Past exam of the relativity course of the University of Cambridge / 2013 / Paper 1 / 1

Use the angular-diameter ratio under a collinear boost, with motion along the common sightline and small angular radius $\vartheta$. For a boost toward the source, relativistic aberration of the incoming light gives $\tan ({\vartheta }^{\prime }/2)=e^{-\eta }\tan (\vartheta /2)$, where $\eta$ is the relative rapidity. Thus the observer seeing the smaller disk has angular diameter reduced by $e^{-\eta }$. The ratio two implies $e^{\eta }=2$, soThe answer is $0.60c$. Both observations occur at one event, so this is a relativistic aberration comparison rather than a change of geometrical distance during observation. Arbitrary transverse relative motion would require additional directional information; the stated motion toward the same distant target selects the collinear interpretation.

For geodesic motion of uncharged dust, use the pressureless stress-energy form, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$ and $u_{\nu }{u}^{\nu }=c^2$. stress-energy conservation expands toContracting with $u_{\nu }$ kills the second term, since $u_{\nu }{\nabla }_{\mu }{u}^{\nu }=\frac{1}{2}{\nabla }_{\mu }(c^2)=0$. Hence ${\nabla }_{\mu }(\rho u^{\mu })=0$, and wherever $\rho >0$,On a flow line $u^{\mu }=d{x}^{\mu }/d\tau$, so this becomes $d^2{x}^{\nu }/d{\tau }^2+{\Gamma }^{\nu }{}_{\alpha \beta }(d{x}^{\alpha }/d\tau )(d{x}^{\beta }/d\tau )=0$. Each relativistic dust element therefore follows a timelike geodesic with proper time as an affine parameter. pressure gradients or external nongravitational forces would add an acceleration term.

The Christoffel symbols of a spatially flat FLRW metric depend on the time-coordinate convention. Here $x^0$ is $t$, not $ct$. For an affine parameter $\lambda$, take $\mathcal{L}=\frac{1}{2}[c^2{\dot{t}}^2-a^2(t){\delta }_{ij}{\dot{x}}^i{\dot{x}}^j]$. Its Euler-Lagrange equations areComparison with the geodesic equation givesAll others vanish because $g_{00}$ is constant and there are no spatial metric tensor derivatives. The lower-index interchange is essential: the printed list specifies one member of each symmetric mixed pair. The factor of two in the spatial geodesic equation comes from those two members.