Prove Vector Quadruple Product with Levi-Civita/Index Notation

  • #1
I'm asked to prove the following using Levi-Civita/index notation:
[itex]
(\mathbf{a \times b} )\mathbf{\times} (\mathbf{c}\times \mathbf{d}) = [\mathbf{a,\ b, \ d}] \mathbf c - [\mathbf{a,\ b, \ c}] \mathbf d \
[/itex]

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 [itex]
(\mathbf{a \times b} )[/itex] into [itex]\varepsilon_{ijk} a^j b^k[/itex] and [itex]
(\mathbf{c \times d} )[/itex] into [itex]\varepsilon_{imn} c^m d^n[/itex] but then I'm just left with [itex](\varepsilon_{ijk} a^j b^k) \mathbf{\times} (\varepsilon_{imn} c^m d^n)[/itex] 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.
 

Answers and Replies

  • #2
95
13
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
Yes, [a,b,c]=aβ‹…(bΓ—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 [itex](\varepsilon_{ijk} a^j b^k) \mathbf{\times} (\varepsilon_{imn} c^m d^n)[/itex] 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
Thought I had a breakthrough after referencing another forum post. Was able to work through this:

[itex]\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*}[/itex]

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:

[itex](\mathbf{A} \cdot \mathbf{B} \times \mathbf{D})\mathbf{C} - (\mathbf{A} \cdot \mathbf{B} \times \mathbf{C})\mathbf{D}[/itex]

So close, yet so far. Anybody see any mistakes? My brain is fried.
 
  • #5
vela
Staff Emeritus
Science Advisor
Homework Helper
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You didn't make a mistake, but to get the identity you want, try contracting ##\varepsilon_{ijk}## with ##\varepsilon_{kpq}## instead.
 
  • #6
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.
 
  • #7
1
0
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
 
  • #8
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.
 
  • #9
Bro, you did it correctly. I verified it. thank you for sharing the solution.
 
  • #10
73
24
I found the mistake in my solution. I don't know if you're interested in knowing where the mistake is.
 
  • #11
6
0
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! πŸ™‚
 
  • #12
73
24
I don't know how and where to start. I'll just think for a minute.
 
  • #13
73
24
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 = πœ€_{ijk} πœ€_{jlm} πœ€_{kst} A_l B_m C_s D_t = πœ€_{ikj} πœ€_{jlm} πœ€_{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 = πœ€_{ijk} πœ€_{jlm} πœ€_{kst} A_l B_m C_s D_t = -πœ€_{ikj} πœ€_{jlm} πœ€_{kst} A_l B_m C_s D_t##
since for a single permutation (an odd permutation) from ##~πœ€_{ijk}~## to ##~πœ€_{ikj}~##, one must use the relation ##~~πœ€_{ikj}/πœ€_{ijk} = -1~\Rightarrow~πœ€_{ijk} = -πœ€_{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 ##πœ€_{ikj} πœ€_{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} \} πœ€_{kst} A_l B_m C_s D_t##
##{~~~~}\qquad\qquad\qquad\qquad\qquad= \{ \delta_{im} \delta_{kl} - \delta_{il} \delta_{km} \} πœ€_{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} \} πœ€_{kst} A_l B_m C_s D_t##
that led to the inequality.
 
  • #14
6
0
Thanks very much,πŸ‘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!
 
  • #15
6
0
Have a great day/week
 

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