Past exam of the short options course of the University of Oxford / 2014 / 7

Use the Minkowski metric with signature $(+---)$ and write $U=(U^0,\mathbf{u})$, $V=(V^0,\mathbf{v})$. Future-pointing timelike means $U^0>\Vert \mathbf{u}\Vert$ and $V^0>\Vert \mathbf{v}\Vert$. Thusso $U+V$ is again future-pointing timelike.

Let $a=\sqrt{g(U,U)}>0$ and $b=\sqrt{g(V,V)}>0$. In a Lorentz frame where $U=(a,0,0,0)$, the future-pointing vector $V$ has $V^0\ge b$. Consequently $g(U,V)=a{V}^0\ge ab$, with equality exactly when $V$ has no spatial component in that frame. HenceTaking positive square roots provesEquality holds exactly when $U,V$ are positive scalar multiples of each other. This is the reverse triangle inequality for future timelike vectors, not the Euclidean triangle inequality.

Take spacetime coordinates $X=(ct,x,y,z)$. Differentiating the initial segment gives the four-velocityThe identity ${\cosh }^{2}r-{\sinh }^{2}r=1$ gives $g(U,U)=c^2$, and $dt/ds=\cosh \omega s>0$. The proper time element therefore satisfies $d{\tau }^2=g(dX,dX)/c^2=d{s}^2$, with the future orientation selecting $d\tau =ds$. Thus $s$ is proper time along this path.

Since $s$ is proper time, the four-acceleration isIts Minkowski norm squared is $g(A,A)=-c^2{\omega }^2$, so the invariant magnitude of proper acceleration isThe vector is spacelike, which accounts for the minus sign under the square root.

Differentiate each segment and express the four-velocity using a continuous rapidity:At $T$ both rapidities are $\omega T$; at $3T$ both are $-\omega T$. Hence the four-velocity is continuous at both junctions. This expression also gives $g(U,U)=c^2$ everywhere, so the parameter remains proper time on the later segments.

Away from the junctions,with ${\theta }^{\prime }$ successively $+\omega ,-\omega ,+\omega$. The rapidity is continuous but its slope reverses. Thus the four-acceleration has a finite jump to its negative at each junction, while its magnitude stays $c\omega$ on either side. There is no velocity jump; the idealized path is continuously differentiable but not twice differentiable at the switching events.

At $s=0$ and $s=4T$, the mission is at the planet's position $x=c/\omega$. The final coordinate time from the last segment isThe planet is at rest in these coordinates, so its proper time elapsed equals this coordinate-time difference. The astronauts' proper time is the parameter interval, $4T$. ThereforeFor proper acceleration $c\omega =10\mathrm{m}{\mathrm{s}}^{-2}$ and $c=3\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, ${\omega }^{-1}=3\times 10^7$ seconds, approximately one year. Thus $\omega T=16$ and $e^{16}\approx 3000^2=9\times 10^6$. Neglecting $e^{-16}$ in the hyperbolic sine,The clocks compare two different timelike paths between the same departure and reunion events.

Put $a=\omega T$. At emission, $s=2T$, the middle segment givesThe planet is at $x_p=c/\omega$, so the inward light signal takes coordinate timeIts arrival time on the planet is therefore $t_r=2(e^a-1)/\omega$. Subtract this from the mission's return time $4\sinh a/\omega$:For large $T$ at fixed acceleration this tends to $2/\omega =2c/(c\omega )$. With the Earth-gravity value above, the wait is approximately two years, despite the millions of planet-years elapsed since departure in part (i).