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?


2 Answers 2


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.)

  • $\begingroup$ Is this similar to, for example, form of del operator in Cartesian co-ordinate system is different than del operator in spherical co-ordinate system ? $\endgroup$
    – atom
    May 2, 2021 at 5:39
  • $\begingroup$ Kind of, except not really. It has more to do with semantics than anything else; the Lagrangian/Hamiltonian approach is just slightly more abstract than Newton's laws, and this makes the formalism more general. $\endgroup$
    – TLDR
    May 2, 2021 at 5:48
  • $\begingroup$ If for a simple pendulum, and for a specific time, value of Lagrangian function is 100 joules in my reference frame which I know (or assume) is inertial one. Next, I know my friend (who is in non-inertial frame w.r.t. me, for example rotating table) AT THE SAME TIME, also measuring value of the Lagrangian function of this same pendulum. So will he measure 100 joules or something else? $\endgroup$
    – atom
    May 2, 2021 at 5:58

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 $\vec v=\vec 0$


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