This is a page of the Peskin-Schroeder QFT textbook,
I am quite baffled by equation 3.16.
Why is it a natural generalization of the former Euclidean 3d case? How to derive 3.17 from 3.16? The problem is, how to interpreter $\partial^\mu$? Is it different from $\partial_\mu$? Namely, is it that $$(\partial^0, \partial^1, \partial^2, \partial^3 ) = (\partial_0, -\partial_1, -\partial_2, -\partial_3)~?$$
If I were to plug the formula 3.16 into 3.17 and do the calculation mechanically, I would have $\partial^\nu x^\rho$. What is it?
For a newbie like me, the biggest barrier is the upper and lower indices.
Best Answer
Since the calculation is straightforward (and a good exercise) I just summarize the main points:
Use the metric $g^{\mu\nu}$ to 'change' between co- (lower) and contravariant (upper indices) components, for more details see dual spaces. This already answer one question: Both are related via
$\partial^\mu = g^{\mu\nu}\partial_\nu$ (sum over $\nu$ is assumed - see Einstein convention).
or the other way around
$\partial_\mu = g_{\mu\nu}\partial^\nu$. (Note $g_{\mu\nu}$ is the inverse of $g^{\mu\nu}$, here (flat spacetime) it is the same matrix. Yes, in general, they are different because they are dual to each other.)
The only 'tricky' point in the calculation is the proper contraction of terms like $\partial^\nu x^\rho$:
Since $\partial^\nu = \frac{\partial}{\partial x_\nu}$ we have to use the metric once to perform the derivative in the "same space":
Only $\partial_\nu = \frac{\partial}{\partial x^\nu}$ can act on $x^\rho$ not $\partial^\nu$ (both $x$-indices are at the same position) with the result $\partial_\nu x^\rho = \frac{\partial x^\rho}{\partial x^\nu} = \delta_\nu^\rho$ (= Kronecker Symbol: only = 1 if $\nu = \rho$ otherwise = 0).
In total: $\partial^\nu x^\rho = g^{\nu\mu}\partial_\mu x^\rho = g^{\nu\mu}\delta_\mu^\rho= g^{\nu\rho}$.
Two last comments: keep in mind that the metric is symmetric, hence $g^{\mu\nu}=g^{\nu\mu}$, and remember that two partial derivatives always commute, i.e., $\partial_\nu \partial_\mu = \partial_\mu \partial_\nu$.