Past exam of the symmetry and relativity course of the University of Oxford / 2022 / 3

For a massive particle, the proper time is determined by the Minkowski interval along its worldline:Choose future-directed time increments. With the Lorentz factor ${\gamma }_u=(1-|\mathbf{u}{|}^2/c^2{)}^{-1/2}$,We use $U^{\alpha }$ for the source's $u^{\alpha }$ to distinguish four-velocity from velocity $\mathbf{u}$. Since proper time is invariant and $d{x}^{\alpha }$ is a four-vector, $U^{\alpha }$ is a four-vector, with $U_{\alpha }{U}^{\alpha }=-c^2$. The rest-frame result $U^{\alpha }=(c,0)$ follows immediately.

The Lorentz boost from A to B givesDivide each spatial displacement by $d{t}^{\prime }={\gamma }_{v}dt(1-v{u}_x/c^2)$. The velocity-addition formula isThe same result follows by transforming the four-velocity and using $w_i=c{U}^{\prime i}/U^{\prime 0}$. The transverse velocities change because the transformed time increment depends on $dx$, even though the transverse spatial coordinates are unchanged.

For $v=|\mathbf{v}|>0$, decompose the velocity as $\mathbf{u}={\mathbf{u}}_{\parallel }+{\mathbf{u}}_{\perp }$, parallel and perpendicular to $\mathbf{v}$. A rotation-free Lorentz boost gives the velocity-addition formulaThese two components are orthogonal, soThe last equality is Lagrange's identity for the cross product. For physical subluminal speeds, $D>0$, givingThe $\mathbf{v}=0$ case follows directly or by continuity. An independent form useful for checks is $1-|\mathbf{w}{|}^2/c^2=(1-u^2/c^2)(1-v^2/c^2)/D^2$, obtained from the transformed four-velocity normalization.

Let ${\mathbf{v}}_1=(c/2,0,0)$ and ${\mathbf{v}}_2=(c/(2\sqrt{2}),c/(2\sqrt{2}),0)$ in A. Each has Lorentz factor $g=2/\sqrt{3}$. The relative Lorentz factor of $S_2$ in the rest frame of $S_1$ follows from their four-velocity tensor contraction:Let $b=\sqrt{1-{\Gamma }^{-2}}$ be the relative speed divided by $c$. In the rest frame of $S_1$, take $\theta$ between the photon's propagation direction and $S_2$'s velocity. The observed angular frequency is the invariant $\omega =-k\cdot U$. Applying it to the emitter, whose rest angular frequency is ${\omega }_0$, gives ${\omega }_0=\Gamma {\omega }_1(1-b\cos \theta )$. Since ${\omega }_1={\omega }_0$,This is a relativistic Doppler effect calculation using both emitters' and receivers' rest clocks; the time delay of reception introduces no extra factor for constant velocities.

For the direction in A, write the wave four-vector as $k^{\mu }=({\omega }_A/c)(1,\mathbf{n})$, where $\mathbf{n}$ is a unit vector. Equality of the emitted and received angular frequencies gives $k\cdot (U_1-U_2)=0$. Since the two laboratory Lorentz factors are equal, their time components cancel:Thus the complete answer in three spatial dimensions isIf propagation is additionally assumed to lie in the stars' motion plane, this reduces to either direction along the internal angle bisector, $22.5^{\circ }$ or $202.5^{\circ }$ from $+x$. Reflection exchanging the stars' laboratory velocities explains the angle bisector by symmetry. The source does not specify coplanar propagation or the stars' positions, so it does not fix the sign or exclude an out-of-plane ray. For example $\mathbf{n}=(0,0,1)$ gives ${\omega }_1={\omega }_2=g{\omega }_A$ and satisfies the same no-shift condition. Initial positions can be chosen to make such a ray connect emission and reception. All these possibilities give the same rest-frame $\theta$ above.