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Why does Lagrangian of free particle depend on the square of the velocity ? For example, $L(v^4)$ also doesn't depend on direction of $v$.

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Your question is answered here: –  nervxxx May 5 '13 at 13:16
Also related: –  Qmechanic Mar 10 '14 at 8:11

2 Answers 2

The Lagrangian should not only be independent of the direction of $\vec{v}$ but it should also change correctly under a Galilean transformation. For instance, if $K$ and $K'$ are two frames of reference with a relative velocity $\vec{V}$ then the two Lagrangians $L$ and $L'$ should differ only by a total time derivative. If $L$ is a function of fourth power of velocity then $v'^4 = (\vec{v} + \vec{V})^4 = v^4 + \{V^4 + (\dot{\vec{x}}\cdot\vec{V})^2 + 2v^2V^2 + 2(\dot{\vec{x}}\cdot\vec{V})v^2 + 2(\dot{\vec{x}}\cdot\vec{V})V^2\}$. The term in the curly brackets cannot be expressed as a total time derivative. If $L$ was $v^2$, it would have been possible.

Galilean invariance forces Lagrangian to be a quadratic function of velocity. You may want to read section 4 of Landau and Lifshitz's Mechanics to understand this point better.

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The written result of raising to 4th power is wrong! –  TMS Jun 5 at 11:54

I think that the reason is merely a matter of convenience to use |v|^2 instead of e.g. |v|. Not until later in the text they prove that The Lagrangian is linearly proportional to |v|^2

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