Symmetric Trace-Free Multipoles

An angular multipole of degree $\ell$ is most economical when represented by a Cartesian tensor that is symmetric in all $\ell$ indices and has every trace removed. These symmetric trace-free (STF) tensors carry exactly the $2\ell+1$ components of one spherical-harmonic sector, while remaining easy to differentiate, contract, and combine with source integrals. This page develops that equivalence and fixes the normalization used in the multipolar waveform construction.

Required background. Weak-Field Gravity supplies the perturbative fields whose angular dependence is decomposed, while Ordinary Differential Equations supplies the differentiation and asymptotic-series operations used on the moment functions.

Helpful background. Einstein Equations explains the field equations from which the exterior moments arise; Waves, Green Functions, and Geometric Optics supplies the retarded-wave setting; and Relativistic Two-Body Dynamics explains how a particular source can provide time-dependent moments. Detailed binary equations of motion and detector inference are deliberately not treated here.

STF projection and multi-index notation

Let $L=i_1\cdots i_\ell$ be a spatial multi-index in three Euclidean dimensions. We write $x_L=x_{i_1}\cdots x_{i_\ell}$, $\partial_L=\partial_{i_1}\cdots\partial_{i_\ell}$, and place angle brackets around indices to mean “symmetrize with unit weight, then subtract every trace.” Thus $A_{\langle L\rangle}$ is symmetric and obeys

$$\delta^{i_1i_2}A_{\langle i_1i_2i_3\cdots i_\ell\rangle}=0.$$

For ranks two and three the operation is explicit:

$$\begin{aligned} A_{\langle ij\rangle} &=A_{(ij)}-\frac13\delta_{ij}A_{(aa)},\\ A_{\langle ijk\rangle} &=A_{(ijk)}-\frac15\left( \delta_{ij}A_{(aak)}+ \delta_{ik}A_{(aaj)}+ \delta_{jk}A_{(aai)} \right). \end{aligned}$$

The coefficient $1/5$ in the second line is fixed, rather than conventional: tracing $i$ with $j$ gives one vector trace from $A_{(ijk)}$, while the three subtraction terms give $3+1+1=5$ copies of that same vector. In particular, for a unit radial vector $n_i=x_i/r$,

$$\widehat n_{ij}=n_i n_j-\frac13\delta_{ij}, \qquad \widehat n_{ijk}=n_i n_j n_k-\frac15 \left(\delta_{ij}n_k+\delta_{ik}n_j+\delta_{jk}n_i\right).$$

Here a hat and angle brackets denote the same STF projection. These low-rank formulas agree with the convention used in the multipolar post-Minkowskian construction Blanchet 2024, §2.1.2, pp. 39–41.

Why one STF rank is one angular multipole

Associate an STF tensor $T_L$ with the homogeneous polynomial

$$P_\ell(\boldsymbol{x})=T_L x_L=r^\ell T_L\widehat n_L.$$

Taking a Laplacian contracts two tensor indices, so trace freedom is exactly the condition

$$\nabla^2P_\ell=0.$$

Homogeneous harmonic polynomials of degree $\ell$, restricted to the unit sphere, are the usual spherical harmonics of degree $\ell$. The STF basis and the $Y_{\ell m}$ basis are therefore two bases for the same irreducible rotation representation, not two approximations to one another. Thorne gives the complete tensor–harmonic dictionary and its normalization Thorne 1980, §§II–III and Table I.

The component count makes the equivalence concrete. A symmetric rank-$\ell$ tensor in three dimensions has $\binom{\ell+2}{2}$ components. Its trace is a symmetric rank-$(\ell-2)$ tensor with $\binom{\ell}{2}$ components. Removing that image leaves

$$\binom{\ell+2}{2}-\binom{\ell}{2}=2\ell+1,$$

precisely the number of values $m=-\ell,\ldots,\ell$. For $\ell=0$ the STF object is a scalar; for $\ell=1$ it is a vector; for $\ell=2$ it is a five-component quadrupole. Radiation in general relativity begins with time-varying $\ell=2$ moments because the isolated-source mass monopole and total-momentum or center-of-mass dipoles are constrained by conservation, not because the STF construction excludes lower ranks.

Orthogonality and reconstruction on the sphere

Let $\widehat\delta_{LK}$ denote the identity on rank-$\ell$ STF tensors. With $d\Omega$ the unit-sphere measure,

$$\int_{S^2}d\Omega\,\widehat n_L\widehat n_K =\delta_{\ell k}\, \frac{4\pi\,\ell!}{(2\ell+1)!!}\, \widehat\delta_{LK},$$

where $K$ has rank $k$ and the identity tensor on the right is used only when $k=\ell$; unequal ranks integrate to zero. The double factorial is $(2\ell+1)!!=(2\ell+1)(2\ell-1)\cdots3\cdot1$. Consequently a square-integrable scalar angular function has the Cartesian multipole expansion below in the $L^2(S^2,d\Omega)$ sense (smooth functions also admit the usual stronger convergence):

$$F(\boldsymbol n)=\sum_{\ell=0}^{\infty}F_L\widehat n_L, \qquad F_L=\frac{(2\ell+1)!!}{4\pi\,\ell!} \int d\Omega\,F(\boldsymbol n)\widehat n_L.$$

This is the STF version of projecting onto orthonormal spherical harmonics. It is especially useful in wave generation because Cartesian differentiation preserves a transparent multi-index form. A generic retarded exterior solution can be organized into terms such as $\partial_L[K_L(t-r)/r]$, and all angular traces can be removed before the radial and time derivatives are evaluated. The procedure avoids carrying reducible tensor pieces that would later cancel under angular projection.

An immediate normalization check is the quadrupole integral. Using isotropy,

$$\int d\Omega\,n_i n_j n_k n_l =\frac{4\pi}{15} \left(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right),$$

and subtracting traces gives the $\ell=2$ case of the STF orthogonality formula. This check detects both an incorrect $1/3$ in $\widehat n_{ij}$ and a missing factor of $\ell!$ in the reconstruction coefficient.

From angular tensors to source moments

For a slow, weak Newtonian source, the leading mass-type moment illustrates the role of the basis:

$$I_L(t)=\int d^3x\,\rho(t,\boldsymbol{x})\,x_{\langle L\rangle}.$$

$I_L$ has dimensions of mass times length$^\ell$. At leading radiative order, $\ell$ time derivatives enter the mass-type radiative moment, and a factor $G/R$ converts it to dimensionless strain. In units $c=1$ with $G$ explicit,

$$\frac{G}{R}I_L^{(\ell)}$$

is dimensionless, as it must be. The corresponding current-type moments use the same STF angular basis but include the source momentum flow and the spatial Levi-Civita tensor. Their detailed source formulas depend on the approximation scheme; the irreducible angular decomposition does not.

This separation is important. STF tensors solve the representation problem—how to encode angular structure without redundant traces. They do not by themselves determine the source dynamics, the retarded boundary condition, or the nonlinear map from source moments to the waveform at infinity. Those steps are developed on the following pages.

Checks and limiting cases

Three quick tests should accompany every STF manipulation.

1. Trace: contract every distinct pair of free indices. A nonzero result means the projection is incomplete.
2. Dimension: projection changes angular content but never physical units; every term in a projected moment must have the same dimension.
3. Stationarity: if an isolated source becomes stationary, its radiative time derivatives vanish. Constant STF moments may describe its exterior Coulombic multipoles, but they do not produce an outgoing $1/R$ waveform.

The last point prevents a common confusion: a non-spherical stationary body can possess a quadrupole moment without emitting gravitational waves. Radiation requires the appropriate retarded-time variation.

Exercise

Let $S_{ijk}=S_{(ijk)}$ be a symmetric rank-three tensor. Show that

$$T_{ijk}=S_{ijk}-\frac15\left( \delta_{ij}S_{aak}+\delta_{ik}S_{aaj}+\delta_{jk}S_{aai} \right)$$

is trace-free, and explain why it has seven independent components.

Solution

Contract the first two indices. Symmetry gives $S_{iik}=S_{aak}$. In the subtraction, $\delta_{ii}S_{aak}=3S_{aak}$, while the contractions of $\delta_{ik}S_{aaj}$ and $\delta_{jk}S_{aai}$ each give another $S_{aak}$. Hence $T_{iik}=S_{aak}-(3+1+1)S_{aak}/5=0$. Symmetry makes every other pair trace vanish as well. A symmetric rank-three tensor in three dimensions has $\binom{5}{2}=10$ components; its trace is an arbitrary three-component vector. Removing those three trace components leaves $10-3=7=2(3)+1$, matching the seven $m$ values of an $\ell=3$ spherical harmonic.

References

• Blanchet, Luc. “Post-Newtonian Theory for Gravitational Waves.” Living Reviews in Relativity 27 (2024) 4. DOI. Open PDF.
• Thorne, Kip S. “Multipole Expansions of Gravitational Radiation.” Reviews of Modern Physics 52 (1980) 299–339. DOI.