Theory of gravitational lensing on a curved cosmic string

Abstract

It is discussed in detail the complete mathematical model of gravitational lensing on a single cosmic string (CS) of general shape and position with respect to the line of sight. CS are one-dimensional extended objects assuredly predicted by modern cosmology. The presence of CS changes the global geometry of the Universe, could clarify the properties of the early Universe, including inflation models, and could serve as a unique proof of higher-dimensional theories. Despite the fact that CS have not yet been reliable detected, there are several strong independent indications of the existence of the CS, based of CMB analysis and search of gravitational lens chains with special properties (The recent new indication of the existence of the CSs is the nHz gravitational waves in the NANOGrav and other PTA Collaborations data.) However, early considered models of straight CS presented only a small fraction of the general CS-configurations to be observed. Now we propose model which could significantly increase the possibilities of CS observational search. It is considered more realistic models have necessarily include the inclinations and bends of the CS. Besides, the recent analysis of observational data on the search for gravitational-lens candidates, shows a large number of pairs that could be explained by the complex geometry of the CS.

1 Introduction

Astronomers and physicists are closely approaching the search for nontrivial structures in the Universe, from topological defects to the consequences of the hidden multidimensionality of space-time. Such studies are actively supported by modern mathematical theory and existing gaps in understanding the unified picture of physical interactions and the structure of hidden sectors of matter: dark matter and dark energy.

Almost 50 years have passed since the prediction of cosmic strings (CS) as cosmic objects by Kibble [1,2,3]. CS were actively studied in subsequent works by Zeldovich [4], Vilenkin and others [5,6,7,8,9,10,11,12,13,14,15]. In particular, the role of CS in the formation of gravitational-lens images was shown in [5], and the mechanism of generation of CMB anisotropy on CS was shown in [16].

The existence of CS does not contradict all currently available cosmological observational data and is widely supported in theory.

Other more speculative stochastic GW sources in the nanohertz frequency range include cosmic strings.

According the recent NANOGrav and other PTA Collaborations data on the gravitational wave (GW) background in the nanohertz frequency range (by pulsar timing method [17]), there are several different physical models to understand this phenomena: primordial GWs from inflation, bursts from networks of CSs, the mergers of SMBHBs. A general prediction of the CS model is the presence of a stochastic gravitational wave background. The GW origin has not been established yet, and CSs do not contradict to the data [18,19,20,21].

CS avoid the problems of the possible topological defects (single monopoles and domain walls), thus being the most interesting candidates from the point of view of observations. Variants of hybrid models (“dumbbells”, “beads”, “necklaces” – CS with monopoles at the ends and conglomerates of such structures) that do not contradict the observational data are also considered. They are particularly interesting, because, firstly, they are preferable from the point of view of theory [22, 23], and secondly, they open up a wider area of space for their search. Indeed, a short dumbbell-type CS could be much closer to the observer than a CS of the same angular size, but “piercing” the surface of the last scattering. Closer CS allow to search for more gravitational lensed galaxies. Hybrid models also appear in superstring theory [24].

The CS-models from superstring theory find support in theory and in recent observational data. Interpretation of the NANOGrav data could be testable by future experiments [25]. The CSs in superstring theory are strongly favored over the classical field strings, [26]. For instance, in the a priori framework of the some superstring scenario (then the annihilation of the brane and antibrane creates FD-strings) it is possible to estimate a lower limit on the cosmic superstring linear density [27].

The search for chains of gravitational lens events that a CS could form seems to be the most perspective astrophysical test. Firstly, it is possible to use data from numerous surveys and carry out such a search in automatic mode [23], and secondly, such a search complements the search for the CMB anisotropy. CS candidates, identified independently both in anisotropy data and by the presence of lens chains, are the most convincing.

The search for CS using gravitational lensed pair of distant objects began in the 1980s with the study of several pairs of quasars, [28]. Numerous further unsuccessful attempts are described in the works [29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45]. The search for observational manifestations of CS loops has been undertaken repeatedly, but also has not yet led to positive results.

Methods of searching for gravitational-lens images (forming chains, the so-called “New Milky Way”), methods of searching for characteristic structures in the CMB anisotropy, as well as methods of gravitational-wave astronomy are applicable to a very wide class of CS predicted by the theory, being almost universal. But for simplicity of calculations, it is usually assumed that the CS is located perpendicular to the line of sight, then two images of distant source are formed in the plane of the CS lens, and are not rotated relative to each other.

The main goal of this paper is to demonstrate that in the case of a CS with a slope or a CS with a bend in the picture plane, the resulting images will be asymmetrical, with different positional angles. This fact significantly expands the search for gravitational lensed pairs.

The article is organized as follows. In the Sect. 2 we provide a brief introduction to gravitational lensing on a single straight CS. This based on recent work by [46]. In the Sect. 3 we present the calculation of gravitation lensing effects due to a CS inclination. In the Sect. 4 we present the model of gravitational lensing on a curved CS. Thus in these two last chapters we consider the CS of general position with respect of the line of sight. We conclude by declaring the importance of consideration of the general position CS and discuss the new search strategy of gravitational lens pairs. After the Conclusion (Sect. 5) there are Appendixes. In the Appendix A we provide the flat approximation for a CS space-time with a conical singularity. In the Appendix B we provide the detailed calculations of energy–momentum tensor for a curved CS. In the Appendix C we describe photon trajectories for small metric perturbation. In the Appendix D we give derivation of a lens equation for a curved CS.

2 Brief introduction to gravitational lensing on a single straight cosmic string

The metric of the cosmic string (here and further CS) in cylindrical coordinates $(t, z, r, \varphi )$ has the well-known form:

$$\begin{aligned} g_{\mu \nu } = \text {diag}\Big (1, -1, -1, -r^2(1 - 4 G\mu )^2\Big ). \end{aligned}$$

(1)

It is a conical metric with a deficit angle $\Delta \theta = 8\pi G\mu$ in a plane perpendicular to the CS (see Appendix A). The CS lensing model in the flat geometric approximation is the following.

$$\begin{aligned} {\left\{ \begin{array}{ll} \phi = -\eta + 4\pi G\mu \bigg (1 - \dfrac{R_s}{R_g}\bigg )\\ \psi = \eta + 4\pi G\mu \bigg (1 - \dfrac{R_s}{R_g}\bigg ) \end{array}\right. } \end{aligned}$$

(2)

where $R_g$ is the distance from an observer to a source (a galaxy), $R_s$ is the distance from an observer to the CS, $\eta$ is first coordinate, an angle between the direction to the CS and the direction to the source ($\xi$ will be the second coordinate, the angle coordinate along the CS).

If in (2) $|\eta | < \theta _E$ we have two images of the source, where

$$\begin{aligned} \theta _E = 8\pi G\mu \bigg (1 - \dfrac{R_s}{R_g}\bigg ). \end{aligned}$$

The first image (shifted to the right of the CS):

$$\begin{aligned} I_1(\eta , \xi ) = {\left\{ \begin{array}{ll} I(\eta - \theta _E / 2, \xi ),&{} \eta > -\theta _E\\ 0,&{}\eta \le -\theta _E. \end{array}\right. } \end{aligned}$$

The second image (shifted to the left of the CS), (Fig. 1):

$$\begin{aligned} I_2(\eta , \xi ) = {\left\{ \begin{array}{ll} I(\eta + \theta _E / 2, \xi ),&{} \eta < \theta _E\\ 0,&{} \eta \ge \theta _E \end{array}\right. } \end{aligned}$$

or

$$\begin{aligned} I_{1+2}(\eta , \xi ) = {\left\{ \begin{array}{ll} I(\eta + \theta _E / 2, \xi ),&{} \eta < -\theta _E\\ I(\eta + \theta _E / 2, \xi ) + I(\eta - \theta _E / 2, \xi ),&{} |\eta | \le \theta _E \\ I(\eta - \theta _E / 2, \xi ),&{} \eta > \theta _E \end{array}\right. } \end{aligned}$$

what coincides with the results obtained earlier [42].

Fig. 1

A simulation of a galaxy lensed by a cosmic string. The double image is clearly seen. Galaxy has been represented by a 2-dimensional gaussian distribution of intensity

3 Effects due to a CS inclination

Let us consider a case when CS has an additional parameter, inclination $i > 0$ (an inclination of $i = 0^{\circ }$ corresponds to an CS located perpendicular to the line of sight, an inclination of $i =90^{\circ }$ corresponds to a CS parallel to one). In this case the lensing parameter $\theta _E$ will be dependent on the position of the source, $\theta _E = \theta _E(i, \xi )$ due to two effects. These effects are the following:

• 1. for each $\xi$ the CS will have different distance to an observer, $R_s = R_s(\xi ),$

• 2. the effective deficit angle $\Delta \theta (i, \xi )$ for $i > 0$ is less than for $i = 0.$

Fig. 2

Geometric framework for inclined straight CS. Blue dotted lines correspond to 2 possible photon paths from the dot in the object plane with coordinates $(\xi , 0)$

Fig. 3

A simulation of a galaxy lensed by an inclined CS. Galaxy has been represented by a 2D gaussian distribution of intensity. The difference of positional angles of two images is visible

Fig. 4

A simulation of a galaxy lensed by a CS. Galaxy has been represented by a 2D gaussian distribution of intensity. The difference of positional angles of two images is visible. The isophots’ cuts are clearly visible

Fig. 5

Geometric framework for bended string

Let us discuss the first effect. For the triangle “observer – point on the CS for $\xi = 0$ – point on the CS from the source” (Fig. 2):

$$\begin{aligned} \frac{R_s(\xi )}{\sin (90^{\circ } - i)} = \frac{R_s(\xi = 0)}{\sin (180^{\circ } - \xi - (90^{\circ } - i))}. \end{aligned}$$

Then

$$\begin{aligned} R_s(\xi ) = \frac{R_s(\xi = 0)}{\cos \xi + \tan i \sin \xi } \approx \frac{R_s(\xi = 0)}{1 + \xi \tan i }, \end{aligned}$$

$$\begin{aligned} \theta _E = \Delta \theta (i) \bigg ( 1 - \frac{R_s(\xi = 0)}{R_g (1 + \xi \tan i)}\bigg ). \end{aligned}$$

We can define the distance to the CS $R_s$ to be $R_s(\xi = 0)$ to shorten the formulae.

The deficit angle depends on the distance to the point of the CS that we observe when passing the source. The conical metric is equivalent to the flat metric with cut, i.e. with two “effective observers” spaced by a distance of L:

$$\begin{aligned} \Delta \theta \cdot h = L, \end{aligned}$$

(3)

where h is the length of the perpendicular from observer to the CS:

$$\begin{aligned} h = R_s \cos i. \end{aligned}$$

In case of inclined CS we can adopt the same geometric framework, but in 3D. For every $\xi$ the picture will be the same except now $\Delta \theta (i, \xi )$ is defined by Eq. (3). Also in this equation instead of h we have $R_s(\xi ).$ Since L is the same for all such $\xi ,$ then

$$\begin{aligned} \Delta \theta R_s \cos i = \Delta \theta (i, \xi ) R_s(\xi ) \end{aligned}$$

and

$$\begin{aligned} \Delta \theta (i, \xi ) = \Delta \theta \, (\cos i + \xi \sin i). \end{aligned}$$

Finally,

$$\begin{aligned} \theta _E(i, \xi ) = \Delta \theta \, (\cos i + \xi \sin i)) \bigg ( 1 - \frac{R_s}{R_g (1 + \xi \tan i)}\bigg ). \end{aligned}$$

All the other steps in constructing a lensing transformation are the same, so we end up with (Figs. 3, 4):

$$\begin{aligned}{} & {} I_{1+2}(\eta , \xi ) \\{} & {} \quad ={\left\{ \begin{array}{ll} I(\eta + \theta _E(i, \xi ) / 2, \xi ),&{} \eta < -\theta _E(i, \xi )\\ I(\eta + \theta _E(i, \xi ) / 2, \xi ) + I(\eta - \theta _E(i, \xi ) / 2, \xi ),&{} |\eta | \le \theta _E(i, \xi ) \\ I(\eta - \theta _E(i, \xi ) / 2, \xi ),&{} \eta > \theta _E(i, \xi ). \end{array}\right. } \end{aligned}$$

Since $\xi \ll 1,$ it is more useful to make an expansion of $\theta _E(i, \xi )$:

$$\begin{aligned} \theta _E(i, \xi ) = \theta _E(i, \xi = 0) + \frac{\partial \theta _E}{\partial \xi }\bigg |_{\xi = 0} \cdot \xi , \end{aligned}$$

where

$$\begin{aligned}{} & {} \theta _E(i, \xi = 0) = \Delta \theta \, \cos i \bigg ( 1 - \frac{R_s}{R_g} \bigg ), \\{} & {} \frac{\partial \theta _E}{\partial \xi }\bigg |_{\xi = 0} = \Delta \theta \sin i. \end{aligned}$$

Note that from here the rate of increase of $\theta _E$ is limited by the value of $\Delta \theta$ in radians. Knowing the limitations on it $(\Delta \theta \lesssim 10^{-5}),$ the tilt limitations during lensing can be neglected in almost all realistic cases.

It should also be borne in mind that such a picture may occur, even if only a part of the CS on the line of sight will have a large slope.

4 Model of gravitational lensing on a curved CS

In the previous paragraph we have discussed the effects of inclination on a picture one can get from an object that lies behind the CS. Now the model can be developed further.

If a CS on a line of sight has a bend with an angle $\theta \ne 0^{\circ },$ the metric does not have the form (1). To get the picture of a galaxy behind the CS one should calculate the geodesics, using the general relativity framework.

Consider a static CS, that consists of 2 straight lines, connected by a Besier curve (see Fig. 10). As a matter of simplicity, the CS is located in the $(x^1, x^2)$ plane, since the effect of inclination (see Sect. 3) could be neglected:

$$\begin{aligned} {\left\{ \begin{array}{ll} X^1(s) = x(s) = {\left\{ \begin{array}{ll} 0, &{} s \in (-\infty , - R)\\ \dfrac{R \sin \theta }{4} \bigg (1 + \dfrac{s}{R} \bigg )^2,&{} s \in [-R, R]\\ s \sin \theta ,&{} s \in (R, +\infty ) \end{array}\right. }\\ X^2(s) = y(s) = {\left\{ \begin{array}{ll} s, &{} s \in (-\infty , - R)\\ \dfrac{R \cos \theta }{4} \bigg (1 + \dfrac{s}{R} \bigg )^2 -\dfrac{R}{4} \bigg (1 - \dfrac{s}{R} \bigg )^2,&{} s \in [-R, R]\\ s \cos \theta ,&{} s \in (R, +\infty ) \end{array}\right. }\\ X^3(s) = z(s) = R_g - R_s. \end{array}\right. } \end{aligned}$$

The angle $\theta$ describes how much the CS changed the positional angle and R is the characteristic length of such bend. The geometry of the problem can be seen on Figs. 5 and 10. After we have defined the CS location, we need to compute the energy–momentum tensor (EMT) of such CS, using the standard formula:

$$\begin{aligned} T_{\mu \nu }({\textbf{x}})= & {} {\mathcal {F}}^{-1}\bigg \{\mu \int _{-\infty }^{+\infty } d\sigma \exp (- i {\textbf{kX}}) \\{} & {} \times \Big ( \partial _{t}X_{\mu } \partial _{t}X_\nu - \partial _{\sigma }X_{\mu }\partial _{\sigma }X_\nu \Big )\bigg \}, \end{aligned}$$

where the ${\mathcal {F}}^{-1}$ is the inverse Fourier transformation from frequency domain ${\textbf{k}}$ to coordinate domain ${\textbf{x}}.$ The complete derivation of EMT can be seen in the Appendix B. If we assume that the bend is small, compared to other parts of the string $(R = 0),$ EMT can be written as:

$$\begin{aligned} T_{\mu \nu } = \mu \left( \begin{array}{cccc} \delta _{\downarrow } + \delta _{\uparrow } &{} 0 &{} 0 &{} 0 \\ 0 &{} - \delta _{\uparrow }\sin ^2\theta &{} - \delta _{\uparrow }\sin \theta \cos \theta &{} 0 \\ 0 &{} - \delta _{\uparrow }\sin \theta \cos \theta &{} -\delta _{\downarrow } - \delta _{\uparrow } \cos ^2\theta &{} 0\\ 0 &{} 0 &{} 0 &{} 0 \\ \end{array} \right) , \end{aligned}$$

where:

$$\begin{aligned}{} & {} \delta _{\downarrow } = \delta (z)\delta (x)\Big (1 - H(y) \Big ),\\{} & {} \delta _{\uparrow } = \delta (z)\delta (x\cos \theta - y\sin \theta )H(x\sin \theta + y\cos \theta ) \end{aligned}$$

and H(x) is a unit Heaviside step-function.

According to the linearized Einstein equation, nonzero components of the EMT give us the nonzero components of the metric perturbation $h_{\mu \nu }.$ The metric itself is divergent if we assume the infinitely thin CS, but the Christoffel symbols and the geodesics equation for photons in such space can be derived. The full derivation can be found in Appendixes C and D. The final result of this section is an initial boundary problem (IBP) for photon trajectory, that should be solved numerically in order to get the lensed picture:

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{d {\textbf{v}}}{da} = -\dfrac{1}{2} \nabla _{\textbf{n}} (h_{\uparrow } + h_{\downarrow }) - \left[ \dfrac{\partial h_{\uparrow }}{\partial a} \left( \begin{array}{cc} \cos ^2\theta &{} \sin \theta \cos \theta \\ \sin \theta \cos \theta &{} \sin ^2\theta \end{array} \right) \right. \\ \quad \quad \left. + \dfrac{\partial h_{\downarrow }}{\partial a} \left( \begin{array}{cc} 1 &{} 0 \\ 0 &{} 0 \end{array} \right) \right] {\textbf{v}}\\ d{\textbf{n}}/da = {\textbf{v}}\\ {\textbf{n}}(a = 0) = {\textbf{n}}_0\\ {\textbf{n}}(a = 1) =0, \end{array}\right. } \end{aligned}$$

where ${\textbf{n}}_0$ is the initial direction to the point source. The definitions for $a, {\textbf{n}}, h_{\uparrow , \downarrow }$ and the partial derivatives of functions $h_{\uparrow , \downarrow }$ can be seen in the Appendix D.

After getting the solution of IBP we can calculate ${\textbf{v}}_f = -{\textbf{v}}(a = 1)$ – the direction under which the light from the point source ${\textbf{n}}_0$ is seen. Since the IBP can have several solutions, we can numerically check the area around ${\textbf{n}}_0$ using the shooting method to formally obtain a map $-{\textbf{v}}_f({\textbf{n}}_0).$ After that the lensing picture can also be computed numerically. The result of this computation can be seen in the Figs. 6 and 7.

Fig. 6

An example of lensing of 16 point sources (blue dots) on a string with a bend $\theta = 10^{\circ },$ tension $G\mu = 7.0 \cdot 10^{-7}$ and $R_s/R_g = 0.5.$ The new positions of the sources are marked by red dots, the string position is drawn by a green dashed line

Fig. 7

An example of lensing of 16 point sources (blue dots) on a CS with a bend $12.5^{\circ },$ tension $G\mu = 7.0 \cdot 10^{-7}$ and $R_s/R_g = 0.5.$ The new positions of the sources are marked by red dots, the CS position is drawn by a green dashed line

Fig. 8

An example of lensing of 16 point sources (blue dots) on a CS with a bend $\theta = 40^{\circ },$ tension $G\mu = 7.0 \cdot 10^{-7}$ and $R_s/R_g = 0.5.$ The new positions of the sources are marked by red dots, the CS position is drawn by a green dashed line

One can also notice, that for big angle of a bend the double image cannot be obtained even for an object that lies behind the CS (see Fig. 8).

Let us discuss the critical angle statement more in details, giving a physical meaning.

The mathematical model is that a double image of a point source located exactly behind the bend of the CS is considered. In the simplest model the critical angle of a curved CS bending is the angle between the first and second straight CS segments, which delimits the moment of transition from the existence of two images of a background source to one image of this source. The existence of a critical angle is possible only in lensing on a curved CS (curved extended object).

According to the classical gravitational lensing theory on a straight CS, a CS in the lens plane has a peculiar so called “strip of influence” depending on its linear density $G\mu .$ If the background source falls into this strip, then two images of this source are formed. If it does not fall, then the observer sees the source alone and shifted relative to the true position. (Lensed images for an extended source are constructed point-by-point.)

When the bending angle is equal to the critical angle, all points of the background source are outside the strip of the second CS segment. Thus, this particular source on the second segment is no longer lensed, does not generate two images.

The numerical search for the value of the critical angle completely follows from its definition. Due to the complexity of the analytical expressions, the number of solutions to the boundary value problem is numerically investigated for a fixed CS linear density $G\mu$ at different bending angles. The angle that corresponds to the transition from two images to one image is considered critical one. The initial position of the background source is selected exactly behind the bend of the CS.

The number of images of a point source (a photon) is the number of solutions to the boundary value problem for a point source. In the presence of a bending, the CS no longer has a cone-shaped metric near the part between the first and the second segments. Accordingly, two images of the source cannot be expected, because they were characteristic of the solution of a straight CS.

When the CS is curved, the deviation from the Euclidean metric will manifest itself from the side in which the second segment is tilted. Accordingly, from the opposite side, the photon will move already outside the band of the second segment.

The photon that the observer sees will have to be closer to the CS in order to have a sufficient deviation of its trajectory. The bending of the photon trajectory with an increase in the bending angle of the CS will be smaller, which means that the distance to the CS will also be smaller. When the critical angle is reached (with an increase in the bending angle), we will observe that the image on the other side of the CS will disappear, because it will be impossible to bend the photon’s trajectory to a value that allows the photon to get to the observer.

From numerical simulations the critical angle for $G\mu = 7.0 \cdot 10^{-7}$ and $R_s/R_g = 0.5$ is $\theta _c \approx 13^{\circ }.$ This result can be an argument for the observational lack of double images of galaxies.

It was also shown in [47, 48] that CS can produce more than 2 images of distant sources, due to their small scale stricture, which can be the case in our model, if one incorporates more than one bend. However, the modern limitations on the deficit angle are of same magnitude, as our angular resolution in visible light. This means that gravitational lensing events on CS with more than 2 images are difficult to find and analyze.

Using the same numerical procedure one can also get the critical angle for variety of CS tensions $\theta _c(G\mu ).$ We have checked this dependence in the interval $G\mu \in [7\times 10^{-8} ,\, 7\times 10^{-6}]$ around the $G\mu \approx 7 \times 10^{-7},$ where the upper bound for CS tension according CSc-1 [23] is located (see Fig. 9).

Fig. 9

Black dots – numerical data for several string tensions $G\mu$ and fixed $R_s/R_g = 0.5.$ Black line – linear extrapolation of $\theta _c(\log G\mu ).$ Blue square – NANOGrav $2\sigma$ bounds for tension [19]

According to our simulations, the critical angle $\theta _c$ decreases sharply with the decrease of $G\mu$ and we argue that it can be near zero for $G\mu \lesssim 10^{-8}.$ Thus, according to the interpretation of recent NANOGrav data [19] (see Blue rectangle on Fig. 9), only straight CS segments can produce several images. (Of course, for the current level of accuracy, this result is more of a theoretical nature.)

5 Conclusions

The model presented in the paper first generalizes the gravitational lensing on a CS of general position. Gravitational lensing on a CS with an inclination in the plane coinciding with the beam of vision is considered. Gravitational lensing on a CS curved in a plane perpendicular to the line of sight is considered. In the work, the curvature of the CS is given by one fracture with a given angle.

A fundamentally new result is that the bending of the CS crucially affects the number of images. So, with a larger value of the bending angle (more than the critical angle, which for $G\mu = 7.0 \cdot 10^{-7}$ and $R_s/R_g = 0.5$ is $\theta _c \approx 13^{\circ }$), the second image disappears, which can serve as an argument to explain the absence of a large number of gravitational-lens chains (new “Milky Ways”).

The simulation results are applied by authors to the analysis of gravitational-lens candidates in the field of the previously found by CMB analysis [49] candidate string CSc-1 (in preparation).

For cosmology, astrophysics, and theoretical physics the discovery of CS is without a doubt a huge step in understanding the structure of the Universe, especially its global properties and the earliest stages of its evolution.

The presented in this paper detailed theoretical study opens up fundamentally new ways to search for gravitational-lens events, which, together with the analysis of CMB anisotropy and gravitational wave observational channel, will allow statistically significant detection of CS.

Appendices

Appendix A: Flat approximation for a CS space-time with a conical singularity

Geodesic trajectory has the form:

$$\begin{aligned} \frac{d^2 x^{\lambda }}{ds^2} = - \Gamma ^{\lambda }_{\mu \nu } \frac{d x^{\mu }}{ds} \frac{d x^{\nu }}{ds} \end{aligned}$$

where

$$\begin{aligned} \Gamma ^{\lambda }_{\mu \nu }=\frac{g^{\lambda \alpha }}{2}\bigg (\frac{\partial g_{\mu \alpha }}{\partial x^{\nu }} + \frac{\partial g_{\nu \alpha }}{\partial x^{\mu }} - \frac{\partial g_{\mu \nu }}{\partial x^{\alpha }}\bigg ). \end{aligned}$$

Nonzero derivatives in the metric of the cosmic string only has $g_{\varphi \varphi }$ component, so only non-zero Christoffel symbols are:

$$\begin{aligned}{} & {} \Gamma ^{\varphi }_{\varphi r} = \frac{g^{\varphi \varphi }}{2} \frac{\partial g_{\varphi \varphi }}{\partial r} = \frac{1}{r}\\{} & {} \Gamma ^{\varphi }_{r \varphi } = \frac{g^{\varphi \varphi }}{2} \frac{\partial g_{\varphi \varphi }}{\partial r} = \frac{1}{r}\\{} & {} \Gamma ^{r}_{\varphi \varphi } = r (1 - 4 G \mu )^2. \end{aligned}$$

The first two characters are the same for a flat space and do not contain string parameters. This means that we can consider the space to be flat. The third character is used in combination with the derivatives $d\varphi /ds.$ If we make a replacement:

$$\begin{aligned} \varphi ' = \varphi (1 - 4 G \mu ). \end{aligned}$$

Then all the equations will take the form as for a flat conical space with deficit angle $\Delta \theta = 8\pi G \mu$ in the plane perpendicular to the string.

Appendix B: Energy–momentum tensor for a curved string

To define the EMT of the curved string, we assume the area on which the galaxy is lensed to be small. So only one segment with nonzero curvature is needed for the model. Thus we approximate the string with two straight lines connected by Besier curve:

$$\begin{aligned} {\left\{ \begin{array}{ll} X^1(s) = x(s) = {\left\{ \begin{array}{ll} 0, &{} s \in (-\infty , - R)\\ \dfrac{R \sin \theta }{4} \bigg (1 + \dfrac{s}{R} \bigg )^2,&{} s \in [-R, R]\\ s \sin \theta ,&{} s \in (R, +\infty ) \end{array}\right. }\\ X^2(s) = y(s) = {\left\{ \begin{array}{ll} s, &{} s \in (-\infty , - R)\\ \dfrac{R \cos \theta }{4} \bigg (1 + \dfrac{s}{R} \bigg )^2 -\dfrac{R}{4} \bigg (1 - \dfrac{s}{R} \bigg )^2,&{} s \in [-R, R]\\ s \cos \theta ,&{} s \in (R, +\infty ) \end{array}\right. }\\ X^3(s) = z(s) = R_g - R_s. \end{array}\right. } \end{aligned}$$

In this article we use the convention $X^0 = t$ and assume the static string (Fig. 10).

Fig. 10

String position in $(x^1, x^2)$ plane

The parameter s in string location ${\textbf{X}}$ is not its natural parameterization $\sigma ,$ which is used in calculation of EMT:

$$\begin{aligned}{} & {} T_{\mu \nu }({\textbf{x}}) \\{} & {} \quad ={\mathcal {F}}^{-1}\bigg \{\mu \int _{-\infty }^{+\infty } d\sigma \exp (- i {\textbf{kX}}) \Big ( \partial _{t}X_{\mu } \partial _{t}X_\nu - \partial _{\sigma }X_{\mu }\partial _{\sigma }X_\nu \Big )\bigg \}. \end{aligned}$$

The connection between s and $\sigma$ is:

$$\begin{aligned}{} & {} \frac{d\sigma }{ds} = \rho (s) = \sqrt{x'(s)^2 + y'(s)^2 + z'(s)^2 }\\{} & {} \quad = {\left\{ \begin{array}{ll} 1,&{} s \in (-\infty , - R)\\ \sqrt{\cos ^2 \theta /2 + (s/R)^2 \sin ^2\theta / 2},&{} s \in [-R, R]\\ 1, &{} s \in (R, +\infty ). \end{array}\right. } \end{aligned}$$

This connection transforms the EMT into:

$$\begin{aligned}{} & {} T^{\mu \nu }({\textbf{x}}) = \mu \int _{-\infty }^{+\infty } \rho (s) ds\, \\{} & {} \quad \times \Big ( \partial _{t}X^{\mu } \partial _{t}X^{\nu } - \partial _{\sigma }X^{\mu }\partial _{\sigma }X^{\nu }\Big ) \delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big ). \end{aligned}$$

The main step in calculation of EMT is to split the integral into 3 parts:

• $\downarrow$ – for $s \in (-\infty , - R),$ where the positional angle is 0;

• $\uparrow$ – for $s \in (R, +\infty ),$ where the positional angle is $\theta ;$

• B – for $s \in [-R, R],$ where the bend is located.

and then find integrals that describe all components of the EMT in some linear combination, for example:

$$\begin{aligned}{} & {} T^{00}({\textbf{x}}) = \mu \int _{-\infty }^{-R} ds\,\delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) \\{} & {} \qquad + \mu \int _{-R}^{R} ds\,\sqrt{\cos ^2 \theta /2 + (s/R)^2 \sin ^2\theta / 2} \times \delta ^{(3)} \\{} & {} \qquad \Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) + \mu \int _{R}^{+\infty } ds\, \delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) \\{} & {} \quad = \mu (\delta _{\downarrow } + \Delta _{00}^{B} + \delta _{\uparrow }) \end{aligned}$$

where the calculated integrals are respectfully:

$$\begin{aligned}{} & {} \delta _{\downarrow } = \delta (z)\delta (x)\Big (1 - H(y + R) \Big )\\{} & {} \Delta _{00}^{B} = \int _{-R}^{R} ds\,\sqrt{\cos ^2 \theta /2 + (s/R)^2 \sin ^2\theta / 2} \times \delta ^{(3)} \\{} & {} \qquad \Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) \\{} & {} \delta _{\uparrow } = \delta (z)\delta (x\cos \theta - y\sin \theta )H(x\sin \theta + y\cos \theta - R). \end{aligned}$$

Some components are trivially zero, such as:

$$\begin{aligned} T^{0i} = T^{i0} = T^{3\mu } = T^{\mu 3} = 0. \end{aligned}$$

The other ones include $\partial _\sigma = \rho (s)^{-1} \partial _s$:

$$\begin{aligned}{} & {} T^{11} ({\textbf{x}}) = 0 \\{} & {} \qquad - \frac{\mu }{4} \int _{-R}^{R} ds\,\frac{\sin ^2\theta \bigg ( 1 + \dfrac{s}{R}\bigg )^2}{\sqrt{\cos ^2 \theta /2 + (s/R)^2 \sin ^2\theta / 2}} \\{} & {} \qquad \times \delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) - \mu \sin ^2\theta \int _{R}^{+\infty } ds\, \delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) \\{} & {} \quad = - \mu ( \Delta _{11}^{B} + \delta _{\uparrow }\sin ^2\theta ) \\{} & {} T^{12} ({\textbf{x}}) = 0\\{} & {} \qquad -\frac{\mu }{4}\int _{-R}^{R} ds\, \frac{\bigg [ \bigg (1 + \dfrac{s}{R}\bigg ) \cos \theta + \bigg (1 - \dfrac{s}{R}\bigg )\bigg ]\bigg (1 + \dfrac{s}{R}\bigg ) \sin \theta }{\sqrt{\cos ^2 \theta /2 + (s/R)^2 \sin ^2\theta / 2}}\\{} & {} \qquad \times \delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big )- \mu \cos \theta \sin \theta \int _{R}^{+\infty } ds\, \delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) \\{} & {} \quad = - \mu ( \Delta _{12}^{B} + \delta _{\uparrow }\sin \theta \cos \theta ) \\{} & {} T^{22} ({\textbf{x}}) = -\mu \int _{-\infty }^{-R} ds\,\delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) \\{} & {} \qquad -\frac{\mu }{4} \int _{-R}^{R} ds\,\frac{\bigg [ \bigg (1 + \dfrac{s}{R}\bigg ) \cos \theta + \bigg (1 - \dfrac{s}{R}\bigg )\bigg ]^2}{\sqrt{\cos ^2 \theta /2 + (s/R)^2 \sin ^2\theta / 2}}\\{} & {} \qquad \times \delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) -\mu \cos ^2\theta \int _{R}^{+\infty } ds\, \delta ^{(3)}\Big ({\textbf{x}} - {\textbf{X}}(s)\Big ) \\{} & {} \quad = -\mu (\delta _{\downarrow } + \Delta _{22}^{B} + \delta _{\uparrow } \cos ^2\theta ). \end{aligned}$$

All the integrals named $\Delta _{ij}^{B}$ are of the form:

$$\begin{aligned}{} & {} \delta (z) \int _{-R}^{R} f(s) \delta \big (x_0 - x(s)\big ) \delta \big (y_0 - y(s)\big ) ds \\{} & {} \quad = {\left\{ \begin{array}{ll} \dfrac{f(s_0)\delta (z)\delta \big (y_0 - y(s_0)\big )}{|x'(s_0)|}\bigg |_{x_0 = x(s_0)}, &{}\text {if}\ y_0 = y(x_0)\\ 0, &{}\text {if}\ y_0 \ne y(x_0). \end{array}\right. } \end{aligned}$$

Thus the exact solutions for $\Delta _{ij}^{B}$ are:

$$\begin{aligned}{} & {} \Delta _{00}^{B} = \delta (z) \delta \bigg (y + x \tan \frac{\theta }{2} - 2\sqrt{\frac{xR}{\sin \theta }} + R \bigg )\\{} & {} \quad \times \sqrt{\dfrac{\cos ^2\dfrac{\theta }{2} + \sin ^2\dfrac{\theta }{2} \bigg (2\sqrt{\dfrac{x}{R\sin \theta }} - 1\bigg )^2}{\dfrac{x\sin \theta }{R}}} \\{} & {} \quad \times (1 - H(x\sin \theta + y\cos \theta - R)) H(y + R) \\{} & {} \Delta _{11}^{B} = \delta (z) \delta \bigg (y + x \tan \frac{\theta }{2} - 2\sqrt{\frac{xR}{\sin \theta }} + R \bigg ) \\{} & {} \quad \times \sqrt{\dfrac{\dfrac{x\sin \theta }{R}}{\cos ^2\dfrac{\theta }{2} + \sin ^2\dfrac{\theta }{2} \bigg (2\sqrt{\dfrac{x}{R\sin \theta }} - 1\bigg )^2}} \\{} & {} \quad \times (1 - H(x\sin \theta + y\cos \theta - R)) H(y + R) \\{} & {} \Delta _{12}^{B} = \delta (z) \delta \bigg (y + x \tan \frac{\theta }{2} - 2\sqrt{\frac{xR}{\sin \theta }} + R \bigg ) \\{} & {} \quad \times \dfrac{1 - \sqrt{\dfrac{x}{R\sin \theta }} (1 - \cos \theta )}{\sqrt{\cos ^2\dfrac{\theta }{2} + \sin ^2\dfrac{\theta }{2} \bigg (2\sqrt{\dfrac{x}{R\sin \theta }} - 1\bigg )^2}} \\{} & {} \quad \times (1 - H(x\sin \theta + y\cos \theta - R)) H(y + R) \\{} & {} \Delta _{22}^{B} = \delta (z) \delta \bigg (y + x \tan \frac{\theta }{2} - 2\sqrt{\frac{xR}{\sin \theta }} + R \bigg )\\{} & {} \quad \times \dfrac{\sqrt{\dfrac{R}{x \sin \theta }}\bigg (1 - \sqrt{\dfrac{x}{R\sin \theta }} (1 - \cos \theta )\bigg )^2}{\sqrt{\cos ^2\dfrac{\theta }{2} + \sin ^2\dfrac{\theta }{2} \bigg (2\sqrt{\dfrac{x}{R\sin \theta }} - 1\bigg )^2}} \\{} & {} \quad \times (1 - H(x\sin \theta + y\cos \theta - R)) H(y + R). \end{aligned}$$

So the full answer is:

$$\begin{aligned}{} & {} T_{\mu \nu } \\{} & {} \quad = \mu \left( \begin{array}{cccc} \delta _{\downarrow } + \Delta _{00}^{B} + \delta _{\uparrow } &{} 0 &{} 0 &{} 0 \\ 0 &{} -\Delta _{11}^{B} - \delta _{\uparrow }\sin ^2\theta &{} -\Delta _{12}^{B} - \delta _{\uparrow }\sin \theta \cos \theta &{} 0 \\ 0 &{} -\Delta _{12}^{B} - \delta _{\uparrow }\sin \theta \cos \theta &{} -\delta _{\downarrow } - \Delta _{22}^{B} - \delta _{\uparrow } \cos ^2\theta &{} 0\\ 0 &{} 0 &{} 0 &{} 0 \\ \end{array} \right) . \end{aligned}$$

To solve the linearized Einstein’s equations we also need the source function. In the approximation of small bend ($R \ll R_s \Delta \theta$ or simply $R = 0$):

$$\begin{aligned} S_{\mu \nu } = \mu \left( \begin{array}{cccc} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} -\delta _{\downarrow } -\delta _{\uparrow }\cos ^2\theta &{} - \delta _{\uparrow }\sin \theta \cos \theta &{} 0 \\ 0 &{} - \delta _{\uparrow }\sin \theta \cos \theta &{} -\delta _{\uparrow } \sin ^2\theta &{} 0\\ 0 &{} 0 &{} 0 &{} \delta _{\downarrow } + \delta _{\uparrow } \\ \end{array} \right) . \end{aligned}$$

Appendix C: Photon trajectories for small metric perturbation

Let $x^1,$ $x^2$ be the axis parallel to the picture plane, $x^3$ be the axis parallel to the line of sight. The source (galaxy) will be at $x^3 = 0,$ the observer will be located at $x^3 = R_g.$ The metric perturbation (in the article it is the curved cosmic string) between the observer and the source will be placed at $x^3 = R_g - R_s$ and it will be in the form:

$$\begin{aligned} h_{\mu \nu } = \left( \begin{array}{cccc} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} h_{11} &{} h_{12} &{} 0 \\ 0 &{} h_{12} &{} h_{22} &{} 0\\ 0 &{} 0 &{} 0 &{} h_{33} \\ \end{array} \right) . \end{aligned}$$

Since photon travel along the null geodesics, it is convenient to choose time $x^0 = t$ as a parameter:

$$\begin{aligned} \frac{d^2 x^i}{dt^2} = - \bigg ( \Gamma ^i_{\mu \nu } - \Gamma ^0_{\mu \nu } \frac{d x^{i}}{dt} \bigg ) \frac{d x^{\mu }}{dt} \frac{d x^{\nu }}{dt},\quad i = 1, 2, 3. \end{aligned}$$

In the weak field approximation one can write:

$$\begin{aligned} \Gamma ^{\lambda }_{\mu \nu } \approx \frac{1}{2}\bigg (\frac{\partial h_{\mu }^{.\,\lambda }}{\partial x^{\nu }} + \frac{\partial h_{\nu }^{.\,\lambda }}{\partial x^{\mu }} - \frac{\partial h_{\mu \nu }}{\partial x_\lambda }\bigg ). \end{aligned}$$

It is easy to see that for this metric perturbation:

$$\begin{aligned} \Gamma ^0_{\mu \nu } = \frac{1}{2}\bigg (\frac{\partial h_{\mu }^{.\,0}}{\partial x^{\nu }} + \frac{\partial h_{\nu }^{.\,0}}{\partial x^{\mu }} - \frac{\partial h_{\mu \nu }}{\partial x_0}\bigg ) = \frac{1}{2}\bigg (0 + 0 - 0 \bigg ) = 0 \end{aligned}$$

thus the equation for photon trajectory is simple:

$$\begin{aligned} \frac{d^2 x^i}{dt^2} = - \Gamma ^i_{\mu \nu } \frac{d x^{\mu }}{dt} \frac{d x^{\nu }}{dt},\quad i = 1, 2, 3. \end{aligned}$$

The complete list of all Christoffel symbols is shown in the Tables 1, 2, 3.

Recall $v_i = dx^i / dt$ and the equations are:

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{d v^1}{dt} = \dfrac{1}{2}\,\dfrac{\partial h_{11}}{\partial x^1} \, \big ( v^1 \big )^2 +\bigg (\dfrac{\partial h_{21}}{\partial x^2}-\dfrac{1}{2}\,\dfrac{\partial h_{22}}{\partial x^1}\bigg ) \big ( v^2 \big )^2 \\ \quad - \dfrac{1}{2}\,\dfrac{\partial h_{33}}{\partial x^1} \big ( v^3 \big )^2+ \dfrac{\partial h_{11}}{\partial x^2} v^1 v^2 + \dfrac{\partial h_{11}}{\partial x^3} v^1 v^3 + \dfrac{\partial h_{12}}{\partial x^3} v^2 v^3\\ \dfrac{d v^2}{dt} = \dfrac{1}{2}\,\dfrac{\partial h_{22}}{\partial x^2} \, \big ( v^2 \big )^2 +\bigg (\dfrac{\partial h_{21}}{\partial x^1}-\dfrac{1}{2}\,\dfrac{\partial h_{11}}{\partial x^2}\bigg ) \big ( v^1 \big )^2 \\ \quad - \dfrac{1}{2}\,\dfrac{\partial h_{33}}{\partial x^2} \big ( v^3 \big )^2 + \dfrac{\partial h_{22}}{\partial x^1} v^1 v^2 + \dfrac{\partial h_{12}}{\partial x^3} v^1 v^3 + \dfrac{\partial h_{22}}{\partial x^3} v^2 v^3\\ \dfrac{d v^3}{dt} = - \dfrac{1}{2}\,\dfrac{\partial h_{33}}{\partial x^3}\big ( v^3 \big )^2 - \dfrac{1}{2}\,\dfrac{\partial h_{11}}{\partial x^3}\big ( v^1 \big )^2 - \dfrac{1}{2}\,\dfrac{\partial h_{22}}{\partial x^3}\big ( v^2 \big )^2 \\ \quad - \dfrac{1}{2}\,\dfrac{\partial h_{33}}{\partial x^1} v^1 v^3 - \dfrac{1}{2}\,\dfrac{\partial h_{33}}{\partial x^2} v^2 v^3 - \dfrac{1}{2}\,\dfrac{\partial h_{12}}{\partial x^3} v^1 v^2. \end{array}\right. } \end{aligned}$$

Table 1 Evaluation of all $\Gamma ^1_{\mu \nu }$

Table 2 Evaluation of all $\Gamma ^2_{\mu \nu }$

Table 3 Evaluation of all $\Gamma ^3_{\mu \nu }$

The lensing effect is small, so we can just adopt the first order expansion in $\mu$ for $v^1$ and $v^2.$ We can use an approximation $v^3 = 1$ since it will be the second order correction at most in the first two equations:

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{d v^1}{dt} = - \dfrac{1}{2}\,\dfrac{\partial h_{33}}{\partial x^1} + \dfrac{\partial h_{11}}{\partial x^3} v^1 + \dfrac{\partial h_{12}}{\partial x^3} v^2\\ \dfrac{d v^2}{dt} = - \dfrac{1}{2}\,\dfrac{\partial h_{33}}{\partial x^2} + \dfrac{\partial h_{12}}{\partial x^3} v^1 + \dfrac{\partial h_{22}}{\partial x^3} v^2\\ v^3 = 1. \end{array}\right. } \end{aligned}$$

(C1)

Using the third equation in (C1), we can replace the d/dt by d/dz and this is the final result before the derivation of lens equation.

Appendix D: Solving a lens equation for a curved CS

Before the derivation we need to solve the linearized Einstein’s equation for a static bended string:

$$\begin{aligned} h_{\mu \nu }({\textbf{x}}) = 4G \int _{\mathbb {R}^3} d{\textbf{x}}'\,\frac{S_{\mu \nu }}{|{\textbf{x}} - {\textbf{x}}'|}. \end{aligned}$$

For the purpose of simplicity we will consider the case $R \ll R_s \Delta \theta$ so the field from bend $s\in [-R, R]$ can be neglected. Recalling:

$$\begin{aligned} h_{\uparrow ,\downarrow } = \frac{\Delta \theta }{2\pi }\int _{\mathbb {R}^3} d{\textbf{x}}'\,\frac{\delta _{\uparrow , \downarrow }}{|{\textbf{x}} - {\textbf{x}}'|} \end{aligned}$$

we can rewrite geodesic equations (C1):

$$\begin{aligned}{} & {} \frac{d {\textbf{v}}}{dz} = -\underbrace{ \frac{1}{2} \left( \begin{array}{c} \partial _x \\ \partial _y \end{array} \right) (h_{\uparrow } + h_{\downarrow }) }_{={\textbf{A}}(x, y, z)} \nonumber \\{} & {} \quad - \underbrace{\bigg [\frac{\partial h_{\uparrow }}{\partial z} \left( \begin{array}{cc} \sin ^2\theta &{} \sin \theta \cos \theta \\ \sin \theta \cos \theta &{} \cos ^2\theta \end{array} \right) + \frac{\partial h_{\downarrow }}{\partial z} \left( \begin{array}{cc} 0 &{} 0 \\ 0 &{} 1 \end{array} \right) \bigg ]}_{=\hat{\mathbf {{B}}}(x, y, z)} {\textbf{v}}\nonumber \\ \end{aligned}$$

(D2)

where ${\textbf{v}} = (v^1, v^2)^T.$ This equation is particularly important since in linear approximation ${\textbf{v}}$ is an angle between the photon path and the z-axis. Suppose that without the string lensing a point is seen from direction ${\textbf{n}},$ ${\textbf{v}}_i$ is the initial direction of photon’s path and $-{\textbf{v}}_f$ is the final direction under which the point is seen with a lens. Thus, the lens equation is:

$$\begin{aligned} {\textbf{n}} = -{\textbf{v}}_f - ({\textbf{v}}_f - {\textbf{v}}_i)\frac{R_g - R_s}{R_g}. \end{aligned}$$

If we know the initial image without the lens, then we know ${\textbf{n}}.$ Our task is to find ${\textbf{v}}_f.$ The connection between ${\textbf{v}}_i$ and ${\textbf{v}}_f$ is presented by the solution of DE (D2) with the initial condition ${\textbf{v}}(R_g {\textbf{n}}, z = 0) = -{\textbf{v}}_i.$ This approximation is valid only in the case, when all the lensing happens at $z = R_g - R_s,$ near the string. But the string is not a compact object, so we propose another scheme.

Suppose the radiation is emitted from the point ${\textbf{r}}(z = 0) = R_g {\textbf{n}}_0.$ ${\textbf{r}} = (x(z), y(z))^T$ is a vector in the picture plane with a fixed value of z. The ray should have the initial conditions ${\textbf{v}}_{\textbf{0}}$ such that it travels to the observer, so ${\textbf{r}}(z = R_g) =0.$ Thus the boundary value problem can be formulated:

$$\begin{aligned} {\left\{ \begin{array}{ll} d{\textbf{v}}/dz = -{\textbf{A}}({\textbf{r}}, z) - \hat{\mathbf {{B}}}({\textbf{r}}, z){\textbf{v}}\\ d{\textbf{r}}/dz = {\textbf{v}}\\ {\textbf{r}}(z = 0) = R_g {\textbf{n}}_0\\ {\textbf{r}}(z = R_g) =0. \end{array}\right. } \end{aligned}$$

(D3)

Once this problem is solved, we can calculate ${\textbf{v}}(z = R_g) = -{\textbf{v}}_f$ and create a map ${\textbf{v}}_f({\textbf{n}}_0),$ which is our lens equation.

To solve this equation numerically, we can treat ${\textbf{v}}_f$ as a parameter in shooting method for IBP (D3). To further simplify the process of numerical integration, we rescale the spatial variables:

$$\begin{aligned} {\left\{ \begin{array}{ll} a = z / R_g \in [0, 1] \\ {\textbf{n}} = {\textbf{r}} / R_g \end{array}\right. } \end{aligned}$$

and the IBP reads:

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{d {\textbf{v}}}{da} = -\dfrac{1}{2} \nabla _{\textbf{n}} (h_{\uparrow } + h_{\downarrow }) - \left[ \dfrac{\partial h_{\uparrow }}{\partial a} \left( \begin{array}{cc} \cos ^2\theta &{} \sin \theta \cos \theta \\ \sin \theta \cos \theta &{} \sin ^2\theta \end{array} \right) \right. \\ \left. + \dfrac{\partial h_{\downarrow }}{\partial a} \left( \begin{array}{cc} 1 &{} 0 \\ 0 &{} 0 \end{array} \right) \right] {\textbf{v}} \\ d{\textbf{n}}/da = {\textbf{v}} \\ {\textbf{n}}(a = 0) = {\textbf{n}}_0 \\ {\textbf{n}}(a = 1) =0. \end{array}\right. } \end{aligned}$$

We also need all the derivatives of string metric: $\nabla _{{\textbf{n}}} h_{\uparrow , \downarrow },$ $\partial h_{\uparrow , \downarrow } / \partial a.$ They are (in terms of scaled variables, $r = R_s / R_g$):

$$\begin{aligned}{} & {} \frac{\partial h_{\uparrow }}{\partial a} = \frac{\Delta \theta }{2\pi } \, \frac{a - 1 + r}{(a - 1 + r)^2 + (n_x \cos \theta - n_y \sin \theta )^2}\\{} & {} \quad \times \bigg ( 1 + \frac{n_x\sin \theta + n_y \cos \theta }{\sqrt{{\textbf{n}}^2 + (a - 1 + r)^2}} \bigg ) \nonumber \\{} & {} \frac{\partial h_{\downarrow }}{\partial a} {=} \frac{\Delta \theta }{2\pi } \, \frac{a {-} 1 {+} r}{(a {-} 1 {+} r)^2 {+} n_x^2}\bigg ( 1 {-} \frac{n_y}{\sqrt{{\textbf{n}}^2 {+} (a {-} 1 {+} r)^2}} \bigg ) \nonumber \\{} & {} \frac{\partial h_{\uparrow }}{\partial n_x} = \frac{\Delta \theta }{2\pi } \, \bigg [-\frac{\sin \theta }{\sqrt{{\textbf{n}}^2 + (a - 1 + r)^2}} \\{} & {} \quad + \cos \theta \frac{n_x \cos \theta - n_y \sin \theta }{(a - 1 + r)^2 + (n_x \cos \theta - n_y \sin \theta )^2}\\{} & {} \quad \times \bigg ( 1 + \frac{n_x\sin \theta + n_y \cos \theta }{\sqrt{{\textbf{n}}^2 + (a - 1 + r)^2}} \bigg )\bigg ] \\{} & {} \frac{\partial h_{\downarrow }}{\partial n_x} {=} \frac{\Delta \theta }{2\pi } \, \frac{n_x}{(a {-} 1 {+} r)^2 {+} n_x^2}\bigg ( 1 {-} \frac{n_y}{\sqrt{{\textbf{n}}^2 {+} (a {-} 1 {+} r)^2}} \bigg ) \\{} & {} \frac{\partial h_{\uparrow }}{\partial n_y} = \frac{\Delta \theta }{2\pi } \, \bigg [-\frac{\cos \theta }{\sqrt{{\textbf{n}}^2 + (a - 1 + r)^2}} \\{} & {} \quad - \sin \theta \frac{n_x \cos \theta - n_y \sin \theta }{(a - 1 + r)^2 + (n_x \cos \theta - n_y \sin \theta )^2}\\{} & {} \quad \times \bigg (1 + \frac{n_x\sin \theta + n_y \cos \theta }{\sqrt{{\textbf{n}}^2 + (a - 1 + r)^2}} \bigg )\bigg ] \\{} & {} \frac{\partial h_{\downarrow }}{\partial n_y} = \frac{\Delta \theta }{2\pi } \, \frac{1}{\sqrt{{\textbf{n}}^2 + (a - 1 + r)^2}}. \end{aligned}$$