inertial frames and Galilean relativity

An inertial frame of reference is a frame in which a body with zero net force moves with constant velocity, that is, a frame in which Newton's first law holds. Any frame moving with a constant velocity $\vec V$ relative to an inertial frame is also inertial; frames that accelerate or rotate are not.

The coordinates of the same event in two inertial frames are related by the Galilean transformation

$$\vec r' = \vec r - \vec V t, \qquad t' = t.$$

Differentiating once gives $\vec v' = \vec v - \vec V$, and differentiating twice gives $\vec a' = \vec a$: acceleration is the same in all inertial frames. Forces in mechanics depend on relative positions and on differences of velocities, which are unchanged by the constant shift $\vec V t$, so $\vec F = m\vec a$ keeps exactly the same form in every inertial frame. This is Galileo's principle of relativity: no mechanical experiment inside a closed laboratory reveals its uniform straight-line motion.

Frame $S'$ moves relative to the inertial frame $S$ with constant velocity $\vec V = 4\ \text{m/s}$ along the common $x$-axis; at $t = 0$ the origins coincide. In $S$ a ball is at $x = 6\ \text{m}$ at $t = 0$ and moves with constant velocity $v_x = 5\ \text{m/s}$. Find $x'$ and $v_x'$ at $t = 3\ \text{s}$.

In frame $S$ a puck of mass $m = 0.5\ \text{kg}$ slides on frictionless ice; its velocity is $\vec v_1 = (6, 0)\ \text{m/s}$ at $t_1 = 1\ \text{s}$ and $\vec v_2 = (6, 4)\ \text{m/s}$ at $t_2 = 3\ \text{s}$. Frame $S'$ moves relative to $S$ with $\vec V = (2, 0)\ \text{m/s}$. Find the acceleration and the net force on the puck in $S$, then repeat the calculation in $S'$ and compare.

A train moves with constant velocity $\vec V = 15\ \text{m/s}$ along the $x$-axis. A passenger throws a ball straight up with initial speed $u = 4\ \text{m/s}$ relative to the train; take $g = 10\ \text{m/s}^2$. Write $x(t)$, $y(t)$ for the ball in the train frame $S'$ and in the ground frame $S$, compare the accelerations, and explain what this says about whether the two frames are inertial.

Self-study plan (no hazardous measurements). (1) Roll a marble slowly along a smooth level table and record it with a phone in slow motion; place a ruler or a paper tape along the table. (2) From the video (or in a video-analysis app) read the marble's coordinate $x$ in the room frame $S$ at several instants. (3) In a spreadsheet build a second frame $S'$ moving with a chosen constant velocity $V$: for every frame compute $x' = x - Vt$. (4) Plot $x(t)$ and $x'(t)$: the slopes (velocities) differ by $V$, but both lines stay straight, so $a = a' = 0$. (5) Optional, only while seated and not standing in a moving vehicle: toss a coin upward in a bus or train moving at constant speed and note that it returns to your hand exactly as in a stationary room; explain the observation with $\vec v' = \vec v - \vec V$.

To show that the same event has different radius vectors and velocities in two inertial frames, while its acceleration (and therefore the force on the body) is the same.

Sketch: two Cartesian frames $S$ and $S'$ with parallel axes, with $S'$ displaced along $x$ by $x_0 = Vt$; the event $P$ with radius vectors $\vec r$ (from $O$) and $\vec r'$ (from $O'$), plus the dashed vector $\vec V t$ closing the triangle $\vec r = \vec r' + \vec V t$. Beside it, a small velocity triangle $\vec v = \vec v' + \vec V$ and two short $v_x(t)$ graphs for $S$ and $S'$: parallel lines with a constant difference $V$ and zero slope, labelled $a = a' = 0$.

A train rolls along a straight track at $20\ \text{m/s}$ ($72\ \text{km/h}$). A passenger throws a ball straight up at $5\ \text{m/s}$ relative to the train. With $g = 9.8\ \text{m/s}^2$ the ball returns to the hand after $t = 2v/g \approx 1.0\ \text{s}$, rising $h = v^2/2g \approx 1.3\ \text{m}$. During that second the ball's horizontal position follows the train: in the ground frame it covers $20\ \text{m}$ and traces a parabola.

Galilean transformation of velocities: the ground-frame velocity is $\vec v = \vec v_{\text{train}} + \vec v_{\text{ball/train}}$. The horizontal component $20\ \text{m/s}$ is shared, so the ball never falls behind; writing $F = ma$ in the train frame gives the same flight time $t = 2v/g$ as in the ground frame.

A $60\ \text{m}$ wide river flows at $2.0\ \text{m/s}$. A kayaker paddles at $3.0\ \text{m/s}$ perpendicular to the bank relative to the water. The crossing takes $t = 60/3.0 = 20\ \text{s}$; the current carries the kayak $2.0 \times 20 = 40\ \text{m}$ downstream, and the ground-frame speed is $\sqrt{3.0^2 + 2.0^2} \approx 3.6\ \text{m/s}$.

Velocity composition between inertial frames: $\vec v_{\text{ground}} = \vec v_{\text{kayak/water}} + \vec v_{\text{water/bank}}$. The two frames move relative to each other with constant velocity, so the acceleration is the same and no extra force enters the crossing time; only the perpendicular component sets $t$.

A bus travelling at $15\ \text{m/s}$ brakes uniformly to rest in $3.0\ \text{s}$, so its acceleration is $-5.0\ \text{m/s}^2$. A $60\ \text{kg}$ passenger keeps moving forward relative to the bus and feels a jolt that the bus-frame description attributes to an inertial force $|F| = ma = 60 \times 5.0 = 300\ \text{N}$. While the bus cruises at constant velocity, a phone resting on the seat reports zero acceleration.

The definition of an inertial frame: Newton's first law holds only in frames moving with constant velocity. In the braking bus the vector sum of real forces on the phone is zero, yet $\vec a \neq 0$ in that frame, so the frame is non-inertial and an inertial force $-m\vec a_{\text{frame}}$ must be added.