Vector algebra using tensorial notation
⎪⎧1 if i = j
Definition of the Kronecker Delta ⇒ δij = ⎨⎪
⎩0 if i ≠ j
and the Levi-Civita tensor ⇒ εijk
εijkεlmn
δil δim δin
= δ jl δ jm δ jn
δkl δkm δkn
⎧0
for i = j or i = k or j = k
⎪
= ⎨-1 for an odd index permutation
⎪1 for an even index permutation
⎩
in particular, note that if i=l then εijkεimn = δ jmδkn − δ jnδ km
RULE: Whenever an index is repeated, a summation is assumed over that index
3
AiBi = ∑ AiBi in 3D
i=1
 
Dot product: A ⋅ B = AiBi
 
Cross product: A ×B = ε ijk A j Bk
(
)
Gradient: ∇ i =
i
∂
∂x i
The k index was
permuted to the
left 2 times, this
gives a (+1)
That’s all we need to know, here are a few examples:
     
  
1) Prove that A × B × C = B(A ⋅ C) − C(A ⋅ B)
  
 
A ×B × C = εijk A j B × C = ε ijkA jεklm BlC m = ε ijkεklm A jBl Cm = εkijε klm A j BlC m
k
i
 
 
= δ ilδ jm − δ imδ jl A j Bl C m = A j Bi C j − A j B j Ci ≡ Bi A ⋅ C − Ci A ⋅ B
(
)
(
(
)
)
(
)
(
)
The i index was
permuted once with j,
thus giving a (–1)

2) Prove that ∇ ⋅ ∇ × A = 0


∂
∂
∂
∂ ∂
∂ ∂
∇⋅∇×A=
∇×A =
εijk
Ak = ε ijk
Ak = −ε jik
A
i
∂x i
∂xi ∂x j
∂ x i ∂x j
∂x i ∂x j k
∂ ∂
∂ ∂
= −ε jik
Ak = −ε ijk
A =0
∂x j ∂x i
∂xi ∂x j k
(
The derivation
order was
inverted, no
harm done since
xj and xi are
independent
variables.
)
No index permutation,
we just decided to
change notation such that
all i’s transform into j’s
and vice versa.
These two expressions
are identical except for
their signs. This
contradiction can only be
avoided if the expression
is equal to zero.
 
 
 
3) Prove that ∇ ⋅ A × B = ∇ × A ⋅ B− ∇ × B ⋅ A
⎛
⎞
 
∂  
∂
∂
∂
∂
∇ ⋅ A× B =
εijk A jBk = εijk
A jBk = εijk ⎜⎜ A j
Bk + Bk
A j ⎟⎟
A× B =
i
∂x i
∂x i
∂x i
∂x i ⎠
⎝ ∂x i
∂
∂
∂
∂
= εijk A j
Bk + ε ijkBk
A j = ε ikj Ak
B j + ε kijBk
A
∂x i
∂x i
∂x i
∂x i j
(
(
)
) ( )
( )
(
)
Again, no index
permutation, we changed
all j’s for k’s and vice
versa.
= −εkij Ak
Two permutations
of k to the left
yield a (+1)
∂
∂
∂
∂
B j + εkij Bk
A j = −A kεkij
B j + Bkεkij
A
∂x i
∂x i
∂x i
∂x i j
A single permutation of k to
the left gives a (-1)

 

= −A ⋅ ∇ × B + B⋅ ∇ × A
(
)
(
)
⎛ B2 ⎞ 



4) Prove that B × ∇ × B = ∇ ⎜⎜ ⎟⎟ − B⋅ ∇ B
⎝2⎠


∂
∂
∂
Bm = εkijε klm B j
Bm = (δilδ jm − δimδ jl )B j
B
B× ∇ × B = ε ijkB jεklm
i
∂x l
∂x l
∂x l m
1 ∂
∂
∂
∂
∂ ⎛⎜ B2 ⎞⎟ 
=Bj
B j − Bj
Bi =
BjBj − B j
Bi =
− B⋅ ∇ Bi
2 ∂x i
∂x i
∂x j
∂x j
∂xi ⎝⎜ 2 ⎠⎟
(
[ (
)
(
)
)]
(
)
∂
∂
B B = 2B j
B
∂x i j j
∂x i j
In this way, we can deal with arbitrarily complicated vector representations by following
simple rules.
As an exercise, try other identities on your own.
2
MIT OpenCourseWare
http://ocw.mit.edu
16.522 Space Propulsion
Spring 2015
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