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$$k^2g^{\mu\nu}-k^\mu k^\nu=k^2P^{\mu\nu}(k)$$

Here 1st term can be written as 2nd term via breaking square term and then raising index.

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  • $\begingroup$ 1st term can be written as 2nd term No, it can’t. You must be confused about some aspect of tensor algebra. Try writing out the two terms when $\mu$ and $\nu$ are, say, zero to see that they are different. $\endgroup$
    – G. Smith
    Jun 23, 2020 at 6:15
  • $\begingroup$ Of course in tensor form there are off diagonal termsin the 2nd term.I see that but rules about raising or lowering index seems not to be working here.is there some restrictions about using these rules? $\endgroup$
    – Roy
    Jun 23, 2020 at 8:14
  • $\begingroup$ $k^2g^{\mu\nu}$ is $k^\lambda k_\lambda g^{\mu\nu}$. You cannot use $g^{\mu\nu}$ to affect the $\lambda$ indices. Writing $k^2g^{\mu\nu}$ as $k^\mu k_\mu g^{\mu\nu}$ would be wrong. You can never have more than two contracted indices. This must be what you are confused about. $\endgroup$
    – G. Smith
    Jun 23, 2020 at 16:53

3 Answers 3

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In both terms the indices $\mu$ and $\nu$ are free indices. That means that when you try to go from the first one to the second $k^2g^{\mu\nu}\rightarrow k^\mu k^\nu$ by breaking the square you can't write $k^2g^{\mu\nu}=k^\mu k_\mu g^{\mu\nu}=k^\mu k^\nu$ because

  1. You would be introducing a dummy index (one that is summed over) with the same letter as one which was previously considered to be free.
  2. In the intermediate step, you would have three indices named the same.

Remember to name with new letters when introducing a new summed over index.

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Your assumption that you can break the square term and raise index is incorrect: The indices will not be the same as the indices in the second term, it is always helpful to name your indices differently each time you "break" a term.

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  • $\begingroup$ But by usual rules this is allowed.one can write 2nd term also as 1st term.i am confused about the applicability of lowering or raising index rules.is it that one always should diversify the name of the indices. $\endgroup$
    – Roy
    Jun 23, 2020 at 9:19
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Elaborating on @shehab 's answer....

Given $k^2g^{\mu\nu}-k^\mu k^\nu = k^2P^{\mu\nu}$,
for an arbitrary co-vector $A_\nu$ we then have

$$\begin{align*} (k^2g^{\mu\nu}-k^\mu k^\nu)A_\nu &=(k^2P^{\mu\nu})A_\nu\\ k^2 A^{\mu} - k^\mu (k^\nu A_\nu) &=k^2 (P^{\mu\nu}A_\nu). \end{align*}$$

Observe that, in general, $$k^2 A^{\mu} \neq k^\mu (k^\nu A_\nu)$$ since $A^\mu$ and $k^\mu$ generally don't point in the same direction.
So, $k^2 P^{\mu\nu} \neq 0$.

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