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.
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1$\begingroup$ Related: Why is there a $\frac 1 2$ in $\frac 1 2 mv^2$? $\endgroup$– Qmechanic ♦Commented Jan 15, 2022 at 15:42
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$\begingroup$ 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: $$ \int_{s_0}^s F ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 $$ $\endgroup$– CleonisCommented Jan 15, 2022 at 16:12
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$\begingroup$ In Landau's Mechanics force has not been introuced yet, and thus, maybe, this answer does not fit with his logical approach $\endgroup$– Paolo SecchiCommented Jan 15, 2022 at 17:24
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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 $\alpha \neq 0$.
answered Jan 15, 2022 at 15:40