Anisotropic and Charged Interior Families

Anisotropy and electric charge can both support more compact static spheres, but by different mechanisms. Anisotropy contributes a force proportional to $p_t-p_r$; charge contributes Maxwell stress, electrostatic repulsion, and exterior field energy. An exact metric is physically interpretable only after these terms are separated, its center is made regular, and its surface is matched to the correct Reissner–Nordström rather than Schwarzschild exterior.

Required background. Hilbert Stress–Energy from Metric Variation supplies the material and Maxwell stress tensors.

Helpful background. The Einstein Equation fixes the coupled equations; The Riemann Curvature Tensor supplies regularity tests; Exact Solutions of Einstein’s Equations places the families; GR Formulations develops constraint data; and Continuum Kinematics and Balance Laws supplies continuum variables.

This page uses geometrized units $G=c=1$. The functions $m$ and $q$ are length parameters; $m_{\mathrm{geom}}=G M_{\mathrm{phys}}$, so the representative metric factor restores as $1-2G M_{\mathrm{phys}}/r+q_{\mathrm{geom}}^2/r^2$ under the stated electromagnetic normalization.

Charged anisotropic equilibrium

Use areal radius and define

$$ds^2=-e^{2\Phi(r)}dt^2 +\frac{dr^2}{F(r)}+r^2d\Omega^2, \qquad F(r)=1-\frac{2m(r)}r+\frac{q(r)^2}{r^2}.$$

Let the material stress tensor in its comoving orthonormal frame be

$$T^{\hat a}{}_{\hat b}\big|_{\mathrm{mat}} =\operatorname{diag}(-\rho,p_r,p_t,p_t).$$

For a radial electric field $E=q/r^2$, the Maxwell contribution has density $E^2/(8\pi)$, radial tension $-E^2/(8\pi)$, and tangential pressure $+E^2/(8\pi)$. Keeping the material and field pieces separate, the Einstein–Maxwell equations give

$$m'=4\pi r^2\rho+\frac{q q'}r, \qquad \Phi'=\frac{m+4\pi r^3p_r-q^2/r} {r^2-2mr+q^2}.$$

If $\rho_e$ is the proper charge density, Maxwell’s equation gives

$$q'=\frac{4\pi r^2\rho_e}{\sqrt{F}}.$$

Material force balance is the charged anisotropic TOV equation,

$$\begin{aligned} p_r'={}&-(\rho+p_r) \frac{m+4\pi r^3p_r-q^2/r}{r^2-2mr+q^2}\\ &+\frac{2(p_t-p_r)}r +\frac{q q'}{4\pi r^4}. \end{aligned}$$

The first term is gravitational and inertial, the second is the anisotropic force, and the third is the electric Lorentz force. Setting $q=0$ and $p_t=p_r$ recovers the ordinary TOV equation. This double limit is the quickest normalization check.

Regularity and surface data

At a smooth center,

$$m=O(r^3), \qquad q=O(r^3), \qquad p_t-p_r=O(r^2).$$

The last condition expresses the absence of a distinguished radial direction at the central point. Merely requiring finite $p_r$ and $p_t$ is not enough: a nonzero constant difference would make $2(p_t-p_r)/r$ singular. The charge condition follows because a finite proper charge density integrated over a small ball gives $q\propto r^3$; it also keeps $E=q/r^2=O(r)$ finite.

Let $R$ be the first zero of the material radial pressure. In the absence of a surface layer or surface charge, matching to Reissner–Nordström requires

$$m(R)=M, \qquad q(R)=Q, \qquad p_r(R)=0,$$

together with lapse normalization. A tangential pressure can remain nonzero at the boundary without directly violating normal-stress matching, although its discontinuity and the matter model still require interpretation. If $q$ jumps, Maxwell’s equation supplies a surface charge; if the extrinsic curvature jumps, the geometry supplies a thin shell.

What an exact compactness gain means

Positive anisotropy $p_t>p_r$ contributes an outward force and can reduce the radial pressure required for a given density profile. Charge likewise supplies outward repulsion. Exact one-parameter interiors can therefore approach compactness values unavailable to isotropic uncharged matter. Bowers and Liang constructed uniform-density anisotropic solutions precisely to isolate this effect Bowers and Liang 1974, pp. 657–665.

That mathematical gain is conditional, not universal. A proposed family should be tested over its full radial interval for

$$\rho\ge0, \qquad \rho+p_r\ge0, \qquad \rho+p_t\ge0,$$

and, when claimed, the dominant inequalities $\rho\ge\lvert p_r\rvert,\lvert p_t\rvert$. One must also state whether the mass $M$ includes exterior electromagnetic energy, whether charge density has one sign, and whether an equation of state or constitutive law actually generates the chosen anisotropy. Satisfying energy conditions does not by itself establish causal sound speeds or radial stability.

There is no single “anisotropic Buchdahl bound” independent of assumptions. Different inequalities follow from different restrictions on $p_r$, $p_t$, charge, and density monotonicity. A metric family that exceeds $2M/R=8/9$ has demonstrated which hypothesis of the isotropic theorem it relaxed; it has not refuted that theorem.

A verification sequence

Given an exact ansatz, first compute $q$ from Maxwell’s equation and $m$ from the $tt$ equation rather than fitting them by analogy with Reissner–Nordström. Next diagonalize the non-electromagnetic stress tensor in an orthonormal frame to identify $\rho,p_r,p_t$. Check the central orders above, the sign and zeros of $F$, and every denominator. Then impose surface matching and take the uncharged and isotropic limits independently. Finally evaluate energy conditions across the whole allowed parameter region, not only at $r=0$ and $r=R$.

Common pitfalls

Calling total radial stress the fluid pressure. The electromagnetic field has negative radial pressure. State whether $p_r$ denotes the material eigenvalue or the total eigenvalue before applying a boundary condition.

Inferring admissibility from compactness alone. Extra support can come with divergent charge density, central anisotropy, energy-condition violation, or an unphysical constitutive law. Those are independent checks.

Exercises

Assume finite central $\rho_e(0)=\rho_{e,c}$ and $F(0)=1$. Find the leading central behavior of $q$ and the electric energy density.

Solution

Maxwell’s equation gives

$$q'=4\pi r^2\rho_{e,c}+O(r^4),$$

so regularity and $q(0)=0$ imply

$$q(r)=\frac{4\pi}{3}\rho_{e,c}r^3+O(r^5).$$

Therefore $E=q/r^2=(4\pi/3)\rho_{e,c}r+O(r^3)$ and

$$\rho_{\mathrm{EM}}=\frac{E^2}{8\pi}=O(r^2).$$

The electromagnetic stress vanishes at the center and does not select a radial direction there.

References

• Richard L. Bowers and E. P. T. Liang, “Anisotropic Spheres in General Relativity,” The Astrophysical Journal 188, 657–665 (1974). DOI.