Lagrangian Function

1. For a free particle It is easy to figure out L is a function of ${\dot{q}}^2$, but what exactly it is? In the Galileo’s relativity prospect:

$$\begin{array}{r}L({\dot{q^{\prime }}}^2)=L({\dot{q}}^2+2\dot{q}v+v^2)\\ L({\dot{q}}^2+2\dot{q}v+v^2)-L({\dot{q}}^2)=\frac{dF}{dt}\end{array}\text{\hspace{1em}(1)(2)}$$

As $\frac{dF}{dt}$ is proportion to $\dot{q}$, $\frac{\partial L}{\partial {\dot{q}}^2}$ is a constant. Thus,

$$L=\frac{1}{2}m{\dot{q}}^2\text{\hspace{1em}(3)}$$

If it is with the Relativity cinerio. $dt$ is not a well defined thing. 2. For a system of particles

$$L=\sum _a\frac{1}{2}{m}_a{v}_a^2-U(r_1,r_2,\dots )$$

time is homogenius and isotropic. Here, the Euler-Lagrange equation could help define the potential.

$$F=\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}-\frac{\partial U}{\partial r_a}$$

While it is not closed, we can consider A and B as a whole.and we want to extract $L_A$ from $L$, we could use $T_b[q_b(t),\dot{q_b}(t)]$ to exclude $q_a$ from the system. So for particle A, the kinetic energy for particle b isonly a time related quantity so it could be exclude from A’s Lagrange function.

Related

• Galilean Transformation
• Lorentz Transformation