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In particle physics, for example, we add gauge fields, look for the covariant derivative and so on. All to find the LI form of the Lagrangian. Why do we need the LI form? My impression is that when quantities are LI we equalize conditions, but I don't see how.

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    $\begingroup$ If by LI you mean Lorentz Invariant, then we look for LI stuff simply because they're invariants. You don't want to build a theory upon something that will change if you apply a Lorentz transformation on it. $\endgroup$ Commented Sep 26, 2016 at 17:28
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    $\begingroup$ Could you elaborate a bit? As it is now, it seems that you are referring to a relation between Lorentz and gauge invariance that is not correct. $\endgroup$
    – DelCrosB
    Commented Sep 26, 2016 at 17:28

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Lorentz invariant quantities do not change from one inertial frame to another. For example, the rest mass of a particle is Lorentz invariant, and since it doesn't change from one inertial frame to another, and therefore is an intrinsic property associated with that particle.

Moreover, we require the Lagrangian (more precisely, the action) to be Lorentz invariant so that the Euler-Lagrange equation (both sides of which has equal number of uncontracted Lorentz indices and transform in the same manner under Lorentz transformation) derived for the fields be Lorentz covariant i.e., do not change in form from one inertial frame to another. Special relativity teaches us that the laws of physics should be covariant.

Edit: We replace ordinary partial derivatives in the Lagrangian by covariant derivatives to ensure gauge invariance (in gauge theories i.e., quantum field theories which are invariant under various local symmetries such as local $U(1)$ symmetry in QED, local $SU(3)$ symmetry in strong interactions and so on.). This has nothing to do with Lorentz invariance. Lorentz invariance is essential for any relativistic quantum field theory. Gauge invariance is required in gauge theories to ensure renormalizability.

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  • $\begingroup$ Nitpick: the Euler-Lagrange equations have as many uncontracted Lorentz indices as the fields they correspond to. For example, if you had a rank-2 tensor field in your Lagrangian, the corresponding EOMs would have two uncontracted indices. This is a consequence of the quotient rule for tensors. $\endgroup$ Commented Sep 26, 2016 at 19:33
  • $\begingroup$ @Michael Seifert You're right. I had vector fields in mind. I'll edit the answer. $\endgroup$
    – SRS
    Commented Sep 26, 2016 at 19:38

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