Past exam of the relativity course of the University of Cambridge / 2014 / Paper 1 / 1

With $\gamma =(1-v^2/c^2{)}^{-1/2}$, four-momentum conservation gives total energy $E=(\gamma m_1+m_2)c^2$ and momentum $\mathbf{p}=\gamma m_1\mathbf{v}$. The compound system therefore hasHere $M^2{c}^4=E^2-c^2{p}^2$. The excess over $m_1+m_2$ is retained internal energy, assuming absorption without escaping radiation.

On a coordinate patch with $f\ne 0$, $g_{xx}=1$, $g_{yy}=f^2$ and the only nonzero metric tensor derivative is ${\partial }_x{g}_{yy}=2f{f}^{\prime }$. ThusAll remaining Christoffel symbols vanish. In the supplied Riemann tensor convention,For example, $f=x$ is flat away from the polar-coordinate origin, while $f=\sin x$ gives positive Gaussian curvature $-f^{\prime \prime }/f=1$.

Use the largest Newtonian tidal-tensor eigenvalue, the radial gravity gradient $2GM/r^3$. Imposing the supplied gradient threshold givesThis is about $1590\mathrm{km}$ from the centre, or $1580\mathrm{km}$ above the stated surface. The tangential eigenvalues have half the radial magnitude. Since $GM/(r_{\min }{c}^2)\simeq 1.4\times 10^{-3}$, the Newtonian tidal estimate is adequate; this calculation uses the given threshold without inferring any additional material-response model.