Past exam of the short options course of the University of Oxford / 2021 / 8 (Special relativity)

Differentiate the worldline with respect to $\theta$:Since $|\tan \theta |\le 1$ and $0<\kappa <1$, this is a future-directed timelike curve. The proper time element satisfies $c^{2}d{\tau }^2=c^{2}d{t}^2-d{x}^2$, henceAlice's total proper time is thereforeThis definite integral is an exact answer; it is an elliptic integral of the second kind. In notation $E(\alpha \mid m)={\int }_0^{\alpha }\sqrt{1-m{\sin }^{2}u}du$, it is $2\sqrt{2}TE(\pi /4\mid 1+{\kappa }^2)$. Although this parameter exceeds one, the integrand is real on the stated interval because its minimum squared value is $(1-{\kappa }^2)/2>0$.

The endpoint events have $(t,x)=(0,0)$ and $(2T,0)$. Sam stays at $x=0$ in his inertial frame, so his elapsed proper time is $2T$. For $\kappa >0$, the integrand is strictly less than $\cos \theta$ except at $\theta =0$, and ${\tau }_A<2T$.

On the fixed parameter interval the integrand tends to $\cos \theta$ as $\kappa \to 0$ and is bounded by $\cos \theta$. The dominated convergence theorem, or uniform convergence here, givesThis equals Sam's time between departure and reunion. Physically, $dx/dt=-c\kappa \tan \theta$ tends uniformly to zero, so Alice's worldline approaches Sam's, and the time dilation disappears.

In the common $ct,x_1$ plane, multiplication of the two Lorentz boost matrices givesThis follows from the addition identities for hyperbolic sine and hyperbolic cosine. The transverse coordinates are unchanged by both transformations. Consequently the rapidity of $\overline{O}$ relative to $O$ isThe signed rapidities add because both boosts use the same spatial axis and the stated matrix convention.

By definition $v/c=\tanh \theta$ and $v^{\prime }/c=\tanh {\theta }^{\prime }$. Using the addition identity for the hyperbolic tangent and the rapidity sum from (i),Thus the velocity-addition formula isThe denominator is positive for the permitted subluminal signed velocities. For the nonnegative speeds in (iii), the composite is also nonnegative and remains below $c$.

When both speeds are much smaller than $c$, expand the denominator of the velocity-addition formula:This is the Galilean transformation limit. For $0\le \beta =v/c<1$ and $0\le {\beta }^{\prime }=v^{\prime }/c<1$,If both $v,v^{\prime }\to c$ from below, the right side tends to zero, soNo sum of these subluminal speeds exceeds the speed of light.

Let $V>0$ be the observer's speed, so its laboratory velocity is $-V{\mathbf{e}}_x$. The particle's laboratory speed along $+y$ follows from the energy–momentum relation:Here $E>m{c}^2$ because the massive particle is moving. Boosting to the observer's frame with velocity $-V$ gives, by the relativistic velocity-addition formula,Equivalently its four-momentum has $p_x^{\prime }=\gamma VE/c^2$, $p_y^{\prime }=p_y$. Thus $\tan \varphi =u/(\gamma V)$. Solving this equation, rather than using the Galilean transverse velocity, yieldsFor a strictly moving observer, $0<\varphi <\pi /2$. At $\varphi =\pi /2$ the limiting observer speed is zero; as $\varphi \downarrow 0$ it tends to $c$. The direction of the observer's velocity remains negative $x$; the displayed $V$ is its positive magnitude.

For a photon, $m=0$ and $u=c$, so the result from (i) reduces toThis is relativistic aberration for a photon initially transverse to the boost. It also follows directly from $p_x^{\prime }=\gamma VE/c^2$ and $E^{\prime }=\gamma E$: the observed direction cosine is $c{p}_x^{\prime }/E^{\prime }=V/c$. The transverse observed speed is $c/\gamma =c\sin \varphi$, and the total photon speed is still $c$. If desired the observed photon energy is $E^{\prime }=E/\sin \varphi$ for $0<\varphi \le \pi /2$.