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For an asymptotically flat stationary solution in its asymptotic rest frame, the ADM mass is the gravitational mass measured by the Arnowitt-Deser-Misner energy. It is fixed by the asymptotic metric falloff, not by a coordinate parameter's name. If spatial ADM momentum is present, distinguish the asymptotic energy component from the invariant mass obtained from the energy-momentum norm. For a static Schwarzschild-Tangherlini metric in $D$ dimensions with $f=1-\mu /r^{D-3}$, the mass is $(D-2){\Omega }_{D-2}\mu /(16\pi G_D)$; the dimensional normalization is also reproduced by the Komar mass normalization in higher dimensions.

For future static-slice normal $u$ and outward radial normal $n$, take $d{\Sigma }_{ab}=(n_a{u}_b-n_b{u}_a)dA$, with both antisymmetric components summed. For the Schwarzschild-Tangherlini metric the flux is $-{\Omega }_{D-2}{r}^{D-2}{f}^{\prime }=-(D-3){\Omega }_{D-2}\mu$. The displayed dimension-dependent coefficient then reproduces the ADM mass. Reversing orientation reverses the flux; including a factor of one-half in the binormal convention also changes its numerical value. A four-dimensional normalization cannot be carried unchanged into other dimensions.

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.