A sphere flies past you at and you photograph it side-on. Length contraction flattens it along its direction of motion by a factor of . Does the photograph show a flattened ellipsoid?
Solution
No — the outline in the photograph is still a perfect circle.
A photograph does not record where the parts of the sphere are at one instant; it records photons arriving at one instant, which left different parts of the sphere at different times. The compensating delay lives across the sphere’s depth: light from the far surface had further to travel, so it left earlier, when the sphere was still further back, and that back-displaced image of the far side fills in exactly what contraction removed. One line makes it checkable: a surface point at depth is photographed where the point was a time ago, at with the rest-frame longitudinal coordinate; maximizing over gives a half-width , the rest radius, exactly. What survives is an apparent rotation — the Terrell–Penrose effect: to a distant camera the surface pattern is that of the sphere turned through at , trailing hemisphere swung into view, never a sphere squashed. (The circular outline is exact at any distance; the rigid-rotation reading holds in the small-angular-size limit.)
The trap is conflating two operations. Measuring length means locating both endpoints simultaneously in your frame — that gives the -contraction. Seeing folds light-travel time into the picture, and aberration maps circles to circles — no accident: a boost acts on the celestial sphere as a conformal (Möbius) transformation, and conformal maps carry circles to circles. Contraction is real; it is just not what a camera measures. What actually betrays the motion in the picture is not shape but light: beaming and Doppler shift make the approaching limb bright and blue, the receding limb dim and red.
Deeper in the notebook: 07. Relativistic Doppler Effect, Aberration, and Beaming · 01. Minkowski Spacetime and the Lorentz Group