Euler-Lagrange equations in QFT and metric signs

I'm having a probably dumb problem with the Euler-Lagrange equations and the dot-product in Minkowski spacetime. I know that some objects are defined naturally with lower-indexes, e.g. $$\partial_{\mu}$$ which in components reads $$\partial_{\mu}=(\partial_0,\partial_1,\partial_2,\partial_3)$$, so that $$\partial^{\mu}=\left(\begin{array}{c} \partial_0\\ -\partial_1\\ -\partial_2\\ -\partial_3\end{array}\right)$$ if $$sgn(\eta)=(+,-,-,-)$$. On the other hand, there are objects that are naturally defined as 4-vectors with upper-indexes, e.g. $$x^{\mu}$$ such that $$x^{\mu}=\left(\begin{array}{c} x^0\\ x^1\\ x^2\\ x^3\end{array}\right)$$ and then $$x_{\mu} = (x^0,-x^1,-x^2,-x^3)$$. Then, of course we will have that $$\partial_{\mu}x^{\mu} = 4$$ and not $$\partial_{\mu}x^{\mu} = 1-3$$. My problem comes now with the Euler-Lagrange equations $$\frac{\partial\mathcal{L}}{\partial\varphi}-\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\varphi)}\right)=0$$. I would say without thinking that:

$$\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\varphi)}\right) = \partial_0\left(\frac{\partial\mathcal{L}}{\partial(\partial_{0}\varphi)}\right)-\partial_i\left(\frac{\partial\mathcal{L}}{\partial(\partial_i\varphi)}\right)$$

But that doesn't seem to be the case. See for example in page 32 in the following link

https://people.phys.ethz.ch/~babis/Teaching/QFT1/qft1.pdf

https://physicspages.com/pdf/Lancaster%20QFT/Lancaster%20Problems%2012.05.pdf

where both authors try to derive the Schrödinger equation from the Lagrangian. If you put the minus sign in the tensor contraction as I did, it is impossible to reach Schrödinger's equation. It looks like the correct calculation would be:

$$\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\varphi)}\right) = \partial_0\left(\frac{\partial\mathcal{L}}{\partial(\partial_{0}\varphi)}\right)+\partial_i\left(\frac{\partial\mathcal{L}}{\partial(\partial_i\varphi)}\right)$$

Does anyone know why this is so?

The correct relation is $$\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\right)= \partial_0 \left(\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}\right)+\partial_i \left(\frac{\partial \mathcal{L}}{\partial(\partial_i \phi)}\right).$$
Essentially you're asking if $$A_\mu B^\mu \overset{?}{=} A_0 B^0 + A_i B^i,$$ or $$A_\mu B^\mu \overset{?}{=} A_0 B^0 - A_i B^i;$$ and the correct answer is, of course, the former.
The way to see it is to write it down, explicitly, with the metric. $$\begin{array} &A_\mu B^\mu &=& \eta_{\mu\nu} A^\mu B^\nu \equiv \displaystyle\sum_{\mu,\nu=0}^{3} \eta_{\mu\nu} A^\mu B^\nu = \eta_{00}A^0B^0 + \eta_{01}A^0B^1 + \cdots + \eta_{33}A^3B^3 \\ &=&+A^0B^0 -A^1B^1-A^2B^2-A^3B^3 \\ &=&+A^0B^0 +A_1B^1+A_2B^2+A_3B^3.\end{array}$$
• Ok, so, if I understand correctly, the thing with the scalar field with Lagrangian $\mathcal{L}=\frac{1}{2}(\partial_{\mu}\varphi)(\partial^{\mu}\varphi)$ is that $\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\varphi)}=\partial^{\mu}\varphi$ and then: – Condereal Feb 10 at 18:28
• Yes, that's the case. The latter you can also write in a more human-readable form as $\ddot{\varphi}-\nabla^2\varphi$. – idiot stroszek Feb 10 at 18:39