Index Notation help

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1. Sep 2, 2015

JordanD

1. The problem is:

( a x b )⋅[( b x c ) x ( c x a )] = [a,b,c]^2 = [ a⋅( b x c )]^2
I am supposed to solve this using index notation.... and I am having some problems.

2. Relevant equations: I guess I just don't understand the finer points of index notation. Every time I think I am getting it down I come back to this problem and get lost in the forest of letters.

I am also still struggling with dummy vs free variables

3. The attempt at a solution: The most logical approach I thought would be to replace

( b x c ) = d and ( c x a ) = e giving me ( a x b )⋅( d x e ) = [a,b,c]^2

(Note: my instinct was to skip ahead to the 'end' but I think for the sake of learning I should show all my steps?)

bold = vector

( a x b )⋅( d x e ) = aibjεijkek ⋅ dlemεlmpep

= aibjdlemεijkεlmp(ekep)

= aibjdlemεijkεlmpδkp

= aibjdlemεijkεlmk

= aibjdlemilδjm - δimδjl)

= aibjdlemδilδjm - aibjdlemδimδjl

= aibjdiej - aibjdjei

Here is where I freeze up. Am I on the right track? Do I plug in for my values of d and e and keep hammering away? Whats the deal?

2. Sep 3, 2015

TSny

Welcome to PF!

I'd say you're on the right track. Looking at the second term of your last expression, it contains bjdj. Keeping in mind the definition of the vector d, can you see by inspection what bjdj reduces to?

For the first term, aibjdiej, plug in the expressions for d and e as you suggest.

3. Sep 3, 2015

JordanD

So I figured it out! I just needed to be more confident in myself haha.

So had I kept going I would have seen that aibjdiej is the result I was looking for. However the second portion should then reduce to zero.... and luckily it does. To prove this you can write out the sum and see that they alternate and cancel out!

4. Sep 3, 2015

TSny

OK. Another way to see that the second portion is zero is to note that d = b x c and therefore d is perpendicular to b. So, the scalar product of b and d (i.e., bjdj) must be zero.