Lagrangian in non-inertial coordinates, example question

Question

I recently had a problem on a quiz for a classical mechanics course that looked something like this:

A bead of mass $m$ is constrained to move on a rod that is tilted at a fixed angle $\theta$ with respect to the $x$-axis. The rod is accelerated along the $y$-axis with constant acceleration $a$, as shown in the diagram. Ignore gravity in your solution.

My question, and confusion while solving this problem, boiled down to what to do with the acceleration along the $y$-axis, and how to set up my Lagrangian to account for this. Defining $\mathscr{L} = T-V$, clearly $V =0$, as we are ignoring gravity in the problem. My issue lies with how to define $T$: is it possible to make $r$ a non-inertial coordinate such that $T = \frac{1}{2}m\dot{r}^2$, or do I have to make sure that my kinetic energy is written in inertial coordinates, that is, $T = \frac{1}{2}m(\dot{x}^2+\dot{y}^2) = \frac{1}{2}m(\dot{r}^2\cos^2(\theta)+(\dot{r}\sin(\theta)+at)^2))$

More generally, how does using non-intertial coordinates affect the standard formulation of the Lagrangian (and Hamiltonian)?

Answer 1

To see how acceleration-induced pseudo forces enter into Lagrangians consider a partcle that has position $x(t)$ measured from a moving reference point $X(t)$ (the floor of a lift say) with $\ddot X=a$. The particle's position in an inertial frame is therefore $x_{\rm inertial}=x+X$ and so it has action integral
$$ S=\int \frac 12m \left(\frac{d(x(t)+X(t))}{dt}\right)^2dt \\ = \int\left( \frac 12 m \dot x^2 + \dot x m \dot X+ \frac 12 m\dot X^2 \right)dt $$ We can ignore the $\dot X^2/2$ as it does not affect Lagrange's eqations for $x$. We can also integrate by parts to shift $m\dot x \dot X\to -m x\ddot X= -mxa $ because total derivatives do not affect anything either. What is left is $$ S= \int\left( \frac 12 m \dot x^2 - x m a\right)dt, $$ which is the action for a particle in a "gravitational" potential $V(x)=max$.