space-time interval and Lorentz transformations

Galilean transformations rest on absolute time and are therefore incompatible with Einstein's second postulate that the speed of light is the same in every inertial frame.

The Lorentz transformations are the linear relations between the coordinates of one and the same event in two inertial frames that preserve this invariance: if $S'$ moves relative to $S$ with speed $v$ along the $x$-axis, then

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

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

These formulas imply the relativity of simultaneity: events that are simultaneous in $S$ are generally not simultaneous in $S'$.

The invariant of the transformation is the spacetime interval

$$\Delta s^2=c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2,$$

which is the same in every inertial frame.

The sign of $\Delta s^2$ classifies a pair of events: a timelike pair ($\Delta s^2>0$) allows a causal connection and has a frame in which the events occur at the same place; a lightlike pair ($\Delta s^2=0$) corresponds to a signal moving at $c$; a spacelike pair ($\Delta s^2<0$) forbids a causal connection but allows a frame in which the events are simultaneous.

In an inertial frame $S$, event A has coordinates $(t=0,\ x=0)$ and event B has coordinates $(t=5.0\ \text{y},\ x=3.0\ \text{ly})$ along the $x$-axis. Compute $\Delta s^2$, find $\Delta s$, and determine the type of the pair of events.

In frame $S$ an event has coordinates $t=3.0$ y and $x=2.0$ ly. Frame $S'$ moves along the $x$-axis with speed $v=0.6c$, and the origins coincide at $t=t'=0$. Find $t'$ and $x'$ and verify the invariance of the interval relative to the event $(0,0)$.

In frame $S$, event E1 is $(t=0,\ x=0)$ and event E2 is $(t=1.0\ \mu\text{s},\ x=600\ \text{m})$. Show that these events cannot be causally connected, find the speed of the frame in which they are simultaneous, and find the distance between them in that frame.

Safe self-study plan (screen and spreadsheet only — no radiation sources, vacuum tubes, or high voltage).

1. Open a Minkowski-diagram simulation or create a spreadsheet.
2. Enter two events with coordinates $(t_1,x_1)$ and $(t_2,x_2)$; in separate cells program $\Delta s^2=c^2\Delta t^2-\Delta x^2$.
3. Add columns for $t'=\gamma(t-vx/c^2)$ and $x'=\gamma(x-vt)$ and output $\Delta s'^2$.
4. Vary $v$ from $0$ to $0.9c$ and confirm that $\Delta s'^2$ stays equal to $\Delta s^2$ up to rounding.
5. For a spacelike pair, set $v=c^2\Delta t/\Delta x$ and check that $\Delta t'=0$.
6. Sketch the hyperbola $c^2t^2-x^2=\Delta s^2$ through the event and note how the $t'$-axis rotates as $v$ changes.

Show that the same two events have different $\Delta t$ and $\Delta x$ in different inertial frames but the same interval $\Delta s^2$, and that the light cone separates timelike pairs from spacelike ones.

Minkowski diagram: the $ct$-axis points up, the $x$-axis points right, with the light cone $x=\pm ct$; two events are marked together with the worldline of an observer and the tilted $ct'$ and $x'$ axes of the primed frame; the hyperbola $c^2t^2-x^2=\Delta s^2$ is drawn through the event, and a dashed line $t'=\text{const}$ illustrates the relativity of simultaneity.

Cosmic rays striking air molecules about $10\ \text{km}$ above sea level create muons. In its own frame a muon lives $\tau = 2.2\ \mu\text{s} = 2.2\times10^{-6}\ \text{s}$. At $v = 0.998c = 2.99\times10^{8}\ \text{m/s}$ the classical estimate gives a flight of $v\tau \approx 660\ \text{m}$, so such a muon should never reach the ground. Ground detectors nevertheless count about one muon per $\text{cm}^2$ per minute.

The muon reaches the ground because the interval $s^2 = c^2\Delta t^2 - \Delta x^2$ takes the same value in every inertial frame. In the Earth frame $\Delta x = 1.0\times10^{4}\ \text{m}$ and $\Delta t = \Delta x/v = 3.3\times10^{-5}\ \text{s}$, giving $s^2 \approx 4.4\times10^{5}\ \text{m}^2$; in the muon frame $\Delta x' = 0$ and $\Delta t' = \tau$, and $c^2\tau^2$ is the same number. The Lorentz transformation $t' = \gamma\left(t - vx/c^2\right)$ with $\gamma = 1/\sqrt{1 - v^2/c^2} = 15.8$ turns the muon's proper time $\tau = 2.2\ \mu\text{s}$ into the Earth-frame time $\Delta t = \gamma\tau = 35\ \mu\text{s}$.

A GPS satellite moves in an orbit at an altitude of $20\,200\ \text{km}$ ($26\,560\ \text{km}$ from the Earth's centre) with speed $v = 3.87\ \text{km/s}$, so $\gamma - 1 = 8.3\times10^{-11}$. Over one day, $86\,400\ \text{s}$, the on-board clock falls behind an Earth clock by about $8.3\times10^{-11}\times 86\,400\ \text{s} = 7.2\ \mu\text{s}$ (the velocity part of the relativistic correction). Without this correction the ranging error would grow by $c \times 7.2\ \mu\text{s} \approx 2.2\ \text{km}$ per day, whereas a phone displays a position to within a few metres.

The Lorentz transformation relates coordinates and time in the Earth frame $S$ and the satellite frame $S'$: $x' = \gamma(x - vt), \qquad t' = \gamma\left(t - \frac{vx}{c^2}\right).$ The satellite's worldline has $x = vt$, hence $t' = \gamma t\left(1 - v^2/c^2\right) = t/\gamma$: the proper time $t'$ read by the satellite clock is shorter than the coordinate time $t$. Invariance of the interval $c^2t^2 - x^2 = c^2t'^2$ means that a receiver on the ground and a satellite agree on the same value of $s^2$, and the navigation solution looks for the position at which the intervals from four satellites coincide.

A probe flies to Proxima Centauri, $d = 4.24\ \text{ly} = 4.01\times10^{16}\ \text{m}$, at $v = 0.8c = 2.40\times10^{8}\ \text{m/s}$, so $\gamma = 5/3 = 1.667$. By Earth clocks the trip lasts $t = d/v = 1.67\times10^{8}\ \text{s} = 5.3$ years. The crew's proper time is $\tau = t/\gamma = 3.2$ years and the distance they measure is $d' = d/\gamma = 2.54\ \text{ly}$.

Invariance of the interval demands that the Earth and the ship obtain the same $s^2$: $c^2t^2 - d^2 = c^2\tau^2, \qquad \tau = t\sqrt{1 - \beta^2} = t/\gamma.$ Substituting the numbers gives $c^2t^2 - d^2 = (5.3\ \text{ly})^2 - (4.24\ \text{ly})^2 = 10.1\ \text{ly}^2$, and this equals $(c\tau)^2 = (3.2\ \text{ly})^2$. The length contraction $d' = d/\gamma$ follows from the same Lorentz transformation $x' = \gamma(x - vt)$ taken at fixed $t'$.