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


or the following link:


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?


1 Answer 1


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. $$\begin{array} &A_\mu B^\mu &=& \eta_{\mu\nu} A^\mu B^\nu \equiv \displaystyle\sum_{\mu,\nu=0}^{3} \eta_{\mu\nu} A^\mu B^\nu = \eta_{00}A^0B^0 + \eta_{01}A^0B^1 + \cdots + \eta_{33}A^3B^3 \\ &=&+A^0B^0 -A^1B^1-A^2B^2-A^3B^3 \\ &=&+A^0B^0 +A_1B^1+A_2B^2+A_3B^3.\end{array}$$

  • $\begingroup$ 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: $\endgroup$
    – Condereal
    Feb 10, 2020 at 18:28
  • $\begingroup$ \begin{align} \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\varphi)}\right)&=\partial_{\mu}(\partial^{\mu}\varphi)\\ &= \partial_0\partial^0\varphi+\partial_1\partial^1\varphi+\partial_2\partial^2\varphi+\partial_3\partial^3\varphi\\ &= \partial_0\partial^0\varphi-\partial_1\partial_1\varphi-\partial_2\partial_2\varphi-\partial_3\partial_3\varphi \end{align} If that's the case, I think I've understood. $\endgroup$
    – Condereal
    Feb 10, 2020 at 18:29
  • $\begingroup$ Yes, that's the case. The latter you can also write in a more human-readable form as $\ddot{\varphi}-\nabla^2\varphi$. $\endgroup$ Feb 10, 2020 at 18:39
  • 1
    $\begingroup$ Yes of course, the notation you propose is less sofisticated (consequently readable) that the one I've written, but for the sake of understanding the procedure, I thought leaving the partial derivatives was instructive. $\endgroup$
    – Condereal
    Feb 10, 2020 at 18:44

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