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Peskin and Schroeder give a brief outline of Lagrangian field theory on page fifteen in their Quantum Field Theory book, where they write:

Lagrangian Field Theory

The fundamental quantity of classical mechanics is the action, $S$, the time integral of the Lagrangian, $L$. In a local field theory the Lagrangian can be written as the spatial integral of a Lagrangian density, denoted by $\mathcal{L}$, which is a function of one or more fields $\phi(x)$ and their derivatives $\partial_\mu\phi$. Thus we have

$$ S = \int Ldt = \int\mathcal{L}(\phi,\partial_\mu\phi)\ d^4x.$$

Since this is a book on field theory, we will refer to $\mathcal{L}$ simply as the Lagrangian. The principle of least action states that when a system evolves from one given configuration to another between times $t_1$ and $t_2$, it does so along the “path” in configuration space for which $S$ is an extremum (normally a minimum).

In relativistic classical field theory, I'd expect the Lagrangian density to be integrated over all space and time so that no point or boundary is more privileged than any other. So what do we assume about the fields, if at all, that enables the Lagrangian density to be integrated between two fixed times, while maintaining a classical field theory that's still relativistic?

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  • $\begingroup$ The general principle is that you need to specify boundary conditions for the field, to define variations at all. For example, you could take any four-dimensional subvolume of spacetime and fix the field on the boundary of that subvolume. In this case, Peskin is choosing $\mathbb{R}^3 \times [t_1, t_2]$. The boundaries are at infinite spatial coordinates (where the boundary condition is just that the field goes to zero) and at the times $t_1$ and $t_2$ (where the field is specified). That's not a Lorentz invariant choice, but it's still a valid choice to make. $\endgroup$
    – knzhou
    Commented Sep 11, 2022 at 4:34
  • $\begingroup$ But the results that Peskin derives later are still Lorentz invariant because either (1) he just uses this as a stepping stone to derive equations of motion, which are Lorentz invariant, or (2) he computes vacuum-to-vacuum transition amplitudes which involve sending $t_1 \to -\infty$, $t_2 \to \infty$. $\endgroup$
    – knzhou
    Commented Sep 11, 2022 at 4:36
  • $\begingroup$ @knzhou "you could take any four-dimensional subvolume of spacetime and fix the field on the boundary of that subvolume" is a key point: and unless I'm mistaken a property of the geometry of flat space time rather than the field in particular? $\endgroup$
    – Physiks lover
    Commented Sep 11, 2022 at 5:24

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The answer to OP's title question is essentially 'Always'. Yes, OP is correct: In a relativistic field theory, the choice of integration region $V\to V^{\prime}$ is transformed by Poincare transformations of spacetime. Poincare transformations can be viewed as quasi-symmetries of the action $S_V=\int_V\!d^4x~{\cal L}$, so Noether's first theorem does still apply.

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