Question bank — Length contraction — derivation

A moving rod is physically squeezed by a force, which is why it measures shorter. ::: False — no force acts. The rod is unstressed in its own frame; the shortening is a geometric consequence of frame-dependent simultaneity, not compression. The proper length $L_0$ is the longest length any observer will ever measure for the object. ::: True — $L_0$ is measured in the rest frame, and every other frame divides it by $\gamma \ge 1$, so all measured lengths satisfy $L\le L_0$. If observer A measures B's rod as shorter, then B must measure A's rod as longer to keep things consistent. ::: False — B also measures A's rod as shorter. There is no contradiction because A and B use different definitions of "at the same time," so they measure different pairs of events. A rod moving perpendicular to its length also contracts along that length. ::: False — only the dimension parallel to the motion contracts. A rod carried sideways past you keeps its full length. At $v=0$ the length contraction formula still applies and gives $L=L_0$. ::: True — with $v=0$, $\gamma =1$, so $L=L_0/1=L_0$. The formula smoothly reduces to "no motion, no contraction." As $v\to c$, the measured length approaches zero. ::: True — $\gamma \to \infty$, so $L=L_0/\gamma \to 0$. A rod at light speed would be measured with zero length along its motion (unreachable for real objects, but that's the limiting trend). Length contraction and time dilation both multiply the proper quantity by $\gamma$. ::: False — length divides by $\gamma$ (gets smaller) while time multiplies by $\gamma$ (gets larger); proper length is the maximum, proper time is the minimum. A photograph of a fast rod would literally show it squished by exactly the factor $1/\gamma$. ::: False — a measurement gives $1/\gamma$, but a photograph records light that left different parts at different times, adding rotation/distortion effects. Contraction is about simultaneous marking, not raw camera images.

"The rod moves in $S$, so I record its front end now and its back end a moment later, then subtract positions." ::: Error — the endpoints must be recorded simultaneously in $S$. If you wait, the rod moves between readings and you get nonsense, not a length. "To contract, I multiply: $L=\gamma L_0$, since $\gamma$ is the relativity factor." ::: Error — that makes $L$ bigger than the rest length. Proper length is the maximum, so you must divide: $L=L_0/\gamma$. "In the Lorentz step the $vt$ terms cancel because $v$ is the same for both ends." ::: Error — they cancel because $t_1=t_2$ (same time), making the two $vt$ terms identical. Equal $v$ alone isn't enough; simultaneity is what does the work. "The muon sees its own decay lifetime shrink, which is how it reaches the ground." ::: Error — in the muon's frame it's the atmosphere that contracts (shorter distance to cross), not the muon's own lifetime. The lifetime shrinking is the Earth-frame view via time dilation — two frames, one physics. "$L_0=\gamma L$ means the rest length is the contracted one." ::: Error — $L_0=\gamma L$ correctly says $L=L_0/\gamma$, so $L$ (the moving-frame measurement) is the shorter one. $L_0$ is still the largest. "Both observers measuring each other's rod as shorter violates the principle of relativity." ::: Error — it upholds the principle: physics looks the same in both frames, so each symmetrically measures the other contracted. The symmetry is the point, not a paradox.

Why does measuring a moving rod's length require simultaneity? ::: Because length is the gap between the two ends at one instant; if the ends are recorded at different times the moving rod shifts between readings, corrupting the gap. Why is the proper length the maximum, not the minimum? ::: In the rest frame the ends never move, so no simultaneity subtlety shrinks the reading. Every moving frame's simultaneity mismatch subtracts a piece, giving a smaller value. Why do length and time scale in opposite directions under $\gamma$? ::: Because proper length is the largest possible measurement while proper time is the smallest, so moving observers see length reduced ($÷\gamma$) but elapsed time increased ($\times \gamma$). Why doesn't the perpendicular dimension contract? ::: A consistency argument: a ring moving through a matching hole must fit in both frames. If perpendicular sizes changed, one frame would see it jam and the other pass — impossible for a single physical event. Why is $\gamma$ always $\ge 1$? ::: Because $v^2/c^2<1$ for real objects, so $1-v^2/c^2$ is between 0 and 1, and one over its square root is at least 1. This guarantees contraction never lengthens a rod. Why is length contraction called "real" if there's no force? ::: "Real" means every observer in that frame consistently measures it and it has physical consequences (e.g. muons reaching the ground). It's a genuine property of measurement, just not a mechanical squeeze. Why can't we simply say the moving rod "looks" shorter, i.e. it's an optical illusion? ::: Because it's a measured coordinate difference recorded by an ideal simultaneous survey, independent of light-travel or viewing angle — it survives careful correction that would remove any optical illusion.

What happens to $L$ when $v$ is a tiny everyday speed, say a car? ::: $v^2/c^2$ is unimaginably small, so $\gamma \approx 1$ and $L\approx L_0$ — the contraction is real but far too small to detect, matching everyday experience. What is the contracted length of a rod pointed exactly perpendicular to its motion? ::: Its full $L_0$ — there is no contraction perpendicular to $v$, only along it, so this rod is measured at its rest length. A rod moves at an angle (partly along, partly across the motion). What happens? ::: Only its component along the motion contracts; the perpendicular component stays fixed, so the rod appears both shortened and rotated relative to its rest orientation (see the angled-rod figure above). If two observers pass at $v$ and each measures the other's identical rod, do they get the same number? ::: Yes — by symmetry each measures the other's rod as $L_0/\gamma$, the same numeric factor, with no contradiction because they're marking different simultaneous events. Does a spherical ball moving fast get measured as a sphere? ::: No — it's measured as an ellipsoid, flattened along the motion by $1/\gamma$ while the two transverse diameters stay full size. What length does a photon-speed ($v=c$) rod have, formally? ::: The formula gives $L\to 0$, but $v=c$ is unreachable for a massive rod, so this is a limiting statement about the trend, not a physical object.