Displacement Memory Effect

A permanent shift in the separation of freely falling test masses after a gravitational wave passes

1. Geodesic Deviation in Bondi Coordinates

Consider a test particle at large $r$ in Bondi coordinates $(u, r, x^A)$. The geodesic equation $\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau} = 0$ for a slowly-moving particle at large $r$ reduces to:

$$\frac{d^2 x^i}{du^2} = -\Gamma^i_{uu} + \mathcal{O}(r^{-2})$$

For the Bondi-Sachs metric, expanding the Christoffel symbol $\Gamma^i_{uu}$ at large $r$:

$$\Gamma^i_{uu} = \frac{1}{2}g^{ij}\left(2\partial_u g_{ju} - \partial_j g_{uu}\right) = \frac{1}{2r}\partial_u C^i{}_j\,x^j + \mathcal{O}(r^{-2})$$

Now consider the geodesic deviation between two nearby test particles with separation vector $\xi^\mu$. The equation of geodesic deviation is $\frac{D^2\xi^\mu}{d\tau^2} = R^\mu{}_{\alpha\beta\gamma}\frac{dx^\alpha}{d\tau}\xi^\beta\frac{dx^\gamma}{d\tau}$. For our slowly-moving particles with $dx^\alpha/d\tau \approx (1, 0, 0, 0)$ in the TT frame:

$$\frac{d^2\xi^i}{du^2} = R^i{}_{u j u}\,\xi^j$$

2. From the Riemann Tensor to $\ddot{h}_{ij}^{TT}$

The Riemann tensor component $R^i{}_{uju}$ in the linearized regime $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ is computed from:

$$R^i{}_{uju} = \partial_u \Gamma^i_{ju} - \partial_j \Gamma^i_{uu} + \mathcal{O}(h^2)$$

Computing each Christoffel symbol in the TT gauge where $h_{0\mu} = 0$ and $h^i{}_i = 0$:

$$\Gamma^i_{ju} = \frac{1}{2}\eta^{ik}\left(\partial_j h_{ku} + \partial_u h_{jk} - \partial_k h_{ju}\right) = \frac{1}{2}\partial_u h^i{}_j$$

$$\Gamma^i_{uu} = \frac{1}{2}\eta^{ik}\left(2\partial_u h_{ku} - \partial_k h_{uu}\right) = 0 \quad \text{(in TT gauge)}$$

Therefore:

$$\boxed{\ddot{\xi}^i = R^i{}_{uju}\,\xi^j = \frac{1}{2}\ddot{h}_{ij}^{TT}\,\xi^j}$$

This is the tidal acceleration due to the gravitational wave, directly proportional to the second time derivative of the metric perturbation.

3. Double Integration: The Permanent Displacement

Starting from $\ddot{\xi}^i = \frac{1}{2}\ddot{h}_{ij}^{TT}\xi^j$, we integrate once from $u = -\infty$ to $u$. At early times $\dot{\xi}^i(-\infty) = 0$ and $\dot{h}_{ij}(-\infty) = 0$ (no radiation before the burst):

$$\dot{\xi}^i(u) = \frac{1}{2}\dot{h}_{ij}^{TT}(u)\,\xi^j_0 + \frac{1}{2}\int_{-\infty}^{u}\ddot{h}_{ij}^{TT}\,(\xi^j - \xi^j_0)\,du'$$

The second term is of order $\mathcal{O}(h^2)$ since $\xi^j - \xi^j_0 = \mathcal{O}(h)$. At linear order:

$$\dot{\xi}^i(u) = \frac{1}{2}\dot{h}_{ij}^{TT}(u)\,\xi^j_0 + \mathcal{O}(h^2)$$

Integrating again from $-\infty$ to $+\infty$, using $\xi^i(-\infty) = \xi^i_0$:

$$\boxed{\Delta\xi^i = \xi^i(+\infty) - \xi^i_0 = \frac{1}{2}\Delta h_{ij}^{TT}\,\xi^j_0 = \frac{1}{2r}\Delta C_{ij}^{TT}\,\xi^j_0}$$

where $\Delta C_{AB} = C_{AB}(+\infty) - C_{AB}(-\infty)$. If $\Delta C_{AB} \neq 0$, the test particles are permanently displaced even after all radiation has passed. This is the displacement memory effect.

4. Bondi Mass Loss: $G_{uu} = 0$ at Order $r^{-1}$

The $uu$-component of the Einstein equation at order $r^{-1}$ gives the evolution of the Bondi mass aspect. Starting from the Einstein tensor:

$$G_{uu}\big|_{r^{-1}} = \partial_u m_B + \frac{1}{8}N_{AB}N^{AB} - \frac{1}{4}D^A D^B N_{AB} = 0$$

Rearranging, this gives the celebrated Bondi mass-loss formula:

$$\boxed{\partial_u m_B = -\frac{1}{8}N_{AB}N^{AB} + \frac{1}{4}D^A D^B N_{AB}}$$

Integrating over $S^2$, the $D^AD^BN_{AB}$ term vanishes by integration by parts (it is a total divergence). This yields the Bondi mass loss:

$$\frac{dM_B}{du} = -\frac{1}{8}\oint_{S^2} N_{AB}N^{AB}\,d^2\Omega \;\leq\; 0$$

The total Bondi mass can only decrease: gravitational waves always carry positive energy to $\mathscr{I}^+$. This is the gravitational analogue of the Larmor formula in electromagnetism.

5. Christodoulou Nonlinear Memory

The key insight of Christodoulou (1991) is that the energy flux of gravitational waves itself acts as a source for a DC shift in the shear. Decompose the Bondi mass-loss equation into a constraint relating $\Delta C_{AB}$ to the energy flux. Start by integrating the mass-loss formula over all retarded time:

$$\Delta m_B = -\frac{1}{8}\int_{-\infty}^{+\infty} N_{AB}N^{AB}\,du + \frac{1}{4}D^A D^B \Delta C_{AB}$$

The total change in shear consists of a linear part (sourced by matter or unbound masses) and a nonlinear part (sourced by the gravitational wave energy flux). Separating $\Delta C_{AB} = \Delta C_{AB}^{\rm lin} + \Delta C_{AB}^{\rm NL}$ and using the constraint equation, the nonlinear memory satisfies:

$$\frac{1}{4}D^AD^B\Delta C_{AB}^{\rm NL} = \frac{1}{8}\int_{-\infty}^{+\infty} N_{AB}N^{AB}\,du$$

Writing the STF tensor in terms of its scalar potential: $C_{AB} = (D_A D_B - \frac{1}{2}\gamma_{AB}D^2)\Phi$, the constraint becomes an elliptic equation on $S^2$ for $\Delta\Phi^{\rm NL}$:

$$\boxed{\frac{1}{4}(D^2 + 2)D^2\,\Delta\Phi^{\rm NL} = \frac{1}{8}\int_{-\infty}^{+\infty} N_{AB}N^{AB}\,du}$$

The operator $(D^2 + 2)D^2$ has eigenvalues $-\ell(\ell+1)[\ell(\ell+1)-2]$ on $S^2$. The $\ell = 0$ and $\ell = 1$ modes are in the kernel (corresponding to supertranslation ambiguity), and the equation is invertible for $\ell \geq 2$.

6. Explicit Solution for the $\ell = 2$ Mode

For a binary merger, the dominant contribution is $\ell = 2, m = 0$. Expand the right-hand side in spherical harmonics: $\int N_{AB}N^{AB}\,du = \sum_{\ell m} \mathcal{E}_{\ell m}\,Y_{\ell m}(\theta, \phi)$. For the $\ell = 2$ mode, the eigenvalue of $(D^2+2)D^2$ is $-6 \times 4 = -24$:

$$\Delta\Phi^{\rm NL}_{20} = -\frac{1}{24}\cdot\frac{1}{2}\,\mathcal{E}_{20} = -\frac{\mathcal{E}_{20}}{48}$$

The memory shear in the $\ell = 2, m = 0$ sector is then:

$$\Delta C_{AB}^{\rm mem}\big|_{\ell=2} = \left(D_A D_B - \frac{1}{2}\gamma_{AB}D^2\right)\Delta\Phi^{\rm NL}_{20}\,Y_{20}$$

7. Memory Strain for a Binary Merger

For a compact binary with total mass $M$, symmetric mass ratio $\eta = m_1 m_2 / M^2$, and total radiated energy $E_{\rm rad}$, the memory strain observed at distance $R$ along the polar axis ($\theta = 0$) is dominated by the $\ell = 2, m = 0$ mode:

$$h_{\rm mem}^+ = \frac{E_{\rm rad}}{24\pi R}\sin^2\!\theta\,(17 + \cos^2\!\theta) + \mathcal{O}(\ell \geq 3)$$

For a face-on equal-mass binary ($\eta = 1/4$) with $E_{\rm rad} \approx 0.05\,Mc^2$ (from numerical relativity), the peak memory strain is:

$$h_{\rm mem}^{\rm peak} \approx \frac{E_{\rm rad}}{12\pi R} \sim \frac{\eta\,M}{R}\times\mathcal{O}(0.05)$$

This is roughly 10-20 times smaller than the peak oscillatory strain $h_{\rm osc} \sim \eta M/R$, making it challenging but potentially detectable by stacking multiple events in LIGO/Virgo/KAGRA or by next-generation detectors like LISA and the Einstein Telescope.

Simulation: Displacement Memory for BBH Mergers

We model the inspiral-merger-ringdown news tensor for several mass ratios, integrate to obtain the shear $C_{AB}(u)$ showing the permanent offset, and compute the Christodoulou memory strain:

Displacement memory for BBH mergers at various mass ratios

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