Index Notation Help: Understanding Swapping Symbols in AxB=-BxA

  • #1

Homework Statement



I was following along with a proof of AxB=-Bxa

it went along the lines of;
Let;
C=AxB=Ciei
D=BxA=Diei
for i=1,2,3
and we know
Ci=eijkAjBk
Di=eijkBjAk
we can manipulate B and A to give
Bj=BsDeltasj
Bk=AmDeltamk

so we find;
Di=eijkDeltasjDeltamkBsAm = eismBsAm

then it says replace s by k and m by j to find;
Di=eikjBkAj=-eijkAjBk=-Ci

I don't understand why you can just swap s to k and m to j, what's stopping you just replacing s to j and m to k and finding that Di=Ci?
(every time I tried to use the delta from the Latex reference I just got a superscript 1 so that's why I write Delta for the Kronecker Deltas)

Thanks in advance :smile:
 
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  • #2
first try this for tex
[tex] C = A \times B = C_i = \epsilon_{ijk} A_j B_k[/tex]
 

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