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## Fictitious forces in linearly accelerated systems

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A non-inertial reference frame is a frame of reference that undergoes acceleration with respect to an inertial frame. Fictitious forces explain movement of bodies in non-inertial reference frames. In this article I discuss linearly accelerated systems.

### Newton's laws of motion

Newton's laws of motion are three laws that describe the relationship between the motion of an object and the forces acting on it. Forces, positions, velocities and accelerations are vectors.

1. Principle of inertia
An object either remains at rest or continues to move at a constant velocity, unless it is acted upon by an external force: $$\pmb F = 0 \iff \frac{d\pmb v}{dt} = 0 \tag {1}$$ $F$ is the "net" force, i.e. the vectorial sum of all forces acting on the body. $v$ is the velocity of the body.
2. Force law
The acceleration $a$ of the body is proportional to the force: $$\pmb F = m \frac{d\pmb v}{dt} = m \pmb a \tag {2}$$ $F$ is the "net" force, i.e. the vectorial sum of all forces acting on the body, $m$ is the mass of the body.
3. Action-reaction law
When one object exerts a force on a second object, that second object exerts a force that is equal in magnitude and opposite in direction on the first object: $$\pmb F_{12} = - \pmb F_{21} \tag {3}$$ This equation is valid for any pairs of forces between the two bodies. The third law is often misinterpreted. Action and reaction are exerted on different objects and so don't cancel. When two bodies interact, they exert equal and opposite force on each other.

### Forces

There are only 4 fundamental forces in the universe:

1. Gravitational force: two masses attract each other over arbitrary distances.
2. Electromagnetic force: electrically charged particles attract or repell each other over arbitrary distances. They are responsible for so called contact forces between objects.
3. Strong nuclear force: holds atoms together despite repelling electrical force. It is effective only in the realm of atoms.
4. Weak nuclear force: is responsible for the radioactive decay of atoms.

In this article I will consider only contact forces and gravitation. As an example of contact forces consider two touching blocks 1 and 2 with the respective masses $m_1$ and $m_2$. The force $F$ is applied to the block 1 and both blocks move to the right.

We can consider both blocks together as one object with the mass $m = m_1 + m_2$. This composite object is moved by the external force $F$ such that the acceleration is $a = F / (m_1 + m_2)$. Now let us look on the individual blocks. The object 1 exerts the force $F_{12}$ on the object 2. Because the object 2 moves with acceleration $a$, the force is $F_{12} = m_2 a$. The object 2 exerts the reaction force $F_{21} = - F_{12} = - m_2 a$ on the object 1. The total force on object 1 is $F_1 = F - F_{21} = (m_1 + m_2) a - m_2 a = m_1 a$ and corresponds to the Newton's law for the object 1 alone.

### Weight

A body of mass $m$ on Earth is subject to the gravitational force $G = - m g, g = 9.81 \gt 0$ that is directed downwards and is vertical to the Earth surface. (The direction upwards is taken positive.)

Since the person in the figure is at rest the surface must exert the same force on the body directed upwards. This force $W = m g$ is considered to be the "weight" of the person. It can be measured by the compression of a string e.g. on a bathroom scale.

Now consider the person on the scale in an elevator that is accelerated upwards with acceleration $a$. The forces acting on the person are unchanged gravity $G$ and the force of the spring (scale) $W_a$. According to Newton's law we have $$W_a + (- m g) = m a$$ and hence the new "weight" of the person is $$W_a = m (g + a)$$ The person gains weight and feels indeed heavier. She feels as if she was in a stronger gravitational field with the gravitational constant $g + a$. In fact, she cannot distinguish between the gravitation and the acceleration. This observation lead Einstein to his ' "equivalence principle'. The acceleration $g + a$ is also called a 'g-force' (it is in fact an acceleration).

The equations above are obviously valid also for accelerating the elevator downwards with $a \lt 0$. If the elevator accelerates downwards the person loses weight. If $a = - g$ then $W_a = 0$ and the person feels weightless, the g-force is 0. This the case of "free fall". We should not think of free fall as a situation without gravity. In fact gravity is the only force exerted on a body in free fall.

### Inertial frames of reference

Newton's laws are valid only in inertial frames of reference. These are systems that are at rest or move linearly at constant velocity. An observer in an inertial frame cannot find out whether they are moving or at rest.

Alice A in a "fixed" system A sees a car moving to the right with constant velocity $v$. She sees all objects inside the car moving with velocity $v$. Bob B in the "moving" system B is sure that he is at rest: nothing is moving around him in the car. The green box can slide on the floor without friction but it remains at rest with respect to the car. Bob assumes that Alice A outside is moving to the left with constant velocity $-v$. The frames A and B are equivalent and respect Newton's laws.

### Linear acceleration

Alice A in an inertial system A sees a car moving to the right with constant acceleration $a$.

The ball hanging on the ceiling has the same acceleration as the car because it does not move with respect to the car. It is declined back by angle $\alpha$ because the only two forces acting on it, the tension $T$ and the weight $G$, must produce the resulting force $\pmb F = \pmb T + \pmb G$ that produces its acceleration $a$. $$\tan \alpha = \frac{F}{G} = \frac{m a}{m g} = \frac{a}{g}$$

The green box remains at the fixed position $x$ in the inertial system A. This corresponds to the Newton's law. The only forces acting on the box are the gravity and the normal force from the floor - these two forces compensate themself and so the box does not move in the A-system.

Bob B in the accelerating system B sees the same situation but explains it differently:

He sees the ball declined but at rest with respect to his car. So the sum of the forces acting on the ball must be zero. He is aware of the tension $T$ and the weight $G$ but their vectorial sum is not zero. He concludes that there is a third force $\pmb H = - (\pmb T + \pmb G)$ such that $\pmb T + \pmb G + \pmb H = 0$. Of cource $\pmb H$ is the same as the accelaration force $\pmb F$ but has opposite direction: $$\pmb H = - \pmb F = - m \pmb a$$ Since this force exists only in the accelerated system it is called the "fictitious force" (German "Scheinkraft"). I find this denotation misleading because the force is rather real in the accelerated system. Other names for this force are: "inertial force" and "pseudo-force". The described accelerated system is an example of a "non-inertial frame of reference". By adding "inertial forces" we can apply the Newton's law of motion in non-inertial systems. Inertial ("fictitious") forces are real forces inside the accelerated system.

The green box moves to the left towards Bob with the acceleration $a$. Bob again concludes that there is force acting on the box. He himself feels this force. He is pressed against the left wall of the car.

#### Motion in inertial system

Let us consider from the perspective of Alice what happens when the rope of the ball breaks or is cut at time 0. Alice uses her fixed inertial system.

The ball has a certain horizontal velocity $v_A(0)$ and a height $h$ but no further horizontal acceleration. It falls to the floor according to the well known motion in the gravitational field of the earth: \begin{align} \frac{d^2 x_A}{d t^2} & = 0 \tag {13a} \\ \frac{d^2 y_A}{d t^2} & = - m g \tag {13a} \\ \end{align} The subscript "A" indicates that the position is measured in the fixed system A. The solution is: \begin{align} x_A(t) & = v_A(0) t \tag {13a} \\ y_A(t) & = h - \frac{1}{2} g t^2 \tag {13b} \end{align} and hits the floor when $y_A(T) = 0$, i.e. $$T = \sqrt{\frac{2 h}{g}}$$ The horizontal displacement measured in the system A is $$d = x_A(T) = v_A(0) T$$ In the same time the car travelled the distance $$D = v_A(0) T + \frac{1}{2} a T^2$$ The displacement inside the car is hence: $$\Delta = d - D = - \frac{1}{2} a T^2 = - \frac{a}{g} h$$

#### Motion in accelerated system

Bob sees the falling of the ball differently. He defines his system $B$ attached to his car.

There are 2 forces acting on the ball: the horizontal inertial force $-m a$ and the vertical gravitation force $-m g$. The equations of motion in the system B are: \begin{align} \frac{d^2 x_B}{d t^2} & = - m a \tag {13a} \\ \frac{d^2 y_B}{d t^2} & = - m g \tag {13a} \\ \end{align} The solution is: \begin{align} x_B(t) & = - \frac{1}{2} a t^2 \tag {13a} \\ y_B(t) & = h - \frac{1}{2} g t^2 \tag {13b} \end{align} The ball hits the floor when $y_B(T) = 0$, i.e. $$T = \sqrt{\frac{2 h}{g}}$$ The horizontal displacement measured in the system B is $$x_B(T) = - \frac{1}{2} a T^2 = - \frac{a}{g} h$$ which coincides with the computation in the inertial system A.

### General translation

Consider the general translational motion.

The system B moves on a straight line with acceleration $\pmb a$ with respect to a fixed system A. The relation between the position of a particle in the respective systems is: $$\pmb x_A(t) = \pmb x_{AB}(t) + \pmb x_B(t)$$ We denote the differentiation with respect to the time by dot. First-order differentiation yields the velocities, second-order differentiation yields the accelerations: \begin{align} \dot{\pmb x}_A (t) & = \dot{\pmb x}_{AB}(t) + \dot{\pmb x}_B(t) \tag {13a} \\ \ddot{\pmb x}_A (t) & = \ddot{\pmb x}_{AB}(t) + \ddot{\pmb x}_B(t) \tag {13a} \\ \end{align} Multiplying the last equation by the mass $m$ and rearranging we get: $$\ddot{\pmb x}_B (t) = \pmb F - m \pmb a$$ So, Newton's law can be used in the accelerated system when the inertial force $- m \pmb a$ is added to the acceleration force $\pmb F$.

### Rotating frames

All rotating frames are non-inertial frames because non-linear motion is always an accelerated motion. I describe rotating frames in another article .