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In Markakis 2012 they derive an equation (Eq.10),

$$\partial^k K^{ij} p_i p_j p_k + (\partial^j K - 2K^{jk} \partial_k \Phi) p_j = 0$$

For some tensor $K^{ij}$ and scalars, $K$, $\Phi$.

From this they state that one of the necessary and sufficient conditions for this equation to be true is,

$$ \partial^{(k}K^{ij)} = 0$$

Can anyone show why this is true? I can see that the first term $\partial^k K^{ij} p_i p_j p_k$ must equal zero, but can't see how this relates to the symmetrization. Also I cannot see how the metric tensor $g^{ij}$ would satisfy this equation?

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  • $\begingroup$ In the equation, is $K$ just the trace of $K^{ij}$? Also, the necessary & sufficient condition cited in the paper is that $\partial^{(k} K^{ij)} = 0$ (note the indices); I assume this was just a typo on your part. $\endgroup$
    – Michael Seifert
    Commented Nov 3, 2017 at 14:55
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    $\begingroup$ As far as the second question goes, the paper is working in flat spacetime, so $g^{ij} = \delta^{ij}$ and $\partial^k \delta^{ij} = 0$ trivially (even without the symmetrization.) $\endgroup$
    – Michael Seifert
    Commented Nov 3, 2017 at 15:05
  • $\begingroup$ Edits made to clarify questions raised in comments $\endgroup$
    – user1887919
    Commented Nov 3, 2017 at 15:19

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You can expand $\partial^{(k}K^{ij)}p_ip_jp_k$, change the ordering of $p_ip_jp_k$ in each term, and re-label the dummy indices to show that $\partial^{(k}K^{ij)}p_ip_jp_k=\partial^{k}K^{ij}p_ip_jp_k$. Therefore, the vanishing of $\partial^{(k}K^{ij)}$ is sufficient for $\partial^{k}K^{ij}p_ip_jp_k$ to vanish.

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    $\begingroup$ While I think this is the right idea, it seems like you're assuming that $\partial^k K^{ij}$ splits into a completely symmetric & a completely antisymmetric part. But this isn't true for tensors of rank 3 and above: in other words, $T^{ab\cdots c} \neq T^{(ab\cdots c)} + T^{[ab\cdots c]}$ unless $T$ is a rank-2 tensor. Can you elaborate on your answer to clarify this? $\endgroup$
    – Michael Seifert
    Commented Nov 3, 2017 at 18:13
  • $\begingroup$ @MichaelSeifert You're absolutely right! I've removed this assumption from the updated answer. $\endgroup$
    – Michael
    Commented Nov 4, 2017 at 0:17
  • $\begingroup$ Explicitly, what is the expression for $\partial^{(k}K^{ij)}$? $\endgroup$
    – user1887919
    Commented Nov 6, 2017 at 17:03
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    $\begingroup$ @user1887919 Using equation (1.67) of these notes, it is $\partial^{(k}K^{ij)}=\frac{1}{6}\left(\partial^k K^{ij} + \partial^k K^{ji} + \partial^i K^{jk} + \partial^i K^{kj} + \partial^j K^{ki} + \partial^j K^{ik}\right)$ $\endgroup$
    – Michael
    Commented Nov 6, 2017 at 21:57

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