Gravitational Rutherford scattering of electrically charged particles from a charged Weyl black hole

Abstract

Considering electrically charged test particles, we continue our study of the exterior dynamics of a charged Weyl black hole which has been previously investigated regarding the motion of mass-less and (neutral) massive particles. In this paper, the deflecting trajectories of charged particles are designated as being gravitationally Rutherford-scattered and detailed discussion of angular and radial particle motions is presented.

Appendices

Appendix A: The method of finding $L_U$ in Eq. (38)

Equation (34) allows for obtaining an expression for $L_U$, by solving

$$\begin{aligned} {\mathfrak {a}}L_{U}^{4}-{\mathfrak {b}} L_{U}^{2}+{\mathfrak {c}}=0, \end{aligned}$$

(A1)

in which

$$\begin{aligned} {\mathfrak {a}}&={(Q^2-2r_U^2)^2\over r_U^6 }, \end{aligned}$$

(A2a)

$$\begin{aligned} {\mathfrak {b}}&={2Q^2(1+q^2)\over r_U^2}-{Q^4(2+q^2)\over 2r_U^4}-{8r_U^2-2Q^2(1-q^2)\over \lambda ^4}, \end{aligned}$$

(A2b)

$$\begin{aligned} {\mathfrak {c}}&={Q^4(1+2q^2)\over 4r_U^2}-2q^2Q^2+{4r_U^6\over \lambda ^4}-{2Q^2r_U^2(1-q^2) \over \lambda ^2}. \end{aligned}$$

(A2c)

Solving Eq. (A1) for $L_{U}^2$ then yields the value in Eq. (38).

Appendix B: Finding the angular equation of motion

Since the closest approach happens at $r_S$, to deal with the integral in Eq. (46), we define the following nonlinear change of variable:

$$\begin{aligned} u \doteq {1\over \frac{r}{r_S}-1}, \end{aligned}$$

(B1)

which reduces Eq. (46) to

$$\begin{aligned} \phi (r) =\kappa _0\left[ \int _{u}^{\infty }{\mathrm {d}u\over \sqrt{{\mathcal {P}}_3(u)} }-u_3\int _{u}^{\infty }{\mathrm {d}u\over (u+u_3)\sqrt{{\mathcal {P}}_3(u)} }\right] , \end{aligned}$$

(B2)

where $u_j\doteq {1\over (r_j/r_S)-1}$, with $j=\{2,3,5,6\}$, and

$$\begin{aligned} {\mathcal {P}}_3(u)\equiv u^3+{\mathbf {a}} u^2+{\mathbf {b}} u+{\mathbf {c}}, \end{aligned}$$

(B3)

with

$$\begin{aligned} {\mathbf {a}}&= u_2+u_5+u_6, \end{aligned}$$

(B4a)

$$\begin{aligned} {\mathbf {b}}&= u_2(u_5+u_6)+u_5 u_6, \end{aligned}$$

(B4b)

$$\begin{aligned} {\mathbf {c}}&= u_2 u_5 u_6. \end{aligned}$$

(B4c)

Defining

$$\begin{aligned} \kappa _0= {\upsilon \over r_S^2} u_3 \sqrt{u_2 u_5 u_6}, \end{aligned}$$

(B5)

and applying another change of variable

$$\begin{aligned} U\doteq \frac{1}{4}\left( u+\frac{{\mathbf {a}}}{3} \right) , \end{aligned}$$

(B6)

we can rewrite Eq. (B2) as

$$\begin{aligned} \phi (r) =\kappa _0\left[ \int _{U}^{\infty }{\mathrm {d}U\over \sqrt{ P _3(U)} }-{u_3\over 4}\int _{U}^{\infty }{\mathrm {d}U\over (U+U_3)\sqrt{ P _3(U)} }\right] , \end{aligned}$$

(B7)

given that $U_3 = \frac{1}{4}\left( u_3+\frac{{\mathbf {a}}}{3}\right)$, and

$$\begin{aligned} P _3(u)\equiv 4 U^3-{\mathbf {g}}_2 U- {\mathbf {g}}_3. \end{aligned}$$

(B8)

Direct integration of Eq. (B7) results in the expression in Eq. (48).

Appendix C: Solving depressed quartic equations

The condition $V_r'(r)=0$ provides the following equation of eighth degree:

$$\begin{aligned} r^8+{\tilde{a}} r^4+{\tilde{b}} r^2+{\tilde{c}}=0. \end{aligned}$$

(C1)

To solve this equation, we firstly make the change of variable $r^2\doteq x$. Afterward, we combine the methods of Ferrari and Cardano to solve a depressed quartic equation of the form (originally studied by Cardano in Ref. [78])

$$\begin{aligned} x^4+{\tilde{a}} x^2+{\tilde{b}} x+{\tilde{c}}=0,~~~~({\tilde{a}},{\tilde{b}},{\tilde{c}}) \in {\mathbb {R}}. \end{aligned}$$

(C2)

This equation can be rewritten as the product of two quadratic equations, as follows:

$$\begin{aligned} x^4+{\tilde{a}} x^2+{\tilde{b}} x+{\tilde{c}} = (x^2-2 {{\tilde{\alpha }}} x+{{\tilde{\beta }}})(x^2+2 {{\tilde{\alpha }}} x+{{\tilde{\gamma }}})=0. \end{aligned}$$

(C3)

Accordingly, we obtain

$$\begin{aligned} {\tilde{a}}&={{\tilde{\beta }}}+{{\tilde{\gamma }}}-4{{\tilde{\alpha }}}^2, \end{aligned}$$

(C4a)

$$\begin{aligned} {\tilde{b}}&=2{{\tilde{\alpha }}}({{\tilde{\beta }}}-{{\tilde{\gamma }}}), \end{aligned}$$

(C4b)

$$\begin{aligned} {\tilde{c}}&={{\tilde{\beta }}} {{\tilde{\gamma }}}. \end{aligned}$$

(C4c)

Solving the first two equations for ${{\tilde{\beta }}}$ and ${{\tilde{\gamma }}}$ yields

$$\begin{aligned} {{\tilde{\beta }}}&=2{{\tilde{\alpha }}}^2 +{{\tilde{a}}\over 2}+{{\tilde{b}}\over 4{{\tilde{\alpha }}}}, \end{aligned}$$

(C5a)

$$\begin{aligned} {{\tilde{\gamma }}}&=2{{\tilde{\alpha }}}^2 +{{\tilde{a}}\over 2}-{{\tilde{b}}\over 4{{\tilde{\alpha }}}}, \end{aligned}$$

(C5b)

which together with Eq. (C4c) results in an equation of sixth degree in ${{\tilde{\alpha }}}$:

$$\begin{aligned} {{\tilde{\alpha }}}^6+{{\tilde{a}}\over 2}{{\tilde{\alpha }}}^4+\left( {{\tilde{a}}^2\over 16}-{{\tilde{c}}\over 4}\right) {{\tilde{\alpha }}}^2-{{{\tilde{b}}}^2\over 64}=0. \end{aligned}$$

(C6)

Applying the change of variable

$$\begin{aligned} {{\tilde{\alpha }}}^2={\tilde{U}}-{{\tilde{a}}\over 6}, \end{aligned}$$

(C7)

we obtain the depressed cubic equation

$$\begin{aligned} {\tilde{U}}^3-{{\tilde{\eta }}}_2 \,{\tilde{U}}-{{\tilde{\eta }}}_3=0, \end{aligned}$$

(C8)

where

$$\begin{aligned} {{\tilde{\eta }}}_2&={{\tilde{a}}^2\over 48}+{{\tilde{c}}\over 4}, \end{aligned}$$

(C9a)

$$\begin{aligned} {{\tilde{\eta }}}_3&={{\tilde{a}}^3\over 864}+{{\tilde{b}}^2\over 64}-{{\tilde{a}} {\tilde{c}}\over 24}. \end{aligned}$$

(C9b)

The real solution to this cubic equation is obtained as [79, 80]

$$\begin{aligned} {\tilde{U}}= 2\sqrt{{{{\tilde{\eta }}}_2\over 3}} \cosh \left( \frac{1}{3}\mathrm {arccosh}\left( {3\over 2}{{\tilde{\eta }}}_3\sqrt{{3\over {{\tilde{\eta }}}_2^3}} \right) \right) . \end{aligned}$$

(C10)

Therefore, the roots of Eq. (C2) are

$$\begin{aligned} x_1&={{\tilde{\alpha }}}+\sqrt{{{\tilde{\alpha }}}^2-{{\tilde{\beta }}}}, \end{aligned}$$

(C11a)

$$\begin{aligned} x_2&={{\tilde{\alpha }}}-\sqrt{{{\tilde{\alpha }}}^2-{{\tilde{\beta }}}}, \end{aligned}$$

(C11b)

$$\begin{aligned} x_3&=-{{\tilde{\alpha }}}+\sqrt{{{\tilde{\alpha }}}^2-{{\tilde{\gamma }}}}, \end{aligned}$$

(C11c)

$$\begin{aligned} x_4&=-{{\tilde{\alpha }}}-\sqrt{{{\tilde{\alpha }}}^2-{{\tilde{\gamma }}}}. \end{aligned}$$

(C11d)

Appendix D: Solving the equation of motion for frontal scattering

Equation (66) can be recast as

$$\begin{aligned} \left( \frac{\mathrm{d}r}{\mathrm{d}\tau }\right) ^{2}=\frac{m^2 \left( r-r_s\right) {\mathfrak {p}}_3(r)}{\lambda ^2 r^2}, \end{aligned}$$

(D1)

in which

$$\begin{aligned} {\mathfrak {p}}_3(r)\equiv r^3+r_s r^2+(r_s^2+{\bar{a}})r+r_s^3+r_s{\bar{a}}+{\bar{b}}. \end{aligned}$$

(D2)

Considering $r_s$ as the initial position, we can rewrite Eq. (D1) as

$$\begin{aligned} \tau (r)={\lambda \over m}\int _{r_s}^{r} {r \mathrm {d}r \over \sqrt{(r-r_s) {\mathfrak {p}}_3(r)}}, \end{aligned}$$

(D3)

which by the linear change of variable

$$\begin{aligned} z \doteq {r \over r_s}-1 \end{aligned}$$

(D4)

reduces to

$$\begin{aligned} \tau (z)={\lambda \over m}\int _{0}^{z} {(z+1) \mathrm {d}z \over \sqrt{z \tilde{{\mathfrak {p}}}_3(z)}}, \end{aligned}$$

(D5)

where

$$\begin{aligned} \tilde{{\mathfrak {p}}}_3(z)\equiv z^3 +4 z^2 +\gamma _1 z+\gamma _0, \end{aligned}$$

(D6)

and

$$\begin{aligned} \gamma _1&=6+\frac{{\bar{a}}}{r_s^2}, \end{aligned}$$

(D7a)

$$\begin{aligned} \gamma _0&=4+\frac{2{\bar{a}}}{r_s^2}+\frac{{\bar{b}}}{r_s^3}. \end{aligned}$$

(D7b)

Now, letting

$$\begin{aligned} u\doteq \frac{1}{z} \end{aligned}$$

(D8)

yields the following reduced integral form of Eq. (D5):

$$\begin{aligned} \tau (u)={-\lambda \over m\sqrt{\gamma _0}}\left( \int _{\infty }^{u} {\mathrm {d}u\over \sqrt{\bar{{\mathfrak {p}}}_3(u)}}+ \int _{\infty }^{u} {\mathrm {d}u \over u \sqrt{\bar{{\mathfrak {p}}}_3(u)}} \right) , \end{aligned}$$

(D9)

in which

$$\begin{aligned} \bar{{\mathfrak {p}}}_3(u)\equiv u^3 +\frac{\gamma _1}{\gamma _0} u^2 +\frac{4}{\gamma _0} u+\frac{1}{\gamma _0}. \end{aligned}$$

(D10)

Applying the last change of variable

$$\begin{aligned} u\doteq 4 \mathrm {U}-\frac{\gamma _1}{3\gamma _0}, \end{aligned}$$

(D11)

we get

$$\begin{aligned} \tau (\mathrm {U})={-\lambda \over m\sqrt{\gamma _0}}\left( \int _{\infty }^{\mathrm {U}} {\mathrm {d}\mathrm {U}\over \sqrt{\bar{{\mathfrak {P}}}_3(\mathrm {U})}}\right. \left. +{1\over 4} \int _{\infty }^{\mathrm {U}} {\mathrm {d}\mathrm {U} \over (\mathrm {U}- {\gamma _1\over 12 \gamma _0}) \sqrt{\bar{{\mathfrak {P}}}_3(\mathrm {U})}} \right) , \end{aligned}$$

(D12)

where we have defined

$$\begin{aligned} \bar{{\mathfrak {P}}}_3(\mathrm {U})\equiv 4\mathrm {U}^3 - {\bar{g}}_2 \mathrm {U}^2 -{\bar{g}}_3. \end{aligned}$$

(D13)

The direct integration of the elliptic integral in Eq. (D12) now results in the expression in Eq. (72).