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The Lagrangian for a relativistic free particle is of the form $L(\vec{v}^2)$ due to the definition of an inertial frame.

We must also demand invariance under Lorentz transformations:

\begin{equation}
v’=\frac{v+V}{1+\frac{vV}{c^2}}
\end{equation}

Keeping to first order in $V$ and using $(1+x)^n \approx 1+nx, x \ll 1$:

\begin{equation}
v’^2=(v+V)^2\left(1+\frac{vV}{c^2}\right)^{-2}=(v^2+2vV+V^2)\left(1-\frac{2vV}{c^2}+…\right)=v^2+2vV\left(1-\frac{v^2}{c^2}\right)
\end{equation}

The new Lagrangian $L(v’^2)$ is given by:

\begin{equation}
L(v’^2)=L(v^2)+\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)
\end{equation}

Applying the Lorentz transformation for new time $t’$:

\begin{equation}
t'(x,t)=\frac{t-\frac{Vx}{c^2}}{\sqrt{1-\frac{V^2}{c^2}}}
\end{equation}
\begin{equation}
\frac{dt’}{dt}=\frac{\partial t’}{\partial x}\frac{dx}{dt}+\frac{\partial t’}{\partial t}=\left(-\frac{\gamma V}{c^2}\right)v+\gamma=\gamma\left(1-\frac{vV}{c^2}\right)=\frac{1-\frac{vV}{c^2}}{\sqrt{1-\frac{V^2}{c^2}}}
\end{equation}

Note that $\frac{dx}{dt}=v$. Keeping to first order in V:

\begin{equation}
\frac{dt’}{dt}=\left(1-\frac{vV}{c^2}\right)\left(1-\frac{V^2}{c^2}\right)^{-1/2}=1-\frac{vV}{c^2}
\end{equation}

The new action is given by:

\begin{equation}S’=\int L(v’^2)\,dt’=\int \left[L(v^2)+\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)\right]\left(1-\frac{vV}{c^2}\right)\,dt
\end{equation}

Keeping to first order in V:

\begin{equation}
S’=\int\left[L(v^2)\left(1-\frac{vV}{c^2}\right)+\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)\right]\,dt
\end{equation}

The change in the integrand of the action should be a total derivative:

\begin{equation}
\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)-L(v^2)\frac{vV}{c^2}=\frac{d}{dt}f(x,t)
\end{equation}

Factoring out $v=\frac{dx}{dt}$, this suggests setting the rest to a constant. Furthermore, the $1-\frac{v^2}{c^2}$ term needs to be cancelled out which indicates $C=0$.

\begin{equation}
\frac{\partial L}{\partial (v^2)}2V\left(1-\frac{v^2}{c^2}\right)-L(v^2)\frac{V}{c^2}=0
\end{equation}

Simplifying:

\begin{equation}
\frac{\partial L}{\partial (v^2)}\left(1-\frac{v^2}{c^2}\right)-\frac{L(v^2)}{2c^2}=0
\end{equation}

The only nontrivial solution is:

\begin{equation}
L(v^2)=C\sqrt{1-\frac{v^2}{c^2}}
\end{equation}

where C is a constant.

We can deduce what $C$ is because in the limit $v \ll c$, we should extract the Lagrangian for a nonrelativistic free particle.

\begin{equation}
L_r(v^2)=C\left(1-\frac{v^2}{c^2}\right)^{-1/2} \approx C\left(1-\frac{v^2}{2c^2}\right)=C-\frac{Cv^2}{2c^2}
\end{equation}

Ignoring the additive constant which doesn’t affect the motion and setting the right term equal to the nonrelativistic Lagrangian, we obtain:

\begin{equation}
-\frac{Cv^2}{2c^2}=L_{nr}(v^2)=\frac{1}{2}mv^2
\end{equation}
\begin{equation}
C=-mc^2
\end{equation}

Thus the form of $L(v^2)$ for a relativistic particle is given by:

\begin{equation}
L=-mc^2\sqrt{1-\frac{v^2}{c^2}}
\end{equation}