# Prove Vector Quadruple Product with Levi-Civita/Index Notation

1. Sep 6, 2014

### johnnydoejr

I'm asked to prove the following using Levi-Civita/index notation:
$(\mathbf{a \times b} )\mathbf{\times} (\mathbf{c}\times \mathbf{d}) = [\mathbf{a,\ b, \ d}] \mathbf c - [\mathbf{a,\ b, \ c}] \mathbf d \$

I'm able to prove it using triple product identities, but I'm completely stuck with the index notation. I was previously able to prove Lagrange's Identity with index notation, but applying similar concepts I just get stuck on the first step with the quadruple product.

Using the same first step of proving Lagrange's identity, I transformed $(\mathbf{a \times b} )$ into $\varepsilon_{ijk} a^j b^k$ and $(\mathbf{c \times d} )$ into $\varepsilon_{imn} c^m d^n$ but then I'm just left with $(\varepsilon_{ijk} a^j b^k) \mathbf{\times} (\varepsilon_{imn} c^m d^n)$ which is seemingly unhelpful.

I also tried letting AxB = W and CxD = Z and transforming WxZ to index notation. Then I tried to 'un-nest' the original cross products in index notation, but it quickly ended up in a place where I couldn't understand what the different indexes represented.

Any help would be appreciated. Thanks.

2. Sep 6, 2014

I am not entirely sure how you defined $[\mathbf{a},\mathbf{b},\mathbf{c}]$. Probably as a determinant? Either way, you must have $[\mathbf{a},\mathbf{b}, \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$. That might be helpful.

There is also an identity that relates $\epsilon_{ijk}\epsilon_{lmk}$ to Kronecker deltas. It might also prove useful.

3. Sep 7, 2014

### johnnydoejr

Yes, [a,b,c]=a⋅(b×c). I'm sure it will be helpful, but only after I can get past the initial steps.

I'm aware of the Kronecker delta identity you refer to, I used it to prove the scalar quadruple product/Lagrange's Identity as part of the same assignment. I don't doubt that that will be involved as well, but again, only after I can get past this initial step.

I don't know how to combine $(\varepsilon_{ijk} a^j b^k) \mathbf{\times} (\varepsilon_{imn} c^m d^n)$ without the index notation blowing up on me to a point where I don't understand it. It could just be that I'm approaching it from the wrong direction in the one step I did make in re-writing the two inner cross products in index notation.

4. Sep 7, 2014

### johnnydoejr

Thought I had a breakthrough after referencing another forum post. Was able to work through this:

\begin{align*} [(A\times B)\times(C\times D)]_{i} &= \varepsilon_{ijk}(A\times B)_{j}(C\times D)_{k} \\ &= \varepsilon_{ijk} \varepsilon_{jmn} A_{m}B_{n} \varepsilon_{kpq} C_{p}D_{q} \\ &= -\varepsilon_{jik} \varepsilon_{jmn} A_{m}B_{n} \varepsilon_{kpq} C_{p}D_{q} \\ &= -(\delta_{im}\delta_{kn} - \delta_{in}\delta_{km})A_{m}B_{n} \varepsilon_{kpq} C_{p}D_{q}\\ &= (-A_{i}B_{k} + A_{k}B_{i}) \varepsilon_{kpq} C_{p}D_{q}\\ &= -A_{i} \varepsilon_{kpq} B_{k}C_{p}D_{q} + B_{i} \varepsilon_{kpq} A_{k}C_{p}D_{q} \\ &= (\mathbf{A} \cdot \mathbf{C} \times \mathbf{D})\mathbf{B} - (\mathbf{B} \cdot \mathbf{C} \times \mathbf{D})\mathbf{A}\end{align*}

This has to be the right approach...too many things went right while I was working through it, but then I came to then end and compared it to the identity:

$(\mathbf{A} \cdot \mathbf{B} \times \mathbf{D})\mathbf{C} - (\mathbf{A} \cdot \mathbf{B} \times \mathbf{C})\mathbf{D}$

So close, yet so far. Anybody see any mistakes? My brain is fried.

5. Sep 7, 2014

### vela

Staff Emeritus
You didn't make a mistake, but to get the identity you want, try contracting $\varepsilon_{ijk}$ with $\varepsilon_{kpq}$ instead.

6. Sep 7, 2014

### johnnydoejr

Thanks, that does it.