Special Relativity

Special relativity rests on two postulates: the laws of physics are the same in every inertial frame, and the speed of light is the same for all observers no matter how fast they move.^[1] That second point sounds harmless but is radical: to keep light's speed fixed for everyone, space and time must themselves bend. The consequences are time dilation (moving clocks run slow) and length contraction (moving objects shorten) — and, famously, the equivalence of mass and energy, $E = mc^2$ . None of it shows up at everyday speeds; it only becomes dramatic as you approach $c$ .

The cleanest way to see why time itself must stretch is a light clock — a pulse bouncing between a floor and a ceiling mirror. Watch the same clock from the train and from the ground:

On the train

the clock is at rest — light goes straight up and down

From the ground

the train moves — the same light traces a longer diagonal

Same light speed, longer path ⇒ the moving clock ticks 1.25× slower. One tick on the train = 1.25 ticks of ground time.

Train speed = 60% of light (γ = 1.25)

Both pulses move at the same speed. On the train the light just goes up and down; to someone watching the train rush past, it has to cover the diagonal — a longer trip for the very same tick. That extra distance is time dilation. Crank the speed up and watch the ground clock fall further behind.

Time dilation falls straight out of that longer diagonal.

Deep dive · Deriving time dilation from the light clock

Give the clock height $L$ . In the train's frame one tick — up and back — is light covering $2L$ at speed $c$ , so the proper time of a tick is

\tau = \frac{2L}{c}.

From the ground the train moves at $v$ , so during ground time $t$ it advances $vt$ while the light travels the two diagonals. Each diagonal has height $L$ and half the horizontal shift, so the light's total path length $ct$ obeys

c\,t = 2\sqrt{L^2 + \left(\tfrac{vt}{2}\right)^2}.

Square and gather the $t$ terms:

c^2 t^2 = 4L^2 + v^2 t^2 \;\Rightarrow\; t^2\,(c^2 - v^2) = 4L^2 \;\Rightarrow\; t = \frac{2L}{\sqrt{c^2 - v^2}}.

Divide inside the root by $c^2$ and use $\tau = 2L/c$ :

t = \frac{2L/c}{\sqrt{1 - v^2/c^2}} = \gamma\,\tau, \qquad \gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}.

So one tick of the moving clock, $\tau$ , is seen from the ground as the longer time $\gamma\tau$ — time dilation, forced by nothing more than "light has the same speed for everyone."

The Lorentz transformation generalises this to all coordinates. For a frame moving at $v$ ,

t' = \gamma\!\left(t - \frac{v x}{c^2}\right), \qquad x' = \gamma\,(x - v t).

Term by term: $x - vt$ is the ordinary shift of a moving origin; the $\gamma$ in front is the stretch we just derived; and the $vx/c^2$ buried in the time equation is the genuinely new piece — it mixes position into time, which is exactly why two events that are simultaneous for one observer need not be for another.