What is the Minkowski metric tensor's trace?

[LCSphysicist]

TL;DR Summary

I would appreciate if you explain to me how to get the Minkowski metric tensor's trace.

I am trying to follow the rule, that is, raising an index and the contract it.
Be $g_{\mu v}$ the metric tensor in Minkowski space.
Raising $n^{v \mu}g_{\mu v}$ and then, we need now to contract it.
Now, in this step i smell a rat (i learned this pun today, hope this mean what i think this means haha)
Can i simply say that $\mu$ is an index using Einstein notation? I am a little confused how to contract this and then reduced it to delta kronecker, which, in the end, will give us the trace equal four.

 

[Ibix]

Be $g_{\mu v}$ the metric tensor in Minkowski space.
Raising $n^{v \mu}g_{\mu v}$ and then, we need now to contract it.

If $g_{\mu\nu}$ is the metric then the inverse metric should be $g^{\mu\nu}$. If you are meaning the metric of flat space, typically that's denoted $\eta_{\mu\nu}$ and the inverse would be denoted $\eta^{\mu\nu}$. There's nothing wrong with using $g$ instead of $\eta$, but you need to use it consistently.

Apart from that, what you've written seems fine. If you want to think of it in several stages, first you would use $g^{\rho\mu}$ to raise an index, giving you $g^{\rho\mu}g_{\mu\nu}$, which is indeed $\delta^\rho_\nu$. Then you can contract over the upper and lower indices - i.e. you needed to set $\rho=\nu$, which (give or take using $g$ or $\eta$) is what you wrote. Writing the sums explicitly (so no summation convention implied) it's $\sum_{\mu=0}^4\sum_{\nu=0}^4g^{\mu\nu}g_{\mu\nu}$.

 

[mpresic3]

Carroll's textbook Spacetime and Gravitation discusses it, I think it is in the first chapter.

 

[mpresic3]

Carroll's textbook Spacetime and Gravitation discusses it, I think it is in the first chapter. And the solution is as you wrote the trace is four. Page 28 in Carroll's textbook