Why can't the Lagrangian for a free point particle depend on distance?

Question

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 $r^2$?

Emilio Pisanty · Accepted Answer

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.

Vladimir Kalitvianski · Answer

Lagrangian of a free point particle may depend on $vec{r}$, as a matter of fact, but this dependence is of no meaning and normally is reduces to a full time derivative of something that contains this dependence.
Lagrangian seves to obtain differential equations supplied with the initial conditions. Through the initial conditions you obtain solutions depending on the initial distance, initial velocity and time.
Free motion is an approximation of a motion in a relatively weak external force, which is just neglected for simplicity within a given period of motion in question.

Why can't the Lagrangian for a free point particle depend on distance?

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