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The Levi-Civita tensor
October 25, 2012
In 3 -dimensions, we define the Levi-Civita tensor, εijk, to be totally antisymmetric, so we get a minus sign under interchange of any pair of indices. We work throughout in Cartesian coordinate. This means that most of the 27 components are zero, since, for example,
ε 212 = −ε 212
if we imagine interchanging the two 2s. This means that the only nonzero components are the ones for which i, j and k all take different value. There are only six of these, and all of their values are determined once we choose any one of them. Define ε 123 ≡ 1
Then by antisymmetry it follows that
ε 123 = ε 231 = ε 312 = + ε 132 = ε 213 = ε 321 = − 1
All other components are zero. Using εijk we can write index expressions for the cross product and curl. The ith^ component of the cross product is given by [u × v]i = εijkuj vk
as we check by simply writing out the sums for each value of i,
[u × v] 1 = ε 1 jkuj vk = ε 123 u 2 v 3 + ε 132 u 3 v 2 + (all other terms are zero) = u 2 v 3 − u 3 v 2 [u × v] 2 = ε 2 jkuj vk = ε 231 u 3 v 1 + ε 213 u 1 v 3 = u 3 v 1 − u 1 v 3 [u × v] 3 = ε 3 jkuj vk = u 1 v 2 − u 2 v 1
We get the curl simply by replacing ui by ∇i = (^) ∂x∂i ,
[∇ × v]i = εijk∇j vk
If we sum these expressions with basis vectors ei, where e 1 = i, e 2 = j, e 3 = k, we may write these as vectors:
u × v = [u × v]i ei = εijkuj vkei ∇ × v = εijkei∇j vk
There are useful identities involving pairs of Levi-Civita tensors. The most general is
εijkεlmn = δilδjmδkn + δimδjnδkl + δinδjlδkm − δilδjnδkm − δinδjmδkl − δimδjlδkn
To check this, first notice that the right side is antisymmetric in i, j, k and antisymmetric in l, m, n. For example, if we interchange i and j, we get
εjikεlmn = δjlδimδkn + δjmδinδkl + δjnδilδkm − δjlδinδkm − δjnδimδkl − δjmδilδkn
Now interchange the first pair of Kronecker deltas in each term, to get i, j, k in the original order, then rearrange terms, then pull out an overall sign,
εjikεlmn = δimδjlδkn + δinδjmδkl + δilδjnδkm − δinδjlδkm − δimδjnδkl − δilδjmδkn = −δilδjmδkn − δimδjnδkl − δinδjlδkm + δilδjnδkm + δinδjmδkl + δimδjlδkn = − (δilδjmδkn + δimδjnδkl + δinδjlδkm − δilδjnδkm − δinδjmδkl − δimδjlδkn) = −εijkεlmn
Total antisymmetry means that if we know one component, the others are all determined uniquely. Therefore, set i = l = 1, j = m = 2, k = n = 3, to see that
ε 123 ε 123 = δ 11 δ 22 δ 33 + δ 12 δ 23 δ 31 + δ 13 δ 21 δ 32 − δ 11 δ 23 δ 32 − δ 13 δ 22 δ 31 − δ 12 δ 21 δ 33 = δ 11 δ 22 δ 33 = 1
Check one more case. Let i = 1, j = 2, k = 3 again, but take l = 3, m = 2, n = 1. Then we have
ε 123 ε 321 = δ 13 δ 22 δ 31 + δ 12 δ 21 δ 33 + δ 11 δ 23 δ 32 − δ 13 δ 21 δ 32 − δ 11 δ 22 δ 33 − δ 12 δ 23 δ 31 = −δ 11 δ 22 δ 33 = − 1
as expected. We get a second identity by setting n = k and summing,
εijkεlmk = δilδjmδkk + δimδjkδkl + δikδjlδkm − δilδjkδkm − δikδjmδkl − δimδjlδkk = 3 δilδjm + δimδjl + δimδjl − δilδjm − δilδjm − 3 δimδjl = (3 − 1 − 1) δilδjm − (3 − 1 − 1) δimδjl = δilδjm − δimδjl
so we have a much simpler, and very useful, relation
εijkεlmk = δilδjm − δimδjl
A second sum gives another identity. Setting m = j and summing again,
εijkεljk = δilδmm − δimδml = 3 δil − δil = 2 δil
Setting the last two indices equal and summing provides a check on our normalization,
εijkεijk = 2 δii = 6
This is correct, since there are only six nonzero components and we are summing their squares.
= εijk∇j (εklmvlwm) = εijkεklm ((∇j vl) wm + vl∇j wm) = (δilδjm − δimδjl) ((∇j vl) wm + vl∇j wm) = δilδjm ((∇j vl) wm + vl∇j wm) − δimδjl ((∇j vl) wm + vl∇j wm) = (∇mvi) wm + vi∇mwm − (∇j vj ) wi − vj ∇j wi
Restoring the vector notation, we have
∇ × (v × w) = (w · ∇) v + (∇ · w) v − (∇ · v) w − (v · ∇) w
If you doubt the advantages here, try to prove these identities by explicitly writing out all of the components!
