Past exam of the general relativity and cosmology course of the University of Oxford / 2018 / 2

In coordinates $(ct,r,\varphi ,z)$ the diagonal metric tensor is $\mathrm{diag}(-1,1,f^2,1)$, and only $g_{\varphi \varphi }$ depends on position. Substitution into the connection formula giveswith all other Christoffel symbols zero away from singular axes.

Use the explicit Ricci tensor formula and the Einstein equation sign printed on the source formula sheet. They giveFor example, in $R_{rr}$ the two terms proportional to $(f^{\prime }/f{)}^2$ cancel. For $R_{\varphi \varphi }$ the derivative of $-f{f}^{\prime }$ contributes $(f^{\prime }{)}^2+f{f}^{\prime \prime }$, the product terms contribute $-2(f^{\prime }{)}^2$, and the determinant term contributes $(f^{\prime }{)}^2$.

Formula-sheet qualification. Directly contracting the separately printed Riemann curvature tensor as instructed on that sheet instead gives $R_{rr}=-f^{\prime \prime }/f$, the opposite sign. For the concrete example $f(r)=\ell \sin (r/\ell )$, that contraction gives $+1/{\ell }^2$, whereas the explicit Ricci tensor formula gives $-1/{\ell }^2$. The two source formulas cannot both hold with the stated contraction. The calculation here uses the explicit Ricci tensor formula, whose pairing with the negative Einstein equation coupling yields the requested stress-energy tensor.

The desired matter tensor assumes $\Lambda =0$; otherwise the cosmological-constant contribution must also be included. With that assumption,and raising both indices givesThe static observer measures energy density $T^{00}=-c^4{f}^{\prime \prime }/(8\pi Gf)$ and equal-magnitude negative axial pressure when that energy is positive. The axial stress is a tension. Keeping the source's curvature sign and its negative Einstein-equation coupling together is essential to this result.

Assume $\varphi$ has period $2\pi$, $f>0$ in the regular bulk, and consider a unit proper length in $z$. The proper transverse area element is $dA=fdrd\varphi$, while the mass density is $T^{00}/c^2$. The axial boundary term in the mass of a cylindrical spacetime is thereforeChoose the usual deficit branch $0\le \lambda <1$, so $f_{\mathrm{out}}=(1-\lambda )r$. Continuity of the induced metric requires $f(R^{-})=(1-\lambda )R$. If there is no distributional thin shell at $R$, continuity of the extrinsic curvature also requires $f^{\prime }(R^{-})=1-\lambda$. A jump in $f^{\prime }$ would add a surface-layer stress and cannot simply be discarded.

Using the printed inner condition $f^{\prime }(0^{+})=0$, these assumptions yieldThis is the literal smooth-bulk answer. It is negative for the ordinary deficit range, so it must not be silently replaced by the usual positive cosmic-string mass formula.

For comparison, a regular axis with proper radial coordinate and angular period $2\pi$ requires $f(0)=0$ and $f^{\prime }(0)=1$, since the circumference-to-radius ratio must approach $2\pi$. Under that different regular-axis condition the same integral givesThe original PDF really prints the derivative limit zero. It describes either a nonregular axis or an inner geometry different from an ordinary smooth string core; any additional singular-axis matter is not included in the bulk integral above and would need its own stress specification. Standard regular-core matching can also require an offset in the exterior radial origin. Thus the regular-axis formula is a qualified physical interpretation, not a derivation from the printed zero. For a general exterior branch replace $1-\lambda$ by $|1-\lambda |$ in the boundary derivative.

Outside the core, $f^{\prime \prime }=0$ and the metric is locally flat: set $\tilde{\varphi }=(1-\lambda )\varphi$. Static test particles have no Newtonian-like attraction because $g_{00}$ is constant and ${\Gamma }^r{}_{00}=0$. Local trajectories are straight lines in locally unwrapped coordinates, so a very thin dark core is hard to identify through ordinary orbital accelerations or local tidal forces.

The global angular range is nevertheless $2\pi (1-\lambda )$ rather than $2\pi$. The missing angle $\delta =2\pi \lambda$ can give gravitational lensing with two nearly identical, essentially undistorted images of a background galaxy, and multiple such pairs aligned along the projected object. For a small deficit, a straight string and favourable geometry, their angular separation is of order $\delta D_{ls}/D_s$, with an additional inclination factor for a tilted string. The matching spectra and common line of image pairs distinguish this from an ordinary compact lens.

A moving cosmic string can also induce a step in the cosmic microwave background temperature across its path. That signature needs motion in addition to the static geometry supplied here. These are consequences of the exterior conical geometry; interpreting the inferred deficit as a positive mass per length requires the regular-axis/core assumptions distinguished in part b.