Einstein's Postulates and the Relativity of Simultaneity

1A clerk in Bern

In 1905 Albert Einstein was 26 and worked as a technical expert at the Swiss patent office in Bern, examining applications for electrical inventions, many of them about synchronising clocks by electrical signals. In June of that year he sent a paper, "On the Electrodynamics of Moving Bodies", to the journal Annalen der Physik. Instead of trying to rescue the ether, he took the experimental facts at face value and raised them to two principles. Everything in this unit follows from them, including the most surprising consequence of all: two events that are simultaneous for one observer are not simultaneous for another.

2Einstein's two postulates

1. The principle of relativity. The laws of physics, all of them, including those of electricity, magnetism and light, are the same in every inertial frame of reference. No experiment can detect absolute uniform motion.
2. The constancy of the speed of light. Light travels through a vacuum at the same speed $c=3.00\times {10}^8\mathrm{m}{\mathrm{s}}^{-1}c\; =\; 3.00\; \backslash times\; 10^\{8\}\backslash ,\backslash mathrm\{m\backslash ,s^\{-1\}\}$ in every inertial frame, whatever the motion of its source or of the observer.

The first postulate extends Galileo's principle from mechanics to all of physics. The second is what the Michelson–Morley experiment found. Together they make the ether unnecessary: there is no special frame "at rest".

Each postulate on its own seems harmless. Together they are revolutionary, because they contradict the Galilean velocity rule $u^{\mathrm{\prime }}=u-vu\text{'}\; =\; u\; -\; v$. If a lamp on a moving train sends a pulse forward, the passenger and the person on the platform must both measure its speed as $cc$. Speed is distance divided by time, so if they agree on the speed of light, they must disagree about distances and times. Space and time are not absolute.

3The relativity of simultaneity

Einstein's own thought experiment: a long train moves at speed $vv$ past a platform. At the instant the middle of the train, where observer O′ sits, passes observer O on the platform, lightning strikes both ends of the train, leaving scorch marks at A and B on the platform, equal distances from O.

For O on the platform: the two flashes travel equal distances at the same speed $cc$ and reach O together. O concludes, correctly, that the strikes were simultaneous.

For O′ on the train: she is moving towards B and away from A, so (as seen from the platform) she meets the flash from B first. In her own frame she is at rest, exactly halfway between the two ends of the train, and by the second postulate both flashes travel towards her at $cc$. Since the flash from B arrived first, she concludes, also correctly, that the strike at B happened before the strike at A.

Neither observer is mistaken. Events that are simultaneous in one frame are not simultaneous in another frame moving relative to it, if the events are separated along the direction of motion. Whether two separate events happen "at the same time" depends on the frame. Only for events at the same place do all observers agree.

The effect is tiny at everyday speeds. In the Lorentz transformation, which replaces the Galilean one, $t^{\mathrm{\prime }}=\gamma (t-\frac{vx}{c^2})t\text{'}\; =\; \backslash gamma\backslash left(t\; -\; \backslash frac\{vx\}\{c^2\}\backslash right)$, where $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\backslash gamma\; =\; \backslash frac\{1\}\{\backslash sqrt\{1\; -\; \backslash frac\{v^2\}\{c^2\}\}\}$ (Lesson 3). Two events simultaneous in S and a distance $LL$ apart along $xx$ are separated in S′ by

4Adding velocities near the speed of light

Because times and distances change between frames, the Galilean rule $u=u^{\mathrm{\prime }}+vu\; =\; u\text{'}\; +\; v$ must also change. If an object moves with velocity $u^{\mathrm{\prime }}u\text{'}$ in frame S′, which itself moves at $vv$ relative to S (both along $xx$), its velocity in S is

At everyday speeds $\frac{u^{\mathrm{\prime }}v}{c^2}\backslash frac\{u\text{'}v\}\{c^2\}$ is negligible and the Galilean rule is recovered. But however the speeds are combined, the result never exceeds $cc$, and if $u^{\mathrm{\prime }}=cu\text{'}\; =\; c$, then $u=\frac{c+v}{1+\frac{v}{c}}=cu\; =\; \backslash frac\{c\; +\; v\}\{1\; +\; \backslash frac\{v\}\{c\}\}\; =\; c$: light has the same speed in both frames, exactly as the second postulate requires.