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}