Deriving the action and the Lagrangian for a free particle in Relativistic mechanics

My question relates to

Landau, Classical Theory of Field, Chapter 2 - Relativistic Mechanics, paragraph 8 - The principle of least action.

As stated there, To determine the action integral for a free material particle, we note that this integral must not depend on our choice of reference system, that is, must be invariant under Lorenz transformations. Then it follows that it must depend on a scalar. Furthermore, it is clear that the integrand must be a differential of first order. But the only scalar of this kind that one can construct for a free particle is the interval $ds$, or $ads$, where a is some constant. So for a free particle the action must have the form $$S = -a\int_a^b ds\,.$$ where $\int_a^b$ is an integral along the world line of a particle between the two particular events of the arrival of the particle at the initial position and at the final position at definite times $t1$ and $t2$, i.e. between two given world points; and $a$ is some constant characterizing the particle.

For me this statements are inaccurate. (maybe it's because I have few knowledge from maths and physics. However.

1. Why should the action be invariant under Lorenz transformations? Is this a postulate or it's known from experiments. If this invariance follows from special theory of relativity, than how? Why the action should have the same value in all inertial frames? The analogue of the action in classical mechanics is not invariant under Galilei transformations, if I am not mistaken.

2. It is stated "But the only scalar of this kind that one can construct for a free particle is the interval." Why? Can't the Lagrangian be for example $$S = -a\int_a^b x^i x_i ds\,.$$ which is also invariant.

The derivation of the Lagrangian of a free particle in classical mechanics was more detailed I think. Does this one need some more detalization? Thanks a lot.

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– Physiks lover Sep 22 '12 at 15:20

1. Yes, the invariance of the action follows from special relativity – and special relativity is right (not only) because it is experimentally verified. All the equations of motion may be derived from the condition $\delta S = 0$, the action is stationary (which usually means it has the minimum value on the allowed trajectory/history among all trajectories/histories with the same initial and final conditions). 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 (note how this Lorentz is spelled; Lorenz also existed but it was a different physicist). Quite generally, you shouldn't think about "derivation of the action". When we work with the action at all, we are doing so because we view the action as the most fundamental expression – and we derive everything else out of it. In that context, we pretty much define a Lorentz-invariant theory as a theory determined by a Lorentz-invariant action.
2. Your integral is Lorentz-invariant but it is not translationally invariant under $x^\mu \to x^\mu + a^\mu$. So it's not Poincaré-invariant (the Poincaré symmetry unifies the Lorentz transformations and spacetime translations) and due to this violation, we also say that it disagrees with the laws of special relativity. You could also create other expressions, e.g. replace $x_\mu x^\mu$ in the integral by some extrinsic curvature invariant of the world line etc. Those terms could be made Poincaré-invariant. So the right claim is that the proper length of the world line is the only Poincaré-invariant functional that doesn't depend on any higher derivatives of the coordinates $x^\mu(\tau)$.
te action is the lenght of the curve in 4 dimension, the lenght of a curve is $ds= \sqrt_{g_{a,b}\dot x _{a} \dotx _{b}}$ – Jose Javier Garcia Sep 22 '12 at 13:58
Dear Lakshya, it's true that non-relativistic actions don't have to be Galileo-invariant while the equations are Galileo-covariant. However, this is only possible because the Galileo group is not simple, in the technical sense. Moreover, the variation of the action under Galilean transformation has a special form, $\Delta\vec v_i\cdot \vec P_i$ where $P$ is the total momentum. It only depends on the Galilean parameter, the velocity change, and conserved quantities, not on general fields, that's why it doesn't spoil the Galilean invariance of the equations. – Luboš Motl Sep 22 '12 at 14:27
Let me say more clearly what's special about the variation of the action under Galilean boosts. If the action is bilinear in velocities, $vTv/2$ where $T$ is a matrix of masses etc., then it transforms to $(v+\Delta v)T(v+\Delta v)/2$. Use the distributive law, you get the original term plus $\Delta v T v$ plus $\Delta v T \Delta v /2$. The last term is a constant - independent of the coordinates and velocities - and a shift of an action by constant disappears in the variations. The middle term is a total derivative. – Luboš Motl Sep 22 '12 at 15:35
Once it's integrated over time to get the whole action, you get $\Delta v T \cdot x|^b_a$, the difference of $x$ between before and after. It only depends on the initial and final conditions that are fixed under the variations, so they won't contribute to the equations of motion. – Luboš Motl Sep 22 '12 at 15:40