# Deriving the action and the Lagrangian for a free massive point particle in Special Relativity

My question relates to

Landau & Lifshitz, 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 Lorentz 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 non-relativistic Newtonian mechanics is not invariant under Galilean transformations, if I am not mistaken. See e.g. this Phys.SE post.

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 for a non-relativistic free point particle was more detailed I think. See e.g. this Phys.SE post. Does the relativistic case need some more detalization?

• – Physiks lover Sep 22 '12 at 15:20
• Best not to use the same symbol $a$ for both the constant outside the integral and the lower integration bound. – WillO Jan 31 '17 at 21:45

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 $x^\mu(\tau)$.

• I have the same doubt as the question poster. In non-relativistic mechanics, the action is not invariant under Galilean group but the equations of motion are invariant. So, why do we require a more stringent requirement of the invariance of action under the symmetry group of the theory in the case of relativistic mechanics? One answer may be that it gives us a valid Lagrangian and that's all we need. But, it turns out too good in general. Even in field theories this stringent requirement gives us the Lagrangian. Is there any other reason for this requirement of Lorentz invariance of action? – user4235 Sep 22 '12 at 13:20
• 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

In SR the equations have the same form in all inertial frames. This implies that the action of a free particle is that shown by L&L.

For convenience, let us take the proper time $s$ of the massive particle as the integration variable in the action, with a Lagrangian given by $L=L(x^\mu,\dot{x}^\mu,s)$, where $\dot{x}^\mu=dx^\mu/ds$. We have to prove that $L=-a$ (constant).

The Lagrangian formalism implies that, by changing the coordinates $x^\mu \rightarrow x'^\mu$, the equations of motion obtained from the Lagrangian $L=L'(x'^\mu,\dot{x}'^\mu,s)$, regarded as a new function of the new coordinates, are equivalent to the equations of motion (the Euler-Lagrange equations) for the old coordinates.

In the case of a Poincarè transformation of coordinates, the new equations describe the motion in a new frame of reference and SR requires that they have the same form as the old equations. As usual in the Lagrangian formalism (as explained in L&L's Mechanics), this is possible if $L'(x'^\mu,\dot{x}'^\mu,s)$ differs from $L(x'^\mu,\dot{x}'^\mu,s)$ by the derivative of a function of the coordinates and the proper time,

$$L(x^\mu,\dot{x}^\mu,s)=L(x'^\mu,\dot{x}'^\mu,s)+\dot{F}'(x'^\mu,s).$$

It is apparent in this equation that in the old frame of reference $\dot{F}(x^\mu,s)=0$. As the old inertial coordinates are not special, SR requires that $\dot{F}'(x'^\mu,s)=0$ in the new (arbitrary) inertial frame. Thus, $L(x^\mu,\dot{x}^\mu,s)=L(x'^\mu,\dot{x}'^\mu,s)$, which means that the Lagrangian and the action are invariant.

As this equation is valid for any constant space-time translation $x^\mu \rightarrow x'^\mu=x^\mu+a^\mu$, $L$ does not depend on $x^\mu$. L&L's argument implies that $L$ does not depend on $\dot{x}^\mu$ (since $\dot{x}_\mu\dot{x}^\mu=-1$ is the only scalar that one can construct). Finally, $L$ does not depend on $s$ because the time is uniform for a free particle. Therefore, $L=-a$ is a constant.

Notwithstanding, note that L&L shows in their book that $\dot{F}(x^\mu)$ does appear in the Lagrangian of a charged particle in the absence of the electric field and the magnetic field (a free particle in the classical sense),

$$\dot{F}(x^\mu)= \frac{e}{c}A_\mu\dot{x}^\mu,$$

where the 4-vector potential $A_\mu$ is a "pure gauge",

$$A_\mu=\frac{\partial \chi}{\partial x^\mu}$$

for a scalar $\chi$.

In fact, gauge transformations introduce $\dot{F}$ in their Lagrangian of the charged particle interacting with the EM field without changing the equations of motion.