Curvature and Detector Observables

A gravitational-wave detector responds to relative motion, phase, or timing, not to a coordinate component of a metric perturbation by itself. The local bridge from field to measurement is the electric part of the Riemann tensor, while the familiar strain formula follows only after a weak-field, freely falling, long-wavelength reduction. Keeping those levels distinct explains both why the response is gauge independent and why a real detector channel is not identical to $h_+$ or $h_\times$. Empirical calibration and inference are outside this page.

Helpful background. Einstein Equations supplies the vacuum equations used to identify radiative curvature. Waves, Green Functions, and Geometric Optics supplies the wavelength and propagation approximations. Gravitational Radiation and Waveforms supplies TT strain and the volume’s curvature convention. Tests of Gravity supplies the later setting in which alternative responses are compared.

The observer’s tidal tensor

Let $u^a$ be the four-velocity of a freely falling reference observer and let $e_i^a$ be an orthonormal spatial triad carried along that worldline. The measured electric curvature is

$$\mathcal E_{ij}=R_{abcd}e_i^a u^b e_j^c u^d.$$

The pair-exchange symmetry of the Riemann tensor makes $\mathcal E_{ij}$ symmetric, including in nonvacuum regions. In vacuum it is also trace free. It has dimensions of inverse time squared. If $\xi^i$ is the small separation to a neighboring geodesic, geodesic deviation gives

$$\frac{D^2\xi^i}{d\tau^2}=-\mathcal E^i{}_j\xi^j.$$

This is an operational statement: an array of ideal test masses reconstructs the tidal map in the observer’s frame. It does not require a TT coordinate system. The derivation and its weak-wave specialization are developed in Flanagan and Hughes 2005, §3, pp. 13–17, PDF.

From curvature to TT strain

On Minkowski spacetime write $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ and retain first order in $h_{\mu\nu}$. In a radiation zone where a TT representative exists, the site’s curvature convention gives

$$\mathcal E_{ij}=R_{0i0j}^{(1)} =-\frac12\ddot h_{ij}^{\mathrm{TT}}.$$

Hence initially comoving masses satisfy

$$\ddot\xi^i=\frac12\ddot h^{\mathrm{TT},i}{}_j\xi^j, \qquad \delta\xi^i=\frac12 h^{\mathrm{TT},i}{}_j\xi_0^j,$$

after choosing the integration constants so that the perturbation and relative velocity vanish before the wave. The second formula is therefore not a separate force law; it is the twice-integrated curvature response with boundary data. Memory precisely exposes why those integration constants matter.

Why the linear response is gauge invariant

An infinitesimal coordinate change generated by $\epsilon_\mu$ sends

$$h_{\mu\nu}\mapsto h_{\mu\nu} -\partial_\mu\epsilon_\nu-\partial_\nu\epsilon_\mu.$$

Substitution into the linearized Riemann tensor produces pairs of commuting partial derivatives that cancel. Equivalently, a first-order curvature perturbation changes by the Lie derivative of the background curvature; that term vanishes on Minkowski spacetime. Thus $R_{\alpha\beta\gamma\delta}^{(1)}$, and therefore $\mathcal E_{ij}$ in a fixed physical tetrad, is unchanged even though the components $h_{\mu\nu}$ change.

This flat-background result must not be overextended. On a curved background the first-order Riemann tensor alone is generally not gauge invariant because the background curvature is nonzero. One then uses invariant combinations, background-adapted master variables, or asymptotic quantities with their residual frame freedom stated explicitly.

An ideal equal-arm response

Let an ideal Michelson have orthogonal unit arm vectors $\boldsymbol a$ and $\boldsymbol b$. If the arm length $L$ is much shorter than the reduced gravitational wavelength, the fractional round-trip differential response is

$$s(t)=D^{ij}h_{ij}^{\mathrm{TT}}(t), \qquad D^{ij}=\frac12(a^ia^j-b^ib^j).$$

Resolving the field in a screen basis gives

$$s=F_+h_++F_\times h_\times, \qquad F_A=D^{ij}e^A_{ij}.$$

The observable is unchanged by a screen rotation: the polarization and antenna-factor columns transform with the same orthogonal matrix $R$, so $(R\boldsymbol F)^{\mathsf T}(R\boldsymbol h)=\boldsymbol F^{\mathsf T}\boldsymbol h$. For one freely suspended arm, the same approximation yields $\delta L/L=\tfrac12 a^ia^jh_{ij}^{\mathrm{TT}}$. These formulas describe an ideal metric-to-length response, not photodiode electronics or calibrated data.

When wavelength and instrument size compete

If $2\pi fL$ is not small (with $c=1$), different parts of a light path sample different wave phases. The response becomes a delayed path integral, or in frequency space

$$\widetilde s(f)=D^{ij}(f,\boldsymbol n) \widetilde h_{ij}^{\mathrm{TT}}(f).$$

The frequency- and direction-dependent tensor $D^{ij}(f,\boldsymbol n)$ contains propagation delays and the instrument geometry. It reduces to the constant tensor above as $fL\to0$. Pulsar timing and Doppler links likewise measure integrated frequency shifts rather than a strain value at one event. The spacetime field, the ideal response, and a calibrated detector channel are three related but distinct objects.

Checks and limitations

Three checks catch most category errors. First, $\mathcal E_{ij}$ must scale as $\ddot h$, so its units are $T^{-2}$. Second, adding a pure-gauge metric perturbation must leave the flat-background linearized curvature zero. Third, a detector response must be invariant when the polarization screen is relabeled. None of these checks supplies detector noise, calibration uncertainty, or a likelihood; those belong to Gravitational-Wave Astronomy.

Exercises

1. A pure gauge carries no tidal field. Let $h_{\mu\nu}=-2\partial_{(\mu}\epsilon_{\nu)}$. Show directly that its linearized Riemann tensor vanishes.

Solution

The linearized tensor can be written

$$R^{(1)}_{\alpha\beta\gamma\delta} =\frac12\left( \partial_\gamma\partial_\beta h_{\alpha\delta} +\partial_\delta\partial_\alpha h_{\beta\gamma} -\partial_\delta\partial_\beta h_{\alpha\gamma} -\partial_\gamma\partial_\alpha h_{\beta\delta} \right).$$

After inserting $h_{\mu\nu}$, every third derivative of $\epsilon_\mu$ occurs twice with opposite signs. Partial derivatives commute on the Minkowski background, so the sum is zero. A coordinate ripple can therefore change $h_{\mu\nu}$ without accelerating neighboring test masses.

2. Michelson normalization. Take $\boldsymbol a=\boldsymbol e_x$, $\boldsymbol b=\boldsymbol e_y$, and a wave along $+z$ in the matching plus-cross screen. Find $F_+$ and $F_\times$.

Solution

With $D^{ij}=\tfrac12\operatorname{diag}(1,-1,0)$ and $e^+_{ij}=\operatorname{diag}(1,-1,0)$,

$$F_+=D^{ij}e^+_{ij}=1.$$

The detector tensor is diagonal whereas $e^\times$ is off diagonal, so $F_\times=0$. Thus $s=h_+$ for this special alignment. Rotating either the detector or the screen changes the two factors but not the final contraction.

References

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