Conserved quantity of a relativistic free Lagrangian for a Lorentz boost

Question

Let $$L~=~-mc^2\sqrt{1- \frac{|\textbf{v}|^2}{c^2} },$$ where $\textbf{v}$ is the usual velocity of the particle in a fixed inertial frame. Then, this is the Lagrangian for a relativistic free particle. Now what does it mean by "the conserved quantity for a Lorentz boost"? Does it mean that the particle is boosted by some fixed velocity and there comes out a quantity that is preserved? I cannot get the exact meaning of the phrase. Could anyone please explain to me?

Answer 1

This is easiest to see from the Hamiltonian formulation, cf. this Phys.SE post. Below we will also give the non-relativistic expressions for comparison, because it is interesting and somewhat subtle, cf. this Phys.SE post, and because OP asked the analogous non-relativistic question earlier.

I) The Hamiltonian is the kinetic energy, i.e the energy minus the rest energy$^1$ $$\begin{align} H~=~&p^0c-mc^2~=~\sqrt{{\bf p}^2c^2+m^2c^4}-mc^2\cr \longrightarrow&\quad \frac{{\bf p}^2}{2m}\quad\text{for}\quad c\to \infty.\end{align} \tag{1} $$

The 3 boost generators $B^i$ are part of the 6 Lorentz generators $$\begin{align}B^i~=~&\frac{J^{0i}}{c} ~=~tp^i-x^i\frac{p^0}{c}\cr \longrightarrow &\quad tp^i-mx^i\quad\text{for}\quad c\to \infty,\cr & i~\in~\{1,2,3\}.\end{align}\tag{2}$$ The pertinent infinitesimal quasi-symmetry transformations are generated by the boosts $$\begin{align}\delta x^i ~=~&\{ x^i , {\bf B}\cdot \delta {\bf v} \} ~=~t~\delta v^i - \frac{p^i}{p^0c}{\bf x}\cdot \delta {\bf v}\cr \longrightarrow &\quad t~\delta v^i\quad\text{for}\quad c\to \infty, \tag{3}\cr \delta p^i ~=~&\{ x^i , {\bf B}\cdot \delta {\bf v} \} ~=~\frac{p^0}{c}\delta v^i\cr \longrightarrow &\quad 0\quad\text{for}\quad c\to \infty,\tag{4}\cr \delta t~=~&0.\tag{5}\end{align}$$ The Hamiltonian Lagrangian $$\begin{align} L_H ~=~&{\bf p}\cdot \dot{\bf x} - H \cr \longrightarrow &\quad {\bf p}\cdot \dot{\bf x} - \frac{{\bf p}^2}{2m} \quad\text{for}\quad c\to \infty \end{align}\tag{6}$$ has a quasi-symmetry $$\begin{align}\delta L_H~=~&\frac{d}{dt}\left( \frac{m^2c}{p^0} {\bf x}\cdot \delta {\bf v} \right)\cr \longrightarrow &\quad \frac{d}{dt}\left( m {\bf x}\cdot \delta {\bf v} \right) \quad\text{for}\quad c\to \infty. \end{align}\tag{7}$$ One may check that the corresponding Noether charges are precisely the boost generators (2).

II) The corresponding Lagrangian formulation$^1$ $$\begin{align}L~=~&mc^2\left(1-\sqrt{1-\frac{\dot{\bf x}^2}{c^2}}\right)\cr \longrightarrow &\quad \frac{1}{2}m\dot{\bf x}^2\quad\text{for}\quad c\to \infty,\end{align}\tag{8}$$ has infinitesimal boost quasi-symmetry $$\begin{align}\delta x^i ~=~&t~\delta v^i - \frac{\dot{x}^i}{c^2}{\bf x}\cdot \delta {\bf v}\cr \longrightarrow &\quad t~\delta v^i\quad\text{for}\quad c\to \infty, \tag{9}\cr \delta t~=~&0,\tag{10}\end{align}$$ and conserved boost charges $$\begin{align}B^i~=~&m\frac{t\dot{x}^i-x^i}{\sqrt{1-\frac{\dot{\bf x}^2}{c^2}}} \cr \longrightarrow &\quad m(t\dot{x}^i-x^i)\quad\text{for}\quad c\to \infty.\tag{11}\end{align}$$ This is left as an exercise to the reader. One way is to integrate out the 3-momentum ${\bf p}$ from the Hamiltonian formulation in section I. See also this related Phys.SE posts and links therein.

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$^1$ We have removed the rest energy, which is a constant, i.e. a total time derivative, in order to be able to go to the non-relativistic limit.