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Rotate any geodesic into an equatorial plane of the magnetically charged dilaton black hole. Conserved energy and angular momentum are $E=V\dot{t}$ and $L=R^2\dot{\varphi }$. Normalizing its tangent to $g(\dot{x},\dot{x})=-ϵ$, with $ϵ=1,0,-1$ for timelike, null and spacelike geodesics, gives ${\dot{r}}^2=E^2-V(ϵ+L^2/R^2)$. Thus its radial equation is ${\dot{r}}^2/2+\tilde{V}=0$. Radial null rays have $\dot{r}=\pm E$ and use the same Schwarzschild tortoise coordinate with $2M$ replaced by $r_{+}$.

For the magnetically charged dilaton black hole, the metric-dual one-form of $k={\partial }_t$ is $k^{♭}=-Vdt$. With orientation $dt\wedge dr\wedge d\theta \wedge d\varphi$, its Hodge star operator gives $\star d{k}^{♭}=-V^{\prime }{R}^2\sin \theta d\theta \wedge d\varphi$. Therefore the Komar mass is $R^2{V}^{\prime }/2=r_{+}/2-P^2/r$. Its radial dependence is allowed because the Ricci tensor is nonzero; the divergence of $d{k}^{♭}$ is sourced by $R_{ab}{k}^b$. At the horizon $M_K(r_{+})=(r_{+}-r_{-})/2=\kappa A_H/(4\pi )$.

Take $M>0$, set $b=Q^2/M$, and denote the squared areal radius by $C(r)=r(r-b)$. The positive-area exterior has $r>b$. The stationary norm is $-f(r)$, where $f(r)=1-2M/r$, so its candidate Killing horizon is at $r=2M$. There it has areaA regular horizon must have $b<2M$. The curvature singularity then lies strictly inside the horizon, and ingoing coordinates extend smoothly across it because $C(2M)>0$. In the future interior $f<0$, the gradient of $r$ is timelike and future causal curves move to decreasing $r$, rather than back to infinity. Thus the regular black hole condition isThe equality case has zero horizon area and a curvature-singular null limiting surface. It is often included in the family as a singular extremal solution, but it is not a regular black hole. If that limiting nomenclature is used, the family condition is $Q^2\le 2M^2$ with this essential qualification. This geometry has the same spherical form as the magnetically charged dilaton black hole; no unprinted field equation is needed for the horizon comparison.

Given an asymptotically flat spacetime with a conformal completion and future null infinity ${\mathcal{I}}^{+}$, its black hole region isThe event horizon is its boundary. For the magnetically charged dilaton black hole, first assume $0\le a=r_{-}<b=r_{+}$. The physical asymptotic component has $r>a$, and the ingoing chart extends regularly across $r=b$. Its future ingoing null direction is $-{\partial }_r$. For $a<r<b$,Hence $\nabla r$ is future timelike. Every nonzero future causal tangent $X$ satisfies $X(r)=g(\nabla r,X)<0$. A future causal curve from this interior can neither cross back to $r=b$ nor reach the asymptotic end $r\to \infty$, so it cannot reach ${\mathcal{I}}^{+}$. It lies in $\mathcal{B}$.

This proof uses the future black-hole extension. The radial interval alone does not distinguish it from the past white hole interior of the maximal extension; that interior has the same interval and can send signals to infinity. Thus the statement needs this future-component qualification if the maximal spacetime is intended.

The Kretschmann scalar confirms that $r=a$ is a genuine curvature singularity: for $0<a<b$, its numerator at $r=a$ is $3a^4(a-b{)}^2$, andFor $a=0$ it instead reduces to $12b^2/r^6$. Since $V<0$ on approaching the singularity from the interior, the singularity is spacelike. At $r=b>a$ the curvature is finite.

For the maximal extension, take $U=-e^{-u/(2b)}$, $W=e^{v/(2b)}$ in the right exterior and continue analytically across the horizons. ThenThe singularity is $UW=(1-a/b)e^{a/b}>0$ and the horizons are $U=0$ and $W=0$. Rescaling the null coordinates by the square root of this positive constant before arctangent compactification produces the usual qualitative Penrose diagram of Schwarzschild spacetime: two exteriors, a future black-hole region, a past white-hole region, and future and past spacelike singularities. The dashed lines below are the horizon branches; only the future interior is the black-hole region.

If $a>b$, the physical domain remains $r>a$, but now $V>0$ throughout it. The apparent zero $r=b$ is beyond the curvature singularity and is not an accessible horizon. The singularity is timelike, and outgoing radial null rays with $\dot{r}=E>0$ escape to infinity from arbitrarily near it. ThusThe excluded equality $r_{-}=r_{+}$ is a singular limiting case, not a regular horizon. These facts describe the causal structure of the magnetic dilaton black hole.