Index Notation Help: Solve [a,b,c]^2

[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. Homework 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 variables3. 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 ) = a_ib_jε_ijke_k ⋅ d_le_mε_lmpe_p

= a_ib_jd_le_mε_ijkε_lmp(e_k ⋅ e_p)

= a_ib_jd_le_mε_ijkε_lmpδ_kp

= a_ib_jd_le_mε_ijkε_lmk

= a_ib_jd_le_m(δ_ilδ_jm - δ_imδ_jl)

= a_ib_jd_le_mδ_ilδ_jm - a_ib_jd_le_mδ_imδ_jl

= a_ib_jd_ie_j - a_ib_jd_je_i

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?

 

[TSny]

Welcome to PF!

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

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

 

[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 a_ib_jd_ie_j 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!

 

[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., b_jd_j) must be zero.

Anyway, your work looks good!