Galilean and Special Relativity

A reference frame is a coordinate system plus synchronized clocks used to record where and when events happen. An inertial reference frame is one that does not accelerate: it is at rest or moves with constant velocity, so Newton's first law holds within it.

Galilean relativity is the classical statement that the laws of mechanics are identical in all inertial frames, meaning no mechanical experiment can detect uniform motion.

If frame $S'$ moves at constant velocity $v$ along the $x$-axis of frame $S$, the Galilean transformations are:

$$x' = x - vt, \qquad t' = t$$

Time is treated as universal: every observer's clock agrees. Velocities then add linearly, $u' = u - v$. For example, a passenger walking forward at $1.0\ \text{m s}^{-1}$ inside a train moving at $30\ \text{m s}^{-1}$ is measured from the platform to move at $31\ \text{m s}^{-1}$.

This scheme works superbly at everyday speeds but fails for light. Galilean addition predicts that observers in relative motion should measure different speeds for the same light beam, yet experiment shows every inertial observer measures the same value $c$.

You must be able to:

• transform positions and velocities between inertial frames using the Galilean equations
• identify whether a frame is inertial
• explain clearly why the Galilean picture breaks down as speeds approach $c$

Einstein resolved the conflict between mechanics and light by building the whole theory on two postulates.

• First postulate: the laws of physics are the same in all inertial reference frames. This extends Galilean relativity from mechanics to all of physics, including electromagnetism, so no experiment of any kind can reveal absolute uniform motion.
• Second postulate: the speed of light in a vacuum, $c = 3.00 \times 10^8\ \text{m s}^{-1}$, is the same for all inertial observers, regardless of the motion of the light source or the observer.

The second postulate is the radical one. If a spacecraft travelling at $0.5c$ switches on its headlights, both the pilot and a stationary observer measure the emitted light moving at exactly $c$, not $1.5c$.

The only way both measurements can be correct is if the observers disagree about distances and time intervals themselves. Every relativistic effect in this topic, including time dilation, length contraction, and the relativity of simultaneity, follows logically from these two statements.

You must be able to state both postulates precisely and use them in explanations. A common exam task is to justify a relativistic effect by arguing from the constancy of $c$ rather than simply quoting a formula, so practise writing two- or three-sentence arguments that begin from the postulates.

The Lorentz transformations replace the Galilean equations and connect the coordinates of a single event as recorded in two inertial frames. If frame $S'$ moves at speed $v$ along the positive $x$-axis of frame $S$, with origins coinciding at $t = t' = 0$, then:

$$x' = \gamma(x - vt), \qquad t' = \gamma\left(t - \frac{vx}{c^2}\right)$$

Here the Lorentz factor is given by:

$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

Coordinates perpendicular to the motion are unchanged ($y' = y$, $z' = z$).

Two features matter physically. First, time is no longer universal: $t'$ depends on position $x$ as well as $t$, which is the mathematical root of the relativity of simultaneity. Second, at everyday speeds $\gamma \approx 1$ and $\frac{vx}{c^2} \approx 0$, so the equations reduce to the familiar Galilean forms, as they must.

Velocities combine through $u' = \frac{u - v}{1 - \frac{uv}{c^2}}$, which never yields a speed greater than $c$. For instance, two spacecraft approaching each other at $0.8c$ each (in the planet frame) measure a relative speed of $0.98c$, not $1.6c$.

You must be able to:

• compute $\gamma$
• transform event coordinates between frames (taking care with signs of $v$ and $x$)
• apply the velocity transformation in both directions
• verify that answers reduce sensibly in the low-speed limit

Two famous consequences follow directly from the Lorentz transformations.

Time dilation: a clock moving relative to an observer runs slow, so:

$$\Delta t = \gamma \Delta t_0$$

Here the proper time $\Delta t_0$ is measured in the frame where the two events happen at the same position (for example, on the moving clock itself).

Length contraction: an object moving relative to an observer is shortened along its direction of motion:

$$L = \frac{L_0}{\gamma}$$

Here the proper length $L_0$ is measured in the object's rest frame. Dimensions perpendicular to the motion are unaffected.

The classic physical evidence is the muon. Muons created high in the atmosphere have such short average lifetimes that, classically, almost none should reach the ground, yet large numbers are detected at sea level.

In the Earth frame the muon's lifetime is dilated by $\gamma$, giving it time to arrive; in the muon's frame the atmosphere's thickness is contracted by the same factor. Both descriptions predict the same detection rate, illustrating that the two effects are one phenomenon viewed from different frames.

The critical skill is identifying which observer measures the proper quantity before substituting numbers: proper time belongs to the frame where the events are co-located, and proper length belongs to the frame where the object is at rest. Choosing these wrongly inverts the factor of $\gamma$ and is the most common error in this topic.

A spacetime (Minkowski) diagram plots $ct$ on the vertical axis against $x$ on the horizontal axis, so a light pulse always has a worldline at $45^\circ$.

The worldline of an observer moving at velocity $v$ defines their time axis $ct'$, since along it their position coordinate is $x' = 0$. Substituting $x' = 0$ into the Lorentz spatial transformation gives $x = vt$, i.e. the line $ct = \left(\frac{c}{v}\right)x$, tilted from the vertical by an angle $\theta$ where $\tan\theta = \frac{v}{c}$.

The $x'$-axis is the set of events with $t' = 0$; setting $t' = 0$ in the time transformation gives $t = \frac{vx}{c^2}$, i.e. $ct = \left(\frac{v}{c}\right)x$. Both primed axes therefore tilt inward toward the $45^\circ$ light line by the same angle, given by:

$$\theta = \arctan\left(\frac{v}{c}\right)$$

To see why simultaneity is relative, consider a train carriage moving at speed $v$ past a platform, with a bulb at its exact centre flashing once. Inside the carriage, the light travels equal distances to the front and back walls and arrives simultaneously.

For the platform observer, the back wall moves toward the emitted signal while the front wall moves away from it; because $c$ is identical for both observers, the light reaches the back wall first. Events simultaneous in one frame are therefore not simultaneous in a frame in relative motion. On the diagram, events simultaneous in $S'$ lie on lines parallel to the $x'$-axis.

You must be able to:

• draw and label tilted axes
• plot events and worldlines
• use the diagram to justify the ordering of events for different observers