Use of the Levi-Civita tensor ('permutation symbol') result

Use of the Levi-Civita tensor ('permutation symbol') result
Call the above equation "eq. 1". This is just eq. 1.66 from the text, but
with (AxB)i now (BxC)i. Then eq. 2 below is This, as we discussed in class, is just from eq. 1.66 in the text, but
instead of (AxB)i, it's A cross product with a second vector that is not
just simply B, but instead is BxC. Also, instead of the i'th component as
in eq. 1.66, it's the l'th component. So this last equation above (eq. 2)
sums over m,n, instead of j,k like in the book. So 'i' (eq. 1.66) is now
'l', and j,k are m,n. Aj (subscript here is 'j' - hope it's readable) from
eq. 1.66 in the book is Am, and Bj (subscript 'j' again) is just (BxC)n.
In the right side of the above equation (eq. 2), substitute from eq. 1,
with the appropriate subscript renaming. (since eq. 2 involves (BxC)n, the
subscripts for the sum involving ( involves n,j,k and not i,j,k as in
eq. 1. Now, collect the sum over j,k together with the sum over m,n and permute
n,j,k twice ('even' permutation) to get j,k,n and rewrite:
Then, collect the factors containing 'n' separate out the sum over 'n' to
get: The sum over 'n' in the parentheses is just (from the starting equation
above) the product of delta functions (have to rename indices since here we
have the sum over 'n' above is a sum over 'k'):
For the first product of delta functions times AmBjCk, j becomes l (so Bj
becomes Bl) and k becomes m (so Ck becomes Cm); for the second product of
delta functions times AmBjCk k becomes l (so Ck becomes Cl) and j becomes
m, so Bj becomes Bm . Thus But we recognize that the two sums over m are just the dot products between
vectors A and C, and A and B, respectively. Thus we have the l'th
component (the left hand side) of AxBxC equal to: Thus, AxBxC = (A dot C)B - (A dot B)C