Why are Lagrangian densities and actions in Quantum Field Theory always Lorentz invariant? Newtons laws of motion are Galilean invariant. But the Newtonian Lagrangian and Newtonian action for a particle are not Galilean invariant. Similarly we want the Euler-Lagrange (EL) equations in quantum field theory to be Lorentz invariant. But how can we say that the Lagrangian (I mean Lagrangian density) should always be Lorentz invariant? Is there any deep reason for that?
One reason I thought of is since we quantize the classical relativistic Lagrangians which are already invariant we get invariant Lagrangians. But then I am not sure why even in classical field theory the Lagrangians are invariant.
I found in some places that Lagrangian must be invariant for the EL equations to be invariant. But that is not obvious. In some places it is said that action must be invariant and since the volume element $d^4x$ is invariant so must Lagrangian densities. But then why should action be invariant?
Edit: This answer explained it using group theory which I haven't studied yet. If there is an answer without using Group theory you can write it otherwise close this question.
 A: Actually, the Lagrangian for Newtonian mechanics is Gallilean invariant. While it does "change" under a tiny boost $x \mapsto x + \epsilon t$, it changes by a total time derivative. When a Lagrangian changes by a total derivative under a transformation, we still say the Lagrangian is "invariant" because adding a total derivative to a Lagrangian doesn't change its dynamics.
Likewise, the Lagrangian density in field theory is similarly invariant under a Lorentz transformation, because it too changes by a total derivative. (This is usually not discussed in intro lectures because they do changes of coordinates which hide this fact.) Just like in the Newtonian mechanics example, changing by a total derivative doesn't affect the dynamics. In other words, if the Lagrangian changes by a total derivative under tiny Lorentz transformations, then you know that Lorentz transforming a solution to the equations of motion will still itself be a solution to the equations of motion.
A: Your point is essentially that, whereas we recognise $\tfrac12m\dot{x}^2-V(x)$ changes under $x\mapsto x+vt$, interest in $\tfrac12\partial^\mu\phi\partial_\mu\phi-V(\phi)$ hinges on its invariance under Lorentz transformations of $x^\mu$. (For now, let's not go to more elaborate alternatives for either.) The comparison here is flawed, though, and the issue isn't that Galilean invariance is inaccurate; we could replace the first Lagrangian with a relativistic alternative, and it wouldn't address the point I'm about to make.
What you've encountered is really the difference between mechanics, for which the action is a time integral of a Lagrangian, and field theory, for which it's a multiple integral of a Lagrangian density. The quantum aspect is irrelevant.
Note that $x^\mu$ in the second example is actually analogous to $t$ in the first; going in the other direction, the equivalent of $x$ is $\phi$. The role of Lorentz transformations is to covariantly transform $\partial_\mu$, which is analogous to $\frac{d}{dt}$, which is invariant under Galilean transformations. By contrast, the $\phi$ Lagrangian isn't invariant under $\phi\mapsto\phi+v_\mu x^\mu$.
A funny addendum you'll appreciate: you can write down theories which do have this invariance; their solutions are called Galileons. (I've attended a conference talk on them, but it's such a little-known term it doesn't have a Wikipedia article, just research articles.)
