Worked examples with Einstein summation notation Victor Liu February 22, 2009
Abstract
Some examples examples of using Einstein Einstein summation summation notation notation are presente presented. d. All manipulatio manipulations ns assume a three-dimensional space with cartesian coordinates (rather, the basic identities and conver conversions sions from vector calculus calculus only work work in cartesian cartesian coordinate coordinates). s). The point here is to use Einstein notation as a way of getting from one vector identity to another.
1
Basi Basics cs from from vecto ector r cal calcu culu luss
ui vi ijk uj vk ∂ i f ∂ i ui ijk ∂ j vk ui ∂ i vj ∂ i ∂ i uj
Iden Identtitie itiess
Antisymmetry of permutation Derivatives of orthonormal basis Derivatives of coordinates Permutations differing by 1 index Permutations with identical indices Permutations differing by 2 indices
3 3.1 3.1 Show
ijk = −jik = jki = . . . ∂ i ej = 0 ∂ i xj = δ ij ij jmn imn = 2δ ij ij ijk ijk = 6 (twice the dimensionality) ijk imn = δ jm jm δ kn kn − δ jn jn δ km km
Exa Examples A vect vector or iden identi titty a × (b × c) =
(a · c)b − (a · b)c. ijk aj (klm bl cm ) = kij klm aj bl cm
= ( δ il il δ jm jm − δ im im δ jl jl )aj bl cm = a j bi cj − aj bj ci = ( aj cj )bi − (aj bj )ci
1
3.2
Proof of vector calculus null identities
Show ∇ × (∇f ) = 0 and ∇ · (∇ × u) = 0. ∇ × (∇f ) = ijk ∂ j ∂ k f = −ikj ∂ j ∂ k f = −ikj ∂ k ∂ j f = −ijk ∂ j ∂ k f =0 This is shown by antisymmetry, since we are free to swap the order of differentiation in the third to last line and switch all instances of j and k in the second to last line. ∇ · (∇ × u) = ∂ i ijk ∂ j uk = ijk ∂ i ∂ j uk = −jik ∂ i ∂ j uk = −jik ∂ j ∂ i uk = −ijk ∂ i ∂ j uk =0 The proof is analogous to the previous case.
3.3
The vector Laplacian identity
Show ∇ × (∇ × u) = ∇(∇ · u) − ∇2 u. ijk ∂ j (klm ∂ l um ) = ijk klm ∂ j ∂ l um
= kij klm ∂ j ∂ l um = ( δ il δ jm − δ im δj l)∂ j ∂ l um = ∂ j ∂ i uj − ∂ j ∂ j ui = ∂ i ∂ j uj − ∂ j ∂ j ui = ∂ i (∂ j uj ) − (∂ j ∂ j )ui
3.4
An example from fluid mechanics
Show v
· ∇v = ∇
va ∂ a vb = ∂ b
va va
2
|v|2 2
+ (∇ × v) × v
+ bac (adf ∂ d vf )vc
1 1 va ∂ b va + (∂ b va )va − abc (adf ∂ d vf )vc 2 2 = v a ∂ b va − (abc adf )(∂ d vf )vc = v a ∂ b va − (δ bd δ cf − δ bf δ cd )(∂ d vf )vc = v a ∂ b va − (∂ b vc )vc + ( ∂ c vb )vc = v a ∂ b va − va (∂ b va ) + va (∂ a vb ) = v a ∂ a vb =
