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Let $r_h=\sqrt{2m}$ and $f(r)=1-r_h^2/r^2$. The geometry is the five-dimensional Schwarzschild-Tangherlini metric. Introduce a tortoise coordinate by $d{r}_{\ast }/dr=f^{-1}$. Decomposing the integrand givesAn irrelevant additive constant has been fixed. Set $v=t+r_{\ast }$ and $u=t-r_{\ast }$. Substituting $dt=dv-dr/f$ or $dt=du+dr/f$ respectively gives the ingoing Eddington-Finkelstein coordinates and outgoing Eddington-Finkelstein coordinates:The radial-time block in each has determinant $-1$, so the apparent singularity at $f=0$ has disappeared. The ingoing chart crosses the future event horizon, and the outgoing chart crosses the past horizon. Radial null geodesics in the ingoing chart obey either $v=\text{constant}$ or $dr/dv=f/2$, displaying how outward light is trapped when $r<r_h$.

For the maximal extension, the exterior radial metric is $-fdudv$. Since the surface gravity is $\kappa =f^{\prime }(r_h)/2=1/r_h$, choosein the right exterior. Their product and the extended metric areThe coefficient at the horizon is finite and nonzero, $-4r_h^2{e}^{-2}$. The product relation therefore extends smoothly through either horizon. For $r>r_h$, $UV<0$ gives two exterior regions. For $0<r<r_h$, $0<UV<1$ gives the future black-hole interior with $U,V>0$ and the past white-hole interior with $U,V<0$. The Kruskal extension of the five-dimensional Schwarzschild-Tangherlini metric has a bifurcation three-sphere at $U=V=0$.

The singularity $r=0$ corresponds to $UV=1$. It is a curvature singularity: for example, its Kretschmann scalar is $72r_h^4/r^8$. Since $g^{ab}{\partial }_{a}r{\partial }_{b}r=f<0$ in the interior, constant-$r$ surfaces are spacelike there, and the singularities are spacelike boundaries. Define $P=\arctan U$, $Q=\arctan V$, then compactified coordinates $T=(Q+P)/2$, $X=(Q-P)/2$. Radial null curves have $T\pm X=\text{constant}$. Moreover, $\tan P\tan Q=1$ gives $T=\pm \pi /4$ at the two singularities. This constructs the Penrose diagram, not just its horizon labels:

Each point away from the singular boundaries represents a three-sphere, not a two-sphere. The diagram has two asymptotically flat exteriors with separate past and future null infinities, a future black hole and a past white hole. The diagram of a collapse spacetime would omit the second exterior and the white-hole region; the maximal extension of the metric given here contains both.

For the Komar integral, normalize the stationary Killing vector field as $k={\partial }_t$, with unit norm at infinity. The Killing equation makes ${\nabla }^a{k}^b$ antisymmetric, and its divergence is proportional to $R^b{}_c{k}^c$. In a vacuum exterior this vanishes, so Stokes theorem makes the flux independent of the enclosing sphere. In the weak-field limit its radial component measures the derivative of the gravitational potential; its large-sphere flux is therefore proportional to the asymptotic gravitational mass. Normalization of $k$ is needed because rescaling it would rescale the integral.

Specify the surface orientation to fix the sign and factor. On a static exterior slice take future unit normal $u^a=f^{-1/2}({\partial }_t{)}^a$, outward radial normal $n^a=f^{1/2}({\partial }_r{)}^a$ andThen $d{\Sigma }^{tr}=-dA$ and $d{\Sigma }^{rt}=dA$, with both antisymmetric components included. Since $k_t=-f$,The three-sphere area is $2{\pi }^2{r}^3$, and $f^{\prime }=2r_h^2/r^3=4m/r^3$. ThusReversing the binormal orientation gives $+8{\pi }^{2}m$; a convention with half the antisymmetric binormal gives half this magnitude. The question does not fix those conventions, so the choice above is explicit.

For comparison with physical mass, the Komar mass normalization in higher dimensions isIn five dimensions this givesThe metric parameter $m$ has dimensions of length squared and is not itself the physical mass. An independent horizon check is $I_K=-2\kappa A_H$ with $A_H=2{\pi }^2{r}_h^3$, which reproduces the same integral. The dimension-adjusted normalization also agrees with the ADM mass; the four-dimensional Komar prefactor should not be reused unchanged.

Write $f=1-r^2/a^2$ and $r_{\ast }=a\mathrm{artanh}(r/a)$. Since $d{r}_{\ast }/dr=f^{-1}$, $dt=du+dr/f$. Substitution gives the outgoing Outgoing Eddington-Finkelstein coordinatesThe radial block has determinant $-1$, independent of $f$, and all coefficients are smooth at $r=a$. This is a coordinate singularity of the static chart, not a spacetime singularity. The transformation was defined inside the horizon, but the new metric itself extends smoothly across it.

At $r=0$ the spherical angular coordinates degenerate as usual. Cartesian spatial coordinates make the original static metric regular there: $f=1+O(r^2)$ and the extra spatial radial correction is $x_i{x}_j/[a^2(1-r^2/a^2)]$. Thus the origin is not a spacetime singularity either.

Define outgoing Eddington-Finkelstein coordinates by $u=t-r_{\ast }$. The radial Schwarzschild metric becomesOutgoing radio rays have constant $u$. Two nearby wave crests emitted at Alice's proper times $\tau$ and $\tau +d\tau$ have separation $du={\dot{u}}_{A}d\tau$. At Bob's fixed radius $r_B$, $d{t}_B=du$ and $d{\tau }_B=\sqrt{f_B}du$, where $f_B=1-2M/r_B$. Thus the exact local redshift isHere the wavelength ratio equals the proper time period ratio in geometric optics. Equivalently, the phase gradient $k_a=-{\omega }_{\infty }(du{)}_a$ gives Alice's measured frequency ${\omega }_A=-k_a{U}_A^a={\omega }_{\infty }{\dot{u}}_A$ and Bob's ${\omega }_B={\omega }_{\infty }/\sqrt{f_B}$.

In the asymptotically distant-observer limit $f_B\to 1$, this becomes ${\lambda }_B/{\lambda }_A=d{u}_A/d{\tau }_A$. “Far away” is the approximation behind the printed formula; at finite $r_B$ the factor $\sqrt{f_B}$ remains.