How to calculate $$\eta_{\mu\nu}\eta^{\mu\nu}$$ Where $$\eta=\begin{bmatrix} -1 \\ &1 \\&&1\\&&&1\end{bmatrix}$$ All other entries are $0$.
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4
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$\begingroup$ I don't understand where "But this is wrong since ..." comes from. Why is the equation following that not compatible with your result (which I believe is correct)? $\endgroup$– jorikiFeb 20 at 3:50
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$\begingroup$ Your last equation will become correct when you set $\gamma$ equal to $\mu.$ Then what is the trace of $\delta?$ $\endgroup$– Kurt G.Feb 20 at 4:53
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$\begingroup$ @joriki yeah, I was following a YT lecture and I had a misunderstanding of what the lecturer was saying, indeed the answer is 4 and not 2 which I assumed it to be, hence the question. I'll edit this post into a Q&A format now. Thank you for your comment. $\endgroup$– GedankenExperimentalistFeb 20 at 12:52
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$\begingroup$ @KurtG. Thank you for pointing out the error, also please refer to my above comment. Thank you. $\endgroup$– GedankenExperimentalistFeb 20 at 12:53
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1 Answer
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4
Writing $\eta^{\mu\nu}$ as $$\eta^{\mu\nu}=\eta^{\mu\alpha}\eta^{\nu\beta}\eta_{\alpha\beta}$$ Thus $$\eta_{\mu\nu}\eta^{\mu\nu}=\eta_{\mu\nu}\eta^{\mu\alpha}\eta^{\nu\beta}\eta_{\alpha\beta}$$ Rearranging $$\eta_{\mu\nu}\eta^{\mu\nu}=\eta^{\mu\alpha}\eta_{\alpha\beta}\eta^{\nu\beta}\eta_{\mu\nu} \\ =\eta^\mu_\beta \eta^\beta_\mu=4$$ Or equivalently since $$\eta^{\mu\nu}\eta_{\mu\chi}={\delta^\nu}_\chi\,$$ Where $\delta$ is Kronecker delta.
Substituting $\chi=\nu$ we get 4. Note that this only works because $\eta$ is symmetrical.
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$\begingroup$ In $\eta^{\mu\nu}\eta_{\mu\chi}=\delta^\chi_\nu$ you cannot have upper $\nu$ on the left and lower on the right. Ditto with $\chi$. Index notation is easy but not forgiving sloppyness. $\endgroup$– Kurt G.Feb 20 at 13:09
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$\begingroup$ Looks OK. Even better would be ${\delta^\nu}_\chi\,.$ You will soon deal with tensors where that matters. $\endgroup$– Kurt G.Feb 20 at 13:13
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$\begingroup$ @KurtG. Done, thank you very much for being so patient. $\endgroup$ Feb 20 at 13:19