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Put $r_s=2\mu$ and $f(r)=1-r_s/r$. Radial null geodesics of the Schwarzschild metric obeyNear the horizon, $\delta =r-r_s$ satisfies$C>0$ describes the exterior and $C<0$ the interior. The exponential approach in Schwarzschild coordinate time resembles part (a), but $\delta$ is not a proper radial distance.

Outside, set $\rho =2\sqrt{r_s\delta }$ and retain the leading near-horizon terms. The radial metric becomes the Rindler limit of a Schwarzschild horizon:It agrees with part (a) for $x=\rho$, $\alpha =c/(2r_s)$. Accordingly $\rho \propto e^{\pm ct/(2r_s)}$, while $\delta \propto {\rho }^2$ explains the factor two in its exponent.

Inside, write $r=r_s-{\rho }^2/(4r_s)$. The leading radial metric is insteadHere the radial coordinate is timelike and the Schwarzschild $t$ coordinate is spacelike. Thus $x<0$ in the Rindler metric does not represent the black-hole interior: $g_{tt}=(\alpha x{)}^2$ stays positive on both Rindler wedges. In the black-hole interior all future-directed radial light rays move towards decreasing $r$; the two signs of $dr/dt$ do not mean opposite future radial directions because $t$ is no longer time.

The horizon is a coordinate singularity in Schwarzschild coordinates and in the accelerated chart, not a curvature singularity. The full Schwarzschild geometry nevertheless has genuine tidal forces; the flat Rindler metric captures its leading local radial near-horizon structure.

For the Reissner-Nordstrom metric with $F=1-2M/r+Q^2/r^2$, the Kretschmann scalar isThe first expression follows by squaring the independent Riemann curvature tensor components in an orthonormal frame. This Kretschmann scalar diverges at $r=0$ for any nonzero $M$ or $Q$, but is finite at every positive root of $F$. Regular Eddington-Finkelstein coordinates show that those positive roots are coordinate singularities of the static chart. In the extremal case $Q^2=M^2>0$, its horizon value is $8/M^4$.

The radial vector field $B=e_r/r^2$ on punctured three-dimensional Euclidean space has flux $4\pi$ through every centered sphere. It has no vector potential smooth on an entire enclosing sphere: the Stokes theorem on a closed surface would force zero flux. Two overlapping patches suffice. In spherical coordinates take $A_{+}=(1-\cos \theta )e_{\varphi }/(r\sin \theta )$, regular away from the negative $z$ axis, and $A_{-}=-(1+\cos \theta )e_{\varphi }/(r\sin \theta )$, regular away from the positive $z$ axis. Their curls both equal $B$.

The Cartesian forms $A_{+}=(-y,x,0)/[r(r+z)]$ and $A_{-}=(y,-x,0)/[r(r-z)]$ show that the first is smooth at the north pole and the second at the south pole. On the overlap their difference is $A_{+}-A_{-}=2\nabla \varphi$, where the azimuth $\varphi$ is defined locally. Its multivaluedness around the axis explains the circulation difference $4\pi$. Splitting the sphere along a latitude, applying the Stokes theorem with $A_{+}$ on the north and $A_{-}$ on the south, and using their opposite boundary orientations recovers the total flux. A coordinate singularity in the tangential spherical basis does not by itself make a vanishing smooth vector field singular at a pole.

The zenith is the point on the celestial sphere in the direction of the observer's local upward vertical. In the horizontal coordinate system, its unit vector is ${\mathbf{e}}_U$ and its altitude is $90^{\circ }$. The zenith belongs to the observer's site; it is not generally the north celestial pole, whose altitude depends on geographical latitude. A direction's zenith distance is its nonnegative angle from the zenith, so $z=90^{\circ }-\gamma$. A direction of altitude $30^{\circ }$ therefore has zenith distance $60^{\circ }$. ^[ref].

At the zenith, every value of azimuth produces $(E,N,U)=(0,0,1)$. There is no horizontal component to identify a bearing: $atan2(0,0)$ cannot provide a geometrically meaningful azimuth, even if software returns a number. This coordinate singularity is a defect of the angular description; the viewing direction itself remains regular. The same ambiguity occurs at the opposite downward vertical. Near the zenith, two directions at fixed altitude separated by a small azimuth increment $d\eta$ have separation $\cos \gamma |d\eta |$ to first order, with angles in radians. At altitude $89^{\circ }$, a $1^{\circ }$ azimuth increment spans only about $0.01745^{\circ }$ on that small arc. An optical telescope can consequently require a large bearing change for a small change of pointing near the zenith; working with unit vectors avoids inventing a preferred bearing at the singular point.