Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 52 / 1

Use geometric units $G=c=1$, metric signature $(-+++)$, and a four-velocity normalized by the physical metric. Write $\varphi =1+\Phi$, where $|\Phi |\ll 1$. For a pressureless perfect fluid, the dust stress-energy tensor is $T_{ij}=\rho U_i{U}_j$ and $g^{ij}{U}_i{U}_j=-1$. Since ${\eta }^{ij}={\varphi }^2{g}^{ij}$, its background trace isConsequently the field equation is $(-{\partial }_t^2+\Delta )\varphi =4\pi \rho {\varphi }^3$. Neglecting time derivatives and higher weak-field orders gives the Newtonian limit of Nordstrom gravity:The constant shift between $\varphi$ and $\Phi$ does not affect the Poisson equation or the force. Here $\rho$ is the physical rest density; if fluid indices are instead defined using the background metric, the intermediate powers of $\varphi$ change but the same leading-order equation results.

For the motion, the Levi-Civita connection of the conformally flat metric isIn particular ${\Gamma }^{\alpha }{}_{00}={\partial }_{\alpha }\log \varphi$. Replacing the affine parameter in the geodesic equation by coordinate time gives, with $w^i=(1,\mathbf{v})$ and $\mathbf{v}=d\mathbf{x}/dt$,At small speeds and for a nearly static field, this becomesThus the Newtonian gravitational potential and freely falling trajectories agree to the stated weak-field, slow-motion order. The comparison is an approximation, not an exact equivalence of relativistic trajectories.

Yes, at the weak-field accuracy of the experiment. The Pound-Rebka experiment measures gravitational redshift, which Nordstrom theory of gravitation predicts even though it does not predict bending of light.

In a static field, a stationary clock has $d\tau =\varphi dt$. A photon has constant Killing energy $E=-p_t$, and an observer at rest has $U={\varphi }^{-1}{\partial }_t$. Its measured frequency is proportional to $-p_i{U}^i=E/\varphi$. Emission at $e$ and reception at $r$ therefore giveAn upward ray has ${\Phi }_r-{\Phi }_e\simeq gH/c^2$ when ordinary units are restored, so $\Delta \nu /\nu \simeq -gH/c^2$. This is the shift tested using the Mössbauer effect. The gravitational redshift follows from the physical clock rates; a claim that straight light paths imply no frequency shift would confuse two different observables.

No: the theory predicts zero gravitational deflection of light. The reason is conformal preservation of null geodesic paths. Along a null geodesic with tangent $k^i$, the geodesic equation for a conformally flat metric reduces tobecause ${\eta }_{ij}{k}^i{k}^j=0$. The acceleration is parallel to the tangent and changes only its parametrization. Equivalently, with ${\ell }^i={\varphi }^2{k}^i$, $d{\ell }^i/d\lambda =0$, so the unparametrized ray is a straight null geodesic of Minkowski spacetime. A positive conformal factor also preserves local angles, so this straight-path result cannot be hidden by changing the local angle measurement. It contradicts the nonzero solar deflection observed in gravitational lensing, while remaining compatible with the gravitational redshift in part (b).