Special Relativity Domain Guide

There is no 'really'. Neither frame is right about whose clock runs slow; both measure the other's clock running slow, and both are correct, because they are measuring different pairs of events. The apparent contradiction always dissolves into the relativity of simultaneity — the two observers do not agree on which distant events happened 'at the same time', so they are not comparing the same thing.

Work in energy units and the algebra collapses. Express masses as rest energies ($0.511\ \mathrm{MeV}$ for an electron, $938\ \mathrm{MeV}$ for a proton) and momenta as $\mathrm{MeV}/c$; then $E = K + mc^{2}$ and $E^{2} = (pc)^{2}+(mc^{2})^{2}$ need no constants at all. Almost every relativity item on this exam is one of those two equations applied twice.

The invariants are the shortcut. $(\Delta s)^{2}$ is the same in every frame, and so is $m^{2}c^{4} = E^{2}-(pc)^{2}$. When a problem looks messy in the lab frame, ask what is invariant and evaluate it in the frame where one term vanishes — in a particle's rest frame $p = 0$, and in the centre-of-momentum frame the total momentum is zero. That single habit turns most collision and decay questions into one line.

ETS lists special relativity as a content area, and our learning model splits it into foundations and Lorentz transformations. The whole structure follows from two postulates: the laws of physics take the same form in all inertial frames, and the speed of light in vacuum is $c$ for every observer regardless of the motion of the source. The second is the radical one, because it forces time and space to be frame-dependent in order to keep one speed fixed.

Time dilation and length contraction are the first consequences, and they are also where most marks are lost — not to misunderstanding but to direction errors. Write both as statements about proper quantities. The proper time $\Delta\tau$ is measured by a clock present at both events; any other frame measures $\Delta t = \gamma\Delta\tau$, which is longer, so the moving clock runs slow. The proper length $L_0$ is measured in the object's rest frame; any other frame measures $L = L_0/\gamma$, which is shorter. Note that these go in opposite directions — one multiplies by $\gamma$ and the other divides — which is precisely why writing 'proper' next to the correct quantity before computing is worth the two seconds it costs. Contraction acts only along the direction of motion; transverse dimensions are untouched.

The Lorentz transformations contain both results and one more. With frame $S'$ moving at speed $v$ along $x$ relative to $S$: $x' = \gamma(x-vt)$ and $t' = \gamma\left(t - vx/c^{2}\right)$, with the inverses obtained by swapping primes and flipping the sign of $v$. The $-vx/c^{2}$ term is the relativity of simultaneity: two events at different $x$ that are simultaneous in $S$ ($\Delta t = 0$) have $\Delta t' = -\gamma v\Delta x/c^{2} \neq 0$ in $S'$. Every classic paradox — the pole and the barn, the twins, the train and the lightning strikes — is resolved by this term and nothing else, so it repays being able to write it from memory.

Velocity addition replaces the Galilean sum. If an object moves at $u$ in frame $S$ and $S'$ moves at $v$ relative to $S$, then $u' = \dfrac{u-v}{1-uv/c^{2}}$. Two features are worth checking once and then trusting: setting $u = c$ gives $u' = c$ for any $v$, which is the second postulate reappearing as an algebraic identity; and combining any two speeds below $c$ always yields a result below $c$, so $0.9c$ plus $0.9c$ gives $0.994c$, not $1.8c$. The same structure governs the relativistic Doppler effect, $f' = f\sqrt{(1-\beta)/(1+\beta)}$ for a receding source, which depends only on the relative speed because light has no medium to be measured against.

Relativistic dynamics is where the exam concentrates its numerical work. Momentum is $p = \gamma mv$, total energy is $E = \gamma mc^{2}$, kinetic energy is $K = (\gamma-1)mc^{2}$, and the three are tied together by $E^{2} = (pc)^{2}+(mc^{2})^{2}$ — an equation whose right-hand side is frame-independent, since $m$ is invariant. Two derived relations save time repeatedly: $\gamma = 1 + K/mc^{2}$, which converts a stated kinetic energy straight into $\gamma$ with no square roots, and $\beta = pc/E$, which extracts the speed from the dynamics without going back through $v$. For a massless particle $m = 0$, so $E = pc$ exactly, and a photon therefore carries momentum $E/c$ despite having no mass.

Conservation laws survive intact, provided you conserve the right things. In any interaction the total energy and the total momentum are conserved, but the total rest mass is not — that is what makes particle creation, annihilation, and nuclear binding possible. The invariant mass of a system, $M^{2}c^{4} = \left(\sum E\right)^{2}-\left(\sum \vec{p}c\right)^{2}$, is conserved and is generally larger than the sum of the constituent masses, the excess being the kinetic energy available in the centre-of-momentum frame. Threshold problems are exactly this quantity evaluated twice: once in the lab and once in the frame where the products sit at rest.