Ordinary Displacement Memory

Ordinary displacement memory is a permanent change in the asymptotic strain produced when the nonradiative state of an isolated system changes between early and late times. The archetypal source is scattering, ejection, or capture that changes the asymptotic velocities of massive bodies. After the burst and its local curvature have passed, freely falling separations need not return to their original values. This page derives that endpoint observable; it does not assess a present detector or event claim.

Helpful background. Einstein Equations supplies the constraints that relate asymptotic fields to matter and radiative flux. Waves, Green Functions, and Geometric Optics supplies retarded propagation and the zero-frequency endpoint limit. Gravitational Radiation and Waveforms supplies TT strain and curvature response. Causality and Global Structure supplies future null infinity, stationary asymptotic eras, and the distinction between null and timelike asymptotic data.

Memory as an endpoint of the Bondi shear

Use the volume’s Bondi convention

$$h_{AB}=q_{AB}+\frac{C_{AB}}r+O(r^{-2}), \qquad N_{AB}=\partial_uC_{AB},$$

with $q^{AB}C_{AB}=0$. For a unit-sphere dyad $m^A\bar m_A=1$, define

$$H=C_{AB}\bar m^A\bar m^B =\lim_{r\to\infty}r(h_+-i h_\times), \qquad \mathcal N=\partial_uH.$$

Suppose news is negligible in stationary or uniformly moving eras before $u_i$ and after $u_f$. The displacement-memory tensor is

$$\Delta C_{AB} =C_{AB}(u_f)-C_{AB}(u_i) =\int_{u_i}^{u_f}N_{AB}\,du,$$

and the complex memory is

$$\Delta H=\int_{u_i}^{u_f}\mathcal N\,du.$$

At large but finite Bondi areal radius $r$, $\Delta h=\Delta H/r+O(r^{-2})$. This is an asymptotically flat relation; replacing $r$ by a cosmological luminosity distance requires the propagation and redshift map. A time-independent supertranslation adds the same electric-parity vacuum shear to both stationary endpoints, so the inhomogeneous shift cancels from their difference when corresponding physical cuts are compared. The Bondi frame and cut matching must still be stated: comparing unrelated cuts can manufacture an apparent offset.

Favata fixes a compatible $h_+-i h_\times$ and spin-weight $-2$ convention before separating ordinary and nonlinear memory in Favata 2010, §§2–3, pp. 3–5, PDF.

The unbound-particle formula

In linearized gravity, let asymptotic bodies be labeled by $A$, with rest mass $m_A$, velocity $\boldsymbol v_A$, Lorentz factor $\gamma_A=(1-v_A^2)^{-1/2}$, and observation direction $\boldsymbol n$. The leading ordinary memory is

$$\Delta h^{\mathrm{TT}}_{ij} =\frac{4G}{r}\, \Delta\!\left[ \sum_A \frac{m_A\gamma_A\,v_{A i}v_{A j}} {1-\boldsymbol n\cdot\boldsymbol v_A} \right]^{\mathrm{TT}}.$$

The symbol $\Delta$ means the late outgoing state minus the early incoming state, with every body and asymptotic momentum accounted for. The TT projection is relative to $\boldsymbol n$. The denominator is a retarded-time Jacobian; it becomes important for relativistic motion. Its smallness alone does not imply divergent on-axis strain, because the transverse projection of $v_iv_j$ also vanishes along the velocity. The full endpoint formula is Favata 2010, pp. 4–5, eq. (11), PDF.

This formula is not obtained by differentiating a bound-orbit quadrupole forever. It follows from the retarded field of asymptotically inertial particles and compares its zero-frequency endpoints. The assumptions are:

• weak fields in the observation zone and leading $1/r$ order;
• asymptotically constant incoming and outgoing velocities;
• isolated-system retarded boundary conditions;
• a common inertial/Bondi frame for the two endpoints;
• no claim that the instantaneous interaction region is described by free particles.

For a system whose bodies remain bound and return to the same asymptotic Coulombic state, this leading ordinary term vanishes. A recoil, ejection, unbound scattering state, or changing matter flux can make it nonzero.

A two-body ejection check

Consider a parent initially at rest that ejects two equal masses $m$ with velocities $\pm v\boldsymbol e_x$. At the order used in this kinematic example, take the parent rest mass to be $2m\gamma$, so energy and momentum are conserved. Corrections from radiated energy belong to the corresponding higher-order source budget. Every initially stationary contribution has zero $v_iv_j$ term. Observe along $\boldsymbol n=\boldsymbol e_z$. Since $\boldsymbol n\cdot\boldsymbol v_A=0$, the late tensor inside the brackets is

$$S_{ij}=2m\gamma v^2\,e_{x i}e_{x j}.$$

The TT projector onto the $x$–$y$ screen gives

$$S^{\mathrm{TT}}_{ij} =m\gamma v^2 \begin{pmatrix} 1&0&0\\ 0&-1&0\\ 0&0&0 \end{pmatrix}.$$

The initial velocity term is zero, so

$$\Delta h_+=\frac{4Gm\gamma v^2}{r}, \qquad \Delta h_\times=0.$$

The offset vanishes continuously as $v\to0$, is even under exchange of the two bodies, and has the expected dimensionless scaling $G(\text{energy})/r$. Rotating the ejection axis rotates the plus/cross components with spin weight $-2$ but does not change the physical tensor.

Permanent displacement from geodesic deviation

For a long-wavelength apparatus, the local tidal equation is

$$\ddot\xi^i=\frac12\ddot h^{\mathrm{TT}}_{ij}\xi_0^j.$$

Choose initially comoving masses, subtract their unperturbed separation, and assume $\dot h^{\mathrm{TT}}_{ij}=0$ in both quiet eras. The first integration then gives zero net relative-velocity change, and the second gives

$$\Delta\xi^i =\frac12\Delta h^{\mathrm{TT}\,i}{}_j\,\xi_0^j.$$

Curvature is nonzero while $\ddot h_{ij}^{\mathrm{TT}}$ acts, then returns to zero. The final separation nevertheless differs because displacement is the twice-integrated tidal history. This is not a contradiction: zero final acceleration does not imply zero accumulated displacement.

The smooth step in the figure makes this distinction visible: after the transition, the ring is again at relative rest but has a different shape. The same response applies to a displacement-memory step produced by ordinary or null sources.

Permanent separation from $h_+(u)=\tfrac12\Delta h_+[1+\tanh(u/T)]$, with $h_\times=0$. The curve shows $h_+/\Delta h_+$ against $u/T$; the early and late rings follow the linear geodesic-deviation response for initially comoving masses. The illustrative $\Delta h_+=0.4$ exaggerates the deformation. As $u\to\pm\infty$, both $\dot h_+$ and $\ddot h_+$ vanish, while the separation changes by $\Delta\boldsymbol\xi=\Delta h^{\mathrm{TT}}\boldsymbol\xi_0/2$. The waveform is a schematic transition, not a source simulation.

Ordinary displacement memory is electric parity on the sphere. It must not be relabeled as spin memory, which is a time-integrated magnetic-parity time-delay observable, or center-of-mass memory, which involves a different subleading electric-parity time moment.

Boundary with observation

The equations define an asymptotic physical signal. Whether a particular instrument can retain a nonoscillatory offset depends on its transfer function, calibration, low-frequency control, observation duration, and noise model. Current searches or event-level conclusions belong to dated Research records and Gravitational-Wave Astronomy. They are not inferred from the existence of the classical memory solution.

Common pitfalls

Equating ordinary and nonlinear memory. Ordinary memory is sourced by a change in asymptotic matter/Coulombic data. Nonlinear memory is sourced by radiative energy flux, including gravitational waves themselves.

Comparing endpoints in different frames. A translation, boost, or supertranslation changes the cut and component representation. Transform both endpoints into one declared Bondi frame before taking $\Delta C_{AB}$.

Expecting final curvature to remain nonzero. Memory is a displacement after an impulse-like tidal history. The late stationary era can have zero news and zero radiative curvature.

Exercises

1. No-change test. Show that a scattering description with identical sets of asymptotic velocities and masses before and after has zero ordinary memory in the particle formula, even if the bodies were strongly accelerated in between.

Solution

The formula depends only on the endpoint difference of

$$\sum_A \frac{m_A\gamma_Av_{Ai}v_{Aj}} {1-\boldsymbol n\cdot\boldsymbol v_A}.$$

If the complete incoming and outgoing sets agree in one inertial frame, the tensors before TT projection are identical, so their difference and its TT projection vanish. The interaction can still emit oscillatory radiation and nonlinear memory; the conclusion concerns only the ordinary endpoint term.

2. Ring response. For the ejection example above, find the final fractional changes of initially orthogonal $x$ and $y$ separations.

Solution

Geodesic deviation gives $\Delta L_a/L=\tfrac12a^ia^j\Delta h_{ij}^{\mathrm{TT}}$. Hence

$$\frac{\Delta L_x}{L} =\frac12\Delta h_+ =\frac{2Gm\gamma v^2}{r}, \qquad \frac{\Delta L_y}{L} =-\frac{2Gm\gamma v^2}{r}.$$

The first-order area change vanishes because the transverse memory tensor is trace free, while the differential arm change equals $\Delta h_+$.

Thus the permanent pattern has the same plus-polarization geometry as an oscillatory wave even though its endpoint offset is nonoscillatory.

References

• Favata, Marc. “The Gravitational-Wave Memory Effect.” Classical and Quantum Gravity 27 (2010): 084036. DOI. Open PDF.