The tensor $t$ is defined in terms of a scalar $\varrho$ and the two vectors $v$ and $u$ (and derivatives) as follows: $$t_{i j}:=\varrho v_i u_j-\frac{1}{2} \varrho \varepsilon_{i lm} v_l (\partial_j v_m)$$ where the Einstein summation convention is employed (and also is henceforth), $\varepsilon$ is the Levi-Civita symbol, and $v$ is a unit vector, so $v_i v_i=1$. Contracting this definition with $v_{i}$ gives the first identity \begin{equation} v_{i}t_{i k}=\varrho v_i v_{i} u_k-\frac{1}{2} \varrho \varepsilon_{i lm} v_{i} v_l (\partial_k v_m)=\varrho u_k \end{equation} by using the properties of $\varepsilon$, whereas contracting it with $\varepsilon_{nmi}v_{m}$ gives the second identity $$ \begin{split} \varepsilon_{nmi}v_{m}t_{i k}&=\varrho \varepsilon_{nmi}v_{m} v_i u_k-\frac{1}{2} \varrho \varepsilon_{nmi} \varepsilon_{i lo} v_{m} v_l (\partial_k v_o)=-\frac{1}{2} \varrho (\delta_{nl}\delta_{mo}-\delta_{no}\delta_{ml})v_{m} v_l (\partial_k v_o)\\ &=-\frac{1}{2} \varrho v_{o} v_n (\partial_k v_o)+\frac{1}{2} \varrho v_{l} v_l (\partial_k v_n)=\frac{1}{2} \varrho (\partial_k v_n) \end{split} $$ where $v_{o} (\partial_k v_o)=0$ because $v$ is a unit vector. Now, I want to show the third tensor identity $$ \varepsilon_{i j k}\left(\varrho^2 (\partial_j u_{k})-\frac{1}{4} \varepsilon_{l m n} \varrho^2 v_l (\partial_jv_m) (\partial_kv_n)\right)=0 $$ but I haven't managed in doing so by algebraic manipulation. Could anyone help me understand how this third equation can be derived? This appeared originally in the paper Hydrodynamic form of the Weyl equation, Acta Physica Polonica B 26, 1201 (1995).
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$\begingroup$ You say that you employed the Einstein summation convention and yet all indices are covariant. By the way I read the first part of the paper, and I cannot find the definition for tensor $T$. $\endgroup$– Ted BlackCommented May 13 at 13:58
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$\begingroup$ Right, I meant that the summation over the indices is dropped (so for example, in the definition I gave of $t$ the indices $l$ and $m$ are summed over). The definition of $T$ appears in the third part of the paper (equation 23) and the expression I wrote is equation 28. $\endgroup$– reverendjamesmCommented May 14 at 12:53
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$\begingroup$ After reading the whole paper (33) seems to be independent of (28). The authors derive (28) from (26) and (27); but for equation (33) they use "yet another identity" which is (32); note that unlike (26), (27) and (28) the stress tensor $\phi^\dagger \sigma_i \overleftrightarrow{\partial}_j \phi$ does not appear in (32). Instead they only have $\phi^\dagger \overleftrightarrow{\partial}_j \phi$ and $\phi^\dagger \sigma_n \phi$ in this identity which points to the fact that it is independent. $\endgroup$– Ted BlackCommented May 14 at 17:22
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$\begingroup$ Thank you. I had hoped (33) could be also reverse-engineered from (28), just like (26) and (27) can be derived from (28) by contracting with $v_i$ and $\varepsilon_{kli}v_l$ respectively. Do you by any chance know how (33) or the spinor identity (32) are then obtained? $\endgroup$– reverendjamesmCommented May 15 at 9:24
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$\begingroup$ Why is $v_n\partial_n v_j = 0$? It does not follow from the "unit vector" condition $v_1^2+v_2^2+v_3^2=1$. Nor is it necessary from a "canonical basis"-perspective. Also, is $\rho$ a scalar function of the coordinates (a scalar in the tensorial sense) or just a constant, ie $\partial_n \rho\equiv 0$? $\endgroup$– ContraKintaCommented May 16 at 9:47
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