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

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.