Past exam of the general relativity II course of the University of Oxford / 2015 / 2

A nonzero vector field has hypersurface orthogonality when it is locally normal to a family of hypersurfaces. In terms of its associated differential one-form $K^{♭}=g_{\mu \nu }{K}^{\nu }d{x}^{\mu }$, the Frobenius theorem gives the equivalent conditionslocally with nonzero $f$ and $dh$; square brackets denote tensor index antisymmetrization.

All coefficients of the Reissner-Nordstrom metric are independent of $t$, and $K={\partial }_t$ has constant coordinate components. Thus the Lie derivative obeys ${\mathcal{L}}_{K}g={\partial }_{t}g=0$: $K$ is a Killing vector field. Moreover,Thus $K$ has hypersurface orthogonality, with the $t=\text{constant}$ hypersurfaces providing the orthogonal family wherever this chart is regular. It is timelike where $F>0$, since $g(K,K)=-F$. Its hypersurface orthogonality also holds in a regular extension across a horizon; singular static coordinates do not invalidate the Frobenius theorem test.

For the three cases, distinguish a curvature singularity from a coordinate singularity. The Reissner-Nordstrom Kretschmann scalar isIt diverges at $r=0$ when either $M$ or $Q$ is nonzero. Any positive root of $F$ instead admits Eddington-Finkelstein coordinates: set $v=t+\int dr/F$ to obtainThe determinant of its $(v,r)$ block is $-1$, including at $F=0$. Hence positive roots of $F$ are removable coordinate singularities of the static chart. The apparent degeneracy at the poles of the angular chart is also a coordinate singularity.

If $M^2<Q^2$, thenThere is no positive root and no horizon hiding $r=0$ in the extended Reissner-Nordstrom spacetime. Its Kretschmann scalar diverges there, so $r=0$ is a naked singularity. It is a timelike curvature singularity, as the normal to nearby $r=\text{constant}$ hypersurfaces has positive squared norm $g^{rr}=F$.

If $M^2=Q^2$ with $M>0$, thenThe curvature singularity remains at $r=0$. At $r=M$ the static chart has a coordinate singularity, removed by the Eddington-Finkelstein coordinates above. The Kretschmann scalar there equals $8/M^4$, which is finite. The coincident horizons form a degenerate Killing horizon, and its future outer branch is the event horizon of the extremal black hole. A double zero of $F$ does not remove the horizon.

For $0<Q^2<M^2$ there are two distinct positive roots:Both roots are coordinate singularities of the static chart. In the standard extension, $r_{+}$ is the outer event horizon and $r_{-}$ is an inner Cauchy horizon. The curvature singularity is $r=0$. Between the horizons $F<0$, so the radial direction is timelike; for $0<r<r_{-}$, $F>0$ again and the central curvature singularity is timelike. These are the three radial regions of the global structure of a charged spherical black hole.^[ref]

For completeness, if $Q=0$ and $M>0$, the Schwarzschild spacetime limit has just the positive horizon $r=2M$ and a spacelike curvature singularity at $r=0$; the formal root $r_{-}=0$ of $r^{2}F$ is not a regular inner horizon. If $M=Q=0$, the metric tensor is flat and the radial origin is regular in Cartesian coordinates.

A regular hypersurface $h=0$, with $dh\ne 0$, is a null hypersurface if its normal is null:The raised normal is then also tangent to the hypersurface, and its induced metric is degenerate. This is a local geometric property.

In an asymptotically flat spacetime, choose its future null infinity ${\mathcal{I}}^{+}$. The black hole region and future event horizon areHere $J^{-}({\mathcal{I}}^{+})$ is the causal past of future null infinity: the region from which a future-directed causal curve can escape to arbitrarily distant observers. An event horizon therefore depends on the global future causal structure, and an arbitrary null hypersurface need not be an event horizon.

In the extremal case, the squared norm of the Killing vector field isFor $M=|Q|>0$, it is negative for every regular $r\ne M$ and vanishes precisely at $r=M$ in the extended spacetime. ThusThe static chart cannot itself test the geometry at its horizon. In the regular Eddington-Finkelstein coordinates, the normal to $h=r-M=0$ is $n=dr$, andAt $r=M$ this gives $n^{♯}={\partial }_v=K$, a nonzero null vector tangent to the surface. Therefore $r=M$ is a null hypersurface, even though $F$ has the same sign on its two sides.

On the equatorial path, the Lagrangian per unit mass isBoth $t$ and $\varphi$ are cyclic coordinates. The Euler-Lagrange equations therefore conserve their canonical momenta:ConsequentlyHere $E$ is the total conserved energy in the specified gauge of the electromagnetic four-potential. The paper's $l$ is the angular momentum per unit mass; the physical angular momentum is $ml$. The equatorial choice is consistent with the $\theta$ Euler-Lagrange equation, since its angular force term contains $\sin \theta \cos \theta =0$ at $\theta =\pi /2$ and the electromagnetic four-potential has no angular component.

With $\tau$ the proper time of the massive test particle, the four-velocity has squared norm $-1$:The conserved quantities in part (i) giveMultiplying the normalization by $F$ and substituting gives the charged-particle radial equation in Reissner-Nordstrom spacetime:At a radial turning point $r_{\ast }>0$, the left side is zero, soThe parenthesis on the left is strictly positive. Hence $F(r_{\ast })\ge 0$. The factorization in part (b) now yieldsThere is no radial turning point between the horizons. Indeed, for $r_{-}<r<r_{+}$, the right-hand side of the radial equation is strictly positive, independently of the electric charge $e$ of the test particle. This is the charged-particle turning-point exclusion in Reissner-Nordstrom spacetime.

The inequalities are necessary, not a claim that a radial turning point exists for every $E,l,e$. At a regular horizon crossing a timelike vector cannot be tangent to a null hypersurface, so a massive test particle actually has $\dot{r}\ne 0$ there. Algebraic equality at $r_{\pm }$ would require $E=eQ/r_{\pm }$ and does not describe a regular finite-proper time radial turning point on the horizon. In the Schwarzschild spacetime limit $Q=0$, the positive-radius inner alternative disappears.