variable of A and B, we have A.(ÑxB) = (Ai+ A2 j + A k .¶B3 − ¶ 2 i− ¶ 3 − ¶ 1 j +¶B2 − ¶B1 k = ¶B3 ¶B2  ¶B3 ¶B1  ¶B2 ¶B  1 ¶y ¶z  2 ¶x ¶z  3 ¶x ¶y  B.(ÑxA) - A.(ÑxB) = B ¶A + A ¶B −B ¶A2 + A ¶B −B2 ¶A + A ¶B2   ¶A ¶B2   ¶A2 ¶B3   ¶A ¶B3   2 ¶z 1 ¶z   3 ¶x 2 ¶x   3 ¶y 1 ¶y  ¶(A3B1) ¶(A2B1) ¶( 3B2 ) ¶( 1B2 ) ¶(A2B3 ) ¶( 1B3 ) ¶y ¶z ¶x ¶z ¶x ¶y ¶(A2B3 − A B2 ) ¶(A B3 − A B1) ¶( 1B2 − A2 1) ¶x ¶y ¶z \ Ñ.(AxB) = B.(ÑxA) - A.(ÑxB) 4 (10) Prove Ñx(AxB) = (B.Ñ)A - B(Ñ.A) - (A.Ñ)B + A(Ñ.B) i Ñx(AxB)= Ñx A B j k A A = Ñx (A2B3 − A B2 i −(AB3 − A B )j + (A B2 − A B k] B2 B3 i j = ¶¶x ¶¶y A2B3 − A B2 A B − A B3 k ¶¶z A B2 − A2B = ¶( 1B2 − 2 1) ¶( 3 1 − 1B3 ) ¶( 1B2 − 2 1) ¶( 2B3 − 3B2 ) ¶( 3 1 − 1B3 ) ¶(A2B3 − 3B2 )  ¶y ¶z   ¶x ¶z   ¶x ¶y  = LHS (B.Ñ)A - B(Ñ.A) = B ¶x + B2 ¶y + B3 ¶z(Ai+ A2 j + A k)− ¶x + ¶y + ¶ 3 ( 1i+ B2 j+ B3k) = B2 ¶A + B3 ¶A − B ¶A − B ¶A i +B ¶A + B3 ¶A − B2 ¶A − B2 ¶A j +B ¶A + B2 ¶A − B3 ¶A − B3 ¶A2 k Similarly, by interchanging the variable of A and B, we have (A.Ñ)B - A(Ñ.B) = A ¶x + A ¶y + A ¶z(B i+ B2 j + B3k)−¶B + ¶y + ¶B3 ( 1i+ 2 j+ 3k) =  ¶ 1 ¶ 1 ¶B2 ¶B3   ¶B2 ¶B2 ¶ 1 ¶B3   ¶B3 ¶B3 ¶ 1 ¶B2   2 ¶y 3 ¶z 1 ¶y 1 ¶z   1 ¶x 3 ¶z 2 ¶x 2 ¶z   1 ¶x 2 ¶y 3 ¶x 3 ¶y  (B.Ñ)A - B(Ñ.A) - (A.Ñ)B + A(Ñ.B) = B2 ¶y + A ¶ y +B3 ¶z + A ¶z − B ¶ y + A2 ¶y −B ¶z + A ¶z i  ¶A2 ¶B2   ¶A2 ¶B3   ¶ 1 ¶ 2   ¶ 3 ¶B2   1 ¶x 1 ¶x   3 ¶z 2 ¶z   2 ¶x 1 ¶x   2 ¶z 3 ¶z  +B ¶ 3 + A ¶ 1 +B2 ¶ 3 + A ¶B2 −B ¶ 1 + A ¶ 3 −B3 ¶ 2 + A ¶B3 k = ¶( 1B2 − 2 1) ¶( 1B3 − 3 1) ¶( 1B2 − A2 1) ¶( 3B2 − 2B3 ) ¶( 3 1 − 1B3 ) ¶( 3B2 − 2B3 )  ¶y ¶z   ¶x ¶z   ¶x ¶y  = ¶( 1B2 − 2 1) ¶( 3 1 − 1B3 ) ¶( 1B2 − 2 1) ¶( 2B3 − 3B2 ) ¶( 3 1 − 1B3 ) ¶(A2B3 − 3B2 )  ¶y ¶z   ¶x ¶z   ¶x ¶y  = RHS RHS = LHS 5 ... - tailieumienphi.vn
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