Non-inertial reference frame and Lagrange's equation of motion

Question

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?

Answer 1

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.)

Answer 2

1. Why does Lagrangian function change?

You are constructing two different Lagrangians because you are using two different coordinate systems to describe the system.

2. Why Lagrange's equation of motion remains valid in non-inertial reference frame?

Lagrangian mechanics is essentially Newtonian mechanics described in a covariant manner. Thus means with arbitrary coordinate systems. Thus the Euler-Lagrange eom remains valid in non-inertial frames (it is covariant) whilst the Newtonian eom is only valid in an inertial frame.

Answer 3

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 $\vec v=\vec 0$