The Non-vanishing Imprint of Gravitational Waves as the Result of Its Nonlinear Evolution in Space

Abstract

This paper focuses on the nonlinear self-interaction of gravitational waves and explores its impact on the spectrum of the resulting gravitational wave. While many authors primarily investigate the nonlinear effects within the framework of "gravitational memory," we take a different approach by conducting a comprehensive analysis of harmonic generation. Theoretical analysis indicates that higher harmonics do not possess suitable conditions for energy accumulation. However, our study presents intriguing evidence supporting the concept of "nonlinear gravitational memory": the conversion and accumulation of gravitational wave energy into a persistent metric deformation in the background space (specifically referred here to as zero harmonics). In simpler terms, a wave leaves a lasting imprint on the background space, even after the gravitational pulse subsides. Furthermore, our study estimates the significance of this effect and demonstrates that it should not be disregarded.

Appendices

Appendix A: The Christoffel symbols the Ricci tensor components for r-spherical waves in curved space with the spherical metric

In this Appendix, we intend to calculate the components of the nonlinear part of the Ricci tensor, i.e., the right-hand side of Eq. (3).

The general relation determining the Ricci tensor of the second-order nonlinearity is

$$\begin{aligned}{} & {} R^{(2)}_{\gamma \tau } = \frac{\partial \varGamma ^{(2)\beta }_{\gamma \tau }}{\partial x^{\beta }}-\frac{\partial \varGamma ^{(2)\beta }_{\gamma \beta }}{\partial x^{\tau }}+\varGamma ^{(0)\beta }_{\gamma \tau }\varGamma ^{(2)\eta }_{\beta \eta }+\varGamma ^{(2)\beta }_{\gamma \tau }\varGamma ^{(0)\eta }_{\beta \eta }+\varGamma ^{(1)\beta }_{\gamma \tau }\varGamma ^{(1)\eta }_{\beta \eta }-\varGamma ^{(2)\alpha }_{\gamma \beta }\varGamma ^{(0)\beta }_{\tau \alpha }\nonumber \\{} & {} \quad -\varGamma ^{(1)\alpha }_{\gamma \beta }\varGamma ^{(1)\beta }_{\tau \alpha }. \end{aligned}$$

(A-1)

Therefore, previously, we have to calculate the Chistoffel symbols of all ranks, determined by the gravitational waves of the source (in the linear approximation), $h^{(1L),\alpha ,\beta }$.¹³

Perform the proper calculations for the Christoffel symbols and the Ricci tensor components step by step.

$$\begin{aligned} \varGamma ^{(0)\beta }_{\gamma \tau }=\frac{1}{2}g^{(0)\beta \beta }\left( \frac{\partial g^{(0)}_{\beta \gamma }}{\partial x^\tau }+\frac{\partial g^{(0)}_{\beta \tau }}{\partial x^\gamma }- \frac{\partial g^{(0)}_{\gamma \tau }}{\partial x^\beta }\right) . \end{aligned}$$

(A-2)

The only nonzero derivatives of $\mathbf {g^{(0)}}$ components (here covariant) are:

$$\begin{aligned}{} & {} \frac{\partial g^{(0)}_{11}}{\partial x^1} = -2r, \\{} & {} \frac{\partial g^{(0)}_{{33}}}{\partial x^1} = -2r\sin ^2\theta , \\{} & {} \frac{\partial g^{(0)}_{33}}{\partial x^2} = -r^2\sin (2\theta ). \end{aligned}$$

Recall that the zero-order Ricci tensor corresponds to the absence of gravitational waves and, thus, is described by the undisturbed metrics and satisfies the stationary equation $R^{(0)}_{\alpha ,\beta }=0$. In the far zone, we have only these non-zero components. In the far zone, we have only these nonzero components

$$\begin{aligned}{} & {} \varGamma ^{(0)2}_{33 }=\sin (2\theta ),\\{} & {} \varGamma ^{(0)3}_{21}= \frac{1}{r},\\{} & {} \varGamma ^{(0)3}_{31}= \frac{1}{r}. \end{aligned}$$

For the right-hand side of the wave equation (only the components of the fundamental harmonic appear) we write

$$\begin{aligned}{} & {} \varGamma ^{(1)\beta }_{\gamma \tau }=\frac{1}{2}h^{(1L)\alpha \beta }\left( \frac{\partial g^{(0)}_{\alpha \gamma }}{\partial x^\tau }+\frac{\partial g^{(0)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial g^{(0)}_{\gamma \tau }}{\partial x^\alpha }\right) \nonumber \\{} & {} \quad - \frac{1}{2}g^{(0)\beta \beta }\left( \frac{\partial h^{(1L)}_{\beta \gamma }}{\partial x^\tau }+\frac{\partial h^{(1L)}_{\beta \tau }}{\partial x^\gamma }- \frac{\partial h^{(1L)}_{\gamma \tau }}{\partial x^\beta }\right) . \end{aligned}$$

(A-3)

For the calculation of the Christoffel symbol components, we need to decide on the polarization of the gravitational wave in the linear approximation. Assuming the radial direction of wave propagation (relative to the well-defined center of the source), we can consider the plus (+) and cross (x) polarizations (see, for instance, [80, 81]), or (as we do), $h_{22}=-h_{33}$, $h_{23}=h_{32}$.

For quasi-monochromatic sources, we can assume that possible time variations of the gravitational wave amplitudes are so small that we neglect them. Thus, we have $\frac{\partial h^{(1\,L)}_{\beta ,\gamma }}{\partial x^0}=i\omega h^{(1\,L)}_{\beta ,\gamma }$. However, when considering general or multiple sources, we need to take into account the time-dependent radiation intensity and, therefore, the amplitude dependence on $x ^ 0$. In this analysis, we restrict ourselves to cases where it is sufficient to consider only the first derivatives and only their linear order.

Since the $r-$order of the main mode of radiated waves $\mathbf {h^{(1L)}}$ is $O(\frac{1}{r})$ (we ignore here the higher modes, such as those in [31], the nonzero components of the first order Christoffel symbols are:

$$\begin{aligned}{} & {} \varGamma ^{(1)2}_{20}= - \frac{1}{2}g^{(0)22}\frac{\partial h^{(1L)}_{22}}{\partial x^0}= e^{i\varPsi } \frac{1}{2}\left( -i\frac{\omega }{c} H^{2(1L)}_{2} + \frac{\partial H^{2(1L)}_{2}}{\partial x^0}\right) \sim O \left( \frac{1}{r}\right) ,\\{} & {} \varGamma ^{(1)2}_{23}= \frac{1}{2}h^{(1L),32}\frac{g^{(0)}_{33}}{\partial x^2} \approx -e^{i\varPsi }\frac{1}{2}\sin (2\theta )\frac{1}{ \sin ^2{\theta }}H^{2(1L)}_3 \sim O \left( \frac{1}{r}\right) ,\\{} & {} \varGamma ^{(1)2}_{33 }= \frac{1}{2}h^{(1L)22}\frac{\partial g^{(0)}_{33}}{\partial x^2} - \frac{1}{2}g^{(0)22} \left( - \frac{\partial h^{(1L)}_{33}}{\partial x^2}\right) = -\frac{1}{2}h^{(1L)22}\frac{\partial g^{(0)}_{33}}{\partial x^2} + \frac{1}{2}g^{(0)22} \frac{\partial h^{(1L)}_{33}}{\partial x^2}\\{} & {} \quad = -\frac{1}{2}\sin (2\theta )h^{2(1L)}_2 - \frac{1}{2}\frac{\partial h^{2(1L)}_2}{\partial x^2} = - \frac{e^{\pm {i}\omega t}}{2}\left( \sin (2\theta )H^{2(1L)}_2 + \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) \\{} & {} \quad \sim O \left( \frac{1}{r}\right) , \\{} & {} \varGamma ^{(1)3}_{20 }= - \frac{1}{2}g^{(0)33}\frac{\partial h^{(1L)}_{32}}{\partial x^0}= - \frac{1}{2}g^{(0)33}g^{(0)}_{22}\frac{\partial h^{2(1L)}_3}{\partial x^0} \\{} & {} \quad = e^{\pm {i}\omega t} \left[ \mp {i}\frac{\omega }{2c}H^{2(1L)}_{3} +\frac{\partial H^{2(1L)}_3}{\partial x^0}\right] . \end{aligned}$$

Similarly,

$$\begin{aligned}{} & {} \varGamma ^{(1)2}_{30 } =e^{i\varPsi }\left[ i\frac{\omega }{2c} H^{2(1L)}_{3} + \frac{\partial H^{(1L)}_{23}}{\partial x^0}\right] ,\\{} & {} \varGamma ^{(1)3}_{22} = \frac{2}{\sin ^2(\theta )}e^{i\varPsi }\frac{\partial H^{2(1L)}_{3}}{\partial x^2},\\{} & {} \varGamma ^{(1)3}_{32} = \frac{1}{2}e^{i\varPsi }\left[ \sin {(2\theta )}H^{2(1L)}_2-\frac{1}{\sin ^2{\theta }}\frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right] . \end{aligned}$$

On the left-hand side, without knowledge about the polarization of the nonlinear product, we have to keep the general formula intact [23].

For the second order Christoffel symbols we have

$$\begin{aligned} \varGamma ^{(2)\beta }_{\gamma \tau }=\frac{1}{2}h^{(1L)\beta \eta }\left( \frac{\partial h^{(1L)}_{\eta \gamma }}{\partial x^\tau }+\frac{\partial h^{(1L)}_{\eta \tau }}{\partial x^\gamma }- \frac{\partial h^{(1L)}_{\gamma \tau }}{\partial x^\eta }\right) . \end{aligned}$$

The possibly nonzero components are

$$\begin{aligned} \varGamma ^{(2)2}_{\gamma \tau }=\frac{1}{2}h^{(1L)22}\left( \frac{\partial h^{(1L)}_{2\gamma }}{\partial x^\tau }+\frac{\partial h^{(1L)}_{2\tau }}{\partial x^\gamma }- \frac{\partial h^{(1L)}_{\gamma \tau }}{\partial x^2}\right) +\frac{1}{2}h^{(1L)23}\left( \frac{\partial h^{(1L)}_{3\gamma }}{\partial x^\tau }+\frac{\partial h^{(1L)}_{3\tau }}{\partial x^\gamma }\right) . \end{aligned}$$

In the more detailed form:

$$\begin{aligned}{} & {} \varGamma ^{(2)2}_{20 }=\frac{1}{2}h^{(1L)22}\frac{\partial h^{(1L)}_{22}}{\partial x^0} + \frac{1}{2}h^{(1L)32}\frac{\partial h^{(1L)}_{32}}{\partial x^0}\\{} & {} \quad = \frac{i\omega }{2c}e^{i\varPsi }\left[ h^{2(1L)}_{2}H^{2(1L)}_{2}+h^{3(1L)}_2H^{2(1L)}_3 \right] \\{} & {} \quad +\,\frac{e^{i\varPsi }}{2}\left[ h^{2(1L)}_3\left( \frac{\partial H^{2(1L)}_2}{\partial x^0}\right) + h^{2(1L)}_3\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^0}\right) \right] \sim O \left( \frac{\omega }{c}\frac{1}{r^2}\right) .\\{} & {} \varGamma ^{(2)2}_{30 }=\frac{1}{2}h^{(1L)22}\frac{\partial h^{(1L)}_{23}}{\partial x^0} + \frac{1}{2}h^{(1L)32}\frac{\partial h^{(1L)}_{33}}{\partial x^0} =\\{} & {} \quad \frac{e^{i\varPsi }}{2}\left[ h^{2(1L)}_2\left( i\omega H^{2(1L)}_3 + \frac{\partial H^{2(1L)}_{3}}{\partial x^0}\right) - h^{2(1L)}_3\left( i\omega H^{2(1L)}_2 +\frac{\partial H^{2(1L)}_{2}}{\partial x^0}\right) \right] \\{} & {} \quad \sim O \left( \frac{\omega }{c}\frac{1}{r^2}\right) . \end{aligned}$$

Following, we get the relations:

$$\begin{aligned}{} & {} \varGamma ^{(2)2}_{21 } =\frac{i\varPsi }{2}H^{2(1L)}_2\frac{\partial h^{2(1L)}_{2}}{\partial x^1}+\frac{e^{i\varPsi }}{r}H^{2(1L)}_2h^{2(1L)}_2+\frac{e^{i\varPsi }}{2}H^{3(1L)}_2\frac{\partial h^{2(1L)}_{3}}{\partial x^1}\\{} & {} \qquad \qquad+ \frac{e^{i\varPsi }}{r}H^{3(1L)}_2h^{2(1L)}_{3} =\\{} & {} \qquad \qquad\frac{e^{i\varPsi }}{2}H^{2(1L)}_2\frac{\partial h^{2(1L)}_{2}}{\partial x^1}+\frac{e^{i\varPsi }}{r}H^{2(1L)}_2h^{2(1L)}_2+\frac{e^{i\varPsi }}{2}H^{3(1L)}_2\frac{\partial h^{2(1L)}_{3}}{\partial x^1}\\{} & {} \qquad \qquad+ \frac{e^{i\varPsi }}{r}H^{3(1L)}_2h^{2(1L)}_{3} \sim O\left( \frac{1}{r^3}\right) \approx \textbf{0},\\{} & {} \varGamma ^{(2)2}_{31} = \frac{e^{i\varPsi }}{2}h^{2(1L)}_2\frac{\partial H^{2(1L)}_{3}}{\partial x^1} + \frac{e^{i\varPsi }}{2}H^{2(1L)}_3\frac{\partial h^{3(1L)}_{3}}{\partial x^1} \\{} & {} \qquad + \frac{e^{i\varPsi }}{r}\left[ H^{2(1L)}_2h^{2(1L)}_{3} + H^{2(1L)}_3h^{3(1L)}_{3}\right] \\{} & {} \qquad \sim O\left( \frac{1}{r^3}\right) \approx \textbf{0},\\{} & {} \varGamma ^{(2)2}_{22 }=\frac{e^{i\varPsi }}{2}H^{2(1L)}_2\frac{\partial h^{2(1L)}_{2}}{\partial x^2} + e^{i\varPsi }H^{3(1L)}_2\frac{\partial h^{2(1L)}_{3}}{\partial x^2} \sim O\left( \frac{1}{r^2}\right) , \\{} & {} \varGamma ^{(2)2}_{23}=\frac{e^{i\varPsi }}{2}H^{2(1L)}_3\frac{\partial h^{3(1L)}_{3}}{\partial x^2} + \sin (2\theta )H^{2(1L)}_3h^{3(1L)}_{3} \sim O\left( \frac{1}{r^2}\right) ,\\{} & {} \varGamma ^{(2)2}_{33}=-\frac{e^{i\varPsi }}{2}H^{2(1L)}_2\left( \sin (2\theta )h^{3(1L)}_3 + \sin ^2\theta \frac{\partial h^{3(1L)}_3}{\partial x_2}\right) \sim O\left( \frac{1}{r^2}\right) . \end{aligned}$$

In the similar manner,

$$\begin{aligned}{} & {} \varGamma ^{(2)3}_{20 }=\frac{e^{i\varPsi }}{2}\left[ h^{3(1L)}_2\left( \pm {i\frac{\omega }{c}} H^{2(1L)}_{3} +\frac{\partial H^{2(1L)}_{3}}{\partial x^0}\right) \right. \\{} & {} \quad \left. + h^{3(1L)}_3\left( \pm {i\frac{\omega }{c}} H^{3(1L)}_{3} +\frac{\partial H^{3(1L)}_{3}}{\partial x^0}\right) \right] \sim O\left( \frac{\omega }{c}\frac{1}{r^2}\right) , \\{} & {} \varGamma ^{(2)3}_{30 }= \frac{e^{i\varPsi }}{2}\left[ h^{3(1L)}_3\left( \pm {i}\frac{\omega }{c} H^{3(1L)}_{3} + \frac{\partial H^{3(1L)}_{3}}{\partial x^0}\right) \right. \\{} & {} \quad \left. + h^{3(1L)}_2\left( \pm {i}\frac{\omega }{c} H^{2(1L)}_{3} + \frac{\partial H^{2(1L)}_{3}}{\partial x^0}\right) \right] \sim O\left( \frac{\omega }{c}\frac{1}{r^2}\right) , \\{} & {} \varGamma ^{(2)3}_{22}=\frac{e^{i\varPsi }}{2}H^{3(1L)}_2\frac{\partial h^{2(1L)}_{2}}{\partial x^2} + \frac{e^{\pm {i}\omega t }}{2}H^{3(1L)}_3\frac{\partial h^{3(1L)}_{2}}{\partial x^2}\\{} & {} \quad +e^{\pm {i}\omega t}\frac{\cos \theta }{\sin \theta }H^{3(1L)}_3h^{3(1L)}_{2} \sim O\left( \frac{\omega }{c}\frac{1}{r^2}\right) , \\{} & {} \varGamma ^{(2)3}_{23}= \frac{e^{i\varPsi }}{2}H^{3(1L)}_3\frac{\partial h^{3(1L)}_{3}}{\partial x^2} + e^{\pm {i}\omega t }\frac{\cos \theta }{\sin \theta }H^{3(1L)}_3h^{3(1L)}_3 \sim O\left( \frac{\omega }{c}\frac{1}{r^2}\right) , \\{} & {} \varGamma ^{(2)3}_{33}=-\frac{e^{i\varPsi }}{2}\left( H^{2(1L)}_3\frac{\partial h^{3(1L)}_{3}}{\partial x_2} + \sin (2\theta ) H^{2(1L)}_3h^{3(1L)}_{3}\right) \sim O\left( \frac{\omega }{c}\frac{1}{r^2}\right) \end{aligned}$$

Now, we calculate the components of the nonlinear (second order) Ricci tensor, i.e., the right-hand side of the differential equation Eq. (3).

Assuming a weak time dependence of the amplitudes on time we drop in the Assuming a weak time dependence of the amplitudes, we can neglect the second-time derivatives as well as the square of the first-time derivatives in the following relations.

One can observe that after integrating over $\theta$ (taking into account that the angular dependence of the nonlinear products is not of interest), we can separate variables and represent the right-hand relations as follows:

$$\begin{aligned} R^{\alpha (20)}_{\beta } = R^{\alpha (0)T}_{\beta }(\tau = t-r/c)R^{\alpha (0)R}_{\beta }(\textbf{r}). \end{aligned}$$

However, the correctness of the integration procedure is not self-evident. It should be carefully checked during the process of solving the equations. Note also that we can assume, that the time dependence is similar for all H-components, i.e.,

$$\begin{aligned}{} & {} R^{\alpha (20)}_{\beta } = R^{(0)T}(\tau )R^{\alpha (0)R}_{\beta }(\textbf{r}). \nonumber \\{} & {} \quad R^{(2)}_{\gamma \tau } = \frac{\partial \varGamma ^{(2)\beta }_{\gamma \tau }}{\partial x^{\beta }}-\frac{\partial \varGamma ^{(2)\beta }_{\gamma \beta }}{\partial x^{\tau }}+\varGamma ^{(0)\beta }_{\gamma \tau }\varGamma ^{(2)\eta }_{\beta \eta }+\varGamma ^{(2)\beta }_{\gamma \tau }\varGamma ^{(0)\eta }_{\beta \eta }+\varGamma ^{(1)\beta }_{\gamma \tau }\varGamma ^{(1)\eta }_{\beta \eta }-\varGamma ^{(2)\alpha }_{\gamma \beta }\varGamma ^{(0)\beta }_{\tau \alpha }\nonumber \\{} & {} \quad -\varGamma ^{(1)\alpha }_{\gamma \beta }\varGamma ^{(1)\beta }_{\tau \alpha }. \end{aligned}$$

(A-4)

Using the results obtained above, we get

for $R^{(2)}_{00}$

$$\begin{aligned} R^{(2)}_{\gamma =0@\tau =0} = -2{\frac{\partial \varGamma ^{(2)2}_{0 2}}{\partial x^0}}-\varGamma ^{(1)2}_{02}\varGamma ^{(1)2}_{02}-2\varGamma ^{(1)3}_{02}\varGamma ^{(1)2}_{03}. \end{aligned}$$

(A-5)

Further, we develop the necessary formulas for the zero harmonic only.

For the zero harmonic, the equations for $\mathbf {R^{(2)}}$ contain only the amplitudes $H^{(1L)}_{\alpha \gamma }.$

Therefore, instead of Eq. (9) we write

$$\begin{aligned} \varGamma ^{(2)\beta }_{\gamma \tau }=\frac{1}{2}H^{(1L)\alpha \beta }\left( \frac{\partial H^{(1L)}_{\alpha \gamma }}{\partial x^\tau }-\frac{\partial H^{(1L)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial H^{(1L)}_{\gamma \tau }}{\partial x^\alpha }\right) . \end{aligned}$$

(A-6)

and obtain

$$\begin{aligned}{} & {} \varGamma ^{(2)3}_{30 }= \frac{1}{2}\left( H^{3(1L)}_3\frac{\partial H^{3(1L)}_{3}}{\partial x^0} + H^{3(1L)}_2 \frac{\partial H^{2(1L)}_{3}}{\partial x^0}\right) \\{} & {} \quad = \frac{1}{4}\left[ \frac{\partial \left( H^{3(1L)}_{3}\right) ^2}{\partial x^0} + \frac{\partial \left( H^{2(1L)}_{3}\right) ^2}{\partial x^0}\right] , \end{aligned}$$

thus,

$$\begin{aligned} \frac{\partial }{\partial x^0}\varGamma ^{(2)3}_{30 }= \frac{1}{4}\left[ \frac{\partial ^2 \left( H^{3(1L)}_{3}\right) ^2}{\partial (x^0)^2} + \frac{\partial ^2 \left( H^{2(1L)}_{3}\right) ^2}{\partial (x^0)^2}\right] . \end{aligned}$$

Let us keep meanwhile this term with the second-order time derivatives.

Finally, we find the second-order Ricci tensor components for the zero harmonics.

$$\begin{aligned}{} & {} R^{(2,0)}_{ 00} \approx -H^{2(1L)}_2\frac{\partial ^2 H^{2(1L)}_{2}}{\partial (x^0)^2} - H^{3(1L)}_2\frac{\partial ^2 H^{2(1L)}_{3}}{\partial (x^0)^2}-\varGamma ^{(1)2}_{0,2}\varGamma ^{*(1)2}_{0,2} \nonumber \\{} & {} \quad - 2\varGamma ^{(1)3}_{02}\varGamma ^{*(1)2}_{03} \sim O\left( \frac{1}{r^2}\right) , \end{aligned}$$

(A-7)

or, after inserting the proper $\varGamma$,

$$\begin{aligned}&\mathbf {R^{(2,0)}_{ 00} =} -H^{2(1L)}_2\frac{\partial ^2 H^{2(1L)}_{2}}{\partial (x^0)^2} - \frac{1}{4}\left[ \frac{\omega ^2}{c^2} \left( H^{2(1L)}_{2}\right) ^2 + \left( \frac{\partial H^{2(1L)}_{2}}{\partial x^0}\right) ^2\right] \nonumber \\&\quad - \frac{1}{2}\left[ \frac{\omega ^2}{c^2}\left( H^{2(1L)}_{3}\right) ^2 +\left( \frac{\partial H^{2(1L)}_3}{\partial x^0}\right) ^2\right] + o\left( \frac{1}{r^2}\right) . \end{aligned}$$

(A-8)

In cases of slow-varying amplitudes of the radiated gravitational wave, one can use the approximated relation

$$\begin{aligned} \textbf{R}^{\mathbf {(2,0)}}_{\textbf{00}} \approx - \frac{1}{4}\frac{\omega ^2}{c^2} \left( H^{2(1L)}_{2}\right) ^2 - \frac{1}{2}\frac{\omega ^2}{c^2}\left( H^{2(1L)}_{3}\right) ^2 + o\left( \frac{1}{r^2}\right) . \end{aligned}$$

(A-9)

This result already justifies omitting the terms of the order $O\left( \frac{1}{r^3}\right)$ and higher (at least, in appropriate cases).

Let us continue our calculations for other $\mathbf {R^{(2,0)}}$ components.

$$\begin{aligned} \mathbf {R^{(2,0)}_{11}} = -\frac{\partial \varGamma ^{(2)2}_{12}}{\partial x^{1}}-\frac{\partial \varGamma ^{(2)2}_{13}}{\partial x^{1}}-\varGamma ^{(2)3}_{13}\varGamma ^{(0)3}_{13} -\varGamma ^{(2)2}_{13}\varGamma ^{(0)3}_{12}, \end{aligned}$$

(A-10)

and using the calculated $\varGamma$-components obtain

$$\begin{aligned} \mathbf {R^{(20)}_{10}}= -\frac{\partial \varGamma ^{(2)2}_{12}}{\partial x^0}. \end{aligned}$$

(A-11)

Supposing that $x^0, x^1$ derivative of the amplitudes are much smaller than the corresponding derivatives of the exponent factors, we get for the zero harmonics

$$\begin{aligned} \varGamma ^{(2)2}_{12 } =\frac{1}{4}\frac{\partial (H^{2(1L)}_{2})^2}{\partial x^1}+\frac{1}{4}\frac{\partial (H^{2(1L)}_{3})^2}{\partial x^1} + \frac{1}{r}\left( H^{2(1L)}_2\right) ^2 + \frac{1}{r}H^{3(1L)}_2H^{2(1L)}_{3} \sim o\left( \frac{1}{r^3}\right) , \end{aligned}$$

therefore, we omit the term $\frac{\partial \varGamma ^{(2)2}_{12}}{\partial x^{1}}$;

also

$$\begin{aligned} \varGamma ^{(2)2}_{13 }\sim o\left( \frac{1}{r^3}\right) , \end{aligned}$$

and thus, we obtain, finally

$$\begin{aligned} \mathbf {R^{(2)}_{11} \sim o\left( \frac{1}{r^3}\right) \approx 0}. \end{aligned}$$

(A-12)

In a similar manner we calculate

$$\begin{aligned} R^{(2)}_{10}= -\frac{1}{4}\frac{\partial }{\partial x^0}\left[ \frac{\partial (H^{2(1L)}_{2})^2}{\partial x^1}+\frac{\partial (H^{2(1L)}_{3})^2}{\partial x^1}\right] \sim o\left( \frac{1}{r^3}\right) . \end{aligned}$$

(A-13)

In comparison with $R^{(2)}_{00}$, we can accept also

$$\begin{aligned} \mathbf {R^{(2)}_{10} \approx 0}. \end{aligned}$$

(A-14)

Pay attention that for monochromatic waves $R^{(2,0)}_{10} = 0$ identically.

Continue the calculations.

$$\begin{aligned} \begin{aligned}&R^{(2,0)}_{\gamma =1@\tau =2} = \frac{\partial \varGamma ^{(2)\beta }_{12}}{\partial x^{\beta }} -\frac{\partial \varGamma ^{(2)\beta }_{1\beta }}{\partial x^2}+\\&\quad \varGamma ^{(0)\beta }_{12}\varGamma ^{(2)\eta }_{\beta \eta }+\varGamma ^{(2)\beta }_{12}\varGamma ^{(0)\eta }_{\beta \eta }+\varGamma ^{(1)\beta }_{12}\varGamma ^{(1)\eta }_{\beta \eta }-\varGamma ^{(2)\alpha }_{1\beta }\varGamma ^{(0)\beta }_{2\alpha }-\varGamma a^{(1)\alpha }_{1\beta }\varGamma ^{(1)\beta }_{2\alpha }.\\ \end{aligned} \end{aligned}$$

(A-15)

Here

$$\begin{aligned} \varGamma ^{(2)2}_{12 }=\frac{1}{2}H^{(1L)22}\frac{\partial H^{(1L)}_{22}}{\partial x^1} + \frac{1}{2}H^{(1L)32}\frac{\partial H^{(1L)}_{32}}{\partial x^1} \end{aligned}$$

is the only nonzero $\varGamma ^{(2)\beta }_{1,2}$ term. Thus,

$$\begin{aligned} \frac{\partial \varGamma ^{(2)\beta }_{12}}{\partial x^{\beta }} = \frac{\partial \varGamma ^{(2)2}_{12}}{\partial x^2}. \end{aligned}$$

As well,

$$\begin{aligned} \frac{\partial \varGamma ^{(2)\beta }_{1\beta }}{\partial x^2} = \frac{\partial \varGamma ^{(2)2}_{12}}{\partial x^2}, \end{aligned}$$

and, thus, we have

$$\begin{aligned} \frac{\partial \varGamma ^{(2)\beta }_{12}}{\partial x^{\beta }} - \frac{\partial \varGamma ^{(2)\beta }_{1\beta }}{\partial x^2} = 0. \end{aligned}$$

Further,

$$\begin{aligned} \varGamma ^{(0)\beta }_{12}\varGamma ^{(2)\eta }_{\beta \eta } = \varGamma ^{(0)3}_{12}\left( \varGamma ^{(2)2}_{32}+\varGamma ^{(2)3}_{33}\right) = \frac{1}{r}\left( \varGamma ^{(2)2}_{32}+\varGamma ^{(2)3}_{33}\right) , \end{aligned}$$

and, thus,

$$\begin{aligned} \varGamma ^{(0)\beta }_{12}\varGamma ^{(2)\eta }_{\beta \eta } = \frac{1}{r}\left( \frac{1}{2}H^{(1L)23*}\frac{\partial H^{(1L)}_{33}}{\partial x^2}-\frac{1}{2}H^{(1L)32*}\frac{\partial H^{(1L)}_{33}}{\partial x^2}\right) = 0. \end{aligned}$$

Proceed further:

$$\begin{aligned}{} & {} \varGamma ^{(2)\beta }_{12}\varGamma ^{(0)\eta }_{\beta \eta } = \varGamma ^{(2)2}_{12}\varGamma ^{(0)\eta }_{2\eta } = 0,\\{} & {} \varGamma ^{(1)\beta }_{12}\varGamma ^{(1)\eta }_{\beta \eta } = 0,\\{} & {} \quad -\varGamma ^{(2)\alpha }_{1\beta }\varGamma ^{(0)\beta }_{2\alpha } = -\frac{1}{r}\varGamma ^{(2)1}_{13} = 0,\\{} & {} \quad -\varGamma ^{(1)\alpha }_{1\beta }\varGamma ^{(1)\beta }_{2\alpha } = 0. \end{aligned}$$

Thus,

$$\begin{aligned} \mathbf {R^{(2,0)}_{12} = 0}. \end{aligned}$$

(A-16)

Now, calculate the $\textbf{R}$-component

$\mathbf {R^{(20)}_{20}}$.

$$\begin{aligned}{} & {} R^{(2)}_{\gamma =0@\tau =2} = \frac{\partial \varGamma ^{(2)\beta }_{02}}{\partial x^{\beta }}-\frac{\partial \varGamma ^{(2)\beta }_{0\beta }}{\partial x^2}\nonumber \\{} & {} \quad +\varGamma ^{(0)\beta }_{02}\varGamma ^{(2)\eta }_{\beta \eta }+\varGamma ^{(2)\beta }_{02}\varGamma ^{(0)\eta }_{\beta \eta }+\varGamma ^{(1)\beta }_{02}\varGamma ^{(1)\eta }_{\beta \eta }-\varGamma ^{(2)\alpha }_{0,\beta }\varGamma ^{(0)\beta }_{2\alpha }-\varGamma ^{(1)\alpha }_{0\beta }\varGamma ^{(1)\beta }_{2\alpha }. \end{aligned}$$

(A-17)

We have

$$\begin{aligned}{} & {} \varGamma ^{(2)\beta }_{02}=\frac{1}{2}h^{(1L)\beta \eta }\left( \frac{\partial h^{(1L)}_{\eta 0}}{\partial x^2}+\frac{\partial h^{(1L)}_{\eta 2}}{\partial x^0}- \frac{\partial h^{(1L)}_{02}}{\partial x^\eta }\right) = \frac{1}{2}h^{(1L)\beta 2}\frac{\partial h^{(1L)}_{22}}{\partial x^0}\\{} & {} \quad + \frac{1}{2}h^{(1L)\beta 3}\frac{\partial h^{(1L)}_{32}}{\partial x^0}. \end{aligned}$$

Thus,

$$\begin{aligned}{} & {} \frac{\partial \varGamma ^{(2)\beta }_{02}}{\partial x^{\beta }} = \frac{1}{2}\frac{\partial h^{(1L)22}}{\partial x^2}\frac{\partial h^{(1L)}_{22}}{\partial x^0} + \frac{1}{2}h^{(1L)22}\frac{\partial ^2 h^{(1L)}_{22}}{\partial x^0 \partial x^2} + \frac{1}{2}\frac{\partial h^{(1L)23}}{\partial x^2}\frac{\partial h^{(1L)}_{32}}{\partial x^0} \\{} & {} \quad + \frac{1}{2}h^{(1L)23}\frac{\partial ^2 h^{(1L)}_{32}}{\partial x^0 \partial x^2}. \end{aligned}$$

and for the zero harmonic

$$\begin{aligned}{} & {} \frac{\partial \varGamma ^{(2)\beta }_{02}}{\partial x^{\beta }} = \frac{1}{2}\frac{\partial H^{(1L)2}_2}{\partial x^2}\frac{\partial H^{(1L)2}_2}{\partial x^0} + \frac{1}{2}H^{(1L)2}_2\frac{\partial ^2 H^{(1L)2}_2}{\partial x^0 \partial x^2} + \frac{1}{2}\frac{\partial H^{(1L)3}_2}{\partial x^2}\frac{\partial H^{(1L)3}_2}{\partial x^0}\\{} & {} \quad + \frac{1}{2}H^{(1L)3}_2\frac{\partial ^2 H^{(1L)3}_2}{\partial x^0 \partial x^2}. \end{aligned}$$

In the same manner

$$\begin{aligned}{} & {} \varGamma ^{(2)\beta }_{\gamma =0@\tau =\beta }=\frac{1}{2}h^{(1L)\beta \eta }\left( \frac{\partial h^{(1L)}_{\eta 0}}{\partial x^\beta }+\frac{\partial h^{(1L)}_{\eta \beta }}{\partial x^0}- \frac{\partial h^{(1L)}_{0\beta }}{\partial x^\eta }\right) = \frac{1}{2}h^{(1L)\beta 2}\frac{\partial h^{(1L)}_{22}}{\partial x^0}\\{} & {} \quad + \frac{1}{2}h^{(1L)\beta 3}\frac{\partial h^{(1L)}_{32}}{\partial x^0}, \end{aligned}$$

or, for the zero harmonic

$$\begin{aligned}{} & {} \varGamma ^{(2)\beta }_{\gamma =0@\tau =\beta }= \frac{1}{2}H^{(1L)22}\frac{\partial H^{(1L)}_{22}}{\partial x^0} +H^{(1L)23}\frac{\partial H^{(1L)}_{23}}{\partial x^0} +\frac{1}{2}H^{(1L)33}\frac{\partial H^{(1L)}_{33}}{\partial x^0} \\{} & {} \quad = H^{(1L)22}\frac{\partial H^{(1L)}_{22}}{\partial x^0} +H^{(1L)23}\frac{\partial H^{(1L)}_{23}}{\partial x^0}. \end{aligned}$$

Therefore,

$$\begin{aligned}{} & {} \frac{\partial \varGamma ^{(2)\beta }_{0\beta }}{\partial x^2} = \frac{\partial }{\partial x^2}\left[ H^{(1L)22}\frac{\partial H^{(1L)}_{22}}{\partial x^0} +H^{(1L)23}\frac{\partial H^{(1L)}_{23}}{\partial x^0} \right] \\{} & {} \quad = \left[ \frac{\partial H^{(1L)2}_2}{\partial x^2}\frac{\partial H^{(1L)2}_2}{\partial x^0} +\frac{\partial H^{(1L)2}_3}{\partial x^2}\frac{\partial H^{(1L)2}_3}{\partial x^0}\right] \\{} & {} \qquad + \left[ H^{(1L)2}_2\frac{\partial ^2 H^{(1L)2}_2}{\partial x^2\partial x^0}+H^{(1L)3}_2\frac{\partial ^2 H^{(1L)2}_{3}}{\partial x^2\partial x^0}\right] . \end{aligned}$$

Now, we find

$$\begin{aligned}{} & {} \frac{\partial \varGamma ^{(2)\beta }_{02}}{\partial x^{\beta }} - \frac{\partial \varGamma ^{(2)\beta }_{0,\beta }}{\partial x^2} = - \frac{1}{2}\left[ \frac{\partial H^{(1L)2}_2}{\partial x^2}\frac{\partial H^{(1L)2}_2}{\partial x^0} +\frac{\partial H^{(1L)2}_3}{\partial x^2}\frac{\partial H^{(1L)2}_3}{\partial x^0}\right] \\{} & {} \quad -\frac{1}{2}\left[ H^{(1L)2}_2\frac{\partial ^2 H^{(1L)2}_2}{\partial x^2\partial x^0} +H^{(1L)3}_3\frac{\partial ^2 H^{(1L)2}_{23}}{\partial x^2\partial x^0}\right] . \end{aligned}$$

Further,

$$\begin{aligned}{} & {} \varGamma ^{(0)\beta }_{02}\varGamma ^{(2)\eta }_{\beta \eta } = 0,\\{} & {} \varGamma ^{(2)\beta }_{02}\varGamma ^{(0)\eta }_{\beta \eta } = \frac{1}{r}\varGamma ^{(2)1}_{02} = 0,\\{} & {} \varGamma ^{(1)\beta }_{02}\varGamma ^{(1)\eta }_{\beta \eta } = 0,\\{} & {} \varGamma ^{(2)\alpha }_{0\beta }\varGamma ^{(0)\beta }_{2\alpha } = 0,\\{} & {} \varGamma ^{(1)\alpha }_{0\beta }\varGamma ^{(1)\beta }_{2\alpha } = 0. \end{aligned}$$

Finally,

$$\begin{aligned}{} & {} \textbf{R}^{(\textbf{20})}_{\textbf{20}} = - \frac{\textbf{1}}{\textbf{2}}\left[ \frac{\partial \textbf{H}^{\mathbf {2(1L)}}_{\textbf{2}}}{\partial \textbf{x}^{\textbf{2}}} \frac{\partial \textbf{H}^{\mathbf {2(1L)}}_{\textbf{2}}}{\partial \textbf{x}^\textbf{0}} +\frac{\partial \textbf{H}^{\mathbf {2(1L)}}_{\textbf{3}}}{\partial \textbf{x}^{\textbf{2}}} \frac{\partial \textbf{H}^{\mathbf {2(1L)}}_{\textbf{3}}}{\partial \textbf{x}^{\textbf{0}}}\right] \nonumber \\{} & {} \quad -\frac{\textbf{1}}{\textbf{2}}\left[ \textbf{H}^{\mathbf {2(1L)}}_{\textbf{2}} \frac{\partial ^\textbf{2} \textbf{H}^{\mathbf {2(1L)}}_{\textbf{2}}}{\partial \textbf{x}^\textbf{2}\partial \textbf{x}^{\textbf{0}}} +\textbf{H}^{\mathbf {2(1L)}}_{\textbf{3}}\frac{\partial ^{\textbf{2}} \textbf{H}^{\mathbf {2(1L)2}}_{\textbf{3}}}{\partial \textbf{x}^{\textbf{2}}\partial \textbf{x}^{\textbf{0}}}\right] . \end{aligned}$$

(A-18)

Or

$$\begin{aligned} \textbf{R}^{\mathbf {(20)}}_{\textbf{20}} = - \frac{\textbf{1}}{\textbf{4}} \frac{\partial ^{\textbf{2}} }{\partial \textbf{x}^{\textbf{0}}\partial \textbf{x}^{\textbf{2}}}\left[ \left( \textbf{H}^{\mathbf {2(1L)}}_{\textbf{2}} \textbf{H}^{\mathbf {2(1L)}}_{\textbf{2}}\right) + \left( \textbf{H}^{\mathbf {2(1L)}}_{\textbf{3}}\textbf{H}^{\mathbf {2(1L)}}_{\textbf{3}}\right) \right] . \end{aligned}$$

(A-19)

As is shown in the main text, we can limit ourselves to the components that have been calculated above.

*******

However, we will continue our calculations formally, mainly for the purpose of illustration.

Consider the component

$\mathbf {R^{(20)}_{32}}$.

$$\begin{aligned}{} & {} R^{(2,0)}_{\gamma =3@\tau =2} = \frac{\partial \varGamma ^{(2)\beta }_{32}}{\partial x^{\beta }}-\frac{\partial \varGamma ^{(2)\beta }_{3\beta }}{\partial x^2}+\varGamma ^{(0)\beta }_{32}\varGamma ^{(2)\eta }_{\beta \eta }+\varGamma ^{(2)\beta }_{32}\varGamma ^{(0)\eta }_{\beta \eta }+\varGamma ^{(1)\beta }_{32}\varGamma ^{(1)\eta }_{\beta ,\eta }\nonumber \\{} & {} \quad -\varGamma ^{(2)\alpha }_{3\beta }\varGamma ^{(0)\beta }_{2\alpha }-\varGamma ^{(1)\alpha }_{3\beta }\varGamma ^{(1)\beta }_{2\alpha }. \end{aligned}$$

(A-20)

We have

$$\begin{aligned}{} & {} \varGamma ^{(2)2}_{23}=\frac{1}{2}H^{(1L)22}\frac{\partial H^{(1L)}_{22}}{\partial x^3}+\frac{1}{2}H^{(1L)23}\frac{\partial H^{(1L)}_{33}}{\partial x^2} = -\frac{1}{2}H^{(1L)23}\frac{\partial H^{(1L)}_{22}}{\partial x^2},\\{} & {} \varGamma ^{(2)3}_{23}=\frac{1}{2}H^{(1L)32}\frac{\partial H^{(1L)}_{22}}{\partial x^3} +\frac{1}{2}H^{(1L)33}\frac{\partial H^{(1L)}_{33}}{\partial x^2} = \frac{1}{2}H^{(1L)22}\frac{\partial H^{(1L)}_{22}}{\partial x^2},\\{} & {} \frac{\partial \varGamma ^{(2)\beta }_{32}}{\partial x^\beta } =^{H_{33} = -H_{22}} \frac{1}{2}\frac{\partial }{\partial x^2}\left( -H^{(1L)23}\frac{\partial H^{(1L)}_{22}}{\partial x^2} +H^{(1L)22}\frac{\partial H^{(1L)}_{22}}{\partial x^2}\right) . \end{aligned}$$

Further,

$$\begin{aligned} \frac{\partial \varGamma ^{(2)\beta }_{3\beta }}{\partial x^2} = \frac{\partial }{\partial x^2}\left[ \frac{1}{2}\frac{\partial }{\partial x^2}\left( -H^{(1L)23}\frac{\partial H^{(1L)}_{22}}{\partial x^2} +H^{(1L)22}\frac{\partial H^{(1L)}_{22}}{\partial x^2}\right) \right] . \end{aligned}$$

Now, we see

$$\begin{aligned} \frac{\partial \varGamma ^{(2)\beta }_{32}}{\partial x^{\beta }}-\frac{\partial \varGamma ^{(2)\beta }_{3\beta }}{\partial x^2} = \textbf{0}. \end{aligned}$$

Continue the calculations.

$$\begin{aligned}{} & {} \varGamma ^{(0)\beta }_{32}\varGamma ^{(2)\eta }_{\beta \eta } = 0,\\{} & {} \varGamma ^{(2)\beta }_{32}\varGamma ^{(0)\eta }_{\beta \eta } = 0,\\{} & {} \varGamma ^{(2)\alpha }_{3\beta }\varGamma ^{(0)\beta }_{2\alpha } = -\frac{1}{r}\varGamma ^{(2)1}_{33} = 0,\\{} & {} \varGamma ^{(1)\alpha }_{3,\beta }\varGamma ^{(1)\beta }_{2\alpha } = \varGamma ^{(1)2}_{32}\varGamma ^{(1)2}_{22} = \frac{1}{r^2}\sin {2\theta }H^{(1L) 33}*0 =0. \end{aligned}$$

Thus, obtain finally

$$\begin{aligned} \mathbf {R^{(20)}_{32} = 0}. \end{aligned}$$

(A-21)

For the rest components we find:

$\mathbf {R^{(20)}_{22}}$

$$\begin{aligned}{} & {} R^{(20)}_{\gamma =2@\tau =2} = \frac{\partial \varGamma ^{(2)\beta }_{22}}{\partial x^{\beta }}-\frac{\partial \varGamma ^{(2)\beta }_{2\beta }}{\partial x^2}+\varGamma ^{(0)\beta }_{22}|_{=0}\varGamma ^{(2)\eta }_{\beta \eta }+\varGamma ^{(2)\beta }_{22}\varGamma ^{(0)\eta }_{\beta \eta }+2\varGamma ^{(1)\beta }_{22}\varGamma ^{(1)\eta }_{\beta \eta }\nonumber \\{} & {} \quad -\varGamma ^{(2)\alpha }_{2\beta }\varGamma ^{(0)\beta }_{2\alpha }-2\varGamma ^{(1)\alpha }_{2\beta }\varGamma ^{(1)\beta }_{2\alpha }.\nonumber \\{} & {} \varGamma ^{(2)\beta }_{\gamma \tau }=\frac{1}{2}H^{(1L)\alpha \beta }\left( \frac{\partial H^{(1L)}_{\alpha \gamma }}{\partial x^\tau }-\frac{\partial H^{(1L)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial H^{(1L)}_{\gamma \tau }}{\partial x^\alpha }\right) \nonumber \\{} & {} \varGamma ^{(2)\beta }_{\gamma =2@\tau =2 }=-\frac{1}{2}H^{(1L)\alpha \beta }\left( \frac{\partial H^{(1L)}_{22}}{\partial x^\alpha }\right) = \frac{1}{2}r^2H^{(1L)2\beta }\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) \nonumber \\{} & {} \frac{\partial }{\partial x^\beta }\varGamma ^{(2)\beta }_{\gamma =2@\tau =2 }=\frac{\partial }{\partial x^{\beta =2}}\varGamma ^{(2)\beta =2}_{\gamma =2@\tau =2 }= -\frac{1}{2}\frac{\partial }{\partial x^2}\left( H^{2(1L)}_2\frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) \nonumber \\{} & {} \quad = -\frac{1}{4}\frac{\partial ^2 \left( H^{2(1L)}_{2}\right) ^2}{\partial (x^2)^2}\nonumber \\{} & {} \frac{\partial \varGamma ^{(2)\beta }_{22}}{\partial x^{\beta }} = \frac{1}{4}\frac{\partial ^2 \left( H^{2(1L)}_{2}\right) ^2}{\partial (x^2)^2}. \end{aligned}$$

(A-22)

In its turn

$$\begin{aligned}{} & {} \varGamma ^{(2)\beta }_{\gamma =2@\tau =\beta }= \frac{1}{2}H^{(1L)\alpha \beta }\left( \frac{\partial H^{(1L)}_{\alpha 2}}{\partial x^\beta }+\frac{\partial H^{(1L)}_{\alpha \beta }}{\partial x^2}- \frac{\partial H^{(1L)}_{2,\beta }}{\partial x^\alpha }\right) =\\{} & {} \quad =\frac{1}{2}H^{(1L)\alpha =2@\beta }\frac{\partial H^{(1L)}_{22}}{\partial x^\beta }+\frac{1}{2}H^{(1L)\alpha =3@\beta }\left( \frac{\partial H^{(1L)}_{32}}{\partial x^\beta }+\frac{\partial H^{(1L)}_{3,\beta }}{\partial x^2}\right) =\\{} & {} \quad =\frac{1}{2}H^{(1L)22}\frac{\partial H^{(1L)}_{22}}{\partial x^2}+\frac{1}{2}H^{(1L)32}\frac{\partial H^{(1L)}_{32}}{\partial x^2}+\frac{1}{2}H^{(1L)33}\frac{\partial H^{(1L)}_{33}}{\partial x^2}=\\{} & {} \quad =\frac{1}{4}\frac{\partial \left( H^{2(1L)}_{2}\right) ^2}{\partial x^2}+\frac{1}{4\sin ^2{\theta }}\frac{\partial \left( H^{2(1L)}_3H^{2(1L)}_{3}\right) }{\partial x^2}+\frac{1}{2}\frac{\partial \left( H^{(1L)33}H^{(1L)}_{33}\right) }{\partial x^2}\\{} & {} \quad =\frac{1}{4}\frac{\partial \left( H^{2(1L)}_{2}\right) ^2}{\partial x^2} + \frac{1}{4\sin ^2{\theta }}\frac{\partial \left( H^{2(1L)}_{3}\right) ^2}{\partial x^2} - \frac{3\cos {\theta }}{4\sin ^3{\theta }}\left( H^{2(1L)}_{3}\right) ^2 +\frac{1}{4}\frac{\partial \left( H^{3(1L)}_{3}\right) ^2}{\partial x^2}. \end{aligned}$$

Thus,

$$\begin{aligned}{} & {} \frac{\partial \varGamma ^{(2)\beta }_{2\beta }}{\partial x^2} = \frac{1}{4}\frac{\partial ^2 \left( H^{2(1L)}_{2}\right) ^2}{\partial (x^2)^2} + \frac{1}{4\sin ^2{\theta }}\frac{\partial ^2 \left( H^{2(1L)}_{3}\right) ^2}{\partial (x^2)^2}\\{} & {} \quad - \frac{3}{4}\frac{\cos {\theta }}{\sin ^3{\theta }}\frac{\partial \left( H^{2(1L)}_{3}\right) ^2}{\partial x^2} +\frac{1}{4}\frac{\partial ^2 \left( H^{3(1L)}_{3}\right) ^2}{\partial (x^2)^2} +\frac{3}{4\sin ^2{\theta }}\left( 1+4\dfrac{\cos ^2{\theta }}{\sin ^2{\theta }}\right) \left( H^{2(1L)}_{3}\right) ^2. \end{aligned}$$

After this,

$$\begin{aligned}{} & {} \frac{\partial \varGamma ^{(2)\beta }_{22}}{\partial x^{\beta }}-\frac{\partial \varGamma ^{(2)\beta }_{2\beta }}{\partial x^2} = -\frac{1}{4}\frac{1}{\sin ^2{\theta }}\frac{\partial ^2 \left( H^{2(1L)}_{3}\right) ^2}{\partial (x^2)^2} +\frac{3}{4}\frac{\cos {\theta }}{\sin ^3{\theta }}\frac{\partial \left( H^{2(1L)}_{3}\right) ^2}{\partial x^2} \\{} & {} \quad -\frac{3}{4\sin ^2{\theta }}\left( 1+4\dfrac{\cos ^2{\theta }}{\sin ^2{\theta }}\right) \left( H^{2(1L)}_{3}\right) ^2-\frac{1}{4}\frac{\partial ^2 \left( H^{3(1L)}_{3}\right) ^2}{\partial (x^2)^2}. \end{aligned}$$

The rest nonzero terms are:

$$\begin{aligned}{} & {} \varGamma ^{(0)2}_{21}=\frac{1}{2}g^{(0)22}\left( \frac{\partial g^{(0)}_{22}}{\partial x^1}\right) = \frac{1}{r}.\\{} & {} \varGamma ^{(0)1}_{22}=-\frac{1}{2}g^{(0)11}\left( \frac{\partial g^{(0)}_{22}}{\partial x^1}\right) = -r.\\{} & {} \varGamma ^{(0)3}_{23}=\frac{1}{2}g^{33}\left( \frac{\partial g^{(0)}_{33}}{\partial x^2}\right) = \frac{\cos {\theta }}{\sin {\theta }}.\\{} & {} \varGamma ^{(2)\beta }_{\gamma \tau }=\frac{1}{2}H^{(1L)\alpha \beta }\left( \frac{\partial H^{(1L)}_{\alpha \gamma }}{\partial x^\tau }-\frac{\partial H^{(1L)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial H^{(1L)}_{\gamma \tau }}{\partial x^\alpha }\right) \\{} & {} \varGamma ^{(2)\beta =1}_{\gamma =2@\tau =3 }=\frac{1}{2}H^{(1L)\alpha 1}\left( \frac{\partial H^{(1L)}_{\alpha \gamma }}{\partial x^\tau }-\frac{\partial H^{(1L)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial H^{(1L)}_{\gamma \tau }}{\partial x^\alpha }\right) =0\\{} & {} \varGamma ^{(2)\beta =2}_{\gamma =2@\tau =1 }=\frac{1}{2}H^{(1L)\alpha 2}\left( \frac{\partial H^{(1L)}_{\alpha 2}}{\partial x^1}\right) = \frac{1}{2}H^{(1L)22}\left( \frac{\partial H^{(1L)}_{22}}{\partial x^1}\right) \\{} & {} \quad +\frac{1}{2}H^{(1L)32}\left( \frac{\partial H^{(1L)}_{32}}{\partial x^1}\right) \\{} & {} \quad =\frac{1}{4}\frac{\left( \partial H^{2(1L)}_{2}\right) ^2}{\partial x^1}+\frac{1}{2}H^{(1L)32}\left( \frac{\partial H^{(1L)}_{32}}{\partial x^1}\right) \\{} & {} \varGamma ^{(2)\beta =1}_{\gamma =2@\tau =2 }=\frac{1}{2}H^{(1L)\alpha 1}\left( \frac{\partial H^{(1L)}_{\alpha 2}}{\partial x^2}-\frac{\partial H^{(1L)}_{\alpha 2}}{\partial x^2}- \frac{\partial H^{(1L)}_{22}}{\partial x^\alpha }\right) = 0,\\{} & {} \varGamma ^{(2)2}_{21} = e^{i\varPsi }\left[ \frac{1}{2}H^{2(1L)}_2\left( \frac{\partial H^{2(1L)}_2}{\partial x^1}\right) + \frac{1}{r}\left( H^{2(1L)}_2\right) ^2+\frac{1}{2}H^{2(1L)}_3\left( \frac{\partial H^{2(1L)}_3}{\partial x^1}\right) \right. \\{} & {} \quad \left. + \frac{1}{r}\left( H^{2(1L)}_3\right) ^2\right] ,\\{} & {} \varGamma ^{(2)3}_{23}=-e^{i\varPsi } \left[ \frac{1}{\sin ^2{\theta }}H^{2(1L)}_3\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^2}\right) + \frac{1}{2}H^{3(1L)}_3\left( \frac{\partial H^{3(1L)}_{3}}{\partial x^2}\right) \right. \\{} & {} \quad \left. + \frac{\cos {\theta }}{\sin {\theta }}H^{3(1L)}_3H^{3(1L)}_3\right] . \end{aligned}$$

Thus,

$$\begin{aligned}{} & {} -\varGamma ^{(2)\alpha }_{2\beta }\varGamma ^{(0)\beta }_{2\alpha }=-\varGamma ^{(0)\beta =2}_{2@\alpha =1}\varGamma ^{(2)1}_{22}-\varGamma ^{(0)\beta =1}_{2@\alpha =2}\varGamma ^{(2)2}_{21}-\varGamma ^{(0)\beta =3}_{2@\alpha =3}\varGamma ^{(2)3}_{23}=\\{} & {} \quad = r\varGamma ^{(2)2}_{21}-\frac{\cos {\theta }}{\sin {\theta }}\varGamma ^{(2)3}_{23}=\\{} & {} \quad = r\left[ \frac{1}{2}H^{2(1L)}_2\left( \frac{\partial H^{2(1L)}_2}{\partial x^1}\right) + \frac{1}{r}\left( H^{2(1L)}_2\right) ^2+\frac{1}{2}H^{2(1L)}_3\left( \frac{\partial H^{2(1L)}_3}{\partial x^1}\right) \right. \\{} & {} \qquad \left. + \frac{1}{r}\left( H^{2(1L)}_3\right) ^2\right] \\{} & {} \qquad +\frac{\cos {\theta }}{\sin {\theta }}\left[ \frac{1}{\sin ^2{\theta }}H^{2(1L)}_3\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^2}\right) + \frac{1}{2}H^{3(1L)}_3\left( \frac{\partial H^{3(1L)}_{3}}{\partial x^2}\right) \right. \\{} & {} \qquad \left. + \frac{\cos {\theta }}{\sin {\theta }}\left( H^{3(1L)}_3\right) ^2 \right] . \end{aligned}$$

Further, the 1-st order nonzero terms (we derive the proper $h_{\alpha \beta }-$functions, but retain the $H_{\alpha \beta }-$ amplitude symbols, anticipating the next multiplication on $h^*_{\alpha \beta }$).

$$\begin{aligned}{} & {} \varGamma ^{(1)\beta }_{\gamma \tau }=\frac{1}{2}H^{(1L)\alpha \beta }\left( \frac{\partial g^{(0)}_{\alpha \gamma }}{\partial x^\tau }+\frac{\partial g^{(0)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial g^{(0)}_{\gamma \tau }}{\partial x^\alpha }\right) \\{} & {} \quad - \frac{1}{2}g^{(0)\beta \beta }\left( \frac{\partial H^{(1L)}_{\beta \gamma }}{\partial x^\tau }+\frac{\partial H^{(1L)}_{\beta \tau }}{\partial x^\gamma }- \frac{\partial H^{(1L)}_{\gamma \tau }}{\partial x^\beta }\right) .\\{} & {} \varGamma ^{(1)\beta =0}_{\gamma =2@\tau =2 }=\frac{1}{2}H^{(1L)\alpha 0}\left( \frac{\partial g^{(0)}_{\alpha \gamma }}{\partial x^\tau }+\frac{\partial g^{(0)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial g^{(0)}_{\gamma \tau }}{\partial x^\alpha }\right) |_{=0} \\{} & {} \quad - \frac{1}{2}g^{(0)00}\left( \frac{\partial H^{(1L)}_{0\gamma }}{\partial x^\tau }+\frac{\partial H^{(1L)}_{\beta \tau }}{\partial x^\gamma }- \frac{\partial H^{(1L)}_{\gamma \tau }}{\partial x^\beta }\right) .\\{} & {} \varGamma ^{(1)0}_{22 }= -e^{i\varPsi }\frac{r^2}{2}\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^0} -\frac{i\omega }{c}H^{2(1L)}_{2}\right) .\\{} & {} \varGamma ^{(1)\beta =0}_{\gamma =2@\tau =3 }=\frac{1}{2}H^{(1L)\alpha 0}\left( \frac{\partial g^{(0)}_{\alpha \gamma }}{\partial x^\tau }+\frac{\partial g^{(0)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial g^{(0)}_{\gamma \tau }}{\partial x^\alpha }\right) |_{=0} \\{} & {} \quad - \frac{1}{2}g^{(0)00}\left( \frac{\partial H^{(1L)}_{0\gamma }}{\partial x^\tau }+\frac{\partial H^{(1L)}_{0\tau }}{\partial x^\gamma }- \frac{\partial H^{(1L)}_{23}}{\partial x^0}\right) \end{aligned}$$

$$\begin{aligned}{} & {} \varGamma ^{(1)0}_{23 }= -e^{i\varPsi }\frac{r^2}{2}\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^0}\mp \frac{i\omega }{c}H^{2(1L)}_{2}\right) .\\{} & {} \varGamma ^{(1)\beta =1}_{\gamma =2@\tau =2 }=\frac{1}{2}H^{(1L)\alpha 1}\left( \frac{\partial g^{(0)}_{\alpha \gamma }}{\partial x^\tau }+\frac{\partial g^{(0)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial g^{(0)}_{\gamma \tau }}{\partial x^\alpha }\right) |_{=0} \\{} & {} \quad - \frac{1}{2}g^{(0)11}\left( \frac{\partial H^{(1L)}_{1\gamma }}{\partial x^\tau }+\frac{\partial H^{(1L)}_{1\tau }}{\partial x^\gamma }- \frac{\partial H^{(1L)}_{22}}{\partial x^1}\right) .\\{} & {} \varGamma ^{(1)1}_{22}= e^{i\varPsi }\left[ \frac{r^2}{2}\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^1}\right) -i\frac{r^2}{2}\frac{\omega }{c}H^{2(1L)}_{2}+rH^{2(1L)}_{2}\right] .\\{} & {} \varGamma ^{(1)1}_{23}= e^{i\varPsi }\left[ \frac{r^2}{2}\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^1}\right) -i\frac{r^2}{2}\frac{\omega }{c}H^{2(1L)}_{3}+rH^{2(1L)}_{3}\right] .\\{} & {} \varGamma ^{(1)\beta =2}_{\gamma =2@\tau =0 }=\frac{1}{2}H^{(1L)\alpha 2}\left( \frac{\partial g^{(0)}_{\alpha 2}}{\partial x^0}+\frac{\partial g^{(0)}_{\alpha 0}}{\partial x^2}- \frac{\partial g^{(0)}_{20}}{\partial x^\alpha }\right) |_{=0} \\{} & {} \quad - \frac{1}{2}g^{(0)22}\left( \frac{\partial H^{(1L)}_{22}}{\partial x^0}+\frac{\partial H^{(1L)}_{20}}{\partial x^2}- \frac{\partial H^{(1L)}_{20}}{\partial x^2}\right) .\\{} & {} \varGamma ^{(1)2}_{20}= - \frac{e^{i\varPsi }}{2}\left[ \frac{\partial H^{2(1L)}_{2}}{\partial x^0}+\frac{i\omega }{c}H^{2(1L)}_{2}\right] .\\{} & {} \varGamma ^{(1)\beta =2}_{\gamma =2@\tau =1 }=\frac{1}{2}H^{(1L)\alpha 2}\left( \frac{\partial g^{(0)}_{\alpha 2}}{\partial x^1}+\frac{\partial g^{(0)}_{\alpha 1}}{\partial x^2}- \frac{\partial g^{(0)}_{21}}{\partial x^\alpha }\right) \\{} & {} \quad - \frac{1}{2}g^{(0)22}\left( \frac{\partial H^{(1L)}_{22}}{\partial x^1}+\frac{\partial H^{(1L)}_{21}}{\partial x^2}- \frac{\partial H^{(1L)}_{21}}{\partial x^2}\right) .\\{} & {} \varGamma ^{(1)2}_{21}= -\frac{e^{i\varPsi }}{2}\left[ \frac{\partial H^{2(1L)}_{2}}{\partial x^1} - \frac{i\omega }{c}H^{2(1L)}_{2}\right] .\\{} & {} \varGamma ^{(1)2}_{22}= -e^{i\varPsi }\left[ \frac{1}{2}\frac{\partial H^{2(1L)}_2}{\partial x^2}\right] .\\{} & {} \varGamma ^{(1)2}_{23}=e^{i\varPsi }\frac{\cos {\theta }}{\sin {\theta }}H^{3(1L)}_3 = -e^{\pm i\omega t}\frac{\cos {\theta }}{\sin ^3{\theta }}H^{2(1L)}_2.\\{} & {} \varGamma ^{(1)3}_{20}= - \frac{e^{i\varPsi }}{2}\left( \frac{\partial H^{3(1L)}_{2}}{\partial x^0}\right) = - \frac{e^{i\varPsi }}{2\sin ^2{\theta }}\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^0}\right) .\\{} & {} \varGamma ^{(1)\beta =3}_{\gamma =2@\tau =1 }= \frac{e^{\pm i\omega t}}{r}h^{2(1L)}_3 - e^{\pm i\omega t}\frac{1}{2\sin ^2{\theta }}\left( e^{\pm i\omega t}\frac{\partial H^{2(1L)}_{3}}{\partial x^1} -\pm \frac{i\omega }{c}h^{2(1L)}_3\right) \\{} & {} \quad - \frac{1}{r\sin ^2{\theta }}h^{2(1L)}_{3}-\frac{1}{2r^2\sin ^2{\theta }}\frac{\partial h^{1(1L)}_{3}}{\partial x^2}\\{} & {} \varGamma ^{(1)3}_{21}= - \frac{e^{i\varPsi }}{2\sin ^2{\theta }}\left[ \left( \frac{\partial H^{2(1L)}_{3}}{\partial x^1} -\frac{i\omega }{c}H^{2(1L)}_3\right) + \frac{2\cos ^2{\theta }}{r}H^{2(1L)}_{3}\right] .\\{} & {} \varGamma ^{(1)3}_{22}= e^{i\varPsi }\left[ -\frac{1}{\sin ^2{\theta }}\frac{\partial H^{2(1L)}_{3}}{\partial x^2} + 2\frac{\cos {\theta }}{\sin ^3{\theta }} H^{2(1L)}_{3}\right] .\\{} & {} \varGamma ^{(1)3}_{23}= - \frac{e^{i\varPsi }}{2}\frac{\partial H^{3(1L)}_{3}}{\partial x^2} = - e^{i\varPsi }\left[ \frac{1}{2\sin ^2{\theta }}\frac{\partial H^{2(1L)}_{2}}{\partial x^2}+ \frac{\cos {\theta }}{\sin ^3{\theta }}H^{2(1L)}_{2}\right] . \end{aligned}$$

Thus, inserting the expressions for the $\varGamma$-symbols, we obtain

$$\begin{aligned}{} & {} \varGamma ^{(1)\beta }_{22}\varGamma ^{(1)*\eta }_{\beta \eta } = \varGamma ^{(1)0}_{22}\varGamma ^{(1)*\eta }_{0\eta }+\varGamma ^{(1)1}_{22}\varGamma ^{(1)*\eta }_{1\eta }|_{=0}+\varGamma ^{(1)2}_{22}\varGamma ^{(1)*\eta }_{2\eta }+\varGamma ^{(1)3}_{22}\varGamma ^{(1)*\eta }_{3\eta } =\\{} & {} \quad \frac{r^2}{4}\left[ \left( \frac{\partial H^{2(1L)}_{2}}{\partial x^0}\right) ^2+\frac{\omega ^2}{c^2}\left( H^{2(1L)}_{2}\right) ^2\right] -\frac{1}{4}\frac{\partial H^{2(1L)}_{2}}{\partial x^2}\\{} & {} \quad \left( \frac{\partial H^{2(1L)}_{2}}{\partial x^2}- 2\frac{\cos {\theta }}{\sin ^3{\theta }}H^{2(1L)}_{2}+ \frac{1}{\sin ^2{\theta }}\frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) \\{} & {} \quad + \frac{\cos {\theta }}{\sin ^5{\theta }}\frac{\partial H^{2(1L)}_{3}}{\partial x^2}H^{2(1L)}_2. \end{aligned}$$

***

$$\begin{aligned}{} & {} \varGamma ^{(1)\alpha }_{2\beta }\varGamma ^{(1)*\beta }_{2\alpha } \\{} & {} \quad = \varGamma ^{(1)\alpha =0}_{2@\beta =2}\varGamma ^{(1)*\beta =2}_{2@\alpha =0} +\varGamma ^{(1)\alpha =1}_{2@\beta =2}\varGamma ^{(1)*\beta =2}_{2@\alpha =1} +\varGamma ^{(1)\alpha =1}_{2@\beta =3}\varGamma ^{(1)*\beta =3}_{2@\alpha =1} + \varGamma ^{(1)\alpha =2}_{2@\beta =2}\varGamma ^{(1)*\beta =2}_{2@\alpha =2} \\{} & {} \qquad + \varGamma ^{(1)\alpha =3}_{2@\beta =3}\varGamma ^{(1)*\beta =3}_{2@\alpha =3}.\\{} & {} \quad = -\frac{1}{4}r^2\left[ \left( \frac{\partial H^{2(1L)}_{2}}{\partial x^0}\right) ^2+\frac{\omega ^2}{c^2}\left( H^{2(1L)}_{2}\right) ^2\right] \\{} & {} \qquad +r^2\left[ -\frac{1}{4}\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^1}\right) ^2-\frac{1}{2r}H^{2(1L)}_{2}\frac{\partial H^{2(1L)}_2}{\partial x^1} -\frac{1}{4}\frac{\omega ^2}{c^2}\left( H^{2(1L)}_{2}\right) ^2\right] \\{} & {} \qquad -\frac{1}{2\sin ^2{\theta }}\left[ \frac{r^2}{2}\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^1}\right) ^2+rH^{2(1L)}_{3}\right] \left[ \frac{\partial H^{2(1L)}_{3}}{\partial x^1} + \frac{2\cos ^2{\theta }}{r}H^{2(1L)}_{3}\right] \\{} & {} \quad -\frac{\omega ^2}{c^2}\frac{r^2}{4\sin ^2{\theta }}\left( H^{2(1L)}_{3}\right) ^2 \\{} & {} \qquad +\frac{1}{4}\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) ^2 +\frac{1}{4\sin ^4{\theta }}\left[ \frac{\partial H^{2(1L)}_{2}}{\partial x^2}+ 2\frac{\cos {\theta }}{\sin {\theta }}H^{2(1L)}_{2}\right] ^2 \end{aligned}$$

After simplifying and ignoring powers of $x^0$ derivatives and higher orders of $\frac{1}{r}$, we obtain:

$$\begin{aligned}{} & {} \varGamma ^{(1)\alpha }_{2\beta }\varGamma ^{(1)* \beta }_{2\alpha } \approx \frac{\omega ^2}{c^2}\left( H^{2(1L)}_{2}\right) ^2\\{} & {} \quad -\frac{r^2}{4}\left[ \left( \frac{\partial H^{2(1L)}_{2}}{\partial x^1}\right) ^2+\frac{2}{r}H^{2(1L)}_{2}\frac{\partial H^{2(1L)}_2}{\partial x^1} +\frac{\omega ^2}{c^2}\left( H^{2(1L)}_{2}\right) ^2\right] \\{} & {} \quad -\frac{1}{2\sin ^2{\theta }}\left[ \frac{r^2}{2}\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^1}\right) ^2+rH^{2(1L)}_{3}\right] \left[ \frac{\partial H^{2(1L)}_{3}}{\partial x^1} + \frac{2\cos ^2{\theta }}{r}H^{2(1L)}_{3}\right] \\{} & {} \quad -\frac{\omega ^2}{c^2}\frac{r^2}{4\sin ^2{\theta }}\left( H^{2(1L)}_{3}\right) ^2 \\{} & {} \quad +\frac{1}{4}\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) ^2 +\frac{1}{4\sin ^4{\theta }}\left[ \frac{\partial H^{2(1L)}_{2}}{\partial x^2}+ 2\frac{\cos {\theta }}{\sin {\theta }}H^{2(1L)}_{2}\right] ^2. \end{aligned}$$

Finally, ignoring the terms containing powers of the $x^0$ derivatives we obtain:

$$\begin{aligned}{} & {} R^{(20)}_{\gamma =2@\tau =2} = \frac{\partial \varGamma ^{(2)\beta }_{22}}{\partial x^{\beta }}-\frac{\partial \varGamma ^{(2)\beta }_{2\beta }}{\partial x^2}+\varGamma ^{(0)\beta }_{22}|_{=0}\varGamma ^{(2)\eta }_{\beta \eta }+\varGamma ^{(2)\beta =1}_{22}|_{=0}\varGamma ^{(0)\eta =3}_{\beta =1@\eta =3}\\{} & {} \quad +2\varGamma ^{(1)\beta }_{22}\varGamma ^{(1)\eta }_{\beta \eta }-\varGamma ^{(2)\alpha =1}_{2@\beta =3}|_{=0}\varGamma ^{(0)\beta =3}_{2@\alpha =1}-2\varGamma ^{(1)\alpha }_{2\beta }\varGamma ^{(1)\beta }_{2,\alpha }. \end{aligned}$$

Thus,

$$\begin{aligned}{} & {} R^{(20)}_{22}\approx -\frac{1}{4}\frac{1}{\sin ^2{\theta }}\frac{\partial ^2 \left( H^{2(1L)}_{3}\right) ^2}{\partial (x^2)^2} +\frac{3}{4}\frac{\cos {\theta }}{\sin ^3{\theta }}\frac{\partial \left( H^{2(1L)}_{3}\right) ^2}{\partial x^2} \\{} & {} \quad +2\frac{\omega ^2}{c^2}\left( H^{2(1L)}_{2}\right) ^2 -2\frac{\partial H^{2(1L)}_{2}}{\partial x^1}\left( \frac{r^2}{2}\frac{\partial H^{2(1L)}_{2}}{\partial x^1}+rH^{2(1L)}_{2}\right) -2\frac{\omega ^2}{c^2}\left( H^{2(1L)}_{2}\right) ^2\\{} & {} \quad -2\left[ \frac{1}{2}\frac{\partial H^{2(1L)}_2}{\partial x^2}\right] ^2 +2\frac{\cos {\theta }}{\sin ^3{\theta }}H^{2(1L)}_2\left[ -\frac{1}{\sin ^2{\theta }}\frac{\partial H^{2(1L)}_{3}}{\partial x^2} + 2\frac{\cos {\theta }}{\sin ^3{\theta }} H^{2(1L)}_{3}\right] \\{} & {} \quad -2\left( \frac{\omega ^2}{c^2}\left( H^{2(1L)}_{2}\right) ^2\right. \\{} & {} \quad \left. -\frac{r^2}{4}\left[ \left( \frac{\partial H^{2(1L)}_{2}}{\partial x^1}\right) ^2+\frac{2}{r}H^{2(1L)}_{2}\frac{\partial H^{2(1L)}_2}{\partial x^1} +\frac{\omega ^2}{c^2}\left( H^{2(1L)}_{2}\right) ^2\right] \right. \\{} & {} \quad \left. -\frac{1}{2\sin ^2{\theta }} \left[ \frac{r^2}{2}\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^1}\right) ^2+rH^{2(1L)}_{3}\right] \left[ \frac{\partial H^{2(1L)}_{3}}{\partial x^1} + \frac{2\cos ^2{\theta }}{r}H^{2(1L)}_{3}\right] \right. \\{} & {} \quad \left. -\frac{\omega ^2}{c^2}\frac{r^2}{4\sin ^2{\theta }}\left( H^{2(1L)}_{3}\right) ^2\right. \\{} & {} \quad \left. +\frac{1}{4}\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) ^2 +\frac{1}{4\sin ^4{\theta }}\left[ \frac{\partial H^{2(1L)}_{2}}{\partial x^2}+ 2\frac{\cos {\theta }}{\sin {\theta }}H^{2(1L)}_{2}\right] ^2\right) . \end{aligned}$$

And

$$\begin{aligned} \begin{aligned} R^{(20)}_{22}&= \frac{1}{4}\left( \frac{\partial ^2 \left( H^{2(1L)}_2H^{2(1L)}_2\right) }{\partial (x^2)^2} +\frac{\partial ^2 \left( H^{3(1L)}_2H^{3(1L)}_3\right) }{\partial (x^2)^2}+\frac{\partial ^2 \left( H^{3(1L)}_2H^{2(1L)}_3\right) }{\partial (x^2)^2}\right) \\&- \frac{1}{r}\left( H^{2(1L)}_2\frac{\partial H^{2(1L)}_2}{\partial x^1}\right) - \frac{1}{r^2}H^{2(1L)}_2H^{2(1L)}_2\\&+\frac{\cos {\theta }}{\sin {\theta }}\left( H^{3(1L)}_2\frac{\partial H^{2(1L)}_{3}}{\partial x^2} + \frac{1}{2}H^{3(1L)}_3\frac{\partial H^{3(1L)}_{3}}{\partial x^2}+ \frac{\cos {\theta }}{\sin {\theta }}H^{3(1L)}_3H^{3(1L)}_3\right) \\&- \left( \sin (2\theta )H^{2(1L)}_2 + \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) \frac{1}{\sin ^2(\theta )}\frac{\partial H^{2(1L)}_{3}}{\partial x^2}\\&+\left[ \frac{1}{4}\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^0}\right) ^2 - \frac{r^2}{2}\left( \frac{\partial H^{2(1L)}_3}{\partial x^1}+\frac{2}{r}H^{2(1L)}_{3}\right) \right. \\&\quad \left. \left( \frac{1}{r}H^{3(1L)}_3 - \frac{1}{r} H^{3(1L)}_2- \frac{1}{2}\frac{\partial H^{3(1L)}_2}{\partial x^1}\right) \right] \\&-\left[ \frac{1}{4}\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) ^2 +\frac{1}{4}\left( \frac{\partial H^{2(1L)}_3}{\partial x^0}\right) ^2+\frac{r^2}{4}\left( \frac{\partial H^{2(1L)}_{3}}{\partial x^1}+\frac{2}{r}H^{2(1L)}_{3}\right) \right. \\&\quad \left. \left( \frac{\partial H^{2(1L)}_{3}}{\partial x^1}+\frac{2}{r}H^{2(1L)}_{3}\right) +\frac{1}{4}\left( \frac{\partial H^{3(1L)}_3}{\partial x^2}\right) ^2\right] .\\ \end{aligned} \end{aligned}$$

(A-23)

First of all, we drop the terms of the order $o\left( \frac{1}{r^2}\right)$ and the second-time derivatives (or the quadratic order of the first-time derivatives). The equation takes the form:

$$\begin{aligned} \begin{aligned} R^{(20)}_{22}&= \frac{1}{4}\left( \frac{\partial ^2 \left( H^{2(1L)}_2H^{2(1L)}_2\right) }{\partial (x^2)^2} +\frac{\partial ^2 \left( H^{3(1L)}_2H^{3(1L)}_3\right) }{\partial (x^2)^2}+\frac{\partial ^2 \left( H^{3(1L)}_2H^{2(1L)}_3\right) }{\partial (x^2)^2}\right) \\&+\frac{\cos {\theta }}{\sin {\theta }}\left( H^{3(1L)}_2\frac{\partial H^{2(1L)}_{3}}{\partial x^2} + \frac{1}{2}H^{3(1L)}_3\frac{\partial H^{3(1L)}_{3}}{\partial x^2}+ \frac{\cos {\theta }}{\sin {\theta }}H^{3(1L)}_3H^{3(1L)}_3\right) \\&- \left( \sin (2\theta )H^{2(1L)}_2 + \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) \frac{1}{\sin ^2(\theta )}\frac{\partial H^{2(1L)}_{3}}{\partial x^2}\\&-\frac{1}{4}\left( \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) ^2\\&-\frac{r}{2}\left( \frac{\partial H^{2(1L)}_3}{\partial x^1}+\frac{2}{r}H^{2(1L)}_{3}\right) H^{3(1L)}_3.\\ \end{aligned} \end{aligned}$$

(A-24)

Let us continue further.

$\mathbf {R^{(20)}_{03}}$

$$\begin{aligned} R^{(2,0)}_{\gamma =0\tau =3} = \frac{\partial \varGamma ^{(2)\beta }_{03}}{\partial x^{\beta }}+\varGamma ^{(0)\beta }_{03}\varGamma ^{(2)\eta }_{\beta ,\eta }+\varGamma ^{(2)\beta }_{03}\varGamma ^{(0)\eta }_{\beta ,\eta }+\varGamma ^{(1)\beta }_{03}\varGamma ^{(1)\eta }_{\beta ,\eta }-\varGamma ^{(2)\alpha }_{0,\beta }\varGamma ^{(0)\beta }_{3,\alpha }-\varGamma ^{(1)\alpha }_{0\beta }\varGamma ^{(1)\beta }_{3,\alpha }.\nonumber \\ \end{aligned}$$

(A-25)

After simplifying according to the previous calculations

$$\begin{aligned}{} & {} R^{(20)}_{03} = \frac{\partial \varGamma ^{(2)\beta }_{03}}{\partial x^{\beta }} +\varGamma ^{(2)1}_{03}\varGamma ^{(0)3}_{13}+\varGamma ^{(1)\beta }_{03}\varGamma ^{(1)\eta }_{\beta ,\eta }-\varGamma ^{(2)1}_{03}\varGamma ^{(0)3}_{31}-\varGamma ^{(1)\alpha }_{0\beta }\varGamma ^{(1)\beta }_{3,\alpha } =\\{} & {} \quad \varGamma ^{(1)2}_{03}\varGamma ^{(1)3}_{23}-\varGamma ^{(1)2}_{02}\varGamma ^{(1)2}_{32}-\varGamma ^{(1)3}_{02}\varGamma ^{(1)2}_{33}-\varGamma ^{(1)2}_{03}\varGamma ^{(1)3}_{32} = -\varGamma ^{(1)2}_{02}\varGamma ^{(1)2}_{32}-\varGamma ^{(1)3}_{02}\varGamma ^{(1)2}_{33}. \end{aligned}$$

We have

$$\begin{aligned} \varGamma ^{(1)2}_{23} \approx -e^{\pm {i}\omega t}\frac{1}{2}\sin (2\theta )H^{3(1L)}_2, \end{aligned}$$

and then

$$\begin{aligned} -\varGamma ^{(1)2}_{02}\varGamma ^{(1)2}_{32} \approx -\frac{1}{4}\sin (2\theta )\frac{\omega }{c} H^{2(1L)}_{2}H^{3(1L)}_2. \end{aligned}$$

Further

$$\begin{aligned}{} & {} \varGamma ^{(1)3}_{02} =\mp {i}\frac{\omega }{2c}e^{\pm {i}\omega t} H^{2(1L)}_{3},\\{} & {} \varGamma ^{(1)2}_{33} = - \frac{e^{\pm {i}\omega t}}{2}\left( \sin (2\theta )H^{2(1L)}_2 + \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) , \end{aligned}$$

and then

$$\begin{aligned} -\varGamma ^{(1)3}_{02}\varGamma ^{(1)2}_{33} = \frac{\omega }{4c}H^{2(1L)}_{3}\left( \sin (2\theta )H^{2(1L)}_2 + \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) . \end{aligned}$$

Finally

$$\begin{aligned} R^{(20)}_{03} = \frac{\omega }{4c}H^{2(1L)}_{3}\frac{\partial H^{2(1L)}_{2}}{\partial x^2}. \end{aligned}$$

(A-26)

This component is not zero automatically but is nullified while averaging over $\theta$.

For $\mathbf {R^{(20)}_{13}}$ we have

$$\begin{aligned}{} & {} R^{(2)}_{\gamma =1,\tau =3} = \frac{\partial \varGamma ^{(2)2}_{13}}{\partial x^2} +\varGamma ^{(0)\beta }_{13}\varGamma ^{(2)\eta }_{\beta ,\eta }+\varGamma ^{(2)\beta }_{13}\varGamma ^{(0)\eta }_{\beta ,\eta }+\varGamma ^{(1)\beta }_{13}\varGamma ^{(1)\eta }_{\beta ,\eta }-\varGamma ^{(2)\alpha }_{1,\beta }\varGamma ^{(0)\beta }_{3,\alpha }\nonumber \\{} & {} \qquad \qquad -\varGamma ^{(1)\alpha }_{1,\beta }\varGamma ^{(1)\beta }_{3,\alpha }. \end{aligned}$$

(A-27)

Keeping only the terms of order not higher than $\frac{1}{r^2}$, we obtain

$$\begin{aligned} R^{(2,0)}_{13} \approx \frac{1}{r}\varGamma ^{(2)\eta }_{3,\eta }+\varGamma ^{(1)\beta }_{13}\varGamma ^{(1)\eta }_{\beta ,\eta }-\varGamma ^{(1)\alpha }_{1,\beta }\varGamma ^{(1)\beta }_{3,\alpha }. \end{aligned}$$

For the zero harmonic

$$\begin{aligned}{} & {} \varGamma ^{(2)2}_{32}=\frac{1}{2}H^{2(1L)}_3\frac{\partial H^{3(1L)}_{3}}{\partial x^2} + \sin (2\theta )H^{2(1L)}_3H^{3(1L)}_{3} \sim O\left( \frac{1}{r^2} \right) ,\\{} & {} \varGamma ^{(2)3}_{33}=-\frac{1}{2}r^2\left( \sin ^2(\theta )H^{3(1L)}_2\frac{\partial H^{3(1L)}_{3}}{\partial x_2} + \sin (2\theta ) H^{3(1L)}_2H^{3(1L)}_{3}\right) \sim O\left( \frac{1}{r^0} \right) . \end{aligned}$$

Thus,

$$\begin{aligned} R^{(20)}_{13} \approx -\frac{1}{2}r\left( \sin ^2(\theta )H^{3(1L)}_2\frac{\partial H^{3(1L)}_{3}}{\partial x_2} + \sin (2\theta ) H^{3(1L)}_2H^{3(1L)}_{3}\right) \sim O\left( \frac{1}{r} \right) . \end{aligned}$$

For $\mathbf {R^{(20)}_{33}}$ we find

$$\begin{aligned}{} & {} R^{(2,0)}_{\gamma =3@\tau =3} = \frac{\partial \varGamma ^{(2)2}_{33}}{\partial x^{2}} +\sin {2\theta }\varGamma ^{(2)\eta }_{2\eta }+\frac{1}{r}\varGamma ^{(2)2}_{33} +\varGamma ^{(1)\beta }_{33}\varGamma ^{(1)\eta }_{\beta \eta }-\frac{1}{r}\varGamma ^{(2)1}_{33}\nonumber \\{} & {} \quad -\sin {2\theta }\varGamma ^{(2)3}_{32}-\varGamma ^{(1)\alpha }_{3\beta }\varGamma ^{(1)\beta }_{3\alpha }. \end{aligned}$$

(A-28)

Or, simplifying and dropping the terms of orders $O\left( \frac{1}{r^3} \right)$ and higher

$$\begin{aligned} R^{(20)}_{33} = \frac{\partial \varGamma ^{(2)2}_{33}}{\partial x^{2}} -\varGamma ^{(1)2}_{33}\varGamma ^{(1)3}_{32}. \end{aligned}$$

Finally, using the relation $H^{3(1L)}_{3} = -H^{2(1L)}_{2}$ we obtain

$$\begin{aligned} \begin{aligned} R^{(2,0)}_{33} =&\frac{\partial }{\partial x^{2}}\left[ \frac{1}{2}H^{2(1L)}_2\left( \sin (2\theta )H^{2(1L)}_2 + \sin ^2\theta \frac{\partial H^{2(1L)}_2}{\partial x_2}\right) \right] \\&+ \frac{1}{4}\left( \sin (2\theta )H^{2(1L)}_2 + \frac{\partial H^{2(1L)}_{2}}{\partial x^2}\right) \sin (2\theta )H^{2(1L)}_2 \sim O\left( \frac{1}{r^2} \right) . \end{aligned} \end{aligned}$$

(A-29)

Appendix B: The "back" nonlinear effect

Emphasize that here (as well as in all sections) we deal, actually, not only with $e^{i(\omega t -kr)}$ terms but also with $e^{i(-\omega t)+kr}$ ones. Therefore, we initially know that all imaginary terms have to be dropped.

Here we consider the fundamental wave (symbolized by the index $^{(1L)}$) more generally, containing possibly different $\omega$ components. The back effect is described by the equation (for our case, where only the "zero" product was taken into account)

$$\begin{aligned} R^L_{\alpha \beta }(\mathbf {h_{\omega }^{(1)N}}) =- R^{NL}_{\alpha \beta }(\mathbf {h_{\omega }^{(1)},h^{(0)}}), \end{aligned}$$

(A-30)

i.e., for each $\omega$-line separately. The zero harmonic effects are considered integrally, meaning that they represent the accumulated contribution of all $\omega$ components.

The results obtained in this Appendix are utilized in Sect. 4.

We consider those equations, which involve $H^{(0)}_{00}$, and $H^{(1)}_{22}$.

On the left-hand side, we have:

$$\begin{aligned} R^L_{00}(\mathbf {h^{(21)}}) = \frac{\partial \varGamma ^{(1)\alpha }_{00}}{\partial x^{\alpha }}-\frac{\partial \varGamma ^{(1)\alpha }_{0\alpha }}{\partial x^0}+\varGamma ^{(0)\alpha }_{00}\varGamma ^{(1)\beta }_{\alpha \beta }+\varGamma ^{(1)\alpha }_{00}\varGamma ^{(0)\beta }_{\alpha \beta }-\varGamma ^{(0)\alpha }_{0\beta }\varGamma ^{(1)\beta }_{0\alpha }-\varGamma ^{(1)\alpha }_{0\beta }\varGamma ^{(0)\beta }_{0\alpha }.\nonumber \\ \end{aligned}$$

(A-31)

Or

$$\begin{aligned} R^{(21)}_{00} = -\frac{\partial \varGamma ^{(2)2}_{0 2}}{\partial x^0}-\frac{\partial \varGamma ^{(2)3}_{0 3}}{\partial x^0}+\varGamma ^{(0)1}_{00}\varGamma ^{(2)2}_{12}-\varGamma ^{(1)2}_{0,2}\varGamma ^{(1)2}_{02}. \end{aligned}$$

(A-32)

These equation should provide the additional $t-$ dependence for $\textbf{H}^{(1)}$ amplitudes. For each $\omega -$line the $r-$ dependence corresponds to the additional time of losses ( $t_{ad} =r/c$).

We do not expect any significant modification in their $r-$ dependence, and thus, we keep them as they were found.

$$\begin{aligned} \varGamma ^{(1)\beta }_{\gamma \tau }=\frac{1}{2}h^{(1L)\alpha \beta }\left( \frac{\partial g^{(0)}_{\alpha \gamma }}{\partial x^\tau }+\frac{\partial g^{(0)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial g^{(0)}_{\gamma \tau }}{\partial x^\alpha }\right) - \frac{1}{2}g^{(0)\alpha \beta }\left( \frac{\partial h^{(1L)}_{\alpha \gamma }}{\partial x^\tau }+\frac{\partial h^{(1L)}_{\alpha \tau }}{\partial x^\gamma }- \frac{\partial h^{(1L)}_{\gamma \tau }}{\partial x^\alpha }\right) , \end{aligned}$$

In the calculations below we (as in section 4) follow only the time variation of $H^{(1)}_{22}$.

$$\begin{aligned} \varGamma ^{(1)\beta }_{\gamma =0@\tau =0} = -\frac{1}{2}h^{(1)1\beta }\frac{\partial g^{(0)}_{00}}{\partial x^1} =-\frac{1}{2r^2}h^{(1)1\beta } = \frac{1}{2r^2}h_1^{\beta (1)} =0. \end{aligned}$$

Continuing we obtain

$$\begin{aligned} \varGamma ^{(1)\beta }_{\gamma =0@\tau =\beta } =\frac{1}{2}h^{(1)\alpha \beta }\left( \frac{\partial g^{(0)}_{\alpha 0}}{\partial x^\beta }- \frac{\partial g^{(0)}_{0\beta }}{\partial x^\alpha }\right) - \frac{1}{2}g^{(0)\beta \beta }\left( \frac{\partial h^{(1)}_{\beta 0}}{\partial x^\beta }+\frac{\partial h^{(1)}_{\beta \beta }}{\partial x^0}- \frac{\partial h^{(1)}_{0\beta }}{\partial x^\beta }\right) , \end{aligned}$$

and,

$$\begin{aligned} \frac{\partial \varGamma ^{(1)\alpha }_{0\alpha }}{\partial x^0} \approx -\frac{1}{2}g^{(0)\beta \beta }\frac{\partial ^2 h^{(1)}_{\beta \beta }}{\partial (x^0)^2}. \end{aligned}$$

Since we saw that just (and only) $h^{(1)}_{22}$ contribute its energy to $H^{(1)}_{2}$ we get for the same accuracy framework

$$\begin{aligned} \frac{\partial \varGamma ^{(1)\alpha }_{0\alpha }}{\partial x^0} \approx -\frac{1}{2}g^{2(0)}_2\frac{\partial ^2 h^{2(1)}_2}{\partial (x^0)^2} = - \frac{1}{2}\frac{\partial ^2 H^{2(1)}_2}{\partial (x^0)^2}. \end{aligned}$$

The imaginary term disappears in this equation since it has to include also the $-\omega$ terms.

Continuing we find

$$\begin{aligned}{} & {} \varGamma ^{(0)\alpha }_{00}\varGamma ^{(1)\beta }_{\alpha \beta } = \varGamma ^{(0)1}_{00}\varGamma ^{(1)\beta }_{1,\beta } = 0.\\{} & {} \varGamma ^{(1)\alpha }_{00}\varGamma ^{(0)\beta }_{\alpha \beta } = 0. \end{aligned}$$

Also

$$\begin{aligned}{} & {} -\varGamma ^{(0)\alpha }_{0,\beta }\varGamma ^{(1)\beta }_{0,\alpha } \approx 0.\\{} & {} -\varGamma ^{(1)\alpha }_{0,\beta }\varGamma ^{(0)\beta }_{0,\alpha } = \left( \right) *\left( \varGamma ^{(1)\alpha }_{0,\beta }\right) \approx 0. \end{aligned}$$

Finally,

$$\begin{aligned} R^L_{00}(\mathbf {h^{(21)}}) = \frac{1}{2}\frac{\partial ^2 H^{2(1)}_2}{\partial (x^0)^2}. \end{aligned}$$

For the right-hand side

$$\begin{aligned} R^{NL(2)}_{ 00} = -\frac{\partial \varGamma ^{(2)2}_{0 2}}{\partial x^0}-\frac{\partial \varGamma ^{(2)3}_{0 3}}{\partial x^0}+\varGamma ^{(0)1}_{00}\varGamma ^{(2)2}_{12}-\varGamma ^{(1)2}_{02}\varGamma ^{(1)2}_{02}, \end{aligned}$$

(A-33)

we obtain as follows.

$$\begin{aligned}{} & {} \varGamma ^{(2)\beta =2}_{\gamma =0@\tau =2} = \frac{1}{2}h^{(1L)2\eta }\left( \frac{\partial h^{(0)}_{\eta 0}}{\partial x^2}+\frac{\partial h^{(0)}_{\eta 2}}{\partial x^0}- \frac{\partial h^{(0)}_{02}}{\partial x^\eta }\right) \\{} & {} \qquad + \frac{1}{2}h^{(0)2\eta }\left( \frac{\partial h^{(1L)}_{\eta 0}}{\partial x^2}+\frac{\partial h^{(1L)}_{\eta 2}}{\partial x^0}- \frac{\partial h^{(1L)}_{02}}{\partial x^\eta }\right) .\\{} & {} \qquad \varGamma ^{(2)\beta =3}_{\gamma =0@\tau =3} = \frac{1}{2}h^{(1L)3\eta }\left( \frac{\partial h^{(0)}_{\eta 0}}{\partial x^3}+\frac{\partial h^{(0)}_{\eta 3}}{\partial x^0}- \frac{\partial h^{(0)}_{03}}{\partial x^\eta }\right) \\{} & {} \qquad + \frac{1}{2}h^{(0)3\eta }\left( \frac{\partial h^{(1L)}_{\eta 0}}{\partial x^3}+\frac{\partial h^{(1L)}_{\eta 3}}{\partial x^0}- \frac{\partial h^{(1L)}_{03}}{\partial x^\eta }\right) \end{aligned}$$

Keeping only the main (in a proper $r-$ region) polarization components, obtain

$$\begin{aligned}{} & {} \varGamma ^{(2)\beta =2}_{\gamma =0@\tau =2} = \frac{1}{2}h^{(1L)23}\frac{\partial h^{(0)}_{32}}{\partial x^0} + \frac{1}{2}h^{(0)23}\frac{\partial h^{(1L)}_{32}}{\partial x^0}.\\{} & {} \varGamma ^{(2)\beta =3}_{\gamma =0@\tau =3} = \frac{1}{2}h^{(1L)23}\frac{\partial h^{(0)}_{32}}{\partial x^0} + \frac{1}{2}h^{(0)23}\frac{\partial h^{(1L)}_{32}}{\partial x^0}.\\{} & {} \varGamma ^{(2)2}_{0 2} = h^{(1L)23}\frac{\partial h^{(0)}_{32}}{\partial x^0} = h^{(1L)23}\frac{\partial h^{(0)}_{32}}{\partial x^0} \sim h^{(1L)22}\frac{\partial h^{(0)}_{00}}{\partial x^0}.\\{} & {} \varGamma ^{(2)3}_{0 3} = h^{(1L)23}\frac{\partial h^{(0)}_{32}}{\partial x^0} \sim h^{(1L)22}\frac{\partial h^{(0)}_{00}}{\partial x^0}. \end{aligned}$$

Thus, we have

$$\begin{aligned} R^{NL(2)}_{ 00} \approx -2\frac{\partial }{\partial x^0}\left( h^{(1L)22}\frac{\partial h^{(0)}_{00}}{\partial x^0}\right) +\varGamma ^{(0)1}_{00}\varGamma ^{(1)2}_{12}-\varGamma ^{(0)2}_{02}\varGamma ^{(1)2}_{02}. \end{aligned}$$

(A-34)

Further,

$$\begin{aligned} \varGamma ^{(0)1}_{00}\approx 0, \end{aligned}$$

and finally,

$$\begin{aligned} R^{NL(2)}_{ 00} \approx -2\frac{\partial }{\partial x^0}\left( h^{(1L)22}\frac{\partial h^{(0)}_{00}}{\partial x^0}\right) = -\frac{1}{r^2}\frac{\partial }{\partial x^0}\left( h^{2(1L)}_2\frac{\partial h^{(0)}_{00}}{\partial x^0}\right) \sim \left( \frac{1}{r^5}\right) \end{aligned}$$

(A-35)

Emphasize, however, that this equation is not completely accurate (as it contains $h^{(1L)22}$ in the right-hand side instead of $h^{(1)22}$), and $H^{(0)}$ represents the static field calculated without "back" corrections. Nevertheless, for our purposes, we accept this inaccuracy.

It is important to note that the right-hand side of this equation exhibits a $\frac{1}{r^3}$ dependence.