Prove Vector Quadruple Product with Levi-Civita/Index Notation

[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.

 

[Quesadilla]

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.

 

[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.

 

[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.

 

[vela]

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

 

[johnnydoejr]

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

Thanks, that does it.

 

[MASTER]

Thanks, that does it.

i didn't get that and yes my mind is also fried after seeing so many symbols. can you please explain how to get the result

 

[maurosauros]

i didn't get that and yes my mind is also fried after seeing so many symbols. can you please explain how to get the result

Hi, I'm new here, but I found the solution to your confusion (I think), and maybe it can help somebody.

So, once you contract $\varepsilon_{ijk}$ with $\varepsilon_{kpq}$, you can rotate them, or do an "even permutation" of them. This means that $\varepsilon_{kpq} = \varepsilon_{pqk} = \varepsilon_{qkp}$. That way you can obtain the identity you desire.

 

[aangavatar]

Bro, you did it correctly. I verified it. thank you for sharing the solution.

 

[kumusta]

I found the mistake in my solution. I don't know if you're interested in knowing where the mistake is.

 

[THAUROS]

Yes of course. But I am afraid that you will need to be specific. I don't have the strong background knowledge you have. But I am determined to learn, I study Neuroscience, so I know the effort will pay off!

 

[kumusta]

I don't know how and where to start. I'll just think for a minute.

 

[kumusta]

The given equation is correct:
$\qquad\qquad(\vec A \times \vec B)\times(\vec C \times \vec D) = [ \vec A \cdot (\vec B \times \vec D) ] \vec C - [ \vec A \cdot (\vec B \times \vec C) ] \vec D$
The result I got was
$\qquad\qquad(\vec A \times \vec B)\times(\vec C \times \vec D) = [ \vec A \cdot (\vec B \times \vec C) ] \vec D - [ \vec A \cdot (\vec B \times \vec D) ] \vec C$
The mistake is in this line of the second component form:
$\qquad[(\vec A \times \vec B)\times(\vec C \times \vec D)]_i = 𝜀_{ijk} 𝜀_{jlm} 𝜀_{kst} A_l B_m C_s D_t = 𝜀_{ikj} 𝜀_{jlm} 𝜀_{kst} A_l B_m C_s D_t$
The correct equation should have been
$\qquad[(\vec A \times \vec B)\times(\vec C \times \vec D)]_i = 𝜀_{ijk} 𝜀_{jlm} 𝜀_{kst} A_l B_m C_s D_t = -𝜀_{ikj} 𝜀_{jlm} 𝜀_{kst} A_l B_m C_s D_t$
since for a single permutation (an odd permutation) from $~𝜀_{ijk}~$ to $~𝜀_{ikj}~$, one must use the relation $~~𝜀_{ikj}/𝜀_{ijk} = -1~\Rightarrow~𝜀_{ijk} = -𝜀_{ikj}~.$ I did not put the minus sign before because I was worried that when I substitute values for the indices later, I would have to put a minus sign again $-$ a second minus sign $-$ for an odd permutation which would, of course, neutralize the effect of the first minus sign. So, using the property $𝜀_{ikj} 𝜀_{jlm} = \{ \delta_{il} \delta_{km} - \delta_{im} \delta_{kl} \}~$ of the Levi-Civita symbol, the previous equation becomes
$\qquad[(\vec A \times \vec B)\times(\vec C \times \vec D)]_i = -\{ \delta_{il} \delta_{km} - \delta_{im} \delta_{kl} \} 𝜀_{kst} A_l B_m C_s D_t$
${~~~~}\qquad\qquad\qquad\qquad\qquad= \{ \delta_{im} \delta_{kl} - \delta_{il} \delta_{km} \} 𝜀_{kst} A_l B_m C_s D_t$
which is the negative of the result that I got in my earlier solution,
$\qquad[(\vec A \times \vec B)\times(\vec C \times \vec D)]_i = \{ \delta_{il} \delta_{km} - \delta_{im} \delta_{kl} \} 𝜀_{kst} A_l B_m C_s D_t$
that led to the inequality.

 

[THAUROS]

Thanks very much,very kind of you to also try and put it simpler terms to help me! I absolutely appreciate your help.
I'll go through it properly with a good physics tutor as soon as I find one. I'm still looking, but I feel confident he/she will come my way!

 

[THAUROS]

Have a great day/week