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I have small question about Classical Mechanics, Lagrangian of a free particle. I just read Deriving the Lagrangian for a free particle blog. So, if I am correct, we have, that the free particle moves with a constant velocity in inertial frame and also that

$$ \vec{0}~=~\frac{d}{dt}\frac{\partial L}{\partial \vec{v}} ~=~\frac{d }{dt} \left(2\vec{v}~\ell^{\prime}\right) $$

$\ell^{\prime} $ means $\frac{\partial L}{\partial v^2}$ .Hence $$ \vec{c}~=~\left(2\vec{v}~\ell^{\prime}\right) $$

So, this two statements mean that $\ell^{\prime}$ is constant, so $$L~=~ \ell(v^2)~=~\alpha v^2+\beta, $$

Isn't this enough to derive the lagrangian of a free particle. If yes (but I'm sure no) why did Landau use the Galilean transformation formulas etc to derive that formula.

Thanks a lot!

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up vote 6 down vote accepted

The answer is No, OP's argument(v1) is not enough to derive the Lagrangian for a non-relativistic free particle. It is true that the constant of motion mentioned by OP

$$\vec{c}~:=~\frac{\partial L}{\partial \vec{v}}~=~2\vec{v}~\ell^{\prime}$$

does not depend on time $t$. (It is in fact the canonical/conjugate momentum, which in general is different from the mechanical/kinetic momentum $m\vec{v}$.) However, $\vec{c}$ could still depend on e.g. the initial velocity of the particle (which is also the velocity $\vec{v}$, since we know that the velocity is constant for a free particle, cf. e.g. the first part of the linked answer).

This is perhaps best illustrated by taking a simple example, say

$$L~=~\ell(v^2)~=~ v^4.$$


$$\vec{c}~=~2\vec{v}~\ell^{\prime} ~=~4\vec{v}~v^2,$$

which is a constant of motion, as it should be.

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Right, other functions of $v^2$ can do it, too. After all, the relativistic Lagrangian proportional to $-mc^2\sqrt{1-v^2/c^2}$ is an example of that. For small $v$, it may be expanded as $mv^2$ plus corrections proportional to any other even power of $v$. – Luboš Motl Apr 6 '12 at 9:13
OK, now I really understood what was the problem! Thank you very much! – achatrch Apr 6 '12 at 16:28

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