Past exam of the relativity course of the University of Cambridge / 2014 / Paper 1 (Relativity; Physics (2) and Physical Sciences: Half Subject Physics (2))

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.

inertial mass sets the response to force, $\mathbf{F}=m_i\mathbf{a}$. Passive gravitational mass sets the response to a given gravitational potential, ${\mathbf{F}}_g=-m_g\nabla \Phi$; active gravitational mass describes a body's gravitational sourcing. Universality of the passive-to-inertial mass ratio makes freely falling trajectories independent of composition. torsion balance comparisons and differential free-fall experiments test this weak equivalence principle by searching for composition-dependent acceleration, with environmental forces and instrumental asymmetries controlled.

The Einstein equivalence principle further says that local nongravitational experiments in a freely falling laboratory obey special relativity, independent of the laboratory's location and velocity. A local inertial coordinate system can remove Christoffel symbols at one event, but cannot remove a nonzero Riemann tensor throughout a finite region. Relative acceleration of separated freely falling particles reveals tidal gravity. Thus a uniformly accelerating laboratory in flat spacetime can mimic a local gravitational field, while extended tidal measurements distinguish genuine curvature. Equality of the measured mass responses does not make every gravitational field globally equivalent to acceleration.

Use coordinates $x^{\mu }=(ct,\mathbf{x})$, signature $(+---)$, and the convention $F^{0i}=-E_i/c$, $F^{ij}=-{ϵ}_{ijk}{B}_k$. The antisymmetric electromagnetic field tensor combines six field components:It transforms as $F^{\prime \mu \nu }={\Lambda }^{\mu }{}_{\alpha }{\Lambda }^{\nu }{}_{\beta }{F}^{\alpha \beta }$. Lowering indices changes the time–space signs: $F_{0i}=E_i/c$, while $F_{ij}=F^{ij}$. The covariant tensor instead uses the inverse coordinate Jacobian on each lower index.

For a frame moving with velocity $\mathbf{v}$ relative to the original one,Electric and magnetic fields therefore mix under a boost; they are frame-dependent parts of one tensor. For example, a purely transverse electric field acquires a magnetic field in a boosted frame. The electromagnetic field invariants $B^2-E^2/c^2$ and $\mathbf{E}\cdot \mathbf{B}$ restrict which fields can be eliminated by a frame change. The covariant force law $d{p}^{\mu }/d\tau =q{F}^{\mu }{}_{\nu }{U}^{\nu }$ guarantees consistent particle dynamics in all inertial frames.

The strongest pre-2014 evidence combines large gravitating mass with extreme compactness. In an X-ray binary, companion-star radial velocities determine a binary mass function; orbital inclination, donor properties and distance complete the mass inference. An unseen component above conventional neutron-star masses is a black-hole candidate. For example, a 2011 Cygnus X-1 analysis combined optical photometry, spectroscopy and distance information to infer a compact-object mass about $14.8M_{\odot }$, with model and observational uncertainties.^[ref]

Near Sagittarius A*, infrared stellar astrometry and radial velocities trace nearly Keplerian orbits around a dark compact mass. The concentration of several million solar masses in a small volume strongly challenges an extended cluster of ordinary remnants; source confusion, reference-frame errors, distance and possible extended mass must be fitted.^[ref] Gas and maser dynamics provide related compact-mass tests in other nuclei.

accretion explains luminous X-rays, rapid variability and broad relativistic disk features. These constrain size and the strong-field environment, but emission alone is not unique to an event horizon. Lack of surface-like signatures is supporting, model-dependent evidence. Mass, compactness and consistent accretion phenomenology together support the black-hole interpretation; large-radius mass measurements alone do not directly demonstrate a horizon.

Boost and velocity transformation. For the moving frame,The inverse has $v\mapsto -v$. Taking differentials of that inverse and dividing by $dt$ gives

Constant proper acceleration. For collinear motion, the instantaneous rest-frame acceleration is $a={\gamma }^{3}dv/dt$. Since $dt/d\tau =\gamma$,Write $v=c\tanh \eta$; then $d\eta /d\tau =a/c$. Starting from rest gives $\eta =a\tau /c$, $dt/d\tau =\cosh \eta$ and $dx/d\tau =c\sinh \eta$. integration with the stated initial origin yields

photon rocket mass ratio. In the instantaneous rest frame, ejecting backward photon energy $d{E}_{\gamma }$ gives, to first order, $c^{2}dM=-d{E}_{\gamma }$ and $Mcd\eta =d{E}_{\gamma }/c$. Hence $dM/M=-d\eta$, andHere $M$ includes all stored fuel and energy. This assumes complete conversion into a perfectly collimated photon exhaust without other losses.

photon-rocket fuel for acceleration and braking. With effectively instantaneous changes of speed, the cruise proper time is $D/(\gamma v)=10\mathrm{yr}$, givingBraking to rest costs another factor $e^{-\eta }$, so $M_f/M_i=e^{-2\eta }=(\sqrt{101}-10{)}^2=0.00248758$. If equal rest masses of matter and antimatter annihilate, the consumed antimatter is half the total consumed fuel:This is the fraction of initial total mass, not the antimatter fraction of the consumed matter–antimatter fuel, which is one half.

finite-acceleration correction to a relativistic cruise. Each constant-proper-acceleration leg has proper duration ${\tau }_a=c\eta /a$ and Earth-frame length $x_a=(c^2/a)(\gamma -1)$. ThusSubtracting the original $D/(\gamma v)$ givesFor $a=10\mathrm{m}{\mathrm{s}}^{-2}$, each accelerating leg takes about $2.85$ proper years and covers $8.60$ light years in the Earth frame, leaving positive cruise distance. The new total is about 13.98 proper years. Adding $2{\tau }_a$ to ten years would overcount the cruise segment replaced by acceleration and braking.

geodesic integrals and affine normalization. Put $F(r)=1-2\mu /r$. Choose the affine parameter to be proper time, so $g_{ab}{\dot{x}}^a{\dot{x}}^b=c^2$. Cyclic $t$ and $\varphi$ in the geodesic Lagrangian give $F\dot{t}=k$ and $r^2\dot{\varphi }=h$. Substituting into the four-velocity normalization giveshenceAn arbitrary constant rescaling of affine parameter changes its normalization; the displayed unit constant assumes the proper-time choice.

circular orbit. Define $V(r)=h^2/r^2-2\mu h^2/r^3-2\mu c^2/r$. The radial geodesic equation is $\ddot{r}=-V^{\prime }/2$. A circular geodesic therefore needs $V^{\prime }(R)=0$, not merely $\dot{r}=0$ at an instant. Solving and substituting into the radial first integral givesExistence requires $R>3\mu$. circular orbits below $6\mu$ are radially unstable, although they still define the instantaneous comparison here.

orthogonal local velocities in Schwarzschild spacetime. Use Alice's static orthonormal frame at the meeting event. Her proper clock is $d{\tau }_A=\sqrt{F}dt$ and tangential proper distance is $Rd\varphi$. Bob's local tangential speed isCarol has $h_C=0$ and $k_C=1$, because she starts from rest at infinity. The radial integral gives ${\dot{r}}_C=-c\sqrt{2\mu /R}$. Her static-frame radial speed is $(dr/\sqrt{F})/(\sqrt{F}dt)=dr/(Fdt)$, and $F{\dot{t}}_C=1$, soThis is inward radial motion, perpendicular to Bob's tangential motion in Alice's frame.

Their Lorentz factors relative to Alice are ${\gamma }_{\mathrm{AB}}=\sqrt{(R-2\mu )/(R-3\mu )}$ and ${\gamma }_{\mathrm{AC}}=\sqrt{R/(R-2\mu )}$. The four-velocity contraction therefore gives

Special-relativistic consistency. Let Bob's tangential velocity in Alice's local frame be $u$ and Carol's inward radial speed be $w$. A boost to Bob's frame gives Carol components $v_{\mathrm{tang}}^{\prime }=-u$ and $v_{\mathrm{rad}}^{\prime }=-w/{\gamma }_u$. ConsequentlyThus the three measured relative speeds agree with local Lorentz velocity addition. Ordinary Euclidean subtraction of the two velocity vectors would omit the transverse-velocity transformation.