Non-inertial reference frame and Lagrange's equation of motion Newton's equation of motion (F=ma) is modified when used from within non-inertial reference frame. I thought same is true for Lagrange's equation of motion also. But I got confused when I read in Landau's Mechanics book that when a system is observed from within non-inertial reference frame, its Lagrangian function changes (as compared to Lagrangian function of the same system in an inertial frame) but Lagrange's equation of motion remains valid.
Lagrangian function is a scalar quantity. And Scalar quantity does not change/modify if either measured from within inertial reference frame or non-inertial reference frame - e.g. temperature of an object.
(1) Why does Lagrangian function change ?
(2) Why Lagrange's equation of motion remains valid in non-inertial reference frame ?
Anybody there to clear my confusion ?
 A: It might help to answer your question by analogy.  In short,
the Euler-Lagrange equations can be likened to the general equation $\nabla f|_M = 0$ used to find (constrained) extrema, while Newton's laws are analogous to a precise form of $\nabla f|_M = 0$, in a specially chosen coordinate system (e.g. the coordinate system where $\nabla f = 0$ has the simplest possible form.)  Using a non-inertial reference frame is analogous to using coordinates for which the equation $\nabla f|_M=0$ fails to have the desired properties, but does not influence the outcome of the optimization (or the underlying algorithm, whether Lagrange multipliers or something else.)
A: Lets say you have coordinate system that rotate about the z-axis with the angle $~\Omega\,\tau~$ where $\Omega~$ is constant, thus the position vector relative to inertial system is
$$\vec R=\left[ \begin {array}{ccc} \cos \left( \Omega\,\tau \right) &-\sin
 \left( \Omega\,\tau \right) &0\\ \sin \left( \Omega
\,\tau \right) &\cos \left( \Omega\,\tau \right) &0
\\ 0&0&1\end {array} \right] \,\begin{bmatrix}
  x \\
  y \\
  z \\
\end{bmatrix}$$
from here you can obtain the kinetic and potential energy
$$T=\frac m2\,\vec{\dot{R}}\cdot \vec{\dot{R}}\\
U=m\,g\,\vec{R}_y$$
and the Lagrange  function $$\mathcal{L}=T-U$$
but if the   coordinate system doesn't rotate $\Omega=0$ the Lagrange  function $\mathcal L$  is not the same as above, so the statement that the Lagrange  function doesn't change  , is in general can't be true ?
if the coordinate system move with constant  velocity vector
$\vec v=\left[v_x~.v_y~,v_z\right]^T~$ then the position vector is now
$$\vec R=\left[ \begin {array}{c} x-v_{{x}}\tau\\ y-v_{{y}}
\tau\\ z-v_{{z}}\tau\end {array} \right] 
$$
in this case the equation of motion are equal to the equations of motion where $\vec v=\vec 0$
