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I lack in understanding of some basic idea regarding 4-vectors and index raising and lowering. From what I understand that: $$ \partial^\mu = \eta^{\mu\nu}\partial_\mu$$ So then, is the following correct? $$ \partial_\mu\psi\partial^\mu\psi^* - \partial_\mu\psi^*\partial^\mu\psi =\partial_\mu\psi\partial^\mu\psi^* - \eta^{\mu\mu}\partial^\mu\psi^*\eta_{\mu\mu}\partial_\mu\psi = \partial_\mu\psi\partial^\mu\psi^* - \partial_\mu\psi\partial^\mu\psi^* = 0$$

EDIT: The line above is apparently wrong. What I think I meant was: $$ \partial_\mu\psi\partial^\mu\psi^* - \partial_\mu\psi^*\partial^\mu\psi =\partial_\mu\psi\partial^\mu\psi^* - \eta_{\mu\lambda}\partial^\mu\psi^*\eta^{\mu\lambda}\partial_\mu\psi = \partial_\mu\psi\partial^\mu\psi^* - \partial_\mu\psi\partial^\mu\psi^* = 0$$ where $*$ is just a complex conjugate.

And if this is correct, is this then true:

$$\eta^{\mu\lambda}\eta_{\mu\lambda} = \delta $$

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  • $\begingroup$ Generally you need to ensure that the only case you have repeated indices in an expression is where there is an implicit sum, which, in turn, can only involve an index repeated exactly twice. That repeated index, however, can be renamed freely to make it unique. So an expression like $\eta_{\mu\lambda}\partial^\mu\psi^*\eta^{\mu\lambda}\partial_\mu\psi$ is at best confusing (which sums are there?) and at worst wrong. You need to uniquify the indices to end up with $\eta_{\mu\lambda}\partial^\mu\psi^*\eta^{\nu\lambda}\partial_\nu\psi$. (Sorry, and earlier version of this comment was bogus.) $\endgroup$
    – user107153
    Commented Jun 12, 2017 at 16:46
  • $\begingroup$ Your very first equation is wrong: it should be instead $\partial^{\mu}=\eta^{\mu\nu}\partial_{\nu}$. $\endgroup$
    – gented
    Commented May 31, 2019 at 12:19
  • $\begingroup$ $\eta^{\mu \nu} \eta_{\mu \rho} = \delta^{\nu}_{\;\rho}$ is the appropriate relation, your last statement is actually equal to $4$ since $\delta^{\mu}_{\;\mu}=d$ where $d$ is the dimension of your spacetime (normally $3+1$ though it depends on your theory). $\endgroup$
    – Triatticus
    Commented May 31, 2019 at 17:47

2 Answers 2

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There are some index issues with your expressions as mentioned by others. The correct way to raise/lower an index is by $$ \partial^\mu = \eta^{\mu\nu}\partial_\nu .$$

Notice how the free index $\mu$ appears in the up position exactly once on each side of the equal sign, and the dummy index $\nu$ is repeated in the up and down position on the same side. The dummy index $\nu$ is summed over, so the expression should be read as: $$ \partial^\mu = \sum_{\nu=0}^3 \eta^{\mu\nu}\partial_\nu. $$

The expression tells you how to get each of the 4 components of $\partial^\mu$.

When we apply that to the relevant part of your second expression we get:

$$ \left(\partial_\mu \psi^*\right) \, \left( \partial^\mu \psi \right) = \left( \eta_{\mu\nu} \partial^\nu\psi^*\right) \, \left( \eta^{\mu\lambda} \partial_\lambda \psi \right). $$

To get the indices correct we make sure that each term in parentheses maintains a single $\mu$ index in the correct position, and that in a given multiplicative statement no index is used more than once in a given up/down position. These means we need a different dummy index for each summation, otherwise we might not know which terms get summed with which.

The final statement is then: \begin{align} \partial_\mu \psi \partial^\mu \psi^* - \partial_\mu \psi^* \partial^\mu \psi &= \partial_\mu \psi \partial^\mu \psi^* - \eta_{\mu\nu} \partial^\nu\psi^* \eta^{\mu\lambda} \partial_\lambda \psi \\ &= \partial_\mu \psi \partial^\mu \psi^* - \left( \eta_{\mu\nu} \eta^{\mu\lambda} \right) \partial^\nu\psi^* \partial_\lambda \psi \\ &= \partial_\mu \psi \partial^\mu \psi^* - \partial_\nu \psi \partial^\nu \psi^* \\ &= \partial_\mu \psi \partial^\mu \psi^* - \partial_\mu \psi \partial^\mu \psi^* \\ &= 0 \end{align}

We used the fact that $$\eta_{\mu\nu} \eta^{\mu\lambda} = {\delta_\nu}^\lambda,$$ and that dummy indices can be renamed arbitrarily ($\nu\rightarrow\mu$).

Finally as a commenter stated, if you fully contract a metric with itself you get: $$\eta_{\mu\nu} \eta^{\mu\nu} = {\delta_\nu}^\nu = 4$$ in a 4-D space.

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No. You cannot have more than one of the same index raised/lowered position. If you want to apply the metric to second set of terms, the second line should read \begin{equation} ....=\partial_\mu \psi \partial^\mu \psi^* - \eta_{\mu\nu} \partial^\nu \psi^* \eta^{\mu\lambda}\partial_\lambda\psi = .... \end{equation}

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  • $\begingroup$ I've edited my post. Is it correct now? $\endgroup$
    – Ilya Lapan
    Commented Jun 12, 2017 at 14:54
  • $\begingroup$ @IlyaLapan You should add your revised solution to the post rather than editing the old one, so people can learn from what you put down first. $\endgroup$
    – Señor O
    Commented Jun 12, 2017 at 14:59
  • $\begingroup$ No. You should still have a $\partial_\mu \psi^*$ term and a $\partial^\mu \psi^*$ term. You can't use index gymnastics to get both lower partial derivatives to act on $\psi$ and both upper partial derivatives to act on $\psi^*$. $\endgroup$
    – Bob
    Commented Jun 12, 2017 at 15:00
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    $\begingroup$ Also, $\eta^{\alpha\mu}\eta_{\mu\beta}=\delta^\alpha_\beta$. (In this case, $\eta^{\mu\nu}\eta_{\mu\nu}=4$....assuming you're working in 3+1 dimensional spacetime). $\endgroup$
    – Bob
    Commented Jun 12, 2017 at 15:03
  • $\begingroup$ I am extremely confused then. Is the LHS of the equation equal to zero at all then? $\endgroup$
    – Ilya Lapan
    Commented Jun 12, 2017 at 15:05

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