Thin-Shell Dynamics

A thin shell turns a tensorial jump condition into an equation for an evolving interface. In spherical symmetry that equation contains square roots with independent orientation signs. Those signs are geometric data: they distinguish an ordinary interior–exterior join from a wormhole throat or another branch. Squaring too early erases precisely that information.

Required background. We use Hypersurface Normals and Tangential Projectors, The Einstein Equation, and Flat-Background Hilbert Stress-Energy from Metric Variation.

Helpful background. The Gibbons-Hawking-York Boundary Term provides the variational counterpart of the nonnull extrinsic-curvature data used here.

Spherical shell geometry and orientation branches

On each side of a timelike spherical shell, take

$$ds_\pm^2=-f_\pm(r)dt_\pm^2+\frac{dr^2}{f_\pm(r)} +r^2d\Omega^2.$$

The shell is $r=R(\tau)$, where $\tau$ is its proper time. Its velocity is $u^\mu_\pm$, and one continuous unit normal $n^\mu_\pm$ points from $M_-$ to $M_+$. After both side data are pulled back to the shell, define

$$[Q]\equiv Q_+-Q_-.$$

Do not replace $n^\mu_\pm$ by the two outward normals of the retained regions. Define instead the radial orientation and signed square root

$$\varepsilon_\pm =\operatorname{sign}\!\left(n^\mu_\pm\nabla_\mu r\right), \qquad \beta_\pm =\varepsilon_\pm\sqrt{\dot R^2+f_\pm(R)}.$$

The signs $\varepsilon_\pm=\pm1$ say whether areal radius increases or decreases in the continuous-normal direction. With the site convention

$$K_{ab}=-e^\mu_a e^\nu_b\nabla_\mu n_\nu,$$

the angular component is

$$K^\theta{}_{\theta\,\pm}=-\frac{\beta_\pm}{R}.$$

This minus sign is essential. Israel’s original paper uses the opposite extrinsic-curvature convention; his spherical dust calculation is given in Israel 1966, §§ 8–9, pp. 10–14.

From the tensor equation to a radial equation

Let the surface stress tensor in the shell’s orthonormal frame be

$$S^a{}_b=\operatorname{diag}(-\sigma,p,p).$$

For a timelike interface the junction equation is

$$8\pi G S_{ab}=[K_{ab}]-h_{ab}[K].$$

Its $\tau$ component implies $[K^\theta{}_\theta]=4\pi G\sigma$. Substituting the angular curvature gives the unsquared shell equation

$$\boxed{\; \beta_- -\beta_+ =4\pi G\,\sigma R \;}$$

or, with $A\equiv4\pi G\sigma R$,

$$\varepsilon_-\sqrt{\dot R^2+f_-} -\varepsilon_+\sqrt{\dot R^2+f_+}=A.$$

This equation, together with an equation of state and surface energy balance, determines the motion. The distributional Bianchi identity gives

$$\dot\sigma +2\frac{\dot R}{R}(\sigma+p) =[T_{\mu\nu}n^\mu u^\nu].$$

In vacuum on both sides, the right-hand side vanishes. For dust, $p=0$, so $m=4\pi R^2\sigma$ is constant.

One may derive an effective potential, but only after recording the original signs. Since $\beta_\pm^2=\dot R^2+f_\pm$,

$$\beta_-+\beta_+=\frac{f_--f_+}{A}.$$

For $A\ne0$, the two signed roots are therefore

$$\beta_-=\frac12\left(A+\frac{f_--f_+}{A}\right), \qquad \beta_+=\frac12\left(\frac{f_--f_+}{A}-A\right).$$

Squaring yields

$$\dot R^2+V(R)=0, \qquad V(R)=\frac{f_++f_-}{2} -\frac{A^2}{4} -\frac{(f_--f_+)^2}{4A^2}.$$

Every solution of the unsquared equation solves this potential equation, but the converse is false. A candidate trajectory must reproduce the declared signs of both $\beta_\pm$ and satisfy $\beta_- -\beta_+=A$.

Checks and common failure modes

For an ordinary ball joined to its exterior, $n^\mu\nabla_\mu r>0$ on both sides, so $\varepsilon_-=\varepsilon_+=+1$ away from trapped regions. At a thin-shell wormhole throat, areal radius decreases toward the throat on one side and increases away on the other; one orientation sign changes. The algebraic equation may look similar after squaring, but the spacetime is not the same.

The radicand need not be interpreted using a static coordinate time. Across a horizon, $f$ can be negative while $\dot R^2+f$ remains nonnegative; the invariant sign is that of $n\cdot\nabla r$. Another common error is dividing by $A$ when the shell density vanishes. The $A=0$ case must be read from the original junction equation. Finally, a surface equation of state does not replace the flux law: bulk matter crossing the shell changes its energy.

Exercises

Take a positive-density dust shell with Minkowski interior, $f_-=1$, and Schwarzschild exterior, $f_+=1-2GM/R$. On the ordinary branch $\varepsilon_-=\varepsilon_+=+1$, suppose the shell is instantaneously static. With $m=4\pi R^2\sigma\gt0$, find the exterior mass $M$ in terms of $m$ and $R$, and interpret the correction.

Solution

At $\dot R=0$, the unsquared equation is

$$1-\sqrt{1-\frac{2GM}{R}}=\frac{Gm}{R}.$$

The right-hand side is nonnegative, consistent with the chosen positive roots. Isolating and then squaring the remaining root gives

$$1-\frac{2GM}{R} =\left(1-\frac{Gm}{R}\right)^2,$$

so

$$M=m-\frac{Gm^2}{2R}.$$

The difference is the negative gravitational self-binding energy of a spherical shell. Substitution into the original equation requires $1-Gm/R\ge0$ algebraically. For an actually timelike shell instantaneously at rest on this ordinary branch, the inequality is strict: $R\gt Gm$, and hence $R\gt2GM$ after the mass relation is used. At equality the exterior radius is a horizon and a timelike tangent cannot have $\dot R=0$ there. Keeping the branch and causal conditions prevents the squared equation from admitting an inadmissible limiting configuration.

References

• Israel, Werner. “Singular Hypersurfaces and Thin Shells in General Relativity.” Il Nuovo Cimento B 44, no. 1 (1966): 1–14; erratum 48, no. 2 (1967): 463. doi:10.1007/BF02710419. Erratum doi:10.1007/BF02712210.