Length Contraction & Simultaneity

Turn the light clock so it moves along its length and a second surprise appears: moving objects are shorter. Then look at two distant events and a third: whether they happen “at the same time” depends on who is watching.

Length contraction#

Take the light clock and move it at speed v parallel to its rod. Call the rod’s length in this frame L′ (we don’t yet know if it equals L). On the way out the pulse chases the front mirror, which runs away at v: it takes L′/(c − v). On the way back the rear mirror rushes to meet it: L′/(c + v). One tick therefore lasts

But the tick must also equal γ(2L/c) by time dilation. Setting the two equal:

Which event happened first?#

A box of rest length L lies on the ground, with doors D₁ (left) and D₂ (right) and a lamp exactly in the middle. Each door opens when light reaches it. Train T₁ moves left and train T₂ moves right, both at speed v relative to the ground. The lamp flashes.

• Ground frame: the pulses travel L/2 each at c, so both doors open at t = L/2c: simultaneously.
• Frame of T₁: the box (length L/γ) moves right at v. D₁ comes to meet its pulse, D₂ runs away from its pulse: Δt₁′ = (L/γ)/2(c + v), Δt₂′ = (L/γ)/2(c − v). D₁ opens first.
• Frame of T₂: by the same reasoning, D₂ opens first.

Moving clocks are out of sync#

To synchronize two clocks, put a lamp midway and set each to zero when its flash arrives. Do that for clocks on the ground at D₁ and D₂ and, separately, for a row of clocks along train T₁. When D₁ opens, a train clock C₁ is opposite it; when D₂ opens, clock C₂ is opposite. On the ground the openings are simultaneous, but the train clocks say D₂ opened Lvγ/c² later. So, judged from the ground, the train clocks disagree at a single instant.

Again this is symmetric: from the train, the ground clocks are the ones out of step.