# Conserved quantity of a relativistic free Lagrangian for a Lorentz boost

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?

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.
$$^1$$ 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.