Analyse the Michelson-Morley experiment, state Einstein's two postulates of special relativity, and apply the consequences of time dilation, length contraction and relativity of simultaneity

What this dot point is asking

NESA wants you to summarise the Michelson-Morley experiment, state Einstein's two postulates, and apply time dilation and length contraction quantitatively. You should also be able to describe the relativity of simultaneity qualitatively.

The answer

The aether problem and Michelson-Morley

By the late nineteenth century, light was understood as a wave. Waves needed a medium, so physicists postulated the luminiferous aether: a hypothetical, all-pervading substance through which light propagated. The aether was assumed stationary (or close to it), so the Earth must move through it at orbital speed (about $30$ km/s).

Michelson and Morley's interferometer aimed to detect this motion. A beam of monochromatic light was split into two perpendicular paths by a half-silvered mirror, reflected off mirrors at equal distance, and recombined. Any difference in transit time between the two paths would produce interference fringes. Rotating the apparatus by $90^\circ$ would swap the "along-aether" and "across-aether" paths and should shift the fringes by a predictable amount (about $0.4$ of a fringe for the 1887 setup).

The result was null: no significant fringe shift was observed in any orientation, season or location. The experiment was repeated many times with increasing precision, always null. The simplest interpretation: there is no aether.

Einstein's two postulates (1905)

Special relativity rests on just two postulates:

1. Principle of relativity. The laws of physics are the same in all inertial (non-accelerating) reference frames. No experiment can identify an absolute rest frame.
2. Constancy of the speed of light. The speed of light in vacuum, $c$, is the same in all inertial frames, regardless of the motion of the source or the observer.

The second postulate is the radical one. It immediately removes the need for an aether and makes the Michelson-Morley null result automatic. But it forces strange consequences for space and time.

The Lorentz factor

Most relativistic formulas use:

At everyday speeds, $\gamma \approx 1$ and relativistic effects are negligible. At $v = 0.5c$, $\gamma \approx 1.155$; at $v = 0.9c$, $\gamma \approx 2.29$; at $v = 0.99c$, $\gamma \approx 7.09$.

Time dilation

A clock moving with speed $v$ relative to an observer ticks slowly compared to a clock at rest with that observer. If the moving clock measures proper time $t_0$ (time between two events at the same place in its own frame), the time interval measured by the stationary observer is:

The classic thought experiment: a light clock bounces a photon between two parallel mirrors a distance $L_0$ apart. In its rest frame the round trip is $t_0 = 2 L_0 / c$. In a frame where the clock moves sideways at speed $v$, the photon traces a longer zig-zag path, but still travels at $c$ (postulate 2). Equating distances gives $t = \gamma t_0$.

Time dilation has been confirmed by atomic clocks flown on aircraft (Hafele-Keating, 1971) and by the increased lifetime of fast-moving muons (covered in evidence-for-special-relativity).

Length contraction

An object moving at speed $v$ along its length is measured to be shorter than its proper length $L_0$ (the length in its rest frame) by:

Contraction is only along the direction of motion; perpendicular dimensions are unchanged. Like time dilation, it is real in the sense that any measurement in the observer's frame, made with synchronised rulers, gives the contracted value. There is no internal stress in the object; it is the geometry of spacetime that differs between frames.

Relativity of simultaneity

Two events that are simultaneous in one inertial frame are generally not simultaneous in another moving relative to the first.

Einstein's train thought experiment: lightning strikes both ends of a train simultaneously according to an observer on the embankment. The flashes reach the embankment observer at the same time. But an observer at the centre of the moving train is travelling toward the front flash and away from the back flash, so the front flash reaches them first. Because the speed of light is the same in both frames (postulate 2), the train observer must conclude the front strike happened earlier than the back strike. The two observers disagree on which events were simultaneous.

This is the deepest consequence of the postulates: there is no universal "now". Time ordering of causally connected events (cause before effect) is preserved, but ordering of spacelike-separated events depends on the frame.

Worked example: muon trip

A muon created at the top of the atmosphere ($15$ km altitude) travels at $0.998c$ toward the ground. Its proper lifetime is $2.2$ $\mu$s.

In the Earth frame: $\gamma = 1 / \sqrt{1 - 0.996} = 1 / 0.0632 = 15.8$.

Lifetime as measured from Earth: $t = \gamma t_0 = 15.8 \times 2.2 \times 10^{-6} = 3.48 \times 10^{-5}$ s.

Distance the muon can cover: $d = v t = 0.998 \times 3.0 \times 10^8 \times 3.48 \times 10^{-5} = 1.04 \times 10^4$ m $= 10.4$ km.

Without dilation, the muon would only travel $660$ m in $2.2$ $\mu$s and almost none would reach the surface. Time dilation explains why many do.

In the muon's frame, the atmosphere is contracted: $L = 15 \text{ km} / 15.8 = 0.95$ km, so it can cover the (now-much-shorter) distance in its proper lifetime. Both frames agree on the outcome (muons reach the ground) via different mechanisms.

Examples in context

Example 1. GPS satellite clocks over Sydney. GPS satellites orbit at $v = 3.87 \text{ km/s}$, so $\beta = v/c = 1.29 \times 10^{-5}$ and $\gamma \approx 1 + \beta^2/2 = 1 + 8.3 \times 10^{-11}$. Per day ($86{,}400 \text{ s}$), special-relativistic time dilation slows the satellite clock by $\Delta t = (\gamma - 1) \times 86{,}400 = 7.2 \mu\text{s}$ relative to a ground clock. (General relativity adds $+45.9 \mu\text{s}$ for altitude, giving net $+38.7 \mu\text{s}/\text{day}$). Without applying this correction, GPS positions over Sydney would drift by $\sim 11 \text{ km}/\text{day}$. The satellite oscillators are tuned to $10.22999999543 \text{ MHz}$ instead of $10.23 \text{ MHz}$ at launch to compensate.

Example 2. Cosmic muon flux measured at sea level in NSW. Muons created at $h = 15 \text{ km}$ altitude live (in their rest frame) $\tau_0 = 2.2 \mu\text{s}$ before decaying. At classical speed $v = 0.998c$, distance travelled in one lifetime is $v \tau_0 = 660 \text{ m}$, far short of $15 \text{ km}$. Yet many muons reach sea level. In our frame, time dilates: $\tau = \gamma \tau_0 = 15.8 \times 2.2 \mu\text{s} = 34.8 \mu\text{s}$, allowing $v \tau = 0.998 c \times 34.8 \mu\text{s} = 10{,}400 \text{ m}$ of travel. In the muon's frame, the atmosphere is length-contracted to $h/\gamma = 950 \text{ m}$. Both viewpoints agree on the observed sea-level flux.

Try this

Q1. State Einstein's two postulates of special relativity. [2 marks]

• Cue. (1) Laws of physics same in all inertial frames. (2) Speed of light $c$ is the same in all inertial frames.

Q2. A spaceship passes Earth at $v = 0.80c$. (a) Find $\gamma$. (b) A ground clock measures $1.0 \text{ s}$. What does the spaceship clock measure during this interval? [3 marks]

• Cue. (a) $\gamma = 1/\sqrt{1 - 0.64} = 1.667$. (b) Spaceship clock (moving relative to Earth) records $t_0 = t/\gamma = 0.60 \text{ s}$.

Q3. A muon traveling at $v = 0.995c$ has a proper lifetime of $\tau_0 = 2.2 \mu\text{s}$. (a) Calculate its lifetime in the lab frame. (b) Calculate the distance it can travel before decaying (lab frame). (c) Show that the same result emerges using length contraction in the muon's frame. [2+2+3 marks]

• Cue. (a) $\gamma = 10.0$; $\tau = 22 \mu\text{s}$. (b) $d = v\tau = 6.57 \text{ km}$. (c) Muon sees Earth approaching; distance contracted by $\gamma$, time = proper, same answer.