# Lagrangian is isotropic in space

In Landau & Lifshitz Mechanics, while deriving the properties of Lagrangian of a free particle in inertial frame, he uses the following points $:$

1. As space is homogeneous in inertial frame, a particle will follow same law of motion at 10 m from origin as it would follow at 5 m. Since equation of motion is contained in $L$, hence $L$ should be independent of radius vector $\vec{r}$.

2. Similarly, since the $L$ of the particle should behave in the same manner regardless of the time I measure it, it is therefore independent of the current time $t$.

3. Now since space is isotropic, $L$ should be independent of velocity $\vec{v}$, and should in fact be a function of $\lvert\vec{v}\rvert^2$.

In point 3) I have a problem in conceptually understanding it. For me isotropic means that all directions are the same, so for a given position of a particle, if I measure it from some angle $\theta$ and again at some other angle $\phi$, then the $L$ obtained in both the cases will be same. But what I think the author is trying to say is that a particle moving in a certain direction given by $\vec{v_1}$ is the same as it's movement in another direction given as $\vec{v_2}$ if $\lvert\vec{v_1}\rvert = \lvert\vec{v_2}\rvert$

First of all, I don't think it should be same, because I don't think it is equivalent to isotropic of space. Morover if I have to say such a thing about $L$ dependence on $\lvert\vec{v}\rvert$, then I think I could argue it with homogeneity of space hence not requiring the isotropic principal (again I think isotropic is contained in homogeneity in general).

In a nutshell I want to understand the application of isotropic of space in inferring the $\lvert\vec{v}\rvert^2$ depedence of $L$.

The general idea is that if the laws governing the system are independent on some parameter that may be

1. distance of experiment from origin $r$ (not $\vec r$ by the way)

2. time of experiment $t$

3. direction of experiment $(θ, φ)$

then the Lagrangian should be independent on that parameter.

In the case of point 3, that means specifically that L might be any function of the magnitude of the velocity $\left| \vec v \right|$ , but must not depend on the direction of $\vec v$.

Of course, "any function of $\left| \vec v \right|$" and "any function of $v^2$" are mathematically equivalent.

As an aside, any such independence on some parameter implies, per Noether's wonderful theorem, to a conservation law : here 1) momentum, 2) energy, 3) angular momentum.

• so are you trying to say that by choice of direction, the the direction of $\vec{v}$ will be precieved differently and hence $L$ can't depend on that. This is acceptable to me, but this is true for every vector and hence the only thing I could say is L should be a function of a scalar. Then why specific scalar v? – Manish Kumar Singh Oct 16 '15 at 14:45
• Absolutely. And yes, that is actually true of other vectors as well. The classic lagrangian with an EM field is a function of $E^2$, $B^2$, and $\vec j \cdot \vec A$ only. – Nicolas Oct 16 '15 at 14:58