Proving Vector Identities Using the Permutation Tensor and Kroenecker Delta

  • #1

Homework Statement



Prove using the Levi-Civita Tensor/Kroenecker Delta that:
(AxB)x(CxD) = (A.BxD).C-(A.BxC).D

Homework Equations



εіјkεimn = δjmδkn – δjnδkm (where δij = +1 when i = j and 0 when i ≠ j)

The Attempt at a Solution



if E = (AxB) then Ei = εіјkAjBk, and
if F = (CxD) then Fm = εimnCnDi

from this point I'm a little confused as I'm not sure if I have to find the cross product of (ExF) using the summation notation, or if I can now relate these via the Kroenecker delta relationship given. I feel I am missing a step as there are 3 cross product relationships and I would greatly appreciate some help here as the only examples I can track down deal with 2 cross product relationships.

Thanks very much in advance
PS
Apologies for the the lack of proper subscripts but that's a problem for another day
 

Answers and Replies

  • #2
vela
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You should use unique letters for the indices to avoid confusion. Right now you have i appearing in the expressions for both E and F.

[tex]\begin{align*}
E_q & = \varepsilon_{qjk}A_j B_k \\
F_r & = \varepsilon_{rmn}C_m D_n \\
(E \times F)_p &= \varepsilon_{pqr} E_q F_r = \varepsilon_{pqr} (\varepsilon_{qjk}A_j B_k) (\varepsilon_{rmn}C_m D_n)
\end{align*}
[/tex]

In the last line, you have a product of two Levi-Civita symbols with a common index, which allows you to write the product it in terms of the delta functions.
 
  • #3
Thank you Vela for your help with this it is very much appreciated as this is my first exposure to the power of the permutation tensor. I was wondering if you could answer one final stupid question on this if you have the time as I'm not sure whether to multiply out of the brackets separately using the 3rd tensor or to multiply one and then the other. I will continue to read on this but if you could help further then I would once again be in your debt.
 
  • #4
vela
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I'm not sure I understand what you're asking. What do you have so far?
 
  • #5
Hi Vela,

When I multiply out the brackets I get the following (please forgive my not having LaTeX):

EpqrEqjkAjBk + EpqrErmnCnDn

If I then cycle the subscripts Epqr becomes Eqrp and the other Epqr becomes Erpq:

EqrpEqjkAjBk + ErpqErmnCnDn

This, hopefully, lets me use the epsilon/delta (I'm using 6 as delta) relationship to give:

6rj6pk - 6rk6pj(AjBk) + 6pm6qn - 6pn6qm(CnDn)

I'm not convinced I've handled these next steps properly and have strayed down a cul-de-sac from which I can't escape, but feel I have made some elementary errors. Sorry to be a pain but if you could possibly point out the error of my ways it would be greatly appreciated.

Best regards,
Stuart
PS
Apologies for the delay in responding to you but Scotland is struggling in the grip of an unusually powerful winter
 
  • #6
vela
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What you wrote would be equal to [itex]\varepsilon_{pqr} (\varepsilon_{qjk}A_j B_k + \varepsilon_{rmn}C_m D_n)[/itex], not (ExF)p. Try again without inserting the extra addition.
 

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