Does ##x_\mu x^\mu## Equal ##(x_\mu)^2## in Einstein Notation?

[Silviu]

Hello! Does $(x_\mu)^2$ actually means $x_\mu x^\mu$ in Einstein notation?

 

[dextercioby]

Drop the book using the $x_{\mu}^2$.

 

[Silviu]

Drop the book using the $x_{\mu}^2$.

So this means they are the same, but $x_{\mu}^2$ is bad notation?

 

[Ibix]

@dextercioby is saying that $(x_\mu)^2$ is an illegal expression in differential geometry, analogous to asking what 1s + 1Hz is. It doesn't mean anything. Unless there's some context we're missing (for example Sean Carroll's lecture notes sstarts his discussion of EM with non-tensor all-lower index notation for Maxwell's equations before re-writing them in tensor form). Where did you see it?

 

[Orodruin]

Drop the book using the $x_{\mu}^2$.

Although I agree that it is really a bastard notation using this for $g_{\mu\nu}x^\mu x^\nu$, I would not categorically advice to drop such a book. You will find that many physics papers use a similar notation in the kinetic term for a field, i.e., $(\partial_\mu\phi)^2$. There really is only one thing that can mean and still be meaningful. Of course, the understanding of this is predicated on first having learned it properly ... In Schwartz's QFT book, he places all indices down with the initial statement that it should be subtextual that one should be considered contravariant and the other covariant.