3+1 Decomposition of Stress-Energy

Observers moving along the future unit normal $n^\mu$ decompose stress-energy into an energy density $\rho$, a spatial momentum density $j_i$, and a spatial stress $S_{ij}$. These are not new matter fields: together they reconstruct the same tensor $T_{\mu\nu}$. Their signs determine the sources of the Hamiltonian constraint, momentum constraint, and extrinsic-curvature evolution equation.

Required background. Hypersurface Normals and Tangential Projectors supplies the normal observers and spatial projection, while The Einstein Equation supplies the stress-energy source and its covariant conservation law.

Helpful background. The Gibbons-Hawking-York Boundary Term supplies the boundary setting in which normal and tangential projections must be kept distinct.

Energy, momentum, and stress seen by normal observers

Fix

$$n^\mu n_\mu=-1, \qquad \gamma_{\mu\nu}=g_{\mu\nu}+n_\mu n_\nu.$$

The three projections of a symmetric stress tensor are

$$\rho=T_{\mu\nu}n^\mu n^\nu,$$ $$j_\mu =-\gamma_\mu{}^\rho n^\sigma T_{\rho\sigma}, \qquad n^\mu j_\mu=0,$$

and

$$S_{\mu\nu} =\gamma_\mu{}^\rho\gamma_\nu{}^\sigma T_{\rho\sigma}, \qquad n^\mu S_{\mu\nu}=0.$$

The minus sign in $j_\mu$ is chosen so that positive $j_i$ is the momentum density in the positive spatial direction measured in the normal observer’s orthonormal frame. Equivalently,

$$T(n,X)=-j(X)$$

for every spatial vector $X^\mu$.

If $e_i{}^\mu$ is an orthonormal spatial triad carried by a normal observer, then $\rho=T(n,n)$, $j_i=-T(n,e_i)$, and $S_{ij}=T(e_i,e_j)$. The diagonal entries of $S_{ij}$ are normal stresses and the off-diagonal entries are shear stresses in that frame.

Reconstructing the four-dimensional tensor

Insert the identity

$$\delta^\mu{}_\nu=\gamma^\mu{}_\nu-n^\mu n_\nu$$

on both indices of $T_{\mu\nu}$. Sorting the normal-normal, mixed, and spatial terms gives

$$T_{\mu\nu} =\rho n_\mu n_\nu +n_\mu j_\nu+j_\mu n_\nu +S_{\mu\nu}.$$

The signs can be checked without memorization. Contracting twice with $n^\mu$ returns $\rho$. Contracting once with $n^\mu$ and once with a spatial vector $X^\nu$ gives $-j(X)$ because $n^\mu n_\mu=-1$. Full spatial projection returns $S_{\mu\nu}$.

Let

$$S=\gamma^{ij}S_{ij}.$$

Taking the spacetime trace of the reconstruction formula yields

$$T\equiv g^{\mu\nu}T_{\mu\nu}=-\rho+S.$$

These relations and their Eulerian-observer interpretation are developed in Gourgoulhon 2007, §4.1.2, pp. 52–53, PDF.

How the projections source Einstein evolution

With $K_{ij}=-\gamma_i{}^\mu\gamma_j{}^\nu\nabla_\mu n_\nu$, the normal-normal Einstein equation gives

$${}^{(3)}R+K^2-K_{ij}K^{ij} =16\pi G\rho+2\Lambda.$$

The mixed projection gives

$$D_jK^j{}_i-D_iK=8\pi G j_i.$$

Finally, the spatial trace reversal

$$\gamma_i{}^\mu\gamma_j{}^\nu \left(T_{\mu\nu}-\frac12Tg_{\mu\nu}\right) =S_{ij}-\frac12\gamma_{ij}(S-\rho)$$

appears in the $K_{ij}$ evolution equation. Energy density therefore sources the scalar constraint, momentum density sources the vector constraint, and the complete spatial stress affects normal evolution.

The split does not replace matter dynamics. Projecting $\nabla_\mu T^{\mu\nu}=0$ along $n_\nu$ gives an energy equation; projecting with $\gamma^\rho{}_\nu$ gives a momentum equation. Their flux, acceleration, and extrinsic-curvature terms depend on the particular matter model and on lapse and shift. Gourgoulhon works out these conservation equations and the perfect-fluid specialization in Gourgoulhon 2007, §§5.2–5.3, pp. 72–82, PDF.

Perfect-fluid projections

Let $e$ and $p$ be rest-frame energy density and pressure, and write

$$T_{\mu\nu}=(e+p)u_\mu u_\nu+p g_{\mu\nu}.$$

Decompose the fluid four-velocity relative to the normal observer as

$$u^\mu=W(n^\mu+v^\mu), \qquad n_\mu v^\mu=0, \qquad W=(1-v^2)^{-1/2},$$

where $v^2=\gamma_{ij}v^iv^j<1$. Direct projection gives

$$\rho=(e+p)W^2-p,$$ $$j_i=(e+p)W^2v_i,$$

and

$$S_{ij}=(e+p)W^2v_iv_j+p\gamma_{ij}.$$

For a comoving normal observer, $v^i=0$ and $W=1$, so

$$\rho=e, \qquad j_i=0, \qquad S_{ij}=p\gamma_{ij}.$$

This recovers the physical rest-frame interpretation and checks the sign of $j_i$. A fluid moving in the positive $x$ direction has $j_x>0$.

Minkowski and Milne checks

In inertial Minkowski coordinates, a perfect fluid at rest with respect to $n=\partial_T$ has exactly the comoving projections above. A Lorentz boost of the matter while retaining the same slicing produces nonzero $j_i$ and anisotropic $S_{ij}$ even though the rest-frame stress is isotropic. The change is observer dependence, not a change in $T_{\mu\nu}$.

For Milne hyperboloids, $n=\partial_\tau$ is a different family of inertial observers. A homogeneous fluid chosen comoving with these normals again has $\rho=e$, $j_i=0$, and $S_{ij}=p\gamma_{ij}$. In vacuum all three projections vanish, and the Milne data satisfy the vacuum constraints despite $K_{ij}\ne0$.

One must separate a kinematic decomposition from a self-consistent solution: placing an arbitrary nonzero test-fluid tensor on the flat Milne metric does not make that pair solve Einstein’s equation. The projections remain well-defined, but the constraints correctly report the source mismatch.

Common pitfalls

Energy density depends on the observer. The scalar $\rho=T(n,n)$ changes when the foliation normal changes; it is not generally the matter rest-frame energy density.

Momentum density carries a conventional minus sign. With signature $(-,+,+,+)$ and the definition used here, $T(n,X)=-j(X)$.

Spatial stress is not determined by $\rho$ and $j_i$. Matter models with the same energy and momentum can have different pressure and anisotropic stress, leading to different $K_{ij}$ evolution.

Exercise: scalar-field projections

For a real scalar field with

$$T_{\mu\nu} =\nabla_\mu\phi\nabla_\nu\phi -\frac12g_{\mu\nu} \left(\nabla_\rho\phi\nabla^\rho\phi+m^2\phi^2\right),$$

define $\Pi=n^\mu\nabla_\mu\phi$ and $\Phi_i=D_i\phi$. Derive $\rho$, $j_i$, and $S_{ij}$.

Solution

The gradient splits as

$$\nabla_\mu\phi=-\Pi n_\mu+\Phi_\mu, \qquad \nabla_\rho\phi\nabla^\rho\phi=-\Pi^2+\Phi_k\Phi^k.$$

Normal-normal projection gives

$$\rho =\frac12\left( \Pi^2+\Phi_k\Phi^k+m^2\phi^2 \right).$$

The mixed metric term vanishes under one normal and one spatial projection, so

$$j_i=-\Pi\Phi_i.$$

Finally,

$$S_{ij} =\Phi_i\Phi_j +\frac12\gamma_{ij} \left( \Pi^2-\Phi_k\Phi^k-m^2\phi^2 \right).$$

If the field is homogeneous on a slice, $\Phi_i=0$, so $j_i=0$. The normal observer measures kinetic-plus-potential energy, while the spatial stress has the standard kinetic-minus-potential combination.

References

• Gourgoulhon, Éric. 3+1 Formalism and Bases of Numerical Relativity. Lecture notes, arXiv:gr-qc/0703035v1 (2007). arXiv.