Lagrangian in Non-inertial Frame

Recall Newtonian mechanics in non-inertial frame, a fictitious force is added in non-inertial frame. Let $\vec{a}$ be the acceleration of the non-inertial frame. The fictitious force is $$\vec{F}_{fictitious} = -m\vec{a}.$$ By Newton's second law, in an inertial frame $$\vec{F}=m\ddot{\vec{x}},$$ where $x$ is the acceleration of a point mass $m$ in the inertial frame. Adding fictitious force on both side, $$\vec{F}+\vec{F}_{fictitious} = m\vec{a} + \vec{F}_{fictitious} = m\vec{a} + (-m\ddot{\vec{x}}) = m\ddot{\vec{x}}',$$ where $x'$ is the acceleration of a point mass $m$ in the non-inertial frame. So, the equation of motion is actually the same. As the Lagrangian comes from the equation of motion, the Lagrangian should be the same.

Example

A ball thrown upward in an elevator accelerating upward. Determine its acceleration in the elevator frame.

Solution: Let $y$ be the height of the ball relative to the ground level that is assumed to be an inertial frame and $y'$ be the height of the ball relative to the elevator floor which is an non-inertial frame. Let $y(0)=0$ and $\dot{y}(0)=u_0$. In the inertial frame, the Lagrangian is $$L = \frac{1}{2}m\dot{y}^2 - mgy.$$ By the Lagrange's equation, the equation of motion is $$\ddot{y}(t)=-g.$$ Hence, we have $$\begin{align} \ddot{y}(t) &= -g\\ \dot{y}(t) &= u_0 - gt\\ y(t) &= u_0t - \frac{1}{2}gt^2. \end{align}$$ In the accelerating frame, from the solution of the inertial frame, we have $$\begin{align} \ddot{y}'(t) &= -(g+a) \\ \dot{y}'(t) &= u_0-(g+a)t = \dot{y}(t)-\frac{1}{2}at\\ y'(t) &= u_0t - \frac{1}{2}(g+a)t^2 = y(t)-\frac{1}{2}at^2. \end{align}$$ The Lagrangian is $$\begin{align} L &= \frac{1}{2}m\dot{y}^2 - mgy\\ &= \frac{1}{2}m(\dot{y}'+at)^2 - mg(y' + \frac{1}{2}at^2) \end{align}$$ Substitute into the Lagrange's equation, we have $$\ddot{y}'(t) = -(g+a),$$ which is the acceleration in the non-inertial frame.

The approach was to write down the Lagrangian using an inertial frame generalized coordinates. Then express the inertial frame generalized coordinates with non-inertial frame coordinate. Then put the Lagrangian into the Lagrange's equation with non-inertial frame generalized coordinates.

Checkpoint

A simple pendulum in which the fixed support is moving vertically upward with position given by $h(t)$. Find the equation of motion in the fixed support frame.

$$\begin{align} x &= l\sin\theta \\ y &= h(t) - l\cos\theta \end{align}$$ Take the potential energy as zero at $y=0$. The Lagrangian is $$\begin{align} L &= \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - mgy \\ &= \frac{1}{2}m(l^2\dot{\theta}^2 + 2l\dot{\theta}\sin\theta\dot{h}(t) + (\dot{h}(t))^2) - mg(h(t)-l\cos\theta)\\ \end{align}$$ $$\begin{align} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) &= \frac{\partial L}{\partial x}\\ \ddot{\theta} &= -\frac{g+\ddot{h(t)}}{l}\sin\theta \end{align}$$