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I have looked thorough the derivation of Lagrange equations in Landau and Lifshitz, Vol 1, $\S 3$, p.5. They argues that the lagrangian of a free particle cannot explicitly depend on position vector $\vec{r}$ because of the isotropy and homogeneity of space. But why can't it depend on $r^2$? (They do not explain why this cannot be possible.) Or say a power of $r^2$?

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    $\begingroup$ The confusion of the OP is that the Lagrangian must depend on the distance only because of isotropy and homogeneity; it cannot depend on the direction $\hat n$ and thus on the vector $\vert r$. $\endgroup$ – ZeroTheHero Mar 16 '18 at 12:36
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Because of the homogeneity of space.

If it depended on $r^2$, then the origin with respect to which you're taking that distance $r$ would have a privileged status in the theory. For a free particle this cannot happen, so you cannot have that dependence.

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