Reissner-Nordström Charged Black Hole

Take the Schwarzschild black hole — the exact vacuum sphere of general relativity — and give it electric charge. That single addition, made almost as soon as Schwarzschild's ink was dry, produces the Reissner-Nordström spacetime, and it is far richer than a mere footnote. The charge carries an electric field, the field carries energy, and by $E=mc^2$ that energy gravitates. But the way it enters the geometry is a surprise: the metric acquires a term $+Q^2/r^2$ that is gravitationally repulsive at small radii, fighting the inward pull of the mass. The competition reshapes the black hole from the inside out. Where Schwarzschild has one horizon, Reissner-Nordström generically has two — an outer event horizon and an inner Cauchy horizon — sitting at the two roots of a quadratic. Squeeze the charge upward and the roots march toward each other until, at the extremal value $|Q|=M$ , they collide: a black hole with zero surface gravity, zero Hawking temperature, and a starring role in supersymmetry and the string-theory count of black-hole microstates. Push past $|Q|=M$ and the horizons vanish entirely, baring a singularity to the outside universe — the possibility that cosmic censorship exists to forbid. Real black holes carry essentially no charge; ambient plasma neutralizes any imbalance in microseconds. Yet Reissner-Nordström is indispensable theory. It is the charged member of the no-hair family, the static limit of Kerr-Newman, the simplest home of a Cauchy horizon and the mass-inflation instability, and the cleanest laboratory for the laws of black-hole mechanics, where charge contributes a work term $\Phi\,\mathrm{d}Q$ to the first law.

Key takeaways

The Reissner-Nordström metric is the unique static, spherically symmetric solution of the coupled Einstein-Maxwell equations — a black hole of mass $M$ and electric charge $Q$ , with metric function $f(r)=1-2M/r+Q^2/r^2$ .
The extra charge term $+Q^2/r^2$ is repulsive at small $r$ , so $f(r)=0$ has two roots $r_\pm=M\pm\sqrt{M^2-Q^2}$ : an outer event horizon $r_+$ and an inner Cauchy horizon $r_-$ .
When $Q^2<M^2$ both horizons exist; at $Q^2=M^2$ they merge into the extremal black hole with surface gravity and Hawking temperature exactly zero; when $Q^2>M^2$ no horizon exists and the singularity is naked.
The $r=0$ singularity is timelike (avoidable), not spacelike — an infalling observer need not hit it, a profound structural break from Schwarzschild.
The inner Cauchy horizon is a boundary of predictability that is violently unstable to perturbations (mass inflation), so the smooth eternal interior is physically untrustworthy.
Charge is astrophysically negligible, but RN grounds the no-hair theorem, the first law $\mathrm{d}M=\tfrac{\kappa}{8\pi}\mathrm{d}A+\Phi\,\mathrm{d}Q$ , extremality, and the string-theory entropy count.
Key idea: adding the field energy of a Coulomb charge to the Schwarzschild geometry inserts a positive $+Q^2/r^2$ term into the metric function, splitting the single horizon into an outer event horizon and an inner Cauchy horizon that merge in the extremal limit $|Q|=M$ .

Where this lives today

Reissner-Nordström is a workhorse of theory rather than of the telescope, and its modern life spans three arenas. In string theory, extremal charged black holes are where microscopic state counting meets gravity: Strominger and Vafa's 1996 derivation of the entropy $S=A/4$ from D-brane microstates was carried out on charged, BPS-saturated cousins of Reissner-Nordström, and the extremal $|Q|=M$ bound is the gravitational face of the supersymmetric BPS bound. In mathematical relativity, the inner Cauchy horizon is the arena for strong cosmic censorship: the mass-inflation program of Poisson, Israel, Ori and Dafermos asks whether $r_-$ is always destroyed by infalling perturbations, protecting determinism — now sharpened into rigorous theorems and actively probed for charged and rotating holes. And as a laboratory, Reissner-Nordström is the simplest testbed for the weak gravity conjecture (which demands that extremal black holes be able to discharge, requiring particles with $q>m$ ) and for overcharging gedanken experiments — every attempt to force $Q>M$ by throwing in charged matter fails, because the electrostatic repulsion needed also prevents capture, quietly upholding cosmic censorship.

The physical picture

Picture the Schwarzschild potential well and then switch on the black hole's electric field. That field stores energy, energy gravitates, so you might expect the well to deepen. It does the opposite where it counts. In the metric function the charge appears as $+Q^2/r^2$ , and because it grows faster than the mass term $-2M/r$ as $r$ shrinks, it eventually wins: the pull of the mass is overwhelmed by the outward push of the field's own gravitating energy. The result is not one horizon but two — the geometry closes over, reopens, and hides a singularity that, uniquely, one can choose to avoid. The two horizons are nothing more mysterious than the two roots of a quadratic. Multiply $f(r)=0$ by $r^2$ and you get $r^2-2Mr+Q^2=0$ , whose solutions $r_\pm=M\pm\sqrt{M^2-Q^2}$ are the outer event horizon and the inner Cauchy horizon. The first figure plots $f(r)$ itself and shows the roots directly: crank the charge up and the two crossings slide toward each other; at $|Q|=M$ the parabola just kisses the axis (a double root at $r=M$ , the extremal case); beyond that it never touches down and there is no horizon at all.

Figure

The metric function f = 1 − 2M/r + Q²/r² at fixed mass M for three charges. Subextremal Q = 0.8M cuts the axis twice, at the inner horizon r₋ = 0.4M and the outer horizon r₊ = 1.6M. Extremal Q = M just touches zero at the degenerate horizon r = M. Superextremal Q = 1.2M never reaches zero, so no horizon exists and the singularity is naked

Between the two horizons the radial coordinate $r$ turns timelike, exactly as inside Schwarzschild, so infall is inescapable there. But below the inner horizon $r_-$ , the charge term flips $f(r)$ positive again, $r$ becomes spacelike once more, and the singularity at $r=0$ sits in the timelike direction — a place, not a moment. An infalling observer can steer around it. In the idealized eternal solution one can even pass through $r_-$ and emerge into a fresh asymptotic universe, an infinite ladder of them. The second figure tracks the two horizon radii as the charge is dialed up, showing precisely how the Schwarzschild horizon at $2M$ and the point $r_-\to0$ evolve into the merged extremal horizon.

Figure

The two horizon radii r₊ and r₋ in units of M as the charge ratio Q/M runs from 0 to 1. At Q = 0 the outer horizon sits at the Schwarzschild value r₊ = 2M while the inner horizon is at r₋ = 0. As charge grows the outer horizon shrinks and the inner one grows until they meet at the extremal point Q/M = 1, r = M. Beyond that the region is superextremal with no horizons

The core equations

The whole solution is one line element with one modified metric function; every horizon, temperature and thermodynamic relation below is read off from it. It has the Schwarzschild form, with the charge entering only through $f(r)$ . Throughout we use geometrized Gaussian units $G=c=1$ , in which mass and charge are both lengths.

\mathrm{d}s^2=-f(r)\,\mathrm{d}t^2+\frac{\mathrm{d}r^2}{f(r)}+r^2\,\mathrm{d}\Omega^2,\qquad f(r)=1-\frac{2M}{r}+\frac{Q^2}{r^2} \qquad (1)

where: $M$ — ADM (asymptotic) mass; $Q$ — electric charge; $f(r)$ — metric/redshift function, with horizons wherever $f=0$ ; $\mathrm{d}\Omega^2=\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2$ — round unit 2-sphere. Reading it: $g_{tt}=-f$ vanishes and $g_{rr}=1/f$ diverges at each horizon, while at $Q=0$ the function collapses to Schwarzschild's $1-2M/r$ . The source is a single static, radial electric field — the curved-space Coulomb field — described by one component of the vector potential. Everything electromagnetic in the solution follows from it.

A_\mu\,\mathrm{d}x^\mu=-\frac{Q}{r}\,\mathrm{d}t,\qquad F_{tr}=\partial_t A_r-\partial_r A_t=-\frac{Q}{r^2},\qquad E=\frac{Q}{r^2} \qquad (2)

where: $A_\mu$ — electromagnetic 4-potential; $F_{\mu\nu}$ — Faraday tensor; $E$ — radial electric field a static observer measures. Reading it: the field is pure Coulomb, $E=Q/r^2$ , and its energy density $\rho=E^2/8\pi=Q^2/8\pi r^4$ is exactly what sources the $+Q^2/r^2$ term in $f$ . Setting $f(r)=0$ is a quadratic in $r$ , so a charged black hole carries two horizons rather than one. Their radii are the object that governs the entire causal structure.

r_\pm=M\pm\sqrt{M^2-Q^2} \qquad (3)

where: $r_+$ — outer event horizon; $r_-$ — inner Cauchy horizon. Reading it: real distinct horizons require $Q^2<M^2$ ; the product $r_+r_-=Q^2$ and sum $r_++r_-=2M$ are handy identities. At $Q=0$ , $r_+=2M$ and $r_-=0$ ; at $|Q|=M$ , $r_+=r_-=M$ . The surface gravity of the outer horizon — the redshifted acceleration needed to hold station just above it — sets the Hawking temperature. Both are fixed by the gap between the two horizons.

\kappa_+=\frac12 f'(r_+)=\frac{r_+-r_-}{2r_+^2}=\frac{\sqrt{M^2-Q^2}}{r_+^2},\qquad T_H=\frac{\kappa_+}{2\pi}=\frac{\sqrt{M^2-Q^2}}{2\pi\,r_+^2} \qquad (4)

where: $\kappa_+$ — surface gravity at $r_+$ ; $f'(r)=2M/r^2-2Q^2/r^3$ ; $T_H$ — Hawking temperature. Reading it: as $|Q|\to M$ the numerator $\sqrt{M^2-Q^2}\to0$ , so an extremal black hole is cold, $T_H\to0$ — a fact with deep thermodynamic consequences. The outer horizon carries area and, through Bekenstein and Hawking, entropy. These are the geometric quantities that enter the black-hole first law.

A=4\pi r_+^2=4\pi\left(M+\sqrt{M^2-Q^2}\right)^2,\qquad S=\frac{A}{4}=\pi r_+^2 \qquad (5)

where: $A$ — area of the outer event horizon; $S$ — Bekenstein-Hawking entropy in geometrized units. Reading it: charging a hole at fixed $M$ shrinks $r_+$ , hence shrinks its area and entropy — the charge stores energy that is not available to enlarge the horizon. Charge enters the first law of black-hole mechanics as an electrostatic work term, exactly parallel to $-p\,\mathrm{d}V$ in ordinary thermodynamics. The horizon behaves like a charged conductor at a definite potential.

\mathrm{d}M=\frac{\kappa_+}{8\pi}\,\mathrm{d}A+\Phi\,\mathrm{d}Q,\qquad \Phi=\frac{Q}{r_+} \qquad (6)

where: $\Phi$ — electrostatic potential of the horizon, conjugate to $Q$ ; $\kappa_+/8\pi=T_H/4$ — the entropy (heat) term. Reading it: $\tfrac{\kappa_+}{8\pi}\mathrm{d}A=T_H\,\mathrm{d}S$ is the thermal term and $\Phi\,\mathrm{d}Q$ the electrical work, so the mass of a charged black hole splits cleanly into heat and work like any thermodynamic energy.

Derivation ADeriving the Reissner-Nordström metric from Einstein-Maxwell

The goal is to show that demanding a static, spherically symmetric spacetime sourced by nothing but a radial electric field forces the metric function to be exactly $f(r)=1-2M/r+Q^2/r^2$ — the Schwarzschild answer plus one new term whose coefficient is the square of the total charge. We write the most general static spherical ansatz, solve Maxwell's equations for the only field the symmetry permits, feed the resulting electromagnetic energy into Einstein's equations, and integrate, checking that both the Schwarzschild and Coulomb limits fall out.

Step 1. Adopt the general static, spherically symmetric line element $\mathrm{d}s^2=-e^{2\alpha(r)}\mathrm{d}t^2+e^{2\beta(r)}\mathrm{d}r^2+r^2\mathrm{d}\Omega^2$ , with two unknown radial functions $\alpha,\beta$ . Why: staticity and spherical symmetry kill every angular and time dependence, so the entire problem reduces to two functions of $r$ alone.

Step 2. Solve the source-free Maxwell equation $\nabla_\mu F^{\mu\nu}=0$ . Spherical symmetry allows only a radial electric field, and integration gives $F^{tr}=Q\,e^{-\alpha-\beta}/r^2$ , with the integration constant $Q$ the total electric charge. Why: Gauss's law in curved space fixes the field of a point charge up to the constant $Q$ , and the Bianchi identity $\mathrm{d}F=0$ is then automatic — there is no freedom left in the electromagnetic sector.

Step 3. Build the electromagnetic stress-energy $T_{\mu\nu}=\tfrac{1}{4\pi}(F_{\mu\alpha}F_\nu{}^{\alpha}-\tfrac14 g_{\mu\nu}F^2)$ . With only $F_{tr}$ nonzero it is traceless and diagonal, $T^\mu{}_\nu=\tfrac{Q^2}{8\pi r^4}\,\mathrm{diag}(-1,-1,+1,+1)$ , an energy density $\rho=Q^2/8\pi r^4$ . Why: this is the gravitating fuel — the Coulomb field's own energy, falling off as $r^{-4}$ , is what will bend the geometry beyond Schwarzschild.

Step 4. Combine the $tt$ and $rr$ Einstein equations. Their difference forces $\alpha+\beta=\text{const}$ , and asymptotic flatness sets the constant to zero, so $\alpha=-\beta$ and $g_{tt}g_{rr}=-1$ just as in Schwarzschild. Why: the equal-and-opposite metric functions are what let us write a single function $f=e^{2\alpha}=e^{-2\beta}$ , collapsing two unknowns into one.

Step 5. The remaining $tt$ equation becomes $\tfrac{1}{r^2}\tfrac{\mathrm{d}}{\mathrm{d}r}[r(1-f)]=Q^2/r^4$ . Integrate once to get $r(1-f)=2M-Q^2/r$ , i.e. $f(r)=1-2M/r+Q^2/r^2$ , with $M$ the integration constant fixed as the ADM mass. Why: the $2M$ piece is the vacuum (Schwarzschild) constant of integration, while the $Q^2/r^2$ piece is the running contribution of the field energy interior to radius $r$ — the two sources cleanly superpose in $f$ . *Verified: setting $Q=0$ recovers Schwarzschild $f=1-2M/r$ ; at large $r$ the field reduces to Coulomb $E=Q/r^2$ ; matches Carroll §6.4 and MTW §31.6.*

Go deeper · Derivation B · 1 minThe two horizons and the surface gravity of the outer one

Derivation BThe two horizons and the surface gravity of the outer one

The aim here is to locate the horizons as the roots of $f(r)=0$ , prove that a real event horizon exists only when $Q^2\le M^2$ , and then compute the surface gravity of the outer horizon in closed form, showing it equals $\sqrt{M^2-Q^2}/r_+^2$ and therefore vanishes precisely at extremality. This ties the causal structure (how many horizons) directly to the thermodynamics (how hot the hole is), since the Hawking temperature is $\kappa_+/2\pi$ .

Step 1. Set $f(r)=0$ and clear the denominator: $1-2M/r+Q^2/r^2=0$ becomes $r^2-2Mr+Q^2=0$ . Why: multiplying by $r^2$ turns the horizon condition into an ordinary quadratic, whose roots are immediately the horizon radii.

Step 2. Apply the quadratic formula: $r_\pm=M\pm\sqrt{M^2-Q^2}$ . Two real, distinct, positive roots exist when $Q^2<M^2$ ; they coincide at $Q^2=M^2$ ; they become complex (no horizon) when $Q^2>M^2$ . Why: the discriminant $M^2-Q^2$ is the single quantity that decides subextremal versus extremal versus naked — the entire trichotomy sits under one square root.

Step 3. Record the root identities from the quadratic: sum $r_++r_-=2M$ and product $r_+r_-=Q^2$ . Why: these let us rewrite everything in terms of the two horizon radii, which is far cleaner than carrying $M$ and $Q$ separately — for instance $M r_+ - Q^2 = \tfrac12 r_+(r_+-r_-)$ .

Step 4. Compute the surface gravity $\kappa_+=\tfrac12 f'(r_+)$ with $f'(r)=2M/r^2-2Q^2/r^3=\tfrac{2}{r^3}(Mr-Q^2)$ . At $r_+$ this gives $\kappa_+=\tfrac{1}{r_+^3}(Mr_+-Q^2)=\tfrac{r_+-r_-}{2r_+^2}$ . Why: surface gravity is half the gradient of $f$ at the horizon — the redshifted force to hold a static observer there — and the root identities turn the messy derivative into the clean horizon gap $r_+-r_-$ .

Step 5. Substitute $r_+-r_-=2\sqrt{M^2-Q^2}$ to obtain $\kappa_+=\sqrt{M^2-Q^2}/r_+^2$ , and hence $T_H=\kappa_+/2\pi=\sqrt{M^2-Q^2}/2\pi r_+^2$ . As $|Q|\to M$ the horizons merge, $r_+-r_-\to0$ , and $\kappa_+\to0$ . Why: this exhibits the extremal black hole as the zero-temperature end of the family — the surface gravity is literally proportional to how far apart the two horizons sit. *Verified: reduces to Schwarzschild $\kappa=1/4M$ at $Q=0$ (since $r_+\to2M$ ); $\kappa_+\to0$ at $|Q|=M$ ; consistent with Wald §12.5 and Poisson Ch. 5.*

Go deeper · Derivation C · 1 minWhy the central singularity is timelike, not spacelike

Derivation CWhy the central singularity is timelike, not spacelike

This derivation establishes the single most striking structural difference between a charged and an uncharged black hole: the $r=0$ singularity of Reissner-Nordström is a genuine curvature singularity, yet it is timelike and therefore avoidable, unlike the spacelike, unavoidable singularity of Schwarzschild. We confirm the curvature really diverges using a coordinate-independent invariant, then read the causal character of $r=0$ off the sign of $f(r)$ in its neighbourhood.

Step 1. Compute the Kretschmann scalar, the fully contracted square of the Riemann tensor, which for Reissner-Nordström is $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}=48M^2/r^6-96MQ^2/r^7+56Q^4/r^8$ . Why: a scalar invariant is chart-independent, so its behaviour certifies whether a blow-up is real geometry or a coordinate artifact — the horizons, by contrast, leave this scalar perfectly finite.

Step 2. Take $r\to0$ : the last term $56Q^4/r^8$ dominates and diverges, so $r=0$ is a true curvature singularity that no change of coordinates can remove. Why: an infinite chart-independent scalar cannot live on a smooth manifold, so the spacetime genuinely ends at $r=0$ — the charge does not soften the singularity, it only reshapes the causal approach to it.

Step 3. Inspect $f(r)$ near the centre: as $r\to0$ the $+Q^2/r^2$ term dominates and $f(r)\to+\infty>0$ . Since $g_{tt}=-f<0$ and $g_{rr}=1/f>0$ there, the coordinate $t$ is timelike and $r$ is spacelike below the inner horizon. Why: the sign of $f$ dictates which coordinate is time; between the horizons $f<0$ makes $r$ timelike (infall inevitable), but below $r_-$ the charge term restores $f>0$ and $r$ is spatial again.

Step 4. Conclude the causal character: because $r$ is spacelike near $r=0$ , the singularity is a timelike world-tube, a place the infalling observer can steer around, rather than a spacelike moment lying unavoidably in the future. Why: this is what makes the Reissner-Nordström interior qualitatively unlike Schwarzschild — determinism and the traversability of the interior become live questions, answered only by the mass-inflation instability of $r_-$ . *Verified: the Kretschmann scalar reduces to Schwarzschild's $48M^2/r^6$ at $Q=0$ ; the $f>0$ sign near $r=0$ gives a timelike singularity; matches Chandrasekhar Ch. 5 and Carroll §6.5.*

Limiting cases

The single dimensionless ratio $|Q|/M$ controls the entire family; the four rows below trace it from neutral to naked.

Case	Result	Meaning
Neutral $Q\to0$	$f(r)=1-\dfrac{2M}{r}$ , $r_+=2M$ , $r_-\to0$	Recovers Schwarzschild: one horizon, spacelike singularity — the charge term switches off.
Subextremal $Q^2<M^2$	$r_\pm=M\pm\sqrt{M^2-Q^2}$	Two distinct horizons, timelike singularity at $r=0$ , nonzero $T_H$ at $r_+$ .
Extremal $Q^2=M^2$	$f(r)=\left(1-\dfrac{M}{r}\right)^2$ , $r_+=r_-=M$	Degenerate double horizon; $\kappa=0$ , $T_H=0$ ; saturates the bound $\|Q\|\le M$ ; central to BPS states.
Superextremal $Q^2>M^2$	$f(r)>0$ for all $r>0$	No horizon; naked timelike singularity — forbidden by weak cosmic censorship.

Common trapThe inner horizon is not a benign second surface

It is tempting to treat the eternal Reissner-Nordström interior at face value — two tidy horizons, an avoidable singularity, and a passage through $r_-$ into new universes. Do not. The inner horizon $r_-$ is a Cauchy horizon: the boundary beyond which initial data on an external slice no longer determine the future, so predictability itself fails there. Worse, $r_-$ is a surface of infinite blueshift for anything that falls in later, and realistic perturbations — even the faint radiative tail of the collapse that formed the hole — pile up on it with diverging energy. This mass-inflation instability converts the smooth Cauchy horizon into a null curvature singularity. Only the region outside the outer horizon $r_+$ is physically robust; the traversable ladder of universes is an artifact of the idealized, unperturbed solution.

Worked examples

1 · A subextremal charged hole, fully worked

A black hole has charge $Q=0.8M$ in geometrized units. Find its two horizon radii, the outer surface gravity, Hawking temperature, horizon area, and electrostatic potential, and compare its temperature with an uncharged hole of the same mass.

Step 1. Discriminant and horizons. With $Q^2=0.64M^2$ , $\sqrt{M^2-Q^2}=\sqrt{0.36}\,M=0.6M$ , so $r_+=M+0.6M=\mathbf{1.6M}$ and $r_-=M-0.6M=\mathbf{0.4M}$ . Check the identities: $r_++r_-=2M$ and $r_+r_-=0.64M^2=Q^2$ . Both hold.

Step 2. Surface gravity and temperature. $\kappa_+=\sqrt{M^2-Q^2}/r_+^2=0.6M/(1.6M)^2=0.6/2.56\,M^{-1}=\mathbf{0.234/M}$ . The Hawking temperature is $T_H=\kappa_+/2\pi=\mathbf{0.0373/M}$ (geometrized units).

Step 3. Area and potential. $A=4\pi r_+^2=4\pi(1.6M)^2=\mathbf{32.2\,M^2}$ , giving entropy $S=A/4=\pi(1.6M)^2=8.04\,M^2$ . The horizon potential is $\Phi=Q/r_+=0.8M/1.6M=\mathbf{0.5}$ .

Step 4. Compare with Schwarzschild. An uncharged hole of the same mass has $r_+=2M$ and $\kappa=1/4M=0.25/M$ . The charged hole is colder ( $0.234/M$ versus $0.25/M$ ) and smaller ( $1.6M$ versus $2M$ ): the charge has drained energy into the electric field and pushed the hole toward its cold, extremal end. Charging at fixed $M$ shrinks the horizon and lowers the temperature.

2 · How much charge would actually matter, fully worked

Estimate the electric charge required to make a solar-mass black hole extremal, and then the far smaller charge at which it would spontaneously discharge — explaining why real black holes are effectively neutral.

Step 1. Extremal charge. Extremality $|Q|=M$ in geometrized units becomes, in SI, $GM^2=Q^2/4\pi\varepsilon_0$ , so $Q_\text{ext}=M\sqrt{4\pi\varepsilon_0 G}=M\sqrt{G/k_e}$ with $k_e=8.99\times10^9$ . Numerically $\sqrt{G/k_e}=8.62\times10^{-11}$ C/kg.

Step 2. For $M=M_\odot=1.99\times10^{30}$ kg: $Q_\text{ext}=1.99\times10^{30}\times8.62\times10^{-11}=\mathbf{1.7\times10^{20}\ C}$ . That is about $1.1\times10^{39}$ elementary charges — an absurd amount of net charge to assemble and hold against its own repulsion.

Step 3. Discharge threshold. Selective accretion stops the moment the electric repulsion on a like-sign proton exceeds gravity: $k_e Q e > G M m_p$ , i.e. $Q/M>G m_p/(k_e e)=7.8\times10^{-29}$ C/kg. Compared with the extremal ratio $8.62\times10^{-11}$ C/kg, this is a fraction $\sim9\times10^{-19}$ of extremality.

Step 4. Interpret. A black hole immersed in ionized plasma repels further like charge and attracts opposite charge once $Q$ reaches roughly $10^{-18}$ of its extremal value — a vanishingly small imbalance. So any astrophysical hole neutralizes almost instantly and sits at $Q\approx0$ . Reissner-Nordström is essential theory but never an astrophysical steady state.

Physics of the portrait

The charged black hole is one of the oldest exact solutions in general relativity — older than the word "black hole" by four decades — and it emerged not from an astrophysicist but from an aeronautical engineer solving Einstein's brand-new equations for a point charge.

Origins · 1 minHans Reissner and how this was discovered

Hans Reissner

1874–1967 · the first charged solution of the Einstein field equations

Hans Jacob Reissner was a German engineer and applied mathematician better known in his lifetime for aeroelasticity and aircraft structures than for gravitation — the Reissner plate and Reissner's variational principle bear his name in engineering. Yet in 1916, only months after Karl Schwarzschild published the vacuum solution and while the field equations were less than a year old, Reissner solved them for the exterior of a spherically symmetric body carrying electric charge, becoming the first person to couple Einstein's gravity to a material field. He treated a point charge; Gunnar Nordström, the Finnish physicist who had built his own competing scalar theory of gravity, independently generalized the result to arbitrary charge in 1918, and Hermann Weyl (1917) and George Barker Jeffery (1921) filled in the full geometry — hence the compound name. None of them could have seen in 1916 how much structure they had uncovered: two horizons, a timelike singularity, an extremal cold limit, and a Cauchy horizon whose instability would occupy relativists for the rest of the century.

History

1916

— Reissner solves the Einstein equations for the field of a charged point mass

1918

— Nordström independently extends the solution to a body of arbitrary charge

1921

— Jeffery works out the full geometry of the charged sphere

1965

— Newman and collaborators rotate Reissner-Nordström into the charged, spinning Kerr-Newman solution

1990

— Poisson and Israel establish mass inflation, showing the inner horizon is generically singular

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §31.6 (Reissner-Nordström geometry) · Wald, General Relativity (1984), §6.4 and §12.5 (charged and extremal black holes) · Carroll, Spacetime and Geometry (2004), §6.4–6.6 (RN metric and causal structure) · Hartle, Gravity (2003), Ch. 15 · Chandrasekhar, The Mathematical Theory of Black Holes (1983), Ch. 5 · Poisson, A Relativist's Toolkit (2004), Ch. 5 (surface gravity, mass inflation).

Provenance. The metric, $f(r)=1-2M/r+Q^2/r^2$ , horizon radii $r_\pm=M\pm\sqrt{M^2-Q^2}$ , surface gravity, Hawking temperature, area/entropy, and first law follow the standard texts: MTW §31.6, Wald §6.4 and §12.5, Carroll §6.4–6.6. A verifier should recompute: (1) the roots of $r^2-2Mr+Q^2=0$ ; (2) the surface gravity $\kappa_+=\tfrac12 f'(r_+)=\sqrt{M^2-Q^2}/r_+^2$ and its vanishing at extremality; (3) the Kretschmann scalar $48M^2/r^6-96MQ^2/r^7+56Q^4/r^8$ , reducing to $48M^2/r^6$ at $Q=0$ ; (4) the electromagnetic stress-energy $T^\mu{}_\nu=(Q^2/8\pi r^4)\,\mathrm{diag}(-1,-1,1,1)$ from $F_{tr}=-Q/r^2$ ; (5) the SI extremal charge $Q_\text{ext}=M\sqrt{G/k_e}\approx1.7\times10^{20}$ C for a solar mass. All quantities in geometrized Gaussian units $G=c=1$ .

Prerequisites. Schwarzschild Black Hole · Event and Apparent Horizons · Killing Horizons and Surface Gravity. Leads to. Kerr-Newman Solution · No-Hair Theorem and Black Hole Uniqueness · Four Laws of Black Hole Mechanics · Trapped Surfaces and Closed Trapped Regions.

Ask the tutor: ask me to dial the charge from zero to extremal and watch the two horizons slide together, the singularity flip from spacelike to timelike, and the Hawking temperature fall to zero.

Do real black holes actually carry any electric charge?

Essentially none. Any black hole sits in a bath of ionized plasma, and the instant it acquires a net charge its electric field selectively pulls in opposite-sign charges and pushes away like ones. The imbalance needed to matter is tiny: repulsion on a proton beats gravity once the charge reaches only about $10^{-18}$ of the extremal value, so a hole discharges to near-neutrality in microseconds. This is why Reissner-Nordström, despite being an exact and beautiful solution, never describes a real astrophysical black hole in steady state. Its value is entirely theoretical — as the charged member of the no-hair family, the static limit of Kerr-Newman, and the simplest home of a Cauchy horizon.

Why is the Reissner-Nordström singularity timelike when Schwarzschild's is spacelike?

It comes down to the sign of the metric function near the centre. In Schwarzschild, $f=1-2M/r\to-\infty$ as $r\to0$ , so $r$ stays timelike all the way in and the singularity is a spacelike surface lying unavoidably in every infaller's future. In Reissner-Nordström the charge term dominates at small $r$ : $f\to+Q^2/r^2\to+\infty$ , which flips $r$ back to spacelike and $t$ to timelike below the inner horizon. A singularity reached along a spatial direction is a place, not a moment — you can accelerate to avoid it. That single sign change is the deepest structural difference between charged and uncharged black holes.

What is the inner Cauchy horizon and why is it dangerous?

The inner horizon $r_-$ is where predictability ends. Initial data specified on a surface outside the black hole determine the physics only up to $r_-$ ; beyond it, what happens also depends on whatever streams in from the timelike singularity, so the future is no longer fixed by the past — the hallmark of a Cauchy horizon. It is also a surface of infinite blueshift: radiation falling in later is compressed to unbounded energy against it. Perturbations therefore pile up and drive the local mass function to diverge, the mass-inflation phenomenon of Poisson and Israel, turning the smooth $r_-$ into a null curvature singularity. This is nature's way of enforcing strong cosmic censorship and rendering the eternal interior physically irrelevant.

What does "extremal" mean, and why does the temperature go to zero?

An extremal Reissner-Nordström black hole has the maximum charge its mass allows, $|Q|=M$ , at which the two horizons merge into one degenerate horizon at $r=M$ and $f(r)=(1-M/r)^2$ has a double root. The surface gravity is $\kappa_+=\sqrt{M^2-Q^2}/r_+^2$ , and at $|Q|=M$ the numerator vanishes, so $\kappa_+=0$ and the Hawking temperature $T_H=\kappa_+/2\pi=0$ . Physically, the two horizons are infinitely far apart in proper distance and the throat between them becomes an infinitely long tube. Extremal holes are thus "cold," cannot radiate thermally, and saturate the bound $|Q|\le M$ — properties that make them the fixed points of supersymmetric (BPS) black-hole physics and the setting for microscopic entropy counting.

Is the bound |Q| ≤ M a law, and can you overcharge a black hole past it?

It is not a theorem of the field equations — a superextremal $Q>M$ metric exists on paper — but crossing it would expose a naked singularity, which the weak cosmic censorship conjecture forbids. Every concrete attempt to overcharge a near-extremal hole by dropping in charged particles fails: to add enough charge you must throw in a particle whose own electrostatic repulsion is so strong it is turned away before it can be captured. When the back-reaction and self-force are included carefully, the hole always stays at or below extremality. So $|Q|\le M$ behaves like an effective law, enforced by the same electromagnetism that defines the solution, and is one of the cleanest pieces of evidence for cosmic censorship.

How does charge change a black hole compared with Schwarzschild at the same mass?

At fixed mass, adding charge shrinks the event horizon and cools the hole. The outer radius drops from $r_+=2M$ toward $r_+=M$ as $|Q|\to M$ , so the area $A=4\pi r_+^2$ and entropy $S=\pi r_+^2$ both fall; the surface gravity and Hawking temperature fall too, reaching zero at extremality. Meanwhile a second, inner horizon appears and grows outward from $r=0$ . Intuitively, energy that would have gone into a bigger horizon is instead locked up in the Coulomb field outside it. Charge also converts the singularity from spacelike to timelike. The one thing charge cannot do is add "hair": the exterior is still fixed entirely by $M$ and $Q$ .