In Landau and Lifschitz Mechanics, p. 99, we have (implicit) the equality $$\Omega_i^2 x_i^2 = \Omega_i \Omega_k \delta_{ik} x_{\ell}^2 $$ written with Einstein summation convention.

The left hand side is clearly: $$\Omega_1^2x_1^2 + \Omega_2^2x_2^2+ \Omega_3^2x_3^2 $$

The answer to this question would indicate that the right hand side should be interpreted this way:

$$\Omega_i\Omega_k\delta_{ik}x_{\ell}^2 = \sum_i\sum_k \left[\Omega_i \Omega_k \delta_{ik} \left[\sum_{\ell} x_{\ell}^2 \right] \right] \\= \Omega_1^2[x_1^2+x_2^2+x_3^2] + \Omega_2^2[x_1^2 + x_2^2+x_3^2] + \Omega_3^2[x_1^2+x_2^2+x_3^2]$$

However, I only get that this is an identity when I interpret it this way:

$$\Omega_i\Omega_k\delta_{ik}x_{\ell}^2 = \sum_i\sum_k\sum_{\ell} \Omega_i\Omega_k\delta_{ik}x_{\ell}^2 = \Omega_1^2x_1^2 + \Omega_2^2x_2^2+ \Omega_3^2x_3^2 $$

So is there a mistake in the textbook? Or is the answer given to the other question incorrect? I've tried googling this, but all of the examples given are single terms, not multiple terms.

EDIT: The full equation in the textbook is (if it helps): $$\Omega_i^2 x_i^2 - \Omega_i x_i \Omega_k x_k = \Omega_i \Omega_k \delta_{ik} x_{\ell}^2 - \Omega_i \Omega_k x_i x_k $$

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    $\begingroup$ Using your latter intrepretation, you still got $$\Omega_1^2[x_1^2+x_2^2+x_3^2] + \Omega_2^2[x_1^2 + x_2^2+x_3^2] + \Omega_3^2[x_1^2+x_2^2+x_3^2]$$ $\endgroup$ – user99914 Jun 5 '16 at 3:39
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    $\begingroup$ Why do you have $x^2_\ell$ in the first equation $\Omega_i^2 x_i^2 = \Omega_i \Omega_k \delta_{ik} x_{\ell}^2$? Is the $\ell$ index a typo? $\endgroup$ – Vlad Jun 5 '16 at 3:43
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    $\begingroup$ I'm voting to close this question as off-topic because the question is based on a false premise. $\endgroup$ – Chill2Macht Jun 5 '16 at 3:44
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    $\begingroup$ No it's not a typo -- that's how it is in the textbook. $\endgroup$ – Chill2Macht Jun 5 '16 at 3:44
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    $\begingroup$ You can delete your own question, you don't need to vote to close. (Actually i am a bit confused by the convention in the book - repeated indices are summed, but what about $\Omega_i^2 x_i^2 = \Omega_i \Omega_i x_i x_i$, which has $4$ i's? $\endgroup$ – user99914 Jun 5 '16 at 3:47

Unfortunately it is unclear what is the expression

$$\Omega_i^2 x_i^2$$

as there are indeed $4$ i's in the expression. But after reading previous page of the book, this term corresponds to

$$\Omega ^2 r^2 = (\Omega \cdot \Omega)(r\cdot r)$$

so the expression should be written as

$$ \Omega_i^2 x_l^2$$



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