Displacement memory for flyby

P.M. Zhang, Q.L. Zhao, J. Balog, P.A. Horvathy

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P.M. Zhang, Q.L. Zhao, J. Balog, P.A. Horvathy. Displacement memory for flyby. Annals of Physics, 2025, 473, pp.169890. 10.1016/j.aop.2024.169890. hal-05070515

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Distributed under a Creative Commons Attribution 4.0 International License
Displacement memory for flyby

P.-M. Zhang¹*, Q.-L. Zhao¹†, J. Balog²‡, P. A. Horvathy³§

¹ *School of Physics and Astronomy,*
*Sun Yat-sen University, Zhuhai, China*

² *Holographic QFT Group,*
*Institute for Particle and Nuclear Physics,*
*HUN-REN Wigner Research Centre for Physics H-1525 Budapest 114,*
*P.O.B. 49, Hungary*

³ *Institut Denis-Poisson CNRS/UMR 7013 - Université
de Tours - Université d'Orléans Parc de Grammont,*
*37200; Tours, FRANCE*

(Dated: May 20, 2025)

Abstract

Zel'dovich and Polnarev, in their seminal paper [1] extending previous work on the Memory
Effect, suggested that a gravitational wave generated by flyby would merely displace the particles.
We confirm their prediction numerically by fine-tuning the wave profile proposed by Gibbons and
Hawking [2], and then analytically for its approximation by a Pöschl-Teller potential. Higher-order
derivative profiles proposed for gravitational collapse, etc [2] are shortly discussed.

Annals Phys. **473** (2025), 169890. doi:10.1016/j.aop.2024.169890 [arXiv:2407.10787 [gr-qc]]

PACS numbers: 04.20.-q Classical general relativity;

04.30.-w Gravitational waves

<footnote>* mailto:zhangpm5@mail.sysu.edu.cn</footnote>

<footnote>† mailto: zhaoqliang@mail2.sysu.edu.cn</footnote>

<footnote>‡ mailto:balog.janos@wigner.hu</footnote>

<footnote>§ mailto:horvathy@univ-tours.fr</footnote>
# Contents

<table><tr><td>I. Introduction</td><td>2</td></tr><tr><td>II. Flyby: transverse motion</td><td>4</td></tr><tr><td>A. Flyby profile: derivative of Gaussian</td><td>4</td></tr><tr><td>III. Pöschl-Teller model for flyby: numerical study</td><td>6</td></tr><tr><td>IV. Pöschl-Teller model for flyby: (semi-) analytic treatment</td><td>9</td></tr><tr><td>A. Results</td><td>11</td></tr><tr><td>B. Asymptotic values</td><td>12</td></tr><tr><td>C. The k<sub>m</sub> ↔ m relation</td><td>13</td></tr><tr><td>D. Coefficient of the ln(1 − z) singularity</td><td>13</td></tr><tr><td>E. Half-waves</td><td>15</td></tr><tr><td>1. Pöschl-Teller</td><td>15</td></tr><tr><td>2. Derived Pöschl-Teller</td><td>16</td></tr><tr><td>V. Higher-derivative profiles</td><td>16</td></tr><tr><td>VI. Longitudinal motion</td><td>20</td></tr><tr><td>VII. Conclusion</td><td>21</td></tr><tr><td>Acknowledgments</td><td>22</td></tr><tr><td>References</td><td>23</td></tr></table>

# I. INTRODUCTION

The Displacement Memory Effect (DM) for gravitational waves generated by flyby was proposed by Zel'dovich and Polnarev [1] (see also [2–6]), extending previous work on the Velocity Memory Effect (VM) [7, 8, 24]. Our previous investigations [9–11] hinted, though, at that for *randomly chosen* parameters particles initially at rest would fly apart with non-zero velocity, as illustrated, e.g. in FIG. # 12. of ref. [11]. However closer recent scrutiny [12–14] indicates that for certain “magical” values of the wave parameters reminiscent of
quantum numbers we do get DM, highlighted by the vanishing of the velocity outside the Wavezone,

$$ \frac{dX}{dU}\bigg|_{(U = -\infty)} = 0 = \frac{dX}{dU}\bigg|_{(U = \infty)}. \qquad (I.1) $$

This paper extends these results to a more physical realm with particular attention at the analytically solvable Pöschl-Teller -version [12, 14–16]. Then we generalize the flyby result to gravitational collapse, etc by studying models with higher-order derivatives of the (Gaussian or Pöschl-Teller profile) proposed in [2–4].

We start with the 4 dimensional Brinkmann metric [17–19],

$$ g_{\mu\nu}dX^{\mu}dX^{\nu} = \delta_{ij}dX^{i}dX^{j} + 2dUdV + \frac{1}{2}\mathcal{A}(U)\left((X^{1})^{2} - (X^{2})^{2}\right)dU^{2}, \qquad (I.2) $$

where $X = (X^i)$ are transverse coordinates and $U, V$ are light-cone coordinates. Its geodesics are given by,

$$ \frac{d^2 X^1}{dU^2} - \frac{1}{2} \mathcal{A} X^1 = 0, \qquad (I.3a) $$

$$ \frac{d^2 X^2}{dU^2} + \frac{1}{2} \mathcal{A} X^2 = 0, \qquad (I.3b) $$

$$ \frac{d^2 V}{dU^2} + \frac{1}{4} \frac{d\mathcal{A}}{dU} \left( (X^1)^2 - (X^2)^2 \right) + \mathcal{A} \left( X^1 \frac{dX^1}{dU} - X^2 \frac{dX^2}{dU} \right) = 0. \qquad (I.3c) $$

The last equation is solved by solving first eqns (I.3a) - (I.3b) for the transverse trajectory $X(U)$ and then lifting it to the gravitational wave spacetime.

Let us recall that geodesic motion posses a conserved quantity referred to as the Jacobi invariant [20–22],

$$ \mathbf{m}^2 = -g_{\mu\nu}\dot{X}^{\mu}\dot{X}^{\nu} = \text{const.} \qquad (I.4) $$

where the dot denotes $d/dU$. Discarding tachions, we shall consider $\mathbf{m}^2 \le 0$.

Postponing the study of the $V$-motion to Sec. VI, we first focus our attention at the motion in transverse space. FIG. 1 suggests that for a random choice of the parameters the outgoing particle has constant nonzero velocity: it exhibits the *velocity effect* (VM). However recent results [12] suggest that for a judicious choice of the amplitude a “miracle” may happen and the outgoing velocity can indeed vanish — letting us wonder if this is also true in a broader and more physical context, and for flyby in particular. Our investigations below confirm that this is indeed what happens.
In our previous paper [12] we studied two cases in detail, namely those whose Brinkmann profile $\mathcal{A}$ is either a simple Gaussian, (III.1), or a Pöschl-Teller potential, (III.2). Both of them have an integer wave number $m$. In Sects. II and III we study what happens for flyby modelled by the *derivative profiles* [2] (II.1) and (III.7), respectively.

## II. FLYBY: TRANSVERSE MOTION

### A. Flyby profile: derivative of Gaussian

The profile for flyby proposed by Gibbons and Hawking in [2] is,

$$ \mathcal{A} \equiv \mathcal{A}^G = \frac{d}{dU} \left( \frac{k}{\sqrt{\pi}} e^{-U^2} \right). \qquad (II.1) $$

Its null geodesics are depicted in FIG.1,

<figure><img src="image_2.png" /><figcaption>Figure 1: The null geodesics for the first derivative of the Gaussian in (II.1) shown in the background exhibit, for randomly chosen amplitude $k$, the velocity effect (VM).</figcaption></figure>

We start with the massless case, $m = 0$. At the first sight, the null geodesics show VM as seen in FIG.1. We study first the transverse motion; that in the $V$-direction will be considered in sect. VI. Numerical fine-tuning then indicates, unexpectedly, that for certain “magical” critical values $k = k_{crit}$ we do get (approximate) DM for both transverse coordinates, as illustrated in FIG.2. In fact we found a well separated discrete series of critical amplitudes, $k_m$, each of which corresponding to a unique DM wave labeled by an integer $m$,

$$ k_1 = 32.6, \quad k_2 = 97.2, \quad k_3 = 194.8, \quad k_4 = 325.5, \quad k_5 = 489.2 \dots \qquad (II.2) $$

Our plots suggest that for each such critical amplitude $k_{crit}$ the Wavezone contains a unique DM trajectory composed approximately of an integer number of half-waves, consistently
with our observations in [12]. The novelty is that DM is now obtained for both coordinates, whereas for the simple Gaussian, shown in FIG.20 of [12], DM could be found in the attractive, but not in the repulsive sector; in [12] we called this “half-DM”.

<figure><img src="image_3.png" /><figcaption>Figure 2: For the flyby profile (II.1) fine-tuning the amplitude yields, for each $k = k_{crit}$, a unique (approximate) DM trajectory for both components, as shown for $m = 1$ and $m = 2$.</figcaption></figure>

The [square-roots of] the critical amplitude, $\sqrt{k_{crit}}$, depend on $m$ approximately linearly, as shown in FIG.3 (to be compared to FIG.5 of [12]).

For $k \neq k_{crit}$ we get instead non-zero outgoing velocity; the trajectory is not composed of an integer number of half-waves: we get VM but no DM, as it is manifest in FIG.1.

For generic wave parameters, the particle absorbs some energy from the passing wave [23–25]. DM implies in turn zero cumulated energy of the solution in FIG.2, as confirmed in FIG.4.

It is instructive to record the energy absorption of the underlying NR model [18–20] for $k_{crit}$ both in the low and high amplitude regimes, shown in FIG.5. We shall return to the relation to the underlying NR system in sec.VI.
<figure><img src="image_4.png" /><figcaption>Figure 3: DM is obtained when the wave number m and the square-root of the amplitude $\sqrt{k}$ are (approximately) linearly correlated.</figcaption></figure>

<figure><img src="image_5.png" /><figcaption>Figure 4: For $k = k_{crit}$ the total energy balance after the flyby is zero, as shown for m = 1 and for m = 2.</figcaption></figure>

### III. PÖSCHL-TELLER MODEL FOR FLYBY: NUMERICAL STUDY

First we revisit the simplest cases. The geodesics for the Gaussian profile proposed in [2]

$$ \mathcal{A}^G(U) = \frac{k}{\sqrt{\pi}} e^{-U^2} \qquad (III.1) $$

can only be found numerically. However its similarity with the Pöschl-Teller profile [15],

$$ \mathcal{A}^{PT}(U) = \frac{k}{2 \cosh^2 U}, \qquad (III.2) $$

allowed us to find analytic solutions by solving the Pöschl-Teller-modified Sturm-Liouville equation [12–14, 16]

$$ \frac{d^2 X}{d U^2} + \frac{1}{2} \mathcal{A}^{PT} X = 0. \qquad (III.3) $$

Putting $t = \tanh U$ and introducing the (at this point real) parameter m by

$$ k = 4m(m+1) \qquad (III.4) $$
<figure><img src="image_6.png" /><figcaption>Figure 5: For $k \neq k_{crit}$ we have VM and the particle extracts some energy from the passing gravitational wave. In the fine-tuned case $k = k_{crit}$, though, the cumulative energy balance is zero: the particle does not absorb any energy.</figcaption></figure>

(III.3) then becomes the Legendre equation,

$$ (1 - t^2) \frac{d^2 X}{dt^2} - 2t \frac{dX}{dt} + m(m + 1) X = 0. \qquad (\text{III.5}) $$

Our clue is that the DM conditions (I.1) [or $X(U = \mp\infty) \to \text{const.}$] require that the solution extends to $t = \mp 1$, — yielding a quantization condition: the parameter $m$ must be a natural number,

$$ m = 1, 2, \dots \qquad (\text{III.6}) $$

Then the Legendre function which solves eqn. (III.5) truncates to a Legendre polynomial of order $m$, $P_m(\tanh U)$, see eqn. (IV.6) and the trajectories exhibit DM, as illustrated in FIGs. 7, 8, 9 of [12]. In a remarkable analogy with what happens for a harmonic oscillator in Quantum Mechanics [12], $m \ge 1$ counts the number of half-waves accommodated in the (approximate) wavezone. The question will be further discussed in sect. IV.

In $D = 2$ transverse dimensions the two components come with a relative minus, therefore the two equations in (I.3a)-(I.3b) differ by the sign of $k$. Bounded solutions arise only in the attractive sector – Legendre polynomials have $m \ge 1$ – and we get “half-DM” (as in the Gaussian case [12]).

Turning to flyby which is our main concern in this paper, we note that the derivative of the Pöschl-Teller potential (dPT),

$$ \mathcal{A} \equiv \mathcal{A}^{\text{dPT}} = \frac{d}{dU} \left( \frac{k}{2 \cosh^2 U} \right) = -k \left( \tanh U - \tanh^3 U \right) \qquad (\text{III.7}) $$
shown in FIG.6 is reminiscent of the derivative-Gaussian proposed in [2], prompting us to study the geodesics of the metric (I.2) with derivative Pöschl-Teller (dPT) profile (III.7) as an alternative model for flyby.

<figure><img src="image_7.png" /><figcaption>Figure 6: The derivative of the Pöschl-Teller potential is a good approximation of that of the Gaussian.</figcaption></figure>

Then we get, instead of (III.5),

$$ (1 - t^2) \frac{d^2 X^i}{dt^2} - 2t \frac{dX^i}{dt} \pm \frac{k}{2} t X^i = 0, \qquad (III.8) $$

with the ± sign referring to the coordinate sector, $i = 1$ or 2. Assuming $k > 0$, $+k$ is for $X^2$ and $-k$ is for $X^1$.

The behavior is reminiscent of but different from that for simple Pöschl-Teller, (III.5), as seen by decomposing (III.8) as,

$$ \underbrace{(1 - t^2) \frac{d^2 X^i}{dt^2} - 2t \frac{dX^i}{dt} + 2m(m + 1) X^i}_{\text{Legendre}} - \underbrace{(1 \mp t) 2m(m + 1) X^i}_{\text{perturbation}} = 0. \qquad (III.9) $$

where $m$ is an a priori real number. Thus (III.8) is a Legendre eqn. with a linear perturbation. For $t \to \pm 1$ both equations become Legendre equations but with different coefficients of the linear term, see FIGs. 7 and 8.

We emphasise, though, that while for the PT profile the analytic calculation is routine-like, the corresponding one for dPT is highly involved, as it will be seen in detail in sect. IV.

We note for completeness that off the critical values, $k \neq k_m$, the trajectories exhibit the velocity effect (VM) as expected and illustrated in FIG.9.
<figure><img src="image_8.png" /><figcaption>Figure 7: For the dPT-flyby profile (III.7), both numerically or analytically obtained trajectories exhibit DM for both components. For convenience, the $X^2$ solutions (in red) are rescaled by a factor 20, while the $X^1$ solutions (in blue) are rescaled, for $m = 1$, $m = 2$, by 0.25 and 0.006, respectively.</figcaption></figure>

<figure><img src="image_9.png" /><figcaption>Figure 8: Analytic dPT trajectories for critical amplitude (i) $k_{crit} = 166...$ and (ii) $k_{crit} = 276...$ in the derivative-Pöschl-Teller approximation, have half-wave numbers $m = 3$ and $m = 4$.</figcaption></figure>

## IV. PÖSCHL-TELLER MODEL FOR FLYBY: (SEMI-) ANALYTIC TREATMENT

Using $t = \tanh U$ the equations for the transversal components are (III.8),

$$ (1 - t^2) \frac{d^2 X^i}{dt^2} - 2t \frac{dX^i}{dt} \pm \frac{k}{2} t X^i = 0, $$

where the $\pm$ sign is for the $i = 1, 2$ components, respectively. We introduce the parameter

$$ p = \mp \frac{k}{2} \qquad (IV.1) $$
<figure><img src="image_10.png" /><figcaption>Figure 9: For $k \neq k_{crit}$ the trajectories exhibit the velocity effect (VM).</figcaption></figure>

and allows us to study the two cases simultaneously. Changing the independent variable to $z = \frac{t+1}{2}$ the equation to be solved becomes,

$$ \frac{d^2 X}{dz^2} + \left( \frac{1}{z} + \frac{1}{z-1} \right) \frac{dX}{dz} + p \left( \frac{1}{z} + \frac{1}{z-1} \right) X = 0, \qquad (IV.2) $$

where $X = X^1$ or $X^2$. Here we recognize the confluent Heun equation with special parameters. The singular points are $z = 0, 1$ (plus the irregular singular point at infinity), corresponding to $t = -1, 1$ and $U = -\infty, +\infty$, respectively.

We are looking for solutions which are regular at both of these points. Assuming regularity at $z = 0$, we can write the solution as a power series

$$ X = \sum_{n=0}^{\infty} s_n z^n. \qquad (IV.3) $$

We normalize the solution by putting $s_0 = 1$. The $s_n$ coefficients can be determined from the recursion relation

$$ s_{n+1} = \frac{[n(n+1)-p]s_n + 2ps_{n-1}}{(n+1)^2}, \quad n = 0, 1, \dots \qquad (IV.4) $$

($s_{-1} = 0$ by convention.)

We look for parameter values such that the power series (IV.3), which converges in the disk $|z| < 1$, defines a function which is regular also at $z = 1$. In similar problems the solution is usually a polynomial. There is no polynomial solution in this case, though.

In the original variable the general solution of the problem behaves as

$$ X \sim 2A_0 U + \text{const.} \qquad (IV.5) $$
for $U \to +\infty$, where $A_0(p)$ is some parameter-dependent coefficient. In the new coordinates this means singular behaviour,

$$X \sim -A_0 \ln \left( \frac{1-t}{2} \right) = -A_0 \ln(1-z) = A_0 \sum_{n=1}^{\infty} \frac{z^n}{n} \qquad (IV.6)$$

for $t \to 1$, $z \to 1$. We introduce

$$x_n = ns_n \qquad (IV.7)$$

for convenience. According to the above analysis, this has asymptotic expansion for large $n$

$$x_n = A_0 + \frac{A_1}{n} + \frac{A_2}{n(n-1)} + \frac{A_3}{n(n-1)(n-2)} + \dots \qquad (IV.8)$$

with $p$-dependent coefficients $A_k$. Actually it follows from the recursion relation that all higher coefficients are proportional to $A_0$:

$$A_k = \frac{a_k}{k!} A_0, \qquad (IV.9)$$

where

$$a_1 = -p, \quad a_2 = p(p-2), \quad p_3 = -p^2(p-8), \quad p_4 = p^3(p-20), \quad \text{etc.} \qquad (IV.10)$$

We look for non-singular solutions with $A_0=0$ for which $X(U)$ goes to a mere constant when $U \to \infty$, as it follows from (IV.5). According to the above considerations in this case all $A_k$ vanish and such solutions are not only regular at $z = 1$ but also analytic there. In fact, these solutions are entire functions.

## A. Results

It remains to find special parameter values $p_m$ such that

$$A_0(p_m) = 0. \qquad (IV.11)$$

(We will also use $k_m = 2p_m$.) The trick, which is often used in the case of series behaving asymptotically like (IV.8), is to apply a series of Robertson transformations. These are defined recursively as

$$x_n^{(0)} = x_n, \qquad (IV.12)$$

$$x_n^{(s)} = \frac{1}{s} \left\{ (n+1-s)x_n^{(s-1)} - (n+1-2s)x_{n-1}^{(s-1)} \right\}, \quad s = 1, 2, \dots \qquad (IV.13)$$
<figure><img src="image_11.png" /><figcaption>Figure 10: The rescaled $A_0(p)$ coefficient</figcaption></figure>

This series of transformations eliminates, step by step, the leading corrections such that for large $n$

$$x_n^{(s)} = A_0 + O\left(\frac{1}{n^{s+1}}\right), \qquad (IV.14)$$

i.e. the $s^{th}$ Robertson transformed series has the same leading term but much smaller corrections to it. Using this method we calculated (see FIG.10)

$$A_0(p) = -\frac{p^2}{6} \left\{ 1 - \frac{p^2}{180} + 4.036062 \cdot 10^{-6} p^4 - 7.801993 \cdot 10^{-10} p^6 + \dots \right\} \qquad (IV.15)$$

The function is even in $p$ which means that the same critical values work for both signs in the original equation, i.e. the DM condition is satisfied simultaneously for both components $X^1, X^2$. The roots of $A_0(p)$ can only be calculated numerically<fnref n="1" />

$$\begin{align*} k_1 &= 29.0560068187, & k_4 &= 276.6156438862, \\ k_2 &= 84.0970923581, & k_5 &= 414.1212560177, \tag{IV.16} \end{align*}$$

$$k_3 = 166.6088726851, \quad k_6 = 579.1267981111.$$

## B. Asymptotic values

We normalize our solution such that $X = 1$ at $U = -\infty$ ($z = 0$). For critical values of $k$ we can calculate $X$ at $U = +\infty$ ($z = 1$). Let us take first $k = k_1$. The solution corresponds to the $X^2$ component and we find

$$X^2(\infty) = -0.0119701374443. \qquad (IV.17)$$

<footnote n="1">¹ The critical values could also be derived readily by requiring that the velocity go to zero when $U \to \infty$, (I.1).</footnote>
which is very nearly zero. For $k = -k_1$ instead, which corresponds to the $X^1$ component,
we find the considerably larger value,

$$X^1(\infty) = -83.5412300512. \qquad (IV.18)$$

Of course, $X^1(\infty)X^2(\infty) = 1$, due to the $U \leftrightarrow -U$ symmetry of the problem and our
normalization. For $k = \pm k_2$ the analogous results are

$$X^1(\infty) = 2176.4880022386, \quad X^2(\infty) = 0.000459455884207. \qquad (IV.19)$$

### C. The $k_m \leftrightarrow m$ relation

The following, purely numerically obtained formula,

$$k_m = b_0 w_m + b_1 + b_2/w_m, \qquad (IV.20)$$

where $w_m = m(m+1)$ and

$$b_0 = 13.750365230964, \quad b_1 = 1.614330366353, \quad b_2 = -0.117986675402. \qquad (IV.21)$$

works with rather good precision.

### D. Coefficient of the $\ln(1-z)$ singularity

Summation of the $A_0$ term gives

$$\sum_{n=1}^{\infty} \frac{z^n}{n} = -\ln(1-z). \qquad (IV.22)$$

The higher terms proportional to $A_1$, $A_2$, etc. all contain terms proportional to $\ln(1-z)$,
but these terms are not singular in the sense that they have finite $z \to 1$ limits. Let us
parametrize these higher terms as

$$\sum_{n=k}^{\infty} \frac{z^n}{n^2(n-1)(n-2)\dots(n-k+1)} = -c_k \ln(1-z) + r_k. \qquad (IV.23)$$

For example,

$$c_1 = -\zeta - \frac{\zeta^2}{2} - \frac{\zeta^3}{3} + \dots \quad r_1 = \frac{\pi^2}{6} - \zeta - \frac{\zeta^2}{4} - \frac{\zeta^3}{9} + \dots \qquad (IV.24)$$
$$c_2 = \frac{\zeta^2}{2} + \frac{\zeta^3}{3} + \dots \qquad r_2 = 2 - \frac{\pi^2}{6} - \zeta + \frac{\zeta^2}{4} + \frac{\zeta^3}{9} + \dots \qquad (\text{IV.25})$$

Here for convenience we introduced the new variable

$$\zeta = 1 - z. \qquad (\text{IV.26})$$

The coefficients $c_k(\zeta)$ start at $O(\zeta^k)$, whereas the remainder (analytic) terms $r_k(\zeta)$ contain all powers of $\zeta$. Let us concentrate on the non-analytic term and, using (IV.10), rearrange the power series as

$$\sum_{k=0}^{\infty} A_k c_k(\zeta) = A_0 \sum_{n=0}^{\infty} u_m \zeta^n. \qquad (\text{IV.27})$$

For the first few coefficients we find

$$u_0 = 1 \qquad u_1 = p \qquad u_2 = \frac{p^2}{4} \qquad u_3 = -\frac{p^2}{18} + \frac{p^3}{36}. \qquad (\text{IV.28})$$

Let us compare this to the first few coefficients of the original power series

$$s_0 = 1 \qquad s_1 = -p \qquad s_2 = \frac{p^2}{4} \qquad s_3 = -\frac{p^2}{18} - \frac{p^3}{36}. \qquad (\text{IV.29})$$

We see that they agree up to a sign change. This is not accidental. First of all we can see that the original differential operator of the confluent Heun equation $\mathcal{D}X = 0$ is

$$\mathcal{D} = z(z-1)\frac{d^2}{dz^2} + (2z-1)\frac{d}{dz} + p(2z-1), \qquad (\text{IV.30})$$

which in the $\zeta$ variable becomes

$$\mathcal{D} = \zeta(\zeta - 1)\frac{d^2}{d\zeta^2} + (2\zeta - 1)\frac{d}{d\zeta} - p(2\zeta - 1) \qquad (\text{IV.31})$$

i.e. it is the same operator with $p \to -p$. Let us write the ansatz

$$X = -c(\zeta) \ln \zeta + r(\zeta), \qquad (\text{IV.32})$$

where we assume that the coefficient functions $c(\zeta)$ and $r(\zeta)$ are analytic at $\zeta = 0$ (power series). Applying the differential operator to this ansatz we get

$$\mathcal{D}c = 0 \qquad (\text{IV.33})$$

and

$$\mathcal{D}r = 2(\zeta - 1)\frac{dc}{d\zeta} + c. \qquad (\text{IV.34})$$

From (IV.33) we then see that $c(\zeta)$ satisfies the $-p$ Heun equation and we find that

$$c(\zeta) = A_0 \text{HeunC}[-p, -2p, 1, 1, 0, \zeta]. \qquad (\text{IV.35})$$

This is an alternative way to see that regularity at $z = 1$ requires $A_0 = 0$.
## E. Half-waves

Of course, the perfect half-waves (HW) are sinusoidal. For $X(U) = \sin U$ the nodes are at

$$ U_1 = 0, \quad U_2 = \pi, \quad U_3 = 2\pi \dots \qquad (IV.36) $$

We know that

$$ \text{between } U_1 \text{ and } U_2: \quad X(U) > 0, \quad X''(U) = -\sin U < 0. \qquad (IV.37) $$

This means that in this interval the function is concave (it bends downwards). In this section we will use convex in the sense of the opposite of concave: meaning that the function is bending upwards. Then we see that

$$ \text{between } U_2 \text{ and } U_3: \quad X(U) < 0, \quad X''(U) = -\sin U > 0. \qquad (IV.38) $$

In this interval the sine function is convex. Then the whole picture is repeated periodically.

### 1. Pöschl-Teller

In the PT case the function satisfies the differential equation

$$ \frac{d^2}{dU^2} X(U) = -\frac{m(m+1)}{\cosh^2 U} X(U) \qquad (IV.39) $$

and we have the familiar solution $X(U) = P_m(\tanh U)$. Let us denote the roots of the Legendre polynomial (for fixed $m$) by $y_1, y_2, \dots, y_m$. Then the nodes of the PT solution are at

$$ U_r = \operatorname{arctanh} y_r, \quad r = 1, 2, \dots, m. \qquad (IV.40) $$

With our normalization $X(-\infty) = 1$ and so for $U < U_1$ $X(U) > 0$, then:

$$ U_1 < U < U_2: \quad X(U) < 0, \quad X''(U) > 0. \qquad (IV.41) $$

Here the solution is convex, bends back upwards until $U_2$. From there on:

$$ U_2 < U < U_3: \quad X(U) > 0, \quad X''(U) < 0. \qquad (IV.42) $$

The solution is concave, bends back downwards until $U_3$. This alternating pattern repeats itself until $U_m$. The solution in each of the intervals is monotonic until the (in this interval
unique) local minimum/maximum is reached; then it is again monotonic towards the next node. This looks like a distorted HW.

The main reason of the HW-behaviour is that $X(U)$ and $X''(U)$ have opposite sign.

Between $U_1$ and $U_m$ there are exactly $m-1$ such distorted HWs. It is easy to see that this $m-1$ counting is correct by plotting

$$m=1 \qquad X(U) = -P_1(\tanh U) = -\tanh U. \qquad (IV.43)$$

This solution is monotonic, there is obviously no HW as seen in FIG.\#7 of [12]. Similarly one can plot

$$m=2 \qquad X(U) = -\frac{1}{2} + \frac{3}{2} \tanh^2 U. \qquad (IV.44)$$

There is a single HW between $U_1 = -\arctanh(1/\sqrt{3})$ and $U_2 = \arctanh(1/\sqrt{3})$, as seen in FIG.\#8 of [12].

Now we turn to flyby, which our main interest in this paper.

## 2. Derived Pöschl-Teller

Here the solution satisfies<fnref n="2" />

$$\frac{d^2}{dU^2} X(U) = -k \tanh U \frac{1}{\cosh^2 U} X(U). \qquad (IV.45)$$

As we have seen, for the HW behaviour it is crucial that $X$ and $X''$ have opposite sign. In this case this is satisfied only for $U > 0$. However, it turns out that this is not a problem. In our normalization $U(-\infty) = 1$ and so the solution starts off positive. But since initially $X''$ is also positive, the solution curve is convex, so it bends upwards and does not vanish for negative $U$. After crossing $U = 0$ the curve becomes concave and now bends downwards until it reaches the first node $U_1$. All nodes of $X^1$ are positive! This means that all our qualitative considerations for the PT case are valid also here: there are exactly $(m-1)$ HWs between $U_1$ and $U_m$.

# V. HIGHER-DERIVATIVE PROFILES

Similar results can be obtained for the higher-derivative profiles proposed in [2, 10, 14].

<footnote n="2">² We take here $X(U) = X^1(U)$. The considerations for the $X^2(U)$ coordinate are analogous.</footnote>
• The system proposed by Braginsky and Thorne [4] corresponds [2] to the 2nd derivative of the Gaussian,

$$ \mathcal{A}^{BT}(U) = \frac{d^2}{dU^2} \left( \frac{k}{\sqrt{\pi}} e^{-U^2} \right). \qquad (V.1) $$

A surprising feature which extends our observation made before in the underived case and illustrated in FIG.⑴(20) of [12] is that DM requires to putting one or the other component to identically zero: and we get *half-DM*, as shown in FIG.11.

<figure><img src="image_12.png" /><figcaption>Figure 11: The 2nd-derivative profile (V.1) proposed by Gibbons and Hawking [2] for the Braginsky and Thorne theory [4] has even profile. We can get DM for one (say $X^2$) coordinate but then the other one should be put to zero, $X^1 \equiv 0$.</figcaption></figure>

This curious “half-sidedness” is explained as follows. Assuming that $k > 0$, DM is obtained for $X^2$ by fine-tuning the amplitude, $k = k_{crit}^{(2)}$. This does not work for $X^1$, though, unless $X^1 \equiv 0$, as confirmed by Figs.11 and FIG.12. However we could proceed also with the

<figure><img src="image_13.png" /><figcaption>Figure 12: Fine-tuning one of the sectors for DM yields VM for the other, when the latter trajectory is identically zero.</figcaption></figure>
opposite casting and fine-tune the amplitude for $X^1$ to get $k = k_{crit}^{(1)}$. But the profiles have opposite signs, therefore

$$k_{crit}^{(1)} = -k_{crit}^{(2)}. \qquad (V.2)$$

However we have just one $k$, and the two choices are consistent only for $k = 0$, – i.e., for no wave at all. The only way out is to turn off one of the components, either $X^1 \equiv 0$, or $X^2 \equiv 0$, depending on the sign of $k$: we get *half-DM*. Taking $k < 0$ the sectors are interchanged.

Not surprisingly, things work along the same lines for the Pöschl-Teller counterpart,

$$\frac{d^2 \mathcal{A}^{PT}}{dU^2} = k \left[ (1 - \tanh^2 U)(1 - 3 \tanh^2 U) \right], \qquad (V.3)$$

for which the coefficient of the linear-in-$X$ term in (III.8) is replaced by $1 - 3t^2$ cf. (III.7), which, for $t$ close to $\pm 1$, multiplies the amplitude by roughly $(-2)$. The trajectories shown in FIG.13 are similar to their Gaussian-based counterparts in FIG.11; the only difference is their slightly higher critical amplitude. Numerical evidence suggests that DM trajectories are composed again of (approximately) integer numbers of half-waves.

<figure><img src="image_14.png" /><figcaption>Figure 13: The trajectories for the 2nd derivative of Pöschl-Teller profile are similar to the Gaussian-based versions proposed by Braginsky and Thorne [2, 4], shown in FIG.11.</figcaption></figure>

* The 3rd derivative of the Gaussian

$$\mathcal{A}^{GC}(U) = \frac{d^3}{dU^3} \left( \frac{k}{\sqrt{\pi}} e^{-U^2} \right) \qquad (V.4)$$

proposed to describe gravitational collapse [2] was studied in some detail in [9, 10] exhibits DM again for both components, as shown in FIG.14. The plot of its Pöschl-
Teller counterpart is very similar up to having somewhat higher critical parameters, $k_1 = 4.7457$ and $k_2 = 31.14958$, respectively. It is therefore omitted.

<figure><img src="image_15.png" /><figcaption>Figure 14: The 3rd derivative of the Gaussian was proposed for gravitational collapse [2]. For $k_{crit}$ the velocity vanishes at $U = \pm\infty$, as required for DM in both components, (I.1).</figcaption></figure>

Models with higher-order derivatives of the Gaussian follow a similar pattern. For the 4th order derivative profile, for example, we would get again half-DM effect, etc. We now explain the reason by studying the behavior under reversing the sign of $U$,

$$U \to -U. \qquad (V.5)$$

We start with an even seed profile,

$$\mathcal{A}^0(-U) = \mathcal{A}^0(U) \qquad (V.6)$$

which could be either the Gaussian or Pöschl-Teller as in [12], for example. Then we consider the profile obtained by deriving $\mathcal{A}^0$ $\ell$ times,

$$\mathcal{A}^{\ell} = \frac{d^{\ell} \mathcal{A}^{0}}{dU^{\ell}}. \qquad (V.7)$$

The sign-reversal (V.5) leaves invariant the 2nd-order $U$-derivative terms in (I.3) but has an $\ell$-dependent effect on the equations of motion:

* for $\ell = 2n + 1$, $n = 0, 1, \dots$ odd, the profile is antisymmetric,

$$\mathcal{A}^{2n+1}(-U) = -\mathcal{A}^{2n+1}(U), \qquad (V.8)$$

implying that the two transverse components in (I.3a)-(I.3b) are merely interchanged under (V.5) (as noticed for flyby in sect.III). Thus they satisfy the same pair of equations up to
permutation. Therefore the DM condition holds either for both or for none of them. When it does, then “full DM” is obtained for both coordinates as illustrated in FIGs. 7-8 for $\ell = 1$, and in FIG.14 for $\ell = 3$.

U-reversal symmetry (V.5) allows us to generalize an observation made in sect.IV: for the chosen normalization the $U \to -U$ symmetry implies

$$X^1(\infty) X^2(\infty) = 1. \qquad (V.9)$$

Thus the larger one of the components is, the smaller the other will be. A higher amplitude improves the precision of DM.

The parity-dependent behavior is in fact quite general, as it could also be illustrated for example, by the sequence (V.7) with odd seed profile, $\mathcal{A}^0 \propto \tanh U$ [30].

• for $\ell = 2n, n = 0, 1, \dots, \text{even}$, the profile is symmetric,

$$\mathcal{A}^{2n}(-U) = \mathcal{A}^{2n}(U). \qquad (V.10)$$

Each of the transverse equations (I.3a) and (I.3b) is left invariant under (V.5)

DM can be obtained by fine-tuning the amplitude in one of the sectors but then the other should be put to zero: we get “half-DM”, as illustrated in FIG.\# 20 of [12] for $\ell = 0$ and in FIG.11 for $\ell = 2$.

## VI. LONGITUDINAL MOTION

So far we studied motion only in the transverse plane. What about the “vertical” coordinate? We summarize the answer given in [12]. The projection of motion to transverse space is independent of $V$ and the parallel-transport eqn. (I.3c) implies, for $\mathfrak{m} = 0$, that the vertical motion is given by,

$$\hat{V}(U) = V_0 - \int_{-\infty}^{U} \mathcal{L}_{NR} du, \qquad (VI.1)$$

where $\mathcal{L}_{NR}$ is the underlying non-relativistic Lagrangian, see \# (V.1) of [12]. We proved also that the DM conditions $X'(U = -\infty) = 0 = X'(U = \infty)$ in (I.1) imply that for $U$ in the Afterzone the integral in (VI.1) vanishes: the classical Hamiltonian action calculated along the underlying NR trajectory should vanish,

$$\int_{-\infty}^{U} \mathcal{L}_{NR} du = 0, \quad \text{for } U \geq U_a \qquad (VI.2)$$
leaving us with $\hat{V}(U) = V_0$. Thus for a massless particle the vertical light-cone coordinate $V$ does not move at all.

Turning to massive geodesics characterized by the Jacobi invariant $m > 0$ in (I.4), the $V$-motion is, see # (VI.5) in [12], a mere linear shifted,

$$V(U) = V_0 - \frac{1}{2} \left( \frac{m}{M} \right)^2 U, \qquad (VI.3)$$

where $M$ is the conserved quantity generated by the Killing vector $\partial_V$, interpreted, in the “Bargmann” framework [18, 19], as the underlying NR mass. The massive trajectories are “tilted”, as shown in FIG. # (13) of [12]. However when the relativistic and non-relativistic masses are identical,

$$m = M, \qquad (VI.4)$$

then, setting

$$Z = V + \frac{1}{2}U \qquad (VI.5)$$

we get,

$$Z(U) = V_0 = \text{const.} \qquad (VI.6)$$

i.e., we get DM also in the longitudinal $Z$-direction, see FIG. # (14) of [12].

## VII. CONCLUSION

Test particles hit by a sandwich gravitational wave exhibit, generically, the velocity effect (VM) [8, 24, 31] : after the wave has passed, the particles fly apart with diverging constant, non-zero velocity [9, 10]. Zel'dovich and Polnarev suggested instead that flyby would approximately generate pure displacement (DM) with vanishing relative velocity [1]. Our main result is to confirm and indeed refine the statement of Zel'dovich and Polnarev. Both numerical and analytical evidence show that a judicious choice of the wave parameters yields pure displacement [12].

We focus our attention at linearly polarized gravitational waves whose profile is a derivative of order $\ell = 0, 1, \dots$ of a seed profile $\mathcal{A}^0$, (V.7). The latter can be, for example, a Gaussian [2] or its analytical Doppelgänger, the Pöschl-Teller potential [12, 14–16]. Other promising candidates (not studied in the paper) could be Rosen-Morse type [37] or inverse-power [38–40] profiles. Analytically solvable models can be built also from square-profile [14, 32].
This paper extends the above results to a more physical context. We find that DM solutions are composed approximately of an integer number of half-waves which is exact for the simple Pöschl-Teller profile (III.2) and holds approximately for $ℓ = 1$, as said in [1]. Particular attention is paid to flyby [1], which might approach observability [41–45].

In the generic VM case the particles pick up some energy, which might lead to the absorption of the wave, as discussed quite some time ago. Sect. 2-5.8 of the seminal review of Ehlers and Kundt [24] anticipates much of later work $^3$. Although they did not identify either the Carroll symmetry [26–30] or DM, their intuitive argument in their subsec.2-5.8 point in this direction:

> ... one finds the effect of a wave pulse on a cloud of resting particles: [their eqn. (2-5.58)] gives the amount of *transverse momentum*, longitudinal momentum, and *kinetic energy* [...] a particle has gained [...] after the passage of the pulse. This consideration suggests convincingly that a cloud of dust is *able to extract energy from a gravitational wave* [24].

Preliminary versions of this research [14] were presented by PAH at the Wigner Institute in Budapest (April 5 2024), at the Workshop *Carrollian Physics and Holography CDFG 2024* organised at the Schrödinger Institute in Vienna (April 17 2024), and at the conference "*Conformal anomalies: theory and applications 2024*, https://indico.math.cnrs.fr/event/10718/Tours (May 7 2024). P-M Zhang presented related results in [14].

Related ideas were put forward also in [7, 8, 23, 31]. Recent results on the Memory Effect and the DM or VM puzzle include refs. [12–14, 32–36].

## Acknowledgments

We are indebted to G. Gibbons for his insights and advices. We benefitted from discussions and correspondence also with P. C. Aichelburg, T. Damour, L. Diósi, M. Elbistan, S. Kar, B. Kocsis, V. Komornik, K. Mitman and S. Silagadze. JB is grateful to SYSU for hospitality extended to him in Zhuhai. PAH thanks the Erwin Schrödinger Institute (ESI, Vienna) for hospitality during the Workshop *Carrollian Physics and Holography CDFG 2024*.

<footnote>$^3$ We are grateful to P.C. Aichelburg for calling our attention at this.</footnote>
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