Monday 20 January 2020

symmetry - 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 $L$.



Answer



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.


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