A x ( B x C )=(A∙C)B-(A∙B)C

1. Sep 1, 2015

Karol

I have to prove the above equation.
I tried:
$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]$$
In order to reach the k'th member of (BxC) but i tested it with i=1 and saw that in the member:
$$\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]=\biggl[ \sum_{1j}B_1C_j\biggr]$$
And it can't be that B1 is inside the sum since it has to receive also other values.
Then i tried:
$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]_k$$
But it doesn't enable me to open the [] brackets, or does it? i am new to this

2. Sep 1, 2015

Staff: Mentor

The first approach is confusing i and k.
The second approach is better, but you still have to rename indices to remove the brackets (j is used twice).

3. Sep 1, 2015

stevendaryl

Staff Emeritus
You have to be careful about distinguishing between bound (or "dummy") indices and free indices. Your expression is wrong because on the left-hand side is a free index, $i$, but on the right-hand side, there is no free index.

I think that you would be better off just brute-forcing this. Just write out explicitly:

$(A \times (B \times C))_1 = A_2 (B \times C)_3 - A_3 (B \times C)_2$
$(B \times C)_2 = B_3 C_1 - B_1 C_3$
$(B \times C)_3 = B_1 C_2 - B_2 C_1$

4. Sep 1, 2015

5. Sep 5, 2015

Karol

$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j \sum_{lm}\varepsilon_{klm}B_lC_m=\sum_{jk} \varepsilon_{ijk}A_j \varepsilon_{klm}\sum_{lm}B_lC_m$$
I try i=1:
$$\bigl[ A\times ( B\times C) \bigr]_i=\varepsilon_{123}\varepsilon_{312}A_2 B_1 C_2$$
And i stopped substituting since i don't know if i am allowed to change the indices outside the second summation symbol. i now want to substitute l=2 and m=1 but $\varepsilon_{klm}$ is outside of $\sum_{lm}\varepsilon_{klm}B_lC_m$, its:
$$....\varepsilon_{klm}\sum_{lm}B_lC_m$$

6. Sep 5, 2015

nasu

The second equation is not right. The component 1 is not a single term but a double sum.
By using the contraction of the epsilons you reduce these sums to two terms.

7. Sep 6, 2015

Karol

Yes, i know that it's not a single term, you didn't understand my question. i could complete the other terms:
$$\bigl[ A\times ( B\times C) \bigr]_i=\varepsilon_{123}\varepsilon_{312}A_2 B_1 C_2+\varepsilon_{132}\varepsilon_{213}A_3 B_1 C_3+\varepsilon_{123}\varepsilon_{321}A_2 B_2 C_1+\varepsilon_{132}\varepsilon_{231}A_3 B_3 C_1$$
The last 3 members have their l and m indices changed while they can be changed only inside the $\sum_{lm}$, no? the member $\varepsilon_{klm}$ is outside the sum

8. Sep 17, 2015

davidmoore63@y

Why is the epsilon klm outside the l-m sum? That's not right. Put the summation signs all the way to the left, then do what nasu says and collapse the epsilons into deltas using the formula in the link.

9. Sep 23, 2015

Karol

$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j \sum_{lm}\varepsilon_{klm}B_lC_m=\sum_{jk}\sum_{lm} \varepsilon_{ijk} \varepsilon_{klm}A_jB_lC_m$$
$$\sum_{k} \varepsilon_{ijk} \varepsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$
But, for example, for i=l:
$$\sum_{k} \varepsilon_{ijk}\varepsilon_{imk}=\varepsilon_{ij1} \varepsilon_{im1}+\varepsilon_{ij2} \varepsilon_{im2}+\varepsilon_{ij3} \varepsilon_{im3}=$$
$$=\varepsilon_{321} \varepsilon_{231}+\varepsilon_{132} \varepsilon_{312}+\varepsilon_{213} \varepsilon_{123}=3$$
While the sum:
$$\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$
Can be at most ±1. i don't know to use this formula, how to fill in the indices on the right side, since they take the values 1,2,3. the left side is a summation while the right is?

10. Sep 24, 2015

stevendaryl

Staff Emeritus
Could you please try working out a concrete example? You have:

$\sum_{k} \varepsilon_{ijk}\varepsilon_{imk}=\varepsilon_{ij1} \varepsilon_{im1}+\varepsilon_{ij2} \varepsilon_{im2}+\varepsilon_{ij3} \varepsilon_{im3}$

A particular example: $i=3, j=2, m=2$:

$\sum_{k} \varepsilon_{32k}\varepsilon_{32k}=\varepsilon_{321} \varepsilon_{321}+\varepsilon_{322} \varepsilon_{322}+\varepsilon_{323} \varepsilon_{323}$

That's equal to 1.

11. Sep 24, 2015

davidmoore63@y

When you have satisfied yourself that the formula is correct, the next step is to apply the delta symbols to the components in the following fashion (apologies no latex):

Sigma (m=1 to 3) Delta(j,m) A(m) = A(j)

It's worth persevering with this. Once mastered, the index notation is very powerful.