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I am reading "An Introduction to the Theory of Piezoelectricity" 2nd edition by Jiashi Yang

I am trying to understand a derivation for an equation that uses multiple Levi-Civita symbols.

The Levi-Civita symbol is defined as: \begin{equation} \varepsilon _{ijk} = \hat i_i \cdot \left( {\hat i_j \times \hat i_k } \right) = \left\{ {\begin{array}{*{20}c} 1 \hfill & {i,j,k = 1,2,3;} \hfill & {2,3,1;} \hfill & {3,1,2,} \hfill \\ { - 1} \hfill & {i,j,k = 3,2,1;} \hfill & {2,1,3;} \hfill & {1,3,2,} \hfill \\ 0 \hfill & {{otherwise}} \hfill & \hfill & \hfill \\ \end{array}} \right. \end{equation}

and the product of two Levi-Civita symbols is given by

\begin{equation} \varepsilon _{ijk} \varepsilon _{pqr} = \left| {\begin{array}{*{20}c} {\delta _{ip} } & {\delta _{iq} } & {\delta _{ir} } \\ {\delta _{jp} } & {\delta _{jq} } & {\delta _{jr} } \\ {\delta _{kp} } & {\delta _{kq} } & {\delta _{kr} } \\ \end{array}} \right| \end{equation} where $\delta$ is the Kronecker delta.

We have a matrix that defines the deformation in three dimensions of

\begin{equation} \left[ {\begin{array}{*{20}c} {y_{1,1} } & {y_{1,2} } & {y_{1,3} } \\ {y_{2,1} } & {y_{2,2} } & {y_{2,3} } \\ {y_{3,1} } & {y_{3,2} } & {y_{3,3} } \\ \end{array}} \right] \end{equation}

where $y_{k,K} = \frac{{\partial y_k }}{{\partial X_K }}$

The Jacobian of the deformation is given as:

$J = \det \left( {y_{k,K} } \right)$

So if I expand it out manually I get

$ J = y_{1,1} \left( {y_{2,2} y_{3,3} - y_{2,3} y_{3,2} } \right) - y_{1,2} \left( {y_{2,1} y_{3,3} - y_{2,3} y_{3,1} } \right) + y_{1,3} \left( {y_{2,1} y_{3,2} - y_{2,2} y_{3,1} } \right) $

The author writes the results (Equation 1.12) using the Levi-Civita symbols and Einstein summation convention notation as

$ \begin{array}{l} J = \det \left( {y_{k,K} } \right) = \varepsilon _{ijk} y_{i,1} y_{j,2} y_{k,3} = \varepsilon _{KLM} y_{1,K} y_{2,L} y_{3,M} = \frac{1}{6}\varepsilon _{klm} \varepsilon _{KLM} y_{k,K} y_{l,L} y_{m,M} \\ \\ \end{array} $

I can see by rearranging (order of terms and factors) my results manually by hand and using the definition of the Levi-Civita symbol that the three versions he gives are consistent with my results (including the 1/6 coefficient in the last one). However I assume there must be an an algebra that can be applied to his three results directly in their form with the Levi-Civita symbols and Einstein notation that can be used to convert from one to the other two.

The issue comes up when I try to follow his derivation for Equation 1.17 for which he says can be verified that for all L, M and N that the following is true:

$\varepsilon _{ijk} y_{i,L} y_{j,M} y_{k,N} = J\varepsilon _{LMN}$

I try to obtain this result by taking his last result in Equation 1.12

$J = \frac{1}{6}\varepsilon _{klm} \varepsilon _{KLM} y_{k,K} y_{l,L} y_{m,M}$

Changing the subscripts klm to ijk and KLM to LMN giving

$J = \frac{1}{6}\varepsilon _{ijk} \varepsilon _{LMN} y_{i,L} y_{j,M} y_{k,N}$

From here I assume I can treat the Levi-Civita symbol as a scalar (which I'm really not sure is correct) and multiply both sides by $\varepsilon _{LMN}$ and I obtain

$J\varepsilon _{LMN} = \frac{1}{6}\varepsilon _{ijk} \varepsilon _{LMN} \varepsilon _{LMN} y_{i,L} y_{j,M} y_{k,N}$

Using the relationship above for the product of two Levi-Civita symbols and the Kronecker delta, $\varepsilon _{LMN}$ multiplied by itself on the right hand side should equal 1 giving

$J\varepsilon _{LMN} = \frac{1}{6}\varepsilon _{ijk} y_{i,L} y_{j,M} y_{k,N}$

which is close to the result of Equation 1.17 except for the 1/6 coefficient.

So it is unclear to me if my calculations are incorrect or if there is a typo in Equation 1.17.

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1 Answer 1

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I'm sure there is a more elegant way to do this, but this is what I came up with. I will assume the Einstein summation convention throughout the following.

Let \begin{align} Y_{LMN} = \varepsilon_{ijk}y_{iL}y_{jM}y_{kN} \end{align} Then we have, by Eq. 1-12. \begin{align} \varepsilon_{LMN}Y_{LMN} = 6\det y \end{align} so that \begin{align} \det y = \frac{1}{6}\varepsilon_{ABC}Y_{ABC} \end{align} where I have changed the dummy indices from $L$, $M$, $N$ to $A$, $B$, $C$ to avoid confusion in the next step. Now take \begin{align} \varepsilon_{LMN} \det y = \frac{1}{6}\varepsilon_{LMN}\varepsilon_{ABC}Y_{ABC} \end{align} and use the determinant expansion of the product of epsilons. I will skip writing it out explicitly and just note that it yields \begin{align} \varepsilon_{LMN} \det y = \frac{1}{6}(&Y_{LMN} - Y_{LNM} - Y_{MLN}\\ &+ Y_{NLM} + Y_{MNL} - Y_{NML}) \tag{1} \end{align} Since the indices $i$, $j$, $k$ are dummy indices in the sum \begin{align} Y_{LMN} = \varepsilon_{ijk}y_{iL}y_{jM}y_{kN} \end{align} we can exchange them freely. For example, \begin{align} Y_{LMN} = \varepsilon_{jki}y_{jL}y_{kM}y_{iN}. \end{align} Then using the definition of the Levi-Civita symbol, we can permute the indices to get, for example, \begin{align} Y_{LMN} = \varepsilon_{ijk}y_{iN}y_{jL}y_{kM} = Y_{NLM} \end{align} In general, we find that $Y_{LMN}$ has the same behavior under index permutations as the Levi-Civita symbol. That is, for cyclic permutations, \begin{align} Y_{LMN} = Y_{MNL} = Y_{NLM} \end{align} and for anti-cyclic permutations \begin{align} Y_{LMN} = -Y_{LNM} = -Y_{MLN} = -Y_{NML} \end{align} Applying this property to $(1)$, we find \begin{align} \varepsilon_{LMN} \det y &= \frac{1}{6}(6Y_{LMN})\\ &= Y_{LMN} \end{align} as claimed.

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  • $\begingroup$ Thanks for the response. I think my main error was thinking that the Levi-Civita times itself "squared" equaled 1 whereas apparently it is equal to 6. I neglected to apply the Einstein summation convention to its subscripts. It is a little confusing for me since with the symbol isolated there is no summation but as soon as it appears times itself you have to account for the summation. I would like to find a table of properties for the symbol such as one would have for something like the Fourier transform. $\endgroup$
    – MikeM
    Feb 11 at 21:00

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