# Does action really have to be Lorentz-invariant in SR?

From Landau & Lifshitz The Classical Theory Of Fields it is said:

To determine the action integral for a free material particle (a particle not under the influence of any external force), we note that this integral must not depend on our choice of reference system, that is, it must be invariant under Lorentz transformations.

This seems understandable. But in comments on this answer Ján Lalinský says that "there is no good physical reason why action should be invariant". Further he suggests another Lagrangian than that given in L&L, namely, if we denote L&L Lagrangian as $L_0$, then the example could be $L_0+Cv_x$, which is clearly anisotropic. Clearly, the equations of motion must not change with this Lagrangian, because $Cv_x$ is a total time derivative (of $Cx$).

On the other hand, in this answer Luboš Motl says that "the invariance of the action follows from special relativity – and special relativity is right (not only) because it is experimentally verified. [snip] If $S$ depended on the inertial system, so would the terms in the equations $\delta S=0$, and these laws of motion couldn't be Lorentz-covariant".

How could I connect L&L and Luboš's arguments with Ján's example? Can both sides be simultaneously right? They seem to contradict each other.

The Lubos answer is correct -- the action must be a Lorentz invariant. The Lagrangian need not be, so long as it can be written as $L = L_{\textrm{lorentz invariant}} + \dot{f}$ or some other such formulation that yields a Lorentz invariant action. This distinction between the action being Lorentz invariant and the Lagrangian being Lorentz invariant is important, since it is frequently a source of the anomalies (http://en.wikipedia.org/wiki/Anomaly_(physics)) that appear in QFT.
• That constant is not a scalar. It depends on the reference system it being $Cx(t_2)-Cx(t_1)$, where the $x$ coordinate is that of the used reference frame. Changing reference frame, you would find: $Cx'(t_2)-Cx'(t_1) \neq Cx(t_2)-Cx(t_1)$. Alternatively you may assume that the constant is a scalar (the same value in every reference frame), but, in this way, you get that the constant has the meaning of difference of coordinates only in a precise reference system. Such non-invariant but trivial boundary terms can always be added to a scalar action preserving the dynamical equations. – Valter Moretti Jan 31 '14 at 9:24
• If $S$ is not a scalar the solutions of $\delta S=0$ may be however covariant equations if $S$ is not a scalar because of the presence of non-scalar boundary terms. This is because the variational procedure is performed keeping fixed the the endpoints of the varied curves and boundary terms act on these endpoints only. However, these anomalous boundary terms can be dropped preserving the equations and making the action completely scalar. The situation is more delicate if the action includes derivatives of order greater than $1$ (like in the gravitational action $\int R \sqrt{g} d^4x$). – Valter Moretti Jan 31 '14 at 9:34