Relativistic Elasticity

Scrollable formula. Use Left and Right arrow keys to reveal hidden content.

Scrollable table. Use Left and Right arrow keys to reveal additional columns.

Scrollable figure. Use Left and Right arrow keys to inspect hidden content.

A relativistic solid remembers which material element occupies each spacetime event and how neighboring elements are deformed relative to a relaxed configuration. That memory requires more structure than a perfect fluid’s density and velocity: a material map and strain-dependent stored energy. Varying that energy with respect to the metric produces anisotropic elastic stress. When shear dependence is removed, the construction reduces to a perfect fluid, providing its most important consistency check.

Required background. Hilbert Stress–Energy from Metric Variation supplies metric variation and stress conservation; Differential Equations supplies hyperbolicity and wave propagation.

Helpful background. The Einstein Equation supplies self-gravitating solids; Relativistic Matter and Compact Objects gives the crust and stellar context; and Computational Relativity supplies evolution formulations.

Material map and strain

Label material elements by coordinates $X^A$ on a three-dimensional matter space. The labels function like names painted on the solid: each stays with its material element even as the solid deforms. A configuration is a rank-three map $X^A(x)$ with timelike one-dimensional kernel, satisfying

u^a\nabla_aX^A=0,

so the labels are carried along material worldlines. If $K_{AB}$ is a relaxed material metric, its pullback

k_{ab}=K_{AB}\nabla_aX^A\nabla_bX^B

is spatial: $k_{ab}u^b=0$ . Comparing $k_{ab}$ with the physical rest-space metric $h_{ab}=g_{ab}+u_au_b$ gives strain. Scalar invariants of the mixed tensor $h^{ac}k_{cb}$ separate volume change from shape change.

The formalism is coordinate independent on matter space: relabeling $X^A$ together with the components of $K_{AB}$ cannot change observable stress. A fixed material volume form assigns the number of particles to each set of labels. Pulling it back makes the particle current conserved identically, so density is determined by deformation rather than varied independently. The material map, current, and stress variation are developed in Karlovini and Samuelsson 2003, § 2, Eqs. (1)–(14).

Stored energy and stress tensor

Assume a local hyperelastic material with rest-frame energy density

\varepsilon=\varepsilon(n,s,I_1,I_2),

where $n$ is particle number density, $s$ is entropy per particle carried by each material element, and $I_1,I_2$ are independent shape invariants normalized to vanish in an unsheared state. “Unsheared” need not mean zero pressure: a crust can be isotropic while strongly compressed. The matter action is

S_{\rm m}=-\int\sqrt{-g}\,\varepsilon\,d^4x.

For an explicit variation, define the positive material-space tensor

C^{AB}=g^{ab}\nabla_aX^A\nabla_bX^B.

Hold the material map, entropy labels, and reference tensors fixed. The energy depends on the metric through $C^{AB}$ , so $\delta C^{AB}=\nabla_aX^A\nabla_bX^B\,\delta g^{ab}$ and $\delta\sqrt{-g}=-\tfrac12\sqrt{-g}g_{ab}\delta g^{ab}$ give

T_{ab}=2\frac{\partial\varepsilon}{\partial C^{AB}} \nabla_aX^A\nabla_bX^B-\varepsilon g_{ab}.

The material derivatives are spatial because $u^a\nabla_aX^A=0$ . Decomposing their trace and trace-free parts therefore gives

T_{ab}=-\frac{2}{\sqrt{-g}}\frac{\delta S_{\rm m}}{\delta g^{ab}} =\varepsilon u_au_b+p h_{ab}+\pi_{ab},

where $p=\tfrac13h^{ab}T_{ab}$ and $\pi_{ab}$ is spatial and trace-free. Its exact expression depends on the chosen invariants, but it is a derivative of the same stored-energy function rather than an independently guessed tensor. This guarantees compatible work and stress. In particular, at fixed particle labels the density variation is $\delta n=\tfrac12n h_{ab}\delta g^{ab}$ .

If $\varepsilon$ depends only on $n$ and $s$ , then all shear derivatives vanish, $\pi_{ab}=0$ , and

p=n\left(\frac{\partial\varepsilon}{\partial n}\right)_s-\varepsilon.

Substituting $2\partial\varepsilon/\partial g^{ab}=n(\partial\varepsilon/\partial n)_s h_{ab}$ into the variation explicitly produces $T_{ab}=\varepsilon u_au_b+p h_{ab}$ . Karlovini and Samuelsson give the invariant stress and fluid reduction in Karlovini and Samuelsson 2003, § 2, Eqs. (26)–(31).

Small strains and elastic waves

Near a homogeneous, unsheared isotropic state, use a local displacement $\xi_i$ and define the ordinary infinitesimal strain and its trace-free part by

e_{ij}=\frac12(\partial_i\xi_j+\partial_j\xi_i), \qquad e^{\rm TF}_{ij}=e_{ij}-\frac13\delta_{ij}\partial_k\xi^k.

Positive $\partial_i\xi^i$ denotes expansion, so $\delta n=-n\partial_i\xi^i$ . With bulk modulus $K=n(\partial p/\partial n)_s$ and shear modulus $\mu\geq0$ , the change in spatial momentum flux is

\delta T^{ij}=-K\delta^{ij}\partial_k\xi^k-2\mu e_{\rm TF}^{ij}.

The minus sign is essential with this extension-positive strain: relativistic $T^{ij}$ is momentum flux, opposite in sign to the tension-positive Cauchy stress often used in elasticity. Momentum conservation now gives

(\varepsilon+p)\partial_t^2\xi_i =\mu\nabla^2\xi_i+\left(K+\frac\mu3\right)\partial_i\partial_j\xi^j.

For a transverse plane wave, $\boldsymbol k\cdot\boldsymbol\xi=0$ , so only the Laplacian term survives. For a longitudinal wave, $\boldsymbol\xi$ is parallel to $\boldsymbol k$ , and both terms contribute. Their speeds are

c_T^2=\frac{\mu}{\varepsilon+p}, \qquad c_L^2=\frac{K+4\mu/3}{\varepsilon+p} =c_s^2+\frac{4\mu}{3(\varepsilon+p)}.

Here $c_s^2=K/(\varepsilon+p)$ is the compressional derivative along the unsheared equation of state. These are exact speeds for the stated linearization, not general formulas for a finitely sheared solid. The unsheared limits are Karlovini and Samuelsson 2003, § 6, Eqs. (133)–(137); their § 5 treats the more general characteristic problem.

Positive moduli and enthalpy give real wave speeds in this homogeneous limit, while relativistic causality requires every characteristic speed to be at most unity. Checking only $c_s$ is insufficient when shear modes exist. The wave cone must be computed for the actual strained state and propagation direction; large shear or crystal anisotropy can change hyperbolicity even when the unsheared material is well behaved.

Crusts, failure, and model limits

In a neutron-star crust, ions form a lattice while degenerate electrons and free neutrons contribute pressure and inertia. An isotropic shear modulus can approximate a polycrystalline average, but a single crystal is anisotropic and a superfluid component can move relative to the lattice. Magnetic stress, defects, temperature, and composition further affect the response.

Elasticity describes reversible deformation. A breaking strain or yield criterion is extra constitutive information, commonly calibrated by microscopic calculation rather than derived from the quadratic elastic energy. Once yielding begins, plastic flow, damage, heating, and possibly magnetic reconnection must be evolved. Extending Hooke’s law beyond its calibrated strain produces unphysical stresses even if the equations remain numerically stable.

The local approximation also assumes deformation varies slowly compared with lattice spacing and that a material map remains meaningful. Melting removes shear rigidity; setting $\mu\to0$ should recover the fluid stress. At a nonmagnetic material interface without a surface layer or phase conversion, normal traction and normal displacement are continuous, while tangential traction vanishes on an inviscid-fluid side. Tangential displacement need not be continuous if slip is permitted. Magnetic, viscous, superfluid, or surface stresses alter these interface conditions.

Exercises

An unsheared material has $c_s^2=0.4$ and $\mu/(\varepsilon+p)=0.5$ . Are its transverse and longitudinal waves causal? What changes as $\mu\to0$ ?

Solution

The transverse speed has $c_T^2=0.5$ , which is causal. But $c_L^2=0.4+(4/3)(0.5)=16/15\gt1$ , so the material is not relativistically admissible despite positive moduli and causal compressional sound speed. As $\mu\to0$ , $c_T\to0$ and $c_L\to c_s$ . If the stored energy loses all shear dependence, its variation gives $\pi_{ab}=0$ and the perfect-fluid stress derived above.

References

Karlovini, Max, and Lars Samuelsson. “Elastic Stars in General Relativity: I. Foundations and Equilibrium Models.” Classical and Quantum Gravity 20, 3613–3648 (2003). DOI. Open preprint.