A gravitational non-radiative memory effect

Abstract

We revisit the issue of memory effects, i.e. effects giving rise to a net cumulative change of the configuration of test particles, using a toy model describing the emission of radiation by a compact source and focusing on the scalar, hence non-radiative, part of the Riemann curvature. Motivated by the well known fact that gravitational radiation is accompanied by a memory effect, i.e. a permanent displacement of the relative separation of test particles, present after radiation has passed, we investigate the existence of an analog effect in the non-radiative part of the gravitational field. While quadrupole and higher multipoles undergo oscillations responsible for gravitational radiation, energy, momentum and angular momentum are conserved charges undergoing non-oscillatory change due to radiation emission. We show how the source re-arrangement due to radiation emission produce time-dependent scalar potentials which induce a time variation in the scalar part of the Riemann curvature tensor. As a result, on general grounds a velocity memory effect appears, depending on the inverse of the square of the distance of the observer from the source, thus making it almost impossible to observe, as shown by comparison to the planned gravitational detector noise spectral densities.

Appendices

Einstein equations in the SVT decomposition

Defining $\psi _e\equiv \psi +1/6\nabla ^2 e$ and $V_i\equiv \beta _i-1/2\dot{F}_i$, one has for the Christoffel coefficients

$$\begin{aligned} \begin{array}{rcl} \varGamma ^0_{00}&{}=&{}{{\dot{\phi }}}\,,\\ \varGamma ^0_{0i}&{}=&{}\phi _{,i}\,,\\ \varGamma ^0_{ij}&{}=&{}\displaystyle -\delta _{ij}{{\dot{\psi }}}_e+\frac{1}{2}\dot{e}_{,ij}-b_{,ij} -\frac{1}{2}\left( V_{i,j}+V_{j,i}\right) +\frac{1}{2} \dot{h}^{TT}_{ji}\,,\\ \varGamma ^k_{00}&{}=&{}\displaystyle \delta ^{km}\left( \phi _{,m}+\dot{b}_{,m}+{{\dot{\beta }}}_m\right) \,,\\ \varGamma ^k_{0i}&{}=&{}\displaystyle -\delta ^k_i{{\dot{\psi }}}_e+\frac{1}{2}\delta ^{km}\dot{e}_{,im} +\frac{1}{2}\left( V_{k,i}-V_{i,k}\right) +\frac{1}{2}\delta ^{km}\dot{h}^{TT}_{mi}\,,\\ \varGamma ^k_{ij}&{}=&{}\displaystyle \delta ^{km}\delta _{ij}\psi _{e,m}-\delta ^k_i\psi _{e,j} -\delta ^k_j\psi _{e,i}+\frac{1}{2}\delta ^{kl}e_{,ijk} \\ &{}&{}+\frac{1}{2}\delta ^{km}F_{l,im}+\frac{1}{2}\delta ^{km}\left( h^{TT}_{im,j}+h^{TT}_{jm,i}-h^{TT}_{ij,m}\right) . \end{array} \end{aligned}$$

(21)

and for Riemann tensor components

$$\begin{aligned} \begin{array}{rcl} R^0_{\ i0j}&{}=&{}\displaystyle -\delta _{ij}\ddot{\psi }_e-\left( \phi +\dot{b}-\frac{1}{2}\ddot{e}\right) _{,ij} -\frac{1}{2}\left( \dot{V}_{i,j}+\dot{V}_{j,i}\right) +\frac{1}{2}\ddot{h}^{TT}_{ij}\,,\\ R^k_{\ ilj}&{}=&{}\displaystyle \delta ^{km}\left( \delta _{ij}\psi _{e,ml}-\delta _{il}\psi _{e,mj}\right) +\delta ^k_l\psi _{e,ij}-\delta ^k_j\psi _{e,il} \\ &{}&{}+\frac{1}{2} \delta ^{km}\left( h^{TT}_{jm,il}+h^{TT}_{il,jm}-h^{TT}_{ij,lm} -h^{TT}_{lm,ij}\right) \,,\\ R^0_{\ ijk}&{}=&{}\displaystyle \delta _{ij}{{\dot{\psi }}}_{e,k}-\delta _{ik}{{\dot{\psi }}}_{e,l} +\frac{1}{2}\left( V_{j,ik}-V_{k,ij}\right) +\frac{1}{2}\left( \dot{h}^{TT}_{ik,j}-\dot{h}^{TT}_{ij,k}\right) . \end{array} \end{aligned}$$

(22)

Now defining $\varPhi \equiv \phi +\dot{b}-\ddot{e}/2$ one has

$$\begin{aligned} \begin{array}{rcl} R_{00}&{}=&{}\displaystyle 3\ddot{\psi }_e+\nabla ^2\varPhi \,,\\ R_{0i}&{}=&{}\displaystyle 2\psi _{e,i}-\frac{1}{2}\nabla ^2 V_i\,,\\ R_{ij}&{}=&{}\displaystyle \delta _{ij}\left( -\ddot{\psi }_e+\nabla ^2\psi _e\right) +\psi _{e,ij}-\varPhi _{,ij}-\frac{1}{2}\left( \dot{V}_{i,j}+\dot{V}_{j,i}\right) +\frac{1}{2}\ddot{h}^{TT}_{ij}-\frac{1}{2}\nabla ^2 h^{TT}_{ij}\,, \end{array} \end{aligned}$$

(23)

hence

$$\begin{aligned} R=-6\ddot{\psi }_e-2\nabla ^2\varPhi +4\nabla ^2\psi _e. \end{aligned}$$

(24)

The Einstein’s equations can be split according to their representation under rotation:

• Scalar

  $$\begin{aligned} \begin{array}{rcl} G_{00}&{}=&{}\displaystyle 2\nabla ^2\psi _e\,,\\ G_{0i}&{}=&{}\displaystyle 2{{\dot{\psi }}}_{e,i}\,,\\ G_{ij}&{}=&{}\delta _{ij}\left[ 2\ddot{\psi }_e+\nabla ^2(\varPhi -\psi _e)\right] +\left( \psi _e-\varPhi \right) _{,ij}. \end{array} \end{aligned}$$

  (25)

• Vector

  $$\begin{aligned} \begin{array}{rcl} G_{00}&{}=&{}0\,,\\ G_{0i}&{}=&{}\displaystyle -\frac{1}{2}\nabla ^2 V_i\,,\\ G_{ij}&{}=&{}\displaystyle -\frac{1}{2}\left[ \dot{V}_{i,j}+\dot{V}_{j,i}\right] . \end{array} \end{aligned}$$

  (26)

• Tensor

  $$\begin{aligned} \begin{array}{rcl} G_{00}&{}=&{}\displaystyle G_{0i}=0\,,\\ G_{ij}&{}=&{}\displaystyle \frac{1}{2}\ddot{h}^{TT}_{ij}-\frac{1}{2}\nabla ^2h^{TT}_{ij}\,, \end{array} \end{aligned}$$

  (27)

The action of a coordinate transformation

$$\begin{aligned} \begin{array}{rclcl} x^0&{}\rightarrow &{} {x'}^0&{}=&{}x^0+\xi ^0\,,\\ x^i&{}\rightarrow &{} {x'}^i&{}=&{}x^i+\xi ^i+a^{,i}\,, \end{array} \end{aligned}$$

(28)

with $\xi ^i_{,i}=0$, has the following effects on SVT fields:

$$\begin{aligned} \begin{array}{rclcl} \phi &{}\rightarrow &{}\phi '&{}=&{}\phi -{{\dot{\xi }}}^0\,,\\ b&{}\rightarrow &{} b'&{}=&{}b+\xi ^0-\dot{a}\,,\\ \beta _i&{}\rightarrow &{}\beta _i'&{}=&{}\beta _i-{{\dot{\xi }}}^i\,,\\ \psi &{}\rightarrow &{}\psi '&{}=&{}\psi +\frac{1}{3}\nabla ^2a.\\ F_i&{}\rightarrow &{} F_i'&{}=&{}F_i-2\xi _i\,,\\ e&{}\rightarrow &{} e'&{}=&{}e-2a\,, \end{array} \end{aligned}$$

(29)

from which one can verify that fields in Eq. (8) are coordinate transformation invariant, at linear order.

Detailed derivation of scalar potentials

The inverse Laplacian operator appears in several equations applied to spherically symmetric functions, and it involves integrals of the type

$$\begin{aligned} -\frac{1}{4\pi }\int d^3x'\frac{1}{|\mathbf {x}-\mathbf {x}'|}f(r)= -\frac{1}{r}\int _0^r f(r'){r'}^2dr'-\int _r^\infty f(r')r'dr'\,, \end{aligned}$$

(30)

where it has been used that

$$\begin{aligned} \frac{1}{|\mathbf {r}(\theta ,\phi )-\mathbf {r}(\iota ,\alpha )|}=\frac{4\pi }{r_>} \sum _{l\ge 0}\sum _{|m|\le l}\frac{1}{2l+1}\left( \frac{r_<}{r_>}\right) ^l Y_{lm}^*(\theta ,\phi )Y_{lm}(\iota ,\alpha )\,, \end{aligned}$$

(31)

where $r\equiv |\mathbf {r}(\theta ,\phi )|$, $r'\equiv |\mathbf {r}'(\iota ,\alpha )|$, $r_{<,>}\equiv max,min(r,r')$ and spherical symmetry has been used.

We complement the results of Sect. 3 by providing the explicit derivation of S from the first of Eqs. (7):

$$\begin{aligned} \begin{array}{rcl} \displaystyle S&{}=&{}\displaystyle -\frac{1}{4\pi }\int d^3 x\frac{{{\dot{\rho }}}}{|\mathbf {x}-\mathbf {x}'|}\\ &{}=&{}\displaystyle \frac{\varDelta (t)}{4\pi }\frac{M}{r}-\frac{M}{4\pi }\left( \frac{1}{r}\int _0^r {{\dot{\varDelta }}}(t-r')dr'+ \int _r^\infty \frac{{{\dot{\varDelta }}}(t-r')}{r'} dr'\right) \\ &{}=&{}\displaystyle \frac{M}{4\pi }\int _r^\infty \frac{\varDelta (t-r')}{{r'}^2}dr'\,, \end{array} \end{aligned}$$

(32)

where integration by parts and the relation ${{\dot{\varDelta }}}(t-r)=-\frac{\mathrm{d}\varDelta (t-r)}{\mathrm{d}r}$ have been used.

Analogously $\sigma$, the remaining scalar component of the energy-momentum tensor, can be found from the second of Eqs. (7) (we remind that ${{\dot{\varTheta }}}(t)=\varDelta (t)$)

$$\begin{aligned} \begin{aligned} \sigma&=\displaystyle \frac{M}{32\pi ^2}\int d^3x'\frac{1}{|\mathbf {x}-\mathbf {x}'|} \left( \frac{\varDelta (t-r')}{{r'}^2}-3\int _{r'}^\infty \frac{{{\dot{\varDelta }}} (t-r'')}{{r''}^2}dr''\right) \\&=\displaystyle \frac{M}{4\pi }\left[ -\frac{1}{r}\int _0^r\varDelta (t-r')dr'-\int _r^\infty \frac{\varDelta (t-r')}{r'}dr' \right. \\&\displaystyle \left. \qquad +\frac{3}{r}\int _0^r dr' {r'}^2\int _{r'}^\infty \frac{\varDelta (t-r'')}{{r''}^3}dr'' +3\int _r^\infty dr' r'\int _{r'}^\infty \frac{\varDelta (t-r'')}{{r''}^3}dr'' \right] \\&=\displaystyle \frac{M}{4\pi }\left[ -\frac{1}{r}\int _0^r\varDelta (t-r')dr'-\int _r^\infty \frac{\varDelta (t-r')}{r'}dr'+ \frac{3}{r}\int _0^rdr''\frac{\varDelta (t-r'')}{{r''}^3} \right. \\&\displaystyle \qquad \left. \int _0^{r''}{r'}^2dr'+3\int _r^\infty dr''\frac{\varDelta (t-r'')}{{r''}^3} \left( \frac{1}{r}\int _0^r{r'}^2dr'+\int _r^{r''}r'dr'\right) \right] \\&=\displaystyle \frac{M}{8\pi }\int _r^\infty \frac{\varDelta (t-r')}{r'} \left( 1-\frac{r^2}{{r'}^2}\right) dr'\,, \end{aligned} \end{aligned}$$

(33)

where integration by parts has been used in the first passage and the integration order over $r',r''$ has been exchanged in the following one (and in the final passage $r''$ has been substitud with $r'$).

The solution for the scalar potential $\psi _e$ can be found by direct integration of the first of Eqs. (9)

$$\begin{aligned} \begin{array}{rcl} \psi _e&{}=&{}\displaystyle -G_N\int d^3x\frac{\rho }{|\mathbf {x}-\mathbf {x}'|}\\ &{}=&{}\displaystyle -\left[ 1-\varTheta (t)\right] \frac{G_NM}{r}-G_NM\left( \frac{1}{r}\int _0^r\varDelta (t-r')dr'+\int _r^\infty \frac{\varDelta (t-r')}{r'}dr'\right) \\ &{}=&{}\displaystyle -\left[ 1-\varTheta (t-r)\right] \frac{G_NM}{r}-G_NM\int _r^\infty \frac{\varDelta (t-r')}{r'}dr'\\ &{}=&{}\displaystyle -\frac{G_NM}{r}+G_NM\int _r^\infty \frac{\varTheta (t-r')}{{r'}^2}dr'\,, \end{array}\nonumber \\ \end{aligned}$$

(34)

where integrations by parts and $\varDelta (t-r)=-\frac{d\varTheta (t-r)}{dr}$ have been used repeatedly.

Taking the difference of the first two of Eqs. (9) and noting that from the first of Eqs. (7) one has $2\nabla ^2\sigma =3(\dot{S}-p)$, one finds $\psi _e-\varPhi =8\pi G_N\sigma$, which using Eq. (34) allows to derive

$$\begin{aligned} \begin{array}{rcl} \varPhi &{}=&{}\displaystyle -G_NM\int _r^\infty \frac{\varDelta (t-r')}{r'}\left( 1-\frac{r^2}{{r'}^2}\right) dr'-\frac{G_NM}{r}+G_NM\int _r^\infty \frac{\varTheta (t-r')}{{r'}^2}dr'\\ &{}=&{}\displaystyle -\frac{G_NM}{r}+G_NM\int _r^\infty \frac{\varTheta (t-r')}{{r'}^2} \left( 2-3\frac{r^2}{{r'}^2}\right) \\ &{}=&{}\displaystyle -\left[ 1-\varTheta (t-r)\right] \frac{G_NM}{r}-G_NM\int _r^\infty \frac{\varDelta (t-r')}{r'} \left( 2-\frac{r^2}{{r'}^2}\right) . \end{array} \end{aligned}$$

(35)

To compute the scalar part of the Riemann tensor entering the geodesic deviation equations one still needs

$$\begin{aligned} \begin{array}{rcl} \displaystyle \ddot{\psi }_e&{}=&{}\displaystyle G_NM\int _r^\infty \frac{{{\dot{\varDelta }}}(t-r)}{{r'}^2}dr'\\ &{}=&{}\displaystyle \varDelta (t-r)\frac{G_NM}{r^2}-2G_NM\int _r^\infty \frac{\varDelta (t-r')}{{r'}^3}dr'\,, \end{array} \end{aligned}$$

(36)

and the gradients of $\varPhi$, which starting from the next to last line in Eq. (35) can be written as

$$\begin{aligned} \displaystyle \varPhi _{,i}&=\displaystyle x_i\left\{ \left[ 1+\varTheta (t-r)\right] \frac{G_NM}{r^3} -6G_NM\int _r^\infty \frac{\varTheta (t-r')}{{r'}^4}dr'\right\} \,,\nonumber \\ \displaystyle \varPhi _{,ij}&=\displaystyle \delta _{ij} \left\{ \left[ 1+\varTheta (t-r)\right] \frac{G_NM}{r^3} -6G_NM\int _r^\infty \frac{\varTheta (t-r')}{{r'}^4}dr'\right\} \nonumber \\&\displaystyle +\frac{x_ix_j}{r^2}\left\{ -\varDelta (t-r)\frac{G_NM}{r^2} -3\left[ 1+\varTheta (t-r)\right] \frac{G_NM}{r^3}+6\varTheta (t-r)\frac{G_NM}{r^3}\right\} \nonumber \\&=\displaystyle \delta _{ij}\left\{ \left[ 1-\varTheta (t-r)\right] \frac{G_NM}{r^3} +2G_NM\int _r^\infty \frac{\varDelta (t-r')}{{r'}^3}dr'\right\} \nonumber \\&\displaystyle +\frac{x_ix_j}{r^2}\left\{ -\varDelta (t-r)\frac{G_NM}{r^2} -3\left[ 1-\varTheta (t-r)\right] \frac{G_NM}{r^3}\right\} . \end{aligned}$$

(37)

Finally combining Eqs. (36) and (37) one gets Eq. (16).