Proving the Vector Triple Product Identity: A x (B x C) = (A x C)B - (A x B)C

[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

 

[mfb]

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).

 

[stevendaryl]

I have to prove the above equation.
I tried:
$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j\biggl[ \sum_{ij}\varepsilon_{ijk}B_iC_j\biggr]$$

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, [itex]i[/itex], 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:

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

 

[nasu]

After you fix the indices, you need to use the so called contraction of the epsilon tensors.
See equation (4)
https://en.wikipedia.org/wiki/Levi-Civita_symbol

 

[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:
$$...\varepsilon_{klm}\sum_{lm}B_lC_m$$

 

[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.

 

[Karol]

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.

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

 

[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.

 

[Karol]

Why is the epsilon klm outside the l-m sum? That's not right. Put the summation signs all the way to the left

$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j \sum_{lm}\varepsilon_{klm}B_lC_m=\sum_{jk}\sum_{lm} \varepsilon_{ijk} \varepsilon_{klm}A_jB_lC_m$$

then do what nasu says and collapse the epsilons into deltas using the formula in the link.

In the link it says:
$$\sum_{k} \varepsilon_{ijk} \varepsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$
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?

 

[stevendaryl]

$$\bigl[ A\times ( B\times C) \bigr]_i=\sum_{jk} \varepsilon_{ijk}A_j \sum_{lm}\varepsilon_{klm}B_lC_m=\sum_{jk}\sum_{lm} \varepsilon_{ijk} \varepsilon_{klm}A_jB_lC_m$$

In the link it says:
$$\sum_{k} \varepsilon_{ijk} \varepsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$
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?

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

[itex]\sum_{k} \varepsilon_{ijk}\varepsilon_{imk}=\varepsilon_{ij1} \varepsilon_{im1}+\varepsilon_{ij2} \varepsilon_{im2}+\varepsilon_{ij3} \varepsilon_{im3}[/itex]

A particular example: [itex]i=3, j=2, m=2[/itex]:

[itex]\sum_{k} \varepsilon_{32k}\varepsilon_{32k}=\varepsilon_{321} \varepsilon_{321}+\varepsilon_{322} \varepsilon_{322}+\varepsilon_{323} \varepsilon_{323}[/itex]

That's equal to 1.

 

[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.