[NOTES/ME-08001]-Motion in Linearly Accelerated Frames

Basically there are two types of non inertial frames that are of interest.

• Frames accelerating with a constant acceleration;
• Rotating frames.

The Newton's Laws are not valid in non inertial frames. I cannot write  Force = Mass $\times$ Acceleration  in a non inertial frame. So the question is how do write the EOM in accelerating and rotating frames? The E OM will look different and what are new consequences of Newton's Laws being different in non inertial frames? The first case is simple. The examples are that of an lift, car or a train accelerating with a constant acceleration. So how do we get the EOM? I will always $K$ for an inertial frame. In most cases it will clear which frame is inertial. If it is not, we must identify an inertial frame. I assume that there is an observer $O$ who is performing an experiment and taking data. Then we have another frame $K{'}$ accelerating with a constant acceleration. Let there be an observer $O{'}$ in the frame $K{'}$. Let $\vec{x}(t)$ and $\vec{x}{'}(t)$ be the position vectors of a point at time $t$ as seen by the two observers $O$ and $O{'}$ respectively. Let the positions of the axes in the two frames coincide at time $t=0$. Also let $\vec{x}_O(t)$ be the position vector of the origin of $K{'}$ at time $t$ as seen by the observer $O$. \FigBelow{10,10}{90}{70}{me-fig-02005} Then it is obvious that \begin{eqnarray} \vec{x}(t)=\vec{x}_0(t) + \vec{x}{'}(t) \end{eqnarray} Differentiating the above equation two times we get \begin{eqnarray} \dd[\vec{x}(t)]{t}&=&\dd[\vec{x}_0(t)]{t} + \dd[\vec{x}{'}(t)]{t}\\ \DD[\vec{x}(t)]{t}&=&\DD[\vec{x}_0(t)]{t} + \DD[\vec{x}{'}(t)]{t}\\ \end{eqnarray} Therefore the accelerations of a body, $\vec{a}$ and $\vec{a}{'}(t)$, in the two frames are related by \begin{equation} \vec{a} = \vec{f}+\vec{a}{'}(t) \end{equation} where $\vec{f}$ denotes the acceleration of the frame $K{'}$ w.r.t. the frame $K$. Multiplying the above equation by the mass of the body gives % \begin{eqnarray} M \vec{a} &=& M \vec{f}+ M \vec{a}{'}(t)\\ \text{or} \qquad \qquad M\vec{a}{'} &=& M \vec{a}(t)- M \vec{f}\\ \end{eqnarray} Using Newton's Laws in the first frame $K$ and replacing $M\vec{a}(t)$ with force $\vec{F}$ we get \begin{equation} \text{or} \qquad \qquad M\vec{a}{'} = \vec{F}- M \vec{f}\\ \end{equation}
Pseudoforces

The second term $- M \vec{f} $ is called pseudo force. The EOM in the non inertial frame look similar but one has to include the pseudo force to the forces acting on the system. The pseudo forces to be included will depend on how the frame is accelerating. The pseudo force is equal to minus of mass times acceleration of the frame with respect to an inertial frame. You may ask how are pseudo forces different from real forces. Can one distinguish between real and pseudo forces experimentally? Before answering this question I have to briefly concepts of gravitational mass and inertial mass. It turns out that it is locally impossible to distinguish between gravitational forces and pseudo forces. This is because the both gravitational forces and pseudo forces in an inertial frame are proportional to the mass of the body. In practical terms it means that effect of gravitational forces in a given inertial frame can be reproduced by that coming from pseudo forces in an non inertial frame. So for example, EOM in uniform gravitational field is same as motion in a lift accelerating upwards without any gravitational forces. In this form it is known as Einstein's equivalence principle, and this statement holds for small regions of space. It should be noted that the pseudo forces on a body are always proportional to the mass of a body.