bio-physics-wiki

In the article on [[Galilean Transformation]] we found that the Lagrangian is a function of $v^2$ only. To find the exact form of the Lagrange function for a free particle, we use the invariance of the Lagrange function under [[Galilean Transformation]]. We transform from [[Inertial System|inertial system]] $K$ to the [[Inertial System|inertial system]] $K'$ where the particle has the velocity $v'=v + \mathbf{\varepsilon}$ for an infinitesimal velocity $\mathbf{\varepsilon}$. We compute the transformation
\begin{align}
L \rightarrow L'
\end{align}
\begin{align}
L'(v'^2)=L'(v + \mathbf{\varepsilon})^2=L(v^2 + 2v\mathbf{\varepsilon} + \mathbf{\varepsilon}^2)
\end{align}
We develop the series for $\mathbf{\varepsilon}$ around $v^2$ we have
\begin{align}
L'(v'^2)=L(v^2)+\frac{\partial L}{\partial v^2}(2v\mathbf{\varepsilon}+\mathbf{\varepsilon}^2) \label{dev}
\end{align}
where the rule for very small epsilon is $\mathbf{\varepsilon}^2=0$. We showed that the [[Galilean Transformation]] allows only constant velocities between different inertial systems, thus $\frac{\partial L}{\partial v^2}$ must be constant since \ref{dev} already contains the velocity $v$. Integrating
\begin{align}
\frac{\partial L}{\partial v^2}=const.
\end{align}
gives
\begin{align}
L(v^2)=a\cdot v^2
\end{align}
This Lagrangian is indeed invariant under the infinitesimal transformation $\mathfrak{v}$
\begin{align}
L'=a\cdot v'^2=a(v+\mathfrak{v})^2=av^2+2av\mathfrak{v}+a\mathfrak{v}^2
\end{align}
or
\begin{align}
L'=L+\frac{d}{dt}(2ar\mathfrak{v}+a\mathfrak{v}^2t)
\end{align}
By the [[Non-uniqueness of the Lagrange Function]] up to the addition of total derivatives $L$ and $L'$ are equal.
<div style="background:#FEF5CA;border:1px solid #797979;border-radius:10px;padding:5px 15px 5px 15px;">
The Lagrangian of a free particle is invariant under Galilean transformation
</div>


----
Further Reading:
* L. D. Landau & E. M. Lifschitz - Lehrbuch der Theoretischen Physik I: Mechanik