advanced30 min studyLesson 817 in Physics

Reissner–Nordström and charged horizons

Classify the roots of the spherical Einstein–Maxwell metric while distinguishing electrovacuum stress from vanishing matter curvature.

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What you’ll understand

  • Convert electric charge into a geometric length with explicit SI factors.
  • Distinguish subextremal, extremal, and superextremal ideal geometries.
  • Compute an outer-horizon surface gravity with a stated normalization.

A charged exterior contains electromagnetic stress even where there is no charged material. Reissner–Nordström geometry is therefore an electrovacuum solution of the coupled Einstein–Maxwell equations, not a Ricci-flat vacuum obtained by attaching a charge label to Schwarzschild. We take positive mass M, zero cosmological constant, spherical symmetry, and a purely electric charge Q. Real astrophysical environments can neutralize charge; the analytic family remains useful for understanding horizon structure without claiming that highly charged ideal black holes are typical observed objects.

Electromagnetic energy modifies the metric

Define mass length m as before and a nonnegative charge length q through Q squared, including the SI vacuum permittivity. The metric retains the spherical Schwarzschild form but gains a positive inverse square radius term in f. Reversing the sign of the electric charge reverses the electric field without changing this metric, because the electromagnetic stress depends quadratically on the field. A charged test particle additionally feels a Lorentz force; a neutral geodesic is a different trajectory problem.

The classical Maxwell stress is trace-free in four dimensions. The traced Einstein equation with zero Lambda therefore gives zero Ricci scalar outside the source. Yet its Ricci tensor does not vanish when the electromagnetic field is nonzero. This is a concrete warning against using a single scalar as a complete vacuum or flatness test. The exact static exterior also idealizes away electromagnetic radiation, external plasma, discharge processes, and any material interior required for a different spacetime completion.

m=GMc2,q2=GQ24πϵ0c4,f(r)=1−2mr+q2r2,ds2=−fc2dt2+f−1dr2+r2dΩ22m=\frac{GM}{c^2},\qquad q^2=\frac{GQ^2}{4\pi\epsilon_0c^4},\qquad f(r)=1-\frac{2m}{r}+\frac{q^2}{r^2},\qquad ds^2=-fc^2dt^2+f^{-1}dr^2+r^2d\Omega_2^2
m and q are lengths, Q is measured in coulombs, and ε₀ is the SI vacuum permittivity. This electric-only convention should not be mixed with rationalized natural-unit charge formulas.↔ Scroll wide equations horizontally.

A discriminant classifies the ideal horizons

Multiplying f by radius squared turns its roots into an elementary quadratic. When q lies strictly between zero and m, there are two distinct positive roots, an outer event horizon in the standard black-hole completion and an inner Cauchy horizon in the ideal extended geometry. At q equal to m they coincide as a double root. When q exceeds m, the ideal metric has no real horizon root. This algebra classifies an exact family; it does not establish which members can form from a specified set of regular initial data.

The outer static region lies beyond the larger root. At distinct roots, regular crossing coordinates remove the static chart singularities. The classical inner horizon has a different causal role: continuation beyond it is not fixed by the exterior Cauchy data in the same way. Perturbations can be strongly amplified there, so the perfectly analytic interior is not a robust prediction for a realistic collapsing system. Such instability statements must not be promoted into an unrestricted proof of every version of cosmic censorship.

r±=m±m2−q2,f(r)=(r−r+)(r−r−)r2(0<q≤m)r_\pm=m\pm\sqrt{m^2-q^2},\qquad f(r)=\frac{(r-r_+)(r-r_-)}{r^2}\quad(0<q\le m)
The double-root case is extremal. At q = 0 the inner algebraic root reaches r = 0, which is a singularity rather than a second regular Schwarzschild horizon.↔ Scroll wide equations horizontally.

Surface gravity depends on the time normalization

For the asymptotically normalized static time, half the radial derivative of f at the outer horizon defines a geometric surface gravity with inverse length units. Multiplication by c squared converts it to acceleration units. Its expression through the root separation makes the extremal limit transparent. The vanishing separation makes this quantity zero, even though the horizon area and mass stay nonzero. Zero surface gravity therefore does not mean absence of a horizon or absence of curvature.

As a nonextremal outer horizon is approached, static proper acceleration diverges while the horizon-normalized quantity stays finite; the redshift factor connecting them is essential. At an extremal charged horizon, the exterior static-acceleration limit is instead the finite value c2/mc^2/m, while κgeo=0\kappa_{\mathrm{geo}}=0. No Hawking-temperature formula is needed for the classical calculation here. If quantum emission is discussed later, it requires additional semiclassical assumptions. The square-root lab can reproduce selected discriminants within its positive control range, but it does not evaluate the Maxwell stress, locate a Cauchy development, or determine whether a horizon is physically formed.

κgeo=12f′(r+)=r+−r−2r+2,gH=c2κgeo,AH=4πr+2\kappa_{\mathrm{geo}}=\frac12f'(r_+)=\frac{r_+-r_-}{2r_+^2},\qquad g_H=c^2\kappa_{\mathrm{geo}},\qquad A_H=4\pi r_+^2
κ_geo has m⁻¹ units and g_H has m/s² units. The time Killing vector is normalized at infinity, and the formula applies to the nonnegative outer-horizon surface gravity of this static family.↔ Scroll wide equations horizontally.

Two charged horizons and their surface gravity

Set q=0.8mq=0.8m for a positive mass length m. Find both horizon radii and the outer geometric surface gravity, then state its conversion to acceleration units.

  1. The discriminant is m2−q2=0.36m2m^2-q^2=0.36m^2, whose positive square root is 0.6m0.6m. Charge is subextremal and nonzero, so the two roots are distinct and positive.

  2. The roots are r+=1.6mr_+=1.6m and r−=0.4mr_-=0.4m. Their separation is 1.2m1.2m, while the squared outer radius is 2.56m22.56m^2.

  3. Substitute into the normalized derivative formula: κgeo=1.2/(2×2.56m)=0.234375/m\kappa_{\mathrm{geo}}=1.2/(2\times2.56m)=0.234375/m. The acceleration-valued surface gravity is this result multiplied by c squared.

The result

The horizons lie at 1.6m and 0.4m; κ_geo = 0.234375/m and g_H = 0.234375c²/m with the stated asymptotic normalization.

Explore the model

Interactive lab

Interactive lab · root-finding

Isolate the positive square root in the horizon algebra

Reset and set target clab=1c_{\mathrm{lab}}=1, upper bound U=2U=2; Step once and predict an exact midpoint root one. Repeat from Reset with both controls clab=0.25c_{\mathrm{lab}}=0.25, U=2U=2; Step twice and predict exact root 0.5. Each Step adds 0.5 to time and one bisection refinement. The lab solves only x2−clab=0x^2-c_{\mathrm{lab}}=0; it can illustrate reachable discriminant square roots but has no charge, metric, or horizon classifier.

Isolate the positive square root in the horizon algebraBisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Current readings: Midpoint after 0 refinements: 1.5; Bracket error bound: 1.5; Computed residual · x² − c: 0.25. All parameters are adjustable with the labeled controls below.Bisection on x squared minus 2: bracket from 0 to 3 after 0 refinements.00.751.52.253−2.360.192.745.297.84x · bracket stays nonnegativef(x) = x² − cn = 0 · [ 0, 3 ]
Quadratic / midpointCurrent bracketPositive root
Midpoint after 0 refinements
1.5
Bracket error bound
1.5
Computed residual · x² − c
0.25
t = 0 s

Bisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe radical in an analytic horizon formula is an algebraic calculation, separate from interpreting the resulting surface globally.

Check your understanding

Why can the Reissner–Nordström electrovacuum have R = 0 without being Ricci-flat?

Try it yourself

Take the extremal case q=m>0q=m>0. Factor f, locate its horizon, and compute geometric surface gravity and area. Explain why zero surface gravity is not zero mass.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The quadratic has a double root; evaluate the derivative or the root-separation formula.

Compare with the solution

The metric factor is f=(1−m/r)2f=(1-m/r)^2, with a double horizon root at r=mr=m. Its derivative vanishes there, so κgeo=0\kappa_{\mathrm{geo}}=0. The area remains 4πm24\pi m^2, and the mass is M=c2m/G>0M=c^2m/G>0. Thus degeneracy of the horizon sets its normalized surface gravity to zero without removing the mass or the finite horizon area. The extremal geometry requires separate care in global limits; it is not obtained by declaring the whole spacetime flat.

Take it with you

  • Electrovacuum includes electromagnetic stress even without local charged material.
  • The charge-to-mass discriminant classifies ideal roots, while formation and interior stability require further analysis.
  • Extremality gives a double horizon and zero normalized surface gravity with finite mass and area.

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