Transverse–Traceless Gauge

Transverse–traceless gauge makes the two radiative amplitudes visible by setting every nondynamical component of a vacuum wave to zero. It is an exceptionally useful wave-zone representative, but it is not a coordinate condition that can be imposed on an arbitrary sourced spacetime everywhere. The construction below separates three statements that are often conflated: reaching Lorenz gauge, using its residual freedom in vacuum, and applying the spatial TT projector to radiation traveling in a specified direction.

Required background. Weak-Field Gravity is used for the linearized metric and its infinitesimal gauge transformation. Waves, Green Functions, and Geometric Optics is used for vacuum wave solutions, Fourier modes, and their characteristic directions.

Helpful background. Einstein Equations explains why Lorenz gauge is compatible with conserved sources. Equivalence Principles and Gravitational Measurement supplies the distinction between a coordinate choice and the local tidal curvature that remains observable.

From Lorenz gauge to a vacuum TT representative

For $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$, define

$$\bar h_{\mu\nu}=h_{\mu\nu}-\frac12\eta_{\mu\nu}h.$$

With the passive-coordinate convention $x'^\mu=x^\mu+\xi^\mu$,

$$\bar h'_{\mu\nu} =\bar h_{\mu\nu}-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu +\eta_{\mu\nu}\partial_\rho\xi^\rho.$$

Taking a divergence gives

$$\partial^\mu\bar h'_{\mu\nu} =\partial^\mu\bar h_{\mu\nu}-\Box\xi_\nu.$$

Thus a solution of $\Box\xi_\nu=\partial^\mu\bar h_{\mu\nu}$ reaches Lorenz gauge, subject to compatible initial and boundary data. Once there, residual gauge vectors satisfy $\Box\xi^\mu=0$.

In a source-free region the field equation is $\Box\bar h_{\mu\nu}=0$. For a nonzero null Fourier mode,

$$\bar h_{\mu\nu}=A_{\mu\nu}e^{ik_\rho x^\rho}, \qquad k^2=0, \qquad k^\mu A_{\mu\nu}=0,$$

the four amplitudes of a residual vector $\xi_\mu=B_\mu e^{ik\cdot x}$ can be used to impose

$$h_{0\mu}^{\mathrm{TT}}=0, \qquad \partial^i h_{ij}^{\mathrm{TT}}=0, \qquad \delta^{ij}h_{ij}^{\mathrm{TT}}=0.$$

These equations say that the perturbation is purely spatial, transverse to the propagation direction, and traceless. For the nonzero plane-wave mode they leave two independent components. Constant coordinate shifts, background Killing fields, or boundary-supported zero modes require separate boundary conditions; they are not extra wave polarizations.

Flanagan and Hughes carry out this residual-gauge construction and relate it to the two physical amplitudes in Flanagan and Hughes 2005, §2.1, eqs. (2.18)–(2.28), pp. 7–8, PDF.

The spatial TT projector

Let $n^i$ be a unit spatial vector pointing along propagation and define the screen projector

$$P_{ij}=\delta_{ij}-n_i n_j.$$

For any symmetric spatial tensor $S_{kl}$, its transverse–traceless part relative to $n^i$ is

$$S^{\mathrm{TT}}_{ij} =\Lambda_{ij,kl}S_{kl}, \qquad \Lambda_{ij,kl} =P_{i(k}P_{l)j}-\frac12P_{ij}P_{kl}.$$

The projector has the required algebraic properties:

$$n^i\Lambda_{ij,kl}=0, \qquad \delta^{ij}\Lambda_{ij,kl}=0, \qquad \Lambda_{ij,mn}\Lambda_{mn,kl}=\Lambda_{ij,kl}.$$

For $n^i=(0,0,1)$, only the $x$–$y$ screen remains. Applying $\Lambda$ gives

$$S^{\mathrm{TT}}_{ij} =\begin{pmatrix} (S_{xx}-S_{yy})/2&S_{xy}&0\\ S_{xy}&-(S_{xx}-S_{yy})/2&0\\ 0&0&0 \end{pmatrix}.$$

The two independent entries multiply the plus and cross tensors. To specify the basis-rotation sign, choose $m^i=(e_x^i+i e_y^i)/\sqrt2$ and change its phase by $m^i\mapsto e^{i\psi}m^i$. This means $e'_x=\cos\psi\,e_x-\sin\psi\,e_y$ and $e'_y=\sin\psi\,e_x+\cos\psi\,e_y$. Since $h_+-ih_\times=h^{\mathrm{TT}}_{ij}\bar m^i\bar m^j$, the complex strain has spin weight $-2$:

$$h_+-i h_\times \longmapsto e^{-2i\psi}(h_+-i h_\times).$$

This mixing changes components, not the physical tidal tensor.

The Fourier-space projector is also nonlocal in position space because $n_i$ becomes a derivative divided by $\sqrt{-\nabla^2}$. Consequently, “take the TT part” can mean either an algebraic projection of a specified far-zone direction or a nonlocal tensor decomposition on a spatial slice. Neither operation by itself proves that a global coordinate system satisfying all TT conditions exists.

Where TT gauge is valid

TT gauge is reliable when the calculation concerns freely propagating radiation in a vacuum region and the boundary conditions admit the required residual transformation. Typical examples are a plane wave, a Fourier superposition of vacuum waves, or the leading $1/r$ field of a localized source in an asymptotically flat radiation zone.

It is generally not a global gauge for a spacetime containing matter. With $T_{\mu\nu}\ne0$, components such as $h_{00}$ and $h_{0i}$ encode constrained gravitational fields sourced by energy and momentum. Setting them all to zero would erase physical near-zone curvature or conflict with the field equations. On a curved background, covariant derivatives do not commute as partial derivatives do, and the clean scalar–vector–tensor split depends on background symmetry and boundary data.

Even in an asymptotically flat problem, a TT metric perturbation at finite coordinate radius need not equal the radiative strain at null infinity. Subleading $1/r^2$ terms, gauge effects, retarded-time choices, and asymptotic-frame transformations must be controlled before that identification is made.

Gauge-invariant check

The local observable is the linearized Riemann tensor. For a TT perturbation,

$$R^{(1)}_{0i0j}=-\frac12\ddot h^{\mathrm{TT}}_{ij}.$$

Adding a pure-gauge perturbation changes $h_{\mu\nu}$ but leaves $R^{(1)}_{\alpha\beta\gamma\delta}$ unchanged. This gives an independent check on any TT construction: compute the curvature before and after the residual transformation. If it changes, the transformation, sign convention, or perturbative order has been handled inconsistently.

Common pitfalls

Using the TT projector as a global coordinate transformation. The projector extracts a direction-dependent tensor part. A coordinate transformation must also solve differential equations and respect source, initial, and boundary data.

Dropping constrained fields near a source. Nonradiative does not mean unphysical. Newtonian and gravitomagnetic fields carry curvature even though they are not free outgoing waves.

Rotating plus into cross and calling it new radiation. Plus and cross are components in an oriented screen basis. Their spin-2 mixing under basis rotation preserves the underlying wave.

Exercises

1. Project a general tensor. For propagation along $+z$, apply the TT projector to

$$S_{ij}= \begin{pmatrix} a&b&c\\ b&d&e\\ c&e&f \end{pmatrix}.$$

Verify transversality, tracelessness, and idempotence.

Solution

Here $P_{ij}=\operatorname{diag}(1,1,0)$. The double screen projection removes $c,e,f$, and subtraction of half the screen trace gives

$$S^{\mathrm{TT}}_{ij} =\begin{pmatrix} (a-d)/2&b&0\\ b&-(a-d)/2&0\\ 0&0&0 \end{pmatrix}.$$

Its contraction with $n^i$ and its three-dimensional trace both vanish. Applying the same operation again changes neither the already transverse entries nor the zero screen trace, so $\Lambda^2S=\Lambda S$.

2. Source obstruction. Why can the weak, static field $h_{00}=2U$, with $U=GM/r$ outside a localized mass, not be globally transformed to a nonzero-frequency TT wave?

Solution

The positive potential $U$ obeys $\nabla^2U=-4\pi G\rho$ and has nonzero tidal curvature $R_{0i0j}=-\partial_i\partial_jU$ at linear order. A smooth gauge transformation cannot remove that curvature. It also has zero temporal frequency and Coulombic spatial falloff rather than nonzero null wave-vector support and outgoing radiative curvature. TT gauge for free waves does not erase sourced constraint fields. In the frequently used Newtonian convention $\Phi=-GM/r$, the same metric has $h_{00}=-2\Phi$.

References

• Flanagan, Éanna É., and Scott A. Hughes. “The Basics of Gravitational Wave Theory.” New Journal of Physics 7 (2005): 204. DOI. Open PDF.