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As an answer to this post, I made an impression that if $\mathscr{L}$ were not a Lorentz scalar in Eq.$(1)$ (see below), then Eq.$(1)$ would not be covariant. But now I think that is wrong! I state the reason below why I think so. If confirmed, I will delete the answer there.


The Euler-Lagrange equation for a field $\phi_a$ ($a=1,2,3,...N$ counts that number of components that mix under Lorentz transformation) given by $$\frac{\partial \mathscr{L}}{\partial\phi_a}=\partial_\mu\Big(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\Big)\tag{1}$$ becomes $$\frac{\partial \mathscr{L}}{\partial\phi^\prime_a}=\partial_\mu\Big(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi^\prime_a)}\Big)\tag{2}$$ where $\phi^\prime=S_{ab}\phi_b$ denotes the transformation of the field $\phi_a$ under Lorentz transformation. As far as the $\mu$ index is concerned, those are contracted and should not change. Also, since $\mathscr{L}$ is a Lorentz scalar, that too doesn't change. Therefore, the covariance of $(1)$ is manifest! So it satisfies the postulate of special relativity.

Now consider a different equation which has nothing to do with anything familiar! Write an equation by replacing $\mathscr{L}$ in $(2)$ with a quantity which has one uncontracted Lorentz index (say, $\mathscr{F}_\mu$) or an index $a$ which has definite transformation under Lorentz transformation (say, $\mathscr{F}_a$): $$\frac{\partial \mathscr{F}_\mu}{\partial\phi_a}=\partial_\mu\Big(\frac{\partial \mathscr{F}_\mu}{\partial(\partial_\mu\phi_a)}\Big)\tag{3}$$ or $$\frac{\partial \mathscr{F}_a}{\partial\phi_b}=\partial_\mu\Big(\frac{\partial \mathscr{F}_a}{\partial(\partial_\mu\phi_b)}\Big)\tag{4}.$$

Question As far as covariance is concerned, wouldn't equations of the form $(3)$ and $(4)$ also be covariant?

If yes, it would be incorrect to say: "if $\mathscr{L}$ were not a Lorentz scalar but a Lorentz vector $\mathscr{F}_\mu$, or something like $\mathscr{F}_a$ equation of motions would not be covariant." In other words, covariance of the equation of motion does not require $\mathscr{L}$ to be a Lorentz scalar. Is this correct?

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    $\begingroup$ I believe the issue is with the action then. I mean, if you integrate the "vector lagrangian" componentwise you would have be a problem since in practice you would be summing up vectors at different tangent spaces, and you would get basis dependent results I believe. Now if you contracted the vector with a covector before integrating, well the the integrand is a scalar and you are back to the old story. That is my point of view on this right now. $\endgroup$ – user1620696 Oct 3 '19 at 23:16
  • $\begingroup$ I see. My point is that the reason why $\mathscr{L}$ cannot be a Lorentz vector cannot be argued by saying that "if it were so, Eq. $(1)$ would not be covariant." Is that part agreable? @user1620696 $\endgroup$ – SRS Oct 3 '19 at 23:33
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    $\begingroup$ That equation is not covariant in curved spacetime. But anyway, in flat spacetime you don't actually care whether $\mathcal{F}$ is a vector; you just have a bunch of actions, and require all of them to be extremized at the same time. $\endgroup$ – Javier Oct 4 '19 at 0:01

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