# Why Lagrangian is $L=\frac{1}{2}mv^2$ and not $mv^2$ for a free particle in an intertial frame? Both are proportional to the square of velocity

Landau writes the Lagrangian of a free particle in a second inertial frame as $$L(v'^{2})=L(v^2)+\frac{\partial L}{\partial v^2}2\textbf{v}\cdot{\epsilon},$$ and then it's written that the Lagrangian is in this case proportional to the square of velocity , and we write it as: $$L=\frac{1}{2}mv^2,$$ my question is: why not just $$L=mv^2$$? The latter case is proportional to the square of velocity as well.

• Jan 15 at 15:42
• As pointed out in the answer to the question why is there a $\tfrac{1}{2} in \tfrac{1}{2}mv^2$: Take F=ma and evaluate - both sides - the integral from position $s_0$ to position $s$ $$\int_{s_0}^sF \ ds = \int_{s_0}^sma \ ds$$ The following is independent of $F=ma$: integration of acceleration with respect to position: $$\int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2$$ Hence: $$\int_{s_0}^s F ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2$$ Jan 15 at 16:12
• In Landau's Mechanics force has not been introuced yet, and thus, maybe, this answer does not fit with his logical approach Jan 15 at 17:24

Both would work. It is just a matter of convention. Notice that if $$x_0$$ is an extremum of the function $$S(x)$$, it is also an extremum of $$\alpha \cdot S(x)$$ for constant $$\alpha \neq 0$$.