# Collisional Penrose process with spinning particles in braneworld black hole

Yongbin Du  Department of Physics, South China University of Technology,
Guangzhou 510641, China  Yunlong Liu  Department of Physics, South China
University of Technology, Guangzhou 510641, China  Xiangdong
Zhang111Corresponding author<EMAIL_ADDRESS>Department of Physics,
South China University of Technology, Guangzhou 510641, China

###### Abstract

The Penrose process of an extremal braneworld black hole is studied. We
analyze the Penrose process by two massive spinning particles collide near the
horizon. By calculating the maximum energy extraction efficiency of this
process, it turns out that the maximal efficiency increases as the tilde
charge parameter $d$ of the braneworld blackhole decreases. Interestingly, for
the negative value of $d$, the efficiency can be even larger than the Kerr
case.

##  I INTRODUCTION

In the ergosphere of Kerr black hole, there exist some time-like orbits whose
energy can be strangely negative due to the Killing vectors in this region
being always spacelike. Penrose first conceived the concept of extracting
energy from these special orbitspenrose . His creative scenario alleged that
if an object plunging into the ergosphere of a rotating black hole and breaks
into two pieces and one of them falls into the event horizon with negative
energy, the other one can escape from the black hole, carrying more energy
than the original object. This mechanism is now well known as the Penrose
process. Although some early work showed Bardeen:1972fi ; Wald:1974kya  the
process may rarely happen in astrophysics. However, soon after that, the
scenario of collisional Penrose process was proposed collision  and was highly
expected at one time to acquire energy from Kerr black holes with greater
efficiency. Nevertheless, some researches such asPiran:1977dm  uncovered that
it may not get higher efficiency as expected, making this interesting topic
gradually recede from the physicists’ arena. It did not get its rebirth until
Banados, Silk and West $($BSW$)$ Patil:2011yb  find that the center-of-mass
energy can be arbitrarily high when the collision happens near the horizon and
the incident particles have tuned angular momentum. Therefore, the collisional
Penrose process based on the BSW mechanism has been constructed and found to
have much higher efficiency than the previous version Harada:2012ap ;
Bejger:2012yb ; Schnittman:2014zsa . These results were soon generalized to
the non-equatorial planeLeiderschneider:2015kwa . Moreover, the BSW effect
including spinning particles are discussed in Zaslavskii:2010jd ;
Deriglazov:2018vwa ; Zaslavskii:2016dfh ; Wei:2010vca ; Harada:2011xz ;
Kimura:2010qy ; Guo:2016vbt . Then, the collisional Penrose process has also
been generalized to the spinning particles caseMaeda:2018hfi ;
Okabayashi:2019wjs , It turns out that when the spinning particles were
considered, the efficiency of contracting energy from the blackhole was
greatly improved compared with the spinless case Maeda:2018hfi . In
particular, Maeda, Okabayashi and Okawa calculate spinning test particles
colliding near the horizon of the Kerr background and obtain their maximal
efficiency, which is defined by $($input energy$)/($output energy$)$, is
around 15.01 for massive particlesMaeda:2018hfi .

On the other hand, most of the previous studies on the Penrose process were
focus on four dimensions. However, whether our spacetime has extra dimensions
is a fundamental question in modern physics. In past one hundred years
different types of extra dimensional models were proposed, such as Kaluza-
Klein $($KK$)$ model kaluza ; klein  and string/M theory which assumes that
the extra dimension is needed but quite tiny(approach Planck scale) in order
to explain why it is not found in experiments. While braneworld modelmodel1
conceives that our visible universe is localized on a 3-brane in high
dimension space-time and the extra dimensions are large. Randall and Sundrum
developed double brane models based on Arkani-Hamed, Dimopoulos and Dvali’s
$($ADD$)$ theory Arkani-Hamed ; Dimopoulos ; Dvali  and tactfully solved the
hierarchy problemmodel1 . Later, they further developed their model in which
all the matter of our universe is localized on the single brane of original
point$(\phi=0)$ and the extra dimension is infinitemodel2 . Interestingly, the
blackhole solution in braneworld model is obtained by Naresh Dadhich $et$
$al.$ gives an effective solution of rotating black hole, which is the form of
Reissner-Nordstrom metric but the real charge is replaced by a new parameter
tidal charge $d$ as an imprint of extra dimensionDadhich:2000am . After their
vital work, Aliev and Gumrukcuoglu acquired stationary and axisymmetric
solutions describing charged rotating black holes on a 3-brane in the Randall-
Sundrum braneworldAliev:2005bi . The Kerr solution then can be viewed as the
limit of the tidal charge $d=0$. The collisional Penrose process of braneworld
Kerr black hole with spinless particles has been discussed inKhan:2019gco .
The present paper aims to study the collisional Penrose process with spinning
particles. The necessities for such extension are two folds, first, the
spinning massive particles are more realistic than the massive spinless
particles. Second, the spinning particles in the Penrose process usually have
much higher maximal energy contraction efficiency compared with the spinless
particles Okabayashi:2019wjs .

This paper is organized as follows. After an introduction, in section 2, we
give the equations of motion of spinning test particles in braneworld Kerr
black hole. In section 3, constraints on the orbit are obtained. Based on
these constraints we discuss the collisional process near the horizon in
section 4. The maximal efficiency of extracting energy in braneworld black
hole with different tidal charge $d$ is given in section 5 and the concluding
remarks are presented in section 6.

##  II EQUATIONS OF MOTION

In this section, we will give the differential equations of motion for
spinning test particles moving in braneworld Kerr spacetime.

###  II.1 Braneworld Kerr space-time

The metric of braneworld Kerr spacetime reads Dadhich:2000am :

$g_{\mu\nu}=\left(\begin{array}[]{cccc}-\frac{\Delta-a^{2}\sin^{2}\theta}{\Sigma}&0&0&-\frac{a\sin^{2}\theta\left(\left(a^{2}+r^{2}\right)-\Delta\right)}{\Sigma}\\\
0&\frac{\Sigma}{\Delta}&0&0\\\ 0&0&\Sigma&0\\\
-\frac{a\sin^{2}\theta\left(\left(a^{2}+r^{2}\right)-\Delta\right)}{\Sigma}&0&0&\frac{\sin^{2}\theta\left(\left(a^{2}+r^{2}\right)^{2}-a^{2}\Delta\sin^{2}\theta\right)}{\Sigma}\\\
\end{array}\right),$ (1)

where

$\displaystyle\Delta$ $\displaystyle=$ $\displaystyle a^{2}+d-2Mr+r^{2},$ (2)
$\displaystyle\Sigma$ $\displaystyle=$ $\displaystyle
a^{2}\cos^{2}\theta+r^{2},$ (3)

where $d$ is the tidal charge parameter which could be either positive or
negative. When this parameter $d=0$, the above metric reduces to the Kerr
metric. We should point out that this metric can be seen as the Kerr-Newman
solution if we replace the tidal charge parameter $d$ with $Q^{2}$. However,
$d$ is not always positive which, as we will show in this paper, plays an
important role in achieving higher efficiency of the collisional Penrose
process. To make sure the horizon exists, the following condition must be
satisfied

$\displaystyle M^{2}\geq a^{2}+d.$ (4)

The equal is taken only in the case that it is extreme braneworld black hole.
Besides, it is evidently that when $d$ is negative in the extreme braneworld
black hole, $a$ can be larger than $M$ to make sure the event horizon exists.
The radius of the outer horizon reads

$\displaystyle r_{H}=\sqrt{-a^{2}-d+M^{2}}+M.$ (5)

###  II.2 Equation of spinning test particles

Due to the existence of spin, the orbit of spinning particles is described by
MPD equations rather than geodesics Papapetrou1951Spinning ;
Dixon1970Dynamics1 ; Dixon1970Dynamics2

$\displaystyle\frac{Dp^{\mu}}{d\tau}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}R^{\mu}_{\nu\rho\sigma}v^{\nu}S^{\rho\sigma},$ (6)
$\displaystyle\frac{DS^{\mu\tau}}{d\tau}$ $\displaystyle=$ $\displaystyle
p^{\mu}v^{\nu}-p^{\nu}v^{\mu},$ (7)

where $p^{\mu}$, $v^{\mu}=dz^{\mu}/d\tau$ and $S^{\mu\nu}$ are the 4-momentum,
4-velocity and the spin tensor respectively. A set of supplementary conditions
are also needed Wald1972Gravitational

$\displaystyle S^{\mu\nu}p_{\nu}=0.$ (8)

Moreover, the mass of particles is defined by

$\mu^{2}=-p^{\mu}p_{\mu}.$ (9)

We use a specific 4-momentum as well, which reads

$u^{\mu}=\frac{p^{\mu}}{\mu}.$ (10)

We should also normalize the affine parameter as

$u^{\mu}v_{\mu}=-1.$ (11)

Hence we could get the difference between 4-momentum and 4-velocity as follows
Motoyuki_1998

$\displaystyle
v^{\mu}-u^{\mu}=\frac{S^{\mu\nu}R_{\nu\rho\lambda}u^{\rho}S^{\sigma\lambda}}{2\mu^{2}+\frac{1}{2}R_{\alpha\beta\gamma\delta}S^{\alpha\beta}S^{\gamma\delta}}.$
(12)

###  II.3 Conservation quantities

As we have said in the preceding text, the world line of spinning particles is
not geodesics anymore, causing the necessary modification of the definition of
conserved quantities Hojman_1978

$Q_{\xi}=p^{\mu}\xi_{\mu}+\frac{1}{2}S^{\mu\nu}\nabla_{\mu}\xi_{\nu}.$ (13)

In braneworld Kerr space-time, we invariably have two Killing vectors. By
introducing a tetrad basis as

$\displaystyle
e^{(0)}_{a}=\frac{\sqrt{\Delta}}{\sqrt{\Sigma}}dt_{a}-\frac{\sqrt{\Delta}\left(a\sin^{2}(\theta)\right)}{\sqrt{\Sigma}}d\varphi_{a},$
(15) $\displaystyle e^{(1)}_{a}=\frac{\sqrt{\Sigma}}{\sqrt{\Delta}}dr_{a},$
$\displaystyle e^{(2)}_{a}=\sqrt{\Sigma}d\theta_{a},$ $\displaystyle
e^{(3)}_{a}=-\frac{a\sin(\theta)}{\sqrt{\Sigma}}dt_{a}+\frac{\left(a^{2}+r^{2}\right)\sin(\theta)}{\sqrt{\Sigma}}d\varphi_{a},$
(16)

the Killing vectors can be written as

$\displaystyle\xi^{(t)}_{\mu}$ $\displaystyle=$
$\displaystyle-\sqrt{\frac{\Delta}{\Sigma}}e^{(0)}_{\mu}-\frac{a\sin{\theta}}{\sqrt{\Sigma}}e^{(3)}_{\mu},$
(17) $\displaystyle\xi^{(\varphi)}_{\mu}$ $\displaystyle=$ $\displaystyle
a\sqrt{\frac{\Delta}{\Sigma}}\sin^{2}{\theta}e^{(0)}_{\mu}+\frac{(r^{2}+a^{2})\sin{\theta}}{\sqrt{\Sigma}}e^{(3)}_{\mu}.$
(18)

From it, we can easily acquire the energy $E$ and the $z$ component of the
angular momentum $J$ in the equatorial plane $(\theta=\pi/2)$

$\displaystyle E$ $\displaystyle=$ $\displaystyle
p^{(0)}\frac{\sqrt{\Delta}}{r}+p^{(3)}\frac{a}{r}+S^{(0)(1)}\frac{(d-Mr)}{r^{3}},$
(19) $\displaystyle J$ $\displaystyle=$ $\displaystyle
p^{(0)}\frac{a\sqrt{\Delta}}{r}+p^{(3)}\frac{\left(a^{2}r^{2}+r^{4}\right)}{r^{3}}$
(20)
$\displaystyle+S^{(1)(3)}\frac{\sqrt{\Delta}}{r}+S^{(0)(1)}\frac{a(d-r(M+r))}{r^{3}}.$

###  II.4 Equations of motion in the equatorial plane

In this paper, we merely discuss the orbit in the equatorial plane, which
means $\theta=\pi/2$ and we introduce a spin vector Maeda:2018hfi

$s^{(a)}=-\frac{1}{2\mu}\epsilon^{(a)}_{(b)(c)(d)}u^{(b)}S^{(c)(d)},$ (21)

where $\epsilon$ is the antisymmetric tensor with $\epsilon_{(0)(1)(2)(3)}=1$.
We also find all the components of the spin vector is trivial except

$s^{(2)}=-s.$ (22)

Therefore it can be obtained that

$S^{(\mu)(\nu)}=\left(\begin{array}[]{cccc}0&-p^{(3)}s&0&p^{(1)}s\\\
p^{(3)}s&0&0&p^{(0)}s\\\ 0&0&0&0\\\ -p^{(1)}s&-p^{(0)}s&0&0\\\
\end{array}\right).$ (23)

Thus the conserved quantities can be written as

$\displaystyle E$ $\displaystyle=$
$\displaystyle\frac{p^{(0)}r^{2}\sqrt{\Delta}+p^{(3)}(ar^{2}-ds+Mrs)}{r^{3}},$
(24) $\displaystyle J$ $\displaystyle=$
$\displaystyle\frac{p^{(0)}(r^{2}s\sqrt{\Delta}+ar^{2}\sqrt{\Delta})}{r^{3}}$
(25) $\displaystyle+\frac{p^{(3)}(a^{2}r^{2}+as(r(M+r)-d)+r^{4})}{r^{3}}.$

By solving these equations we obtain

$\displaystyle u^{(0)}$ $\displaystyle=$
$\displaystyle\frac{E\left(-adrs+aMr^{2}s+ar^{3}(a+s)+r^{5}\right)-J\left(r^{2}(ar+Ms)-drs\right)}{\mu\sqrt{a^{2}+d+r(r-2M)}\left(ds^{2}-Mrs^{2}+r^{4}\right)},$
(26) $\displaystyle u^{(3)}$ $\displaystyle=$
$\displaystyle\frac{r^{3}(J-E(a+s))}{\mu\left(ds^{2}-Mrs^{2}+r^{4}\right)}.$
(27)

For the reason of normalization condition

$-(u^{(0)})^{2}+(u^{(1)})^{2}+(u^{(3)})^{2}=-1,$ (28)

we have

$u^{(1)}=\sigma\sqrt{(u^{(0)})^{2}-(u^{(3)})^{2}-1},$ (29)

where $\sigma=\pm 1$. The value is positive when the particle is outgoing and
negtive when it is ingoing respectively. It is obtained hereinabove that

$\displaystyle v^{(0)}$ $\displaystyle=$
$\displaystyle\frac{\lambda_{2}u^{(0)}}{\lambda_{1}},$ (30) $\displaystyle
v^{(1)}$ $\displaystyle=$
$\displaystyle\frac{\lambda_{2}u^{(1)}}{\lambda_{1}},$ (31) $\displaystyle
v^{(3)}$ $\displaystyle=$
$\displaystyle\frac{u^{(3)}(\lambda_{2}+\lambda_{3})}{\lambda_{1}},$ (32)

where

$\displaystyle\lambda_{1}$ $\displaystyle=$ $\displaystyle
ds^{2}\left(4(u^{(3)})^{2}+1\right)-Mrs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4},$
(33) $\displaystyle\lambda_{2}$ $\displaystyle=$ $\displaystyle
ds^{2}-Mrs^{2}+r^{4},$ (34) $\displaystyle\lambda_{3}$ $\displaystyle=$
$\displaystyle 3Mrs^{2}-4ds^{2}.$ (35)

By using the transformation of coordinates Eqs.(15)-(16), we finally get the
differential equations

$\displaystyle\frac{\text{d}t}{\text{d$\tau$}}$ $\displaystyle=$
$\displaystyle\frac{u^{(0)}\left(a^{2}\sqrt{\Delta}\left(ds^{2}-Mrs^{2}+r^{4}\right)+r^{2}\sqrt{\Delta}\left(ds^{2}-Mrs^{2}+r^{4}\right)\right)}{r\Delta\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-Mrs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)}$
(36)
$\displaystyle+\frac{u^{(3)}\left(a^{3}\left(-3ds^{2}+2Mrs^{2}+r^{4}\right)-ar(2M-r)\left(-3ds^{2}+2Mrs^{2}+r^{4}\right)\right)}{r\Delta\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-Mrs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)},$
$\displaystyle\frac{\text{d}r}{\text{d$\tau$}}$ $\displaystyle=$
$\displaystyle\frac{u^{(1)}\sqrt{\Delta}\left(ds^{2}-Mrs^{2}+r^{4}\right)}{r\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-Mrs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)},$
(37) $\displaystyle\frac{\text{d$\varphi$}}{\text{d$\tau$}}$ $\displaystyle=$
$\displaystyle\frac{au^{(0)}\sqrt{\Delta}\left(ds^{2}-Mrs^{2}+r^{4}\right)}{r\Delta\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-Mrs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)}$
(38)
$\displaystyle+\frac{u^{(3)}\left(a^{2}\left(-3ds^{2}+2Mrs^{2}+r^{4}\right)-r(2M-r)\left(-3ds^{2}+2Mrs^{2}+r^{4}\right)\right)}{r\Delta\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-Mrs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)}.$

In what follows, for convenience of calculation, we rescale the variables to
the dimensionless variables as

$\displaystyle\bar{E}=\frac{E}{\mu},\bar{J}=\frac{J}{\mu
M},\bar{s}=\frac{s}{\mu M},\bar{t}=\frac{t}{\mu
M},\quad\bar{r}=\frac{r}{M},\quad\bar{a}=\frac{a}{\mu
M},\quad\bar{\tau}=\frac{\tau}{\mu M}.$

For simplicity, we omit the sign $"-"$ in the following discussion. Now the
differential equations can be rewritten as

$\displaystyle\frac{\text{d}t}{\text{d$\tau$}}$ $\displaystyle=$
$\displaystyle\frac{u^{(0)}\left(a^{2}\sqrt{\Delta}\left(ds^{2}-rs^{2}+r^{4}\right)+r^{2}\sqrt{\Delta}\left(ds^{2}-rs^{2}+r^{4}\right)\right)}{r\Delta\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-rs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)}$
(39)
$\displaystyle+\frac{u^{(3)}\left(a^{3}\left(-3ds^{2}+2rs^{2}+r^{4}\right)-ar(2-r)\left(-3ds^{2}+2rs^{2}+r^{4}\right)\right)}{r\Delta\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-rs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)},$
$\displaystyle\frac{\text{d}r}{\text{d$\tau$}}$ $\displaystyle=$
$\displaystyle\frac{u^{(1)}\sqrt{\Delta}\left(ds^{2}-rs^{2}+r^{4}\right)}{r\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-rs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)},$
(40) $\displaystyle\frac{\text{d$\varphi$}}{\text{d$\tau$}}$ $\displaystyle=$
$\displaystyle\frac{au^{(0)}\sqrt{\Delta}\left(ds^{2}-rs^{2}+r^{4}\right)}{r\Delta\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-rs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)}$
(41)
$\displaystyle+\frac{u^{(3)}\left(a^{2}\left(-3ds^{2}+2rs^{2}+r^{4}\right)-r(2-r)\left(-3ds^{2}+2rs^{2}+r^{4}\right)\right)}{r\Delta\left(ds^{2}\left(4(u^{(3)})^{2}+1\right)-rs^{2}\left(3(u^{(3)})^{2}+1\right)+r^{4}\right)}.$

##  III CONSTRAINS OF THE ORBIT

###  III.1 Different types of orbits

Given we are exploring the collisional Penrose process where certain particles
should escape with more energy after the collision in the ergosphere, it is
essential to find the radial turning point, which means
$\text{d}r/\text{d}t=0$. The differential equation with dimensionless
variables makes it clear that the radial velocity of the particle depends on
$E,J,s,r,a$ and $d$. Thus by taking distinct value of the turning point
$r_{0}$, we could classify different types of orbits and obtain the relations
between these quantities. By introducing the impact parameter $b=J/E$. When
$\text{d}r/\text{d}t=0$, the radial function can be reduced to

$\displaystyle\left[E\left(-\frac{ads}{r}+ar(a+s)+as+r^{3}\right)-bE\left(ar-\frac{ds}{r}+s\right)\right]^{2}$
$\displaystyle-\Delta r^{2}(bE-E(a+s))^{2}+\Delta\left(-P^{2}\right)r^{4}=0,$
(42)

where

$P=\frac{ds^{2}}{r^{4}}-\frac{s^{2}}{r^{3}}+1.$ (43)

We focus on the orbits whose turning points are exactly on the horizon $r_{H}$
which corresponding to the highest energy contraction efficiency Maeda:2018hfi
; r6 . These orbits are usually called critical orbits Maeda:2018hfi ; r6 .
Then the critical value of impact parameter $b=\frac{J}{E}$ reads

$\displaystyle b_{cr}$ $\displaystyle=$
$\displaystyle\frac{as\left(\left(\sqrt{-a^{2}-d+1}+1\right)^{2}+\sqrt{-a^{2}-d+1}-d+1\right)}{s\left(\sqrt{-a^{2}-d+1}-d+1\right)+a\left(\sqrt{-a^{2}-d+1}+1\right)^{2}}$
(44)
$\displaystyle+\frac{\left(\sqrt{-a^{2}-d+1}+1\right)^{4}+a^{2}\left(\sqrt{-a^{2}-d+1}+1\right)^{2}}{s\left(\sqrt{-a^{2}-d+1}-d+1\right)+a\left(\sqrt{-a^{2}-d+1}+1\right)^{2}}.$

In order to reach the horizon, the particles must have $b<b_{cr}$. Moreover,
we can also see that if $d=0$, $b_{cr}$ comes back to the form in Kerr black
hole Maeda:2018hfi . In extreme braneworld black hole case, the above relation
can be further reduced to

$b_{cr}=\frac{a^{2}+1}{a}=\frac{2-d}{\sqrt{1-d}},$ (45)

which does not depend on $s$. The plot of $b_{cr}$ depends on parameter $d$ in
extreme braneworld black hole is showed in Fig. 1. We found that when $d$ is
negative, we can see that the rate of change is not very large. While when
$d=0$, $b_{cr}$ approaches the minimum value$(b_{cr}=2)$.

Figure 1:  The trend of $b_{cr}$ with parameter $d$ in extreme braneworld Kerr
black hole.

If the orbit is near critical$(b\approx b_{cr})$ but $b>b_{cr}$, the particle
will bounce off near the horizon. In this paper, we consider the positive $J$
for it gives higher efficiency Maeda:2018hfi . Moreover, we can also find that
for non-critical orbit, the particle has no turning point and would plunge
into the horizon Maeda:2018hfi ; r6 .

###  III.2 Time-like condition of the orbits

To make sure the orbit exists near the horizon, the time-like condition should
be imposed

$-(v_{0})^{2}+(v_{1})^{2}+(v_{3})^{2}<0.$ (46)

Follow the same line of Maeda:2018hfi , the above timelike condition can be
reduced to

$s^{2}(aE+Es-J)^{2}<G,$ (47)

where

$G=\frac{\left(ds^{2}+r^{4}-rs^{2}\right)^{4}}{r^{6}\left(8d^{2}s^{2}-2d\left(4r^{4}+5rs^{2}\right)+3r^{2}\left(2r^{3}+s^{2}\right)\right)}.$
(48)

The right of the inequality is a function of $r,d$ and $s$, and we show the
image of the function $G$ in Fig. 2.

Figure 2:  The image of the function $G$, where $d=-0.28$ is chosen.

In the following, we focus on the extreme braneworld black hole corresponding
to the highest energy extraction efficiency. In this case, we have
$d=1-a^{2}$. Firstly, we shall take a view of the time-like constraints on the
critical orbits$(b=b_{cr})$ under this background. We have

$E^{2}<\frac{\left(\left(a^{2}-1\right)s^{2}-r^{4}+rs^{2}\right)^{4}}{r^{6}s^{2}\left(4a^{2}+3r-4\right)(a-b+s)^{2}\left(2\left(a^{2}-1\right)s^{2}+2r^{4}+rs^{2}\right)}.$
(49)

If the particle comes from infinity, $E\geqslant 1$, we could get allowed
region of spin $s$ which shown in Fig. 3.

Figure 3: Allowed region of spin $s$ when $d=-0.24$$($ red $)$ and $d=0.08$$($
gray $)$.

If the orbit is non-critical, the time-like condition Eq.(47) gives
constraints on $b$

$a-\frac{Y}{E}+s<b<a+\frac{Y}{E}+s,$ (50)

where

$Y=\frac{\left(ds^{2}+r^{4}-rs^{2}\right)^{2}}{r^{3}\sqrt{s^{2}\left(8d^{2}s^{2}-8dr^{4}-10drs^{2}+6r^{5}+3r^{2}s^{2}\right)}}.$
(51)

We set $b=b_{cr}(1+\zeta)$ with $\zeta<0$, then we have

$\frac{a-b_{cr}-\frac{Y}{E}+s}{b_{cr}}<\zeta<\frac{a-b_{cr}+\frac{Y}{E}+s}{b_{cr}},$
(52)

which gives the constraints on the impact parameter.

##  IV COLLISION OF MASSIVE SPINNING PAITICLES

Now we discuss the collision of massive spinning particles in the extreme
braneworld black hole in order to get higher efficiency. The mass of the two
colliding particles are chosen to be identical for simplicity and they are
both from infinity. We use 1 and 2 to label them respectively. After their
plastic collision, particle 3, whose 4-momenta is $p_{3}^{\mu}$, will move to
infinity, while the particle 4 with $p_{4}^{\mu}$ is going to fall into the
black hole Maeda:2018hfi ; r6 . By assuming

$\displaystyle p^{\mu}_{1}+p^{\mu}_{2}=p^{\mu}_{3}+p^{\mu}_{4},$ (53)
$\displaystyle S^{\mu\nu}_{1}+S^{\mu\nu}_{2}=S^{\mu\nu}_{3}+S^{\mu\nu}_{4}.$
(54)

which means the total momentum and spins are invariable, and the total energy
and angular momentum are conserved during the collision,

$\displaystyle E_{1}+E_{2}=E_{3}+E_{4},$ (55) $\displaystyle
J_{1}+J_{2}=J_{3}+J_{4}.$ (56)

To make it easier to discuss the quantities of the orbits at the plastic
collision, we simplify the conservative relations as follow

$\displaystyle E_{1}+E_{2}=E_{3}+E_{4},$ (57) $\displaystyle
J_{1}+J_{2}=J_{3}+J_{4},$ (58) $\displaystyle s_{1}+s_{2}=s_{3}+s_{4},$ (59)
$\displaystyle u^{(1)}_{1}+u^{(1)}_{2}=u^{(1)}_{3}+u^{(1)}_{4}.$ (60)

We also assume the collision point is quite close to the
horizon$(r_{c}=\frac{1}{(1-\epsilon)},0<\epsilon<<1)$, as is expected to
acquire higher energy extraction efficiency. Thus the radial velocity can be
expanded as

$u^{(1)}=\sigma\frac{E+a^{2}E-aJ}{\left|as-1\right|\epsilon}+O(\epsilon^{0}).$
(61)

Given the conservation equation of radial velocity (60), we have

$\displaystyle\sigma_{1}\frac{E_{1}+a^{2}E_{1}-aJ_{1}}{\left|as_{1}-1\right|\epsilon}+\sigma_{2}\frac{E_{2}+a^{2}E_{2}-aJ_{2}}{\left|as_{2}-1\right|\epsilon}$
$\displaystyle=$
$\displaystyle\sigma_{3}\frac{E_{3}+a^{2}E_{3}-aJ_{3}}{\left|as_{3}-1\right|\epsilon}+\sigma_{4}\frac{E_{4}+a^{2}E_{4}-aJ_{4}}{\left|as_{4}-1\right|\epsilon}+O(\epsilon).$
(62)

As we have discussed in the above section, we could classify different orbits
by the value of $b$. If $b=b_{cr}$, the orbit is critical which means the
turning point of radial velocity is just on the horizon. We also have near
critical orbit whose $b=b_{cr}+O(\epsilon)$ and non-critical orbit whose
$b=b_{cr}+O(\epsilon^{0})$. In the following analysis, without loss of
generality, along the same line of Maeda:2018hfi ; r6 , the calculation is
carried out in case that particle 1 is critical ($J_{1}=b_{cr}E_{1}$), while
particle 3 is near-critical($J_{3}=b_{cr}E_{3}+O(\epsilon)$) and particle 2 is
non-critical($J_{2}<b_{cr}E_{2}$) Maeda:2018hfi ; r6 . Note that in our case
$b_{cr}=\frac{a^{2}+1}{a}$, then we have

$\displaystyle E_{1}+a^{2}E_{1}-aJ_{1}=0,$ (63)
$\displaystyle(\sigma_{4}\frac{1}{1-as_{4}}-\sigma_{2}\frac{1}{1-as_{2}})(aJ_{2}-a^{2}E_{2}-E_{2})=O(\epsilon).$
(64)

Due to $J_{2}<b_{cr}E_{2}$, the particle 2 must be ingoing before the
collision, which means $\sigma_{2}=-1$. We assume that the particle 4 is
ingoing as well. Thus we should set $s_{1}=s_{3}$. To facility the
calculations, we introduce parameters $\zeta<0,\alpha_{3}>0$ and $\beta_{3}$
as

$\displaystyle J_{2}$ $\displaystyle=$ $\displaystyle b_{cr}E_{2}(1+\zeta),$
(65) $\displaystyle J_{3}$ $\displaystyle=$ $\displaystyle
b_{cr}E_{3}\left(1+\alpha_{3}\epsilon+\beta_{3}\epsilon^{2}+O(\epsilon^{3})\right),$
(66)

where $\alpha_{3},\beta_{3}$ are the parameters of $O(\epsilon^{0})$. Next we
shall calculate the expression of the energy of particle 3 and 2 through the
conservative relations. Firstly, we expand the radial velocity of the four
particles:

$\displaystyle u^{(1)}_{1}$ $\displaystyle=$
$\displaystyle\frac{\sigma_{1}f(E_{1},s_{1},a,0)}{a\left(1-a^{2}s_{1}^{2}\right)}-\frac{E_{1}^{2}\sigma_{1}\epsilon
h(s_{1},a)}{a\left(1-a^{2}s_{1}^{2}\right)^{2}f(E_{1},s_{1},a,0)}+O(\epsilon^{2}),$
(67) $\displaystyle u^{(1)}_{2}$ $\displaystyle=$
$\displaystyle-\frac{E_{2}\left(2a^{2}s_{2}+2a-s_{2}\right)\left(a^{2}\zeta-
as_{2}+\zeta+1\right)}{a(as_{2}-1)^{2}(as_{2}+1)}+\frac{\left(a^{2}+1\right)E_{2}\zeta}{\epsilon(1-as_{2})}+\epsilon\Pi+O(\epsilon^{2}),$
(68) $\displaystyle u^{(1)}_{3}$ $\displaystyle=$
$\displaystyle\frac{\sigma_{3}f(E_{3},s_{1},a,\alpha_{3})}{a\left(1-a^{2}s_{1}^{2}\right)}-\frac{E_{3}^{2}\sigma_{3}\epsilon(\beta_{3}g_{1}(s_{1},a,\alpha_{3})-g_{2}(s_{1},a,\alpha_{3})+h(s_{1},a))}{a\left(a^{2}s_{1}^{2}-1\right)^{2}f(E_{3},s_{1},a,\alpha_{3})}+O(\epsilon^{2}),$
$\displaystyle u^{(1)}_{4}$ $\displaystyle=$
$\displaystyle\frac{\left(2a^{3}s_{2}^{2}-a\left(s_{2}^{2}+2\right)+s_{2}\right)(E_{1}+E_{2}-E_{3})-\left(a^{2}+1\right)E_{2}\zeta(2a(as_{2}+1)-s_{2})}{a(as_{2}-1)^{2}(as_{2}+1)}$
(69)
$\displaystyle+\frac{\left(a^{2}+1\right)E_{2}\zeta}{\epsilon(1-as_{2})}+\frac{\left(a^{3}+a\right)\alpha_{3}E_{3}\left(a^{2}s_{2}^{2}-1\right)}{a(as_{2}-1)^{2}(as_{2}+1)}+\epsilon\Omega+O(\epsilon^{2}),$

where $\Omega$ and $\Pi$ are functions relevant to $E_{3},E_{2},s_{1},s_{2},b$
and $a$, which can be found in the appendix, and

$\displaystyle f(E,\text{s},a,\alpha_{3})$ $\displaystyle=$
$\displaystyle\sqrt{E^{2}\left(\left(a^{3}+a\right)\alpha_{3}(as+1)-a(2as+s+2)+s+1\right)}$
(70)
$\displaystyle\times\sqrt{\left(a\left(\left(a^{2}+1\right)\alpha_{3}(as+1)-2as+s-2\right)+s-1\right)-a^{2}\left(a^{2}s^{2}-1\right)^{2}},$
$\displaystyle h(\text{s},a)$ $\displaystyle=$
$\displaystyle\left(6a^{2}\left(a^{2}-1\right)^{2}-1\right)s^{4}+a\left(12a^{2}-5\right)s+2a^{2}+a\left(20a^{4}-11a^{2}+2\right)s^{3}$
(71) $\displaystyle+\left(24a^{4}-19a^{2}+4\right)s^{2}-1,$ $\displaystyle
g_{1}(\text{s},a,\alpha_{3})$ $\displaystyle=$ $\displaystyle
a\left(a^{2}+1\right)(as-1)(as+1)^{2}\left(a\left(a^{2}\alpha_{3}+\alpha_{3}-2\right)(as+1)+s\right),$
(72) $\displaystyle g_{2}(\text{s},a,\alpha_{3})$ $\displaystyle=$
$\displaystyle\left(a^{2}+1\right)\alpha_{3}(as_{1}+1)^{2}\left(3a^{4}s_{1}^{2}+8a^{3}s_{1}+a^{2}\left(5-3s_{1}^{2}\right)-4as_{1}+s_{1}^{2}-1\right)$
(73)
$\displaystyle-a\left(a^{2}+1\right)^{2}\alpha_{3}^{2}(as_{1}+1)^{2}(2a(as_{1}+1)-s_{1}).$

The conservation equation demands

$u^{(1)}_{1}+u^{(1)}_{2}=u^{(1)}_{3}+u^{(1)}_{4}.$ (74)

For $\zeta$ is negative, we expand Eq. (74) order by order, and found that the
identity relation is always satisfied for the order of $\epsilon^{-1}$. Then
we consider the order of $\epsilon^{0}$ and get

$\frac{\sigma_{1}f(E_{1},s_{1},a,0)}{a\left(1-a^{2}s_{1}^{2}\right)}-\frac{E_{3}g_{1}(s_{2},a,\alpha_{3})}{a\left(1-a^{2}s_{2}^{2}\right)}-\frac{\sigma_{3}f(s_{1},E_{3},\alpha_{3})}{a\left(1-a^{2}s_{1}^{2}\right)}+\frac{E_{1}\left(2a^{2}s_{2}+2a-s_{2}\right)}{a\left(1-a^{2}s_{2}^{2}\right)}=0.$
(75)

It can be seen as a quadratic equation of $E_{3}$ as

$A_{m}E_{3}^{2}-2B_{m}E_{3}+C_{m}=0,$ (76)

where the detailed expression $A_{m},B_{m}$ and $C_{m}$ can be found in the
appendix.

The conservation equation of the order of $\epsilon^{0}$ can also be written
as

$\displaystyle
E_{3}+\frac{\sigma_{3}f(E_{3},s_{1},a,\alpha_{3})a(1-a^{2}s_{2}^{2})}{a(1-a^{2}s_{1}^{2})g_{1}(s_{2},a,\alpha_{3})}$
$\displaystyle=$
$\displaystyle\frac{1}{g_{1}(s_{2},a,\alpha_{3})}\left[E_{1}\left(2a^{2}s_{2}+2a-s_{2}\right)+\frac{\sigma_{1}f(E_{1},s_{1},a,0)a(1-a^{2}s_{2}^{2})}{a\left(1-a^{2}s_{1}^{2}\right)}\right].$
(77)

The right side of the equation can be considered as a constant irrelevant to
$E_{3}$. Hence the range of $E_{3}$ is affected by $\sigma_{3}$. When
$\sigma_{3}$ is negative, $E_{3}$ must be larger than the value of the right
side. The larger solution of $E_{3}$ gives larger efficiency because the
efficiency depends on the value of $E_{3}$ Maeda:2018hfi . Therefore, it is
sufficient to consider the case of the negative value of $\sigma_{3}$.

We can solve the equation (76) and only retain the larger one

$E_{3}=\frac{\sqrt{B_{m}^{2}-A_{m}C_{m}}+B_{m}}{A_{m}}.$ (78)

While $E_{3}$ is fixed, we can also get the value of $E_{2}$ for the given
$E_{1},a,s_{1},s_{2},\alpha_{3}$ and $\beta_{3}$ through the conservation
equation of radial momentum in the order of $\epsilon^{1}$. The efficiency of
energy contraction is defined by

$\eta=\frac{E_{3}}{E_{1}+E_{2}}.$ (79)

Thus we need only care about whether $E_{2}$ can be taken the smallest value
of 1. That is what we are going to analyze in the next section.

##  V THE MAXIMAL EFFICIENCY

The solution to Eq.(76) already gives the value of $E_{3}$, which depends on
the parameter $E_{1},a,s_{1},s_{2}$ and $\alpha_{3}$. We normalize $E_{1}=1$
and $\sigma_{1}=1$, instead of $\sigma_{1}=-1$ which can not get higher
efficiency Leiderschneider:2015kwa . Hence, if we know the value of $d$ (or
$a)$, we can find a maximal value of $E_{3}$ in terms of $s_{1},s_{2}$ and
$\alpha_{3}$ under the time-like condition.

We plot the contour map of $E_{3}$ in Fig. 4 for distinct $d$ and we find that
$E_{3}$ can always achieves the maximal value in the condition $\alpha_{3}=0$.

$($a$)$ $d=0.08$

$($b$)$ $d=-0.12$

$($c$)$ $d=-0.24$

Figure 4:  From the contour map of $E_{3}$ we find for distinct $d$.

We also plot the contour map of $E_{3}$ as the function of $s_{1}$ and $s_{2}$
under the time-like condition in Fig. 5 for different values of $d$.

$($a$)$ $d=0.12$

$($b$)$ $d=-0.04$

$($c$)$ $d=-0.20$

Figure 5: The point with maximal $E_{3}$ as the function of $s_{1}$ and
$s_{2}$ under the time-like condition is marked in red in the contour map.

For each $d$, we could find a maximal value of $E_{3}$. Note that $E_{2}\geq
1$ when the particle 2 falling from infinity, if $E_{2}=1$ is possible, then
the maximal value of $E_{3}$ gives the maximal efficiency. Note that $E_{3}$
is decoupled with parameters $\beta_{3}$ and $\xi$. So our target is
simplified to sure $E_{2}=1$ is possible for some admissible values of
$\beta_{3}$ and $\xi$. In what follows we should focus on whether $E_{2}$ can
be 1 when $E_{3}$ is taken the maximal value. Note that Eq.(52) gives the
constraints on $\zeta$, then we set $E_{2}=1$ and observe whether there exists
some value of $\beta_{3}$ in the given range of $\zeta$. We show the results
in Fig. 6 when $d=-0.24$. The result for other values of $d$ is similar, and
we found that $E_{2}=1$ can always be achieved.

Figure 6: The relationship between $\beta_{3}$ and $\zeta$ for $d=-0.24$ when
$E_{3}$ is fixed $($the maximal value$)$ under timelike condition. It reflects
that $E_{2}=1$ can be satisfied in this case.

Therefore we list the maximal efficiency for different $d$ in Table 1. In
particular, when $d=0$, the spacetime metric reduces to the Kerr black hole,
the maximal efficiency also goes back to 15.01 which is completely the same as
inMaeda:2018hfi . Moreover, for the negative value of $d$, the efficiency
$\eta$ is larger than the Kerr case. The maximal efficiency $\eta$ as a
function of $d$ is plotted in Fig. 7. The result shows that $\eta$ increases
as the parameter $d$ decreases.

$d$ | $a$ | $s_{1}$ | $s_{2}$ | $E_{3}$ | $E_{2}$ | $\eta$
---|---|---|---|---|---|---
-0.24 | 1.11355 | 0.0019225 | -0.254649 | 206.255 | 1 | 103.127
-0.20 | 1.09545 | 0.00351892 | 0.256994 | 113.359 | 1 | 56.6796
-0.12 | 1.0583 | 0.00710832 | -0.262068 | 56.8549 | 1 | 28.4274
0.00 | 1 | 0.0137838 | -0.268102 | 30.0203 | 1 | 15.0101
0.12 | 0.938083 | 0.0224733 | -0.242029 | 18.976 | 1 | 9.48799
0.24 | 0.87178 | 0.0341456 | -0.202261 | 13.0032 | 1 | 6.50159
Table 1: The maximal efficiency and its corresponding value of $s_{1},s_{2}$
and $E_{2}$ for different $d$ Figure 7: The maximal efficiency of $\eta$ as a
function of $d$.

##  VI CONCLUDING REMARK

In this paper, we study the collision of two massive spinning particles around
an extreme braneworld black hole and calculate the corresponding maximal
efficiency of the energy extraction from the black hole which is usually
referred to as the collisional Penrose process. Compared with Kerr spacetime,
the braneworld background possesses an extra tilde charge parameter $d$. It
reduces to the Kerr black hole when the parameter $d=0$. We consider the
particles freely falling from infinity and collide near the horizon of the
braneworld black hole. After a collision, one particle escapes to infinity
with larger energy than the incident particles.

By employing the MPD equations which describe the motion of the spinning
particles in the braneworld spacetime, the momentum and the velocity of the
particles have been obtained explicitly. Moreover, by imposing timelike
constraint which comes from the causality requirement, we found a restriction
on the angular momentum of the particles. Then, with the help of the
conservation of energy and momentum during the collision, we calculate the
corresponding maximum energy extraction efficiency $\eta_{max}$ of the
collisional Penrose process, and the detailed calculation shows the
$\eta_{max}$ does depend on the tilde charge parameter $d$ of the braneworld
spacetime. When $d=0$, the corresponding maximum energy extraction efficiency
$\eta_{max}=15.01$ goes back to the Kerr case Maeda:2018hfi . This of course
can be viewed as a consistency check. Moreover, for the negative value of $d$,
the efficiency $\eta$ can be larger than the Kerr case. The result shows that
$\eta_{max}$ increases as the parameter $d$ decreases.

##  VII ACKNOWLEDGEMENT

This work is supported by NSFC with No.11775082. Y. Du would like to thank
Tianyi Li for helpful discussions.

##  Appendix A appendix

$\displaystyle A_{m}$ $\displaystyle=$
$\displaystyle\frac{g_{1}(s_{2},a,\alpha_{3})^{2}}{a^{2}\left(a^{2}s_{2}^{2}-1\right)^{2}}+\frac{-4a^{4}s_{1}^{2}-8a^{3}s_{1}+5a^{2}s_{1}^{2}-\left(a^{2}+1\right)^{2}a^{2}\alpha_{3}^{2}(as_{1}+1)^{2}}{a^{2}\left(1-a^{2}s_{1}^{2}\right)^{2}}$
(80)
$\displaystyle+\frac{2\left(a^{2}+1\right)a\alpha_{3}(as_{1}+1)(2a(as_{1}+1)-s_{1})-4a^{2}+2as_{1}-s_{1}^{2}+1}{a^{2}\left(1-a^{2}s_{1}^{2}\right)^{2}},$
$\displaystyle B_{m}$ $\displaystyle=$
$\displaystyle\frac{g_{1}(s_{2},a,\alpha_{3})E_{1}\left(2a^{2}s_{2}+2a-s_{2}\right)}{a^{2}\left(1-a^{2}s_{2}^{2}\right)^{2}}+\frac{\sigma_{1}\sqrt{\mathcal{P}}}{a^{2}\left(1-a^{2}s_{1}^{2}\right)\left(1-a^{2}s_{2}^{2}\right)},$
(81) $\displaystyle C_{m}$ $\displaystyle=$
$\displaystyle\left(\frac{E_{1}\left(2a^{2}s_{2}+2a-s_{2}\right)}{a\left(1-a^{2}s_{2}^{2}\right)}\right)^{2}+\left(\frac{\sigma_{1}\sqrt{\mathcal{P}}}{a\left(1-a^{2}s_{1}^{2}\right)}\right)^{2}$
(82)
$\displaystyle+\frac{\sqrt{\mathcal{P}}E_{1}\left(2a^{2}s_{2}+2a-s_{2}\right)\sigma_{1}}{a^{2}\left(1-a^{2}s_{1}^{2}\right)\left(1-a^{2}s_{2}^{2}\right)}+1.$

where

$\displaystyle\mathcal{P}$ $\displaystyle=$ $\displaystyle
a^{6}\left(-s_{1}^{4}\right)+4a^{4}E_{1}^{2}s_{1}^{2}+2a^{4}s_{1}^{2}+8a^{3}E_{1}^{2}s_{1}-5a^{2}E_{1}^{2}s_{1}^{2}+4a^{2}E_{1}^{2}-a^{2}-2aE_{1}^{2}s_{1}$
(83) $\displaystyle+E_{1}^{2}s_{1}^{2}-E_{1}^{2}$

$\displaystyle\Omega$ $\displaystyle=$ (84)
$\displaystyle\frac{\sigma_{4}}{\mathcal{H}_{1}}\Big{(}a^{6}s_{2}^{2}\big{(}-12s_{2}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})+6s_{2}^{2}(E_{1}+E_{2}-E_{3})^{2}$
$\displaystyle+14(E_{1}+E_{2}-E_{3})^{2}-s_{2}^{4}\big{)}+2a^{5}s_{2}\big{(}-3s_{2}^{3}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})$
$\displaystyle+s_{2}^{2}\left(7(E_{1}+E_{2}-E_{3})^{2}+3(J_{1}+J_{2}-J_{3})^{2}\right)-14s_{2}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})$
$\displaystyle+5(E_{1}+E_{2}-E_{3})^{2}+s_{2}^{4}\big{)}+a^{4}\big{(}-8s_{2}^{3}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})$
$\displaystyle+s_{2}^{2}\left(19(E_{1}+E_{2}-E_{3})^{2}+14(J_{1}+J_{2}-J_{3})^{2}\right)-20s_{2}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})$
$\displaystyle+2(E_{1}+E_{2}-E_{3})^{2}+s_{2}^{4}\big{)}+2a^{3}\big{(}3s_{2}^{4}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})$
$\displaystyle+s_{2}^{3}\left(2(E_{1}+E_{2}-E_{3})^{2}-3(J_{1}+J_{2}-J_{3})^{2}-2\right)-7s_{2}^{2}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})$
$\displaystyle+s_{2}\left(4(E_{1}+E_{2}-E_{3})^{2}+5(J_{1}+J_{2}-J_{3})^{2}\right)-2(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})\big{)}$
$\displaystyle+a^{2}\big{(}8s_{2}^{3}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})+s_{2}^{2}\left(4(E_{1}+E_{2}-E_{3})^{2}-5(J_{1}+J_{2}-J_{3})^{2}+1\right)$
$\displaystyle-4s_{2}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})-5s_{2}^{4}(E_{1}+E_{2}-E_{3})^{2}+(E_{1}+E_{2}-E_{3})^{2}$
$\displaystyle+2(J_{1}+J_{2}-J_{3})^{2}\big{)}+6a^{7}s_{2}^{3}(E_{1}+E_{2}-E_{3})^{2}+2as_{2}\big{(}s_{2}^{3}(-(E_{1}+E_{2}-E_{3}))(J_{1}+J_{2}-J_{3})$
$\displaystyle+3s_{2}(E_{1}+E_{2}-E_{3})(J_{1}+J_{2}-J_{3})-3(E_{1}+E_{2}-E_{3})^{2}+s_{2}^{2}(J_{1}+J_{2}-J_{3})^{2}$
$\displaystyle-2(J_{1}+J_{2}-J_{3})^{2}+1\big{)}-\left(s_{2}\left(2s_{2}^{2}-7\right)(E_{1}+E_{2}-E_{3})+J_{1}+J_{2}-J_{3}\right)$
$\displaystyle\times(-s_{2}(E_{1}+E_{2}-E_{3})+J_{1}+J_{2}-J_{3})-1\Big{)}$

where

$\displaystyle\mathcal{H}_{1}=2\sqrt{\left(E_{1}a^{2}+E_{2}a^{2}-E_{3}a^{2}-J_{1}a-J_{2}a+J_{3}a+E_{1}+E_{2}-E_{3}\right)^{2}}(as_{2}-1)(as_{2}+1)^{2}$
(85)

$\displaystyle\Pi$ $\displaystyle=$ (86)
$\displaystyle\frac{\sigma_{2}}{\mathcal{H}_{2}}\big{(}\left(a^{2}+1\right)^{2}E_{2}^{2}\zeta^{2}\left(a\left(2\left(3a^{4}-3a^{2}+1\right)s_{2}^{3}+a\left(14a^{2}-5\right)s_{2}^{2}+2\left(5a^{2}-2\right)s_{2}+2a\right)-1\right)$
$\displaystyle-2\left(a^{2}+1\right)E_{2}^{2}\zeta\left(a\left(\left(3a^{4}-3a^{2}+1\right)s_{2}^{3}+a\left(7a^{2}-3\right)s_{2}^{2}+\left(5a^{2}-1\right)s_{2}+a\right)-1\right)(as_{2}-1)$
$\displaystyle-(as_{2}-1)^{4}\left(a^{2}(as_{2}+1)^{2}+E_{2}^{2}\right)\big{)}$

where

$\displaystyle\mathcal{H}_{2}=2a^{2}(as_{2}-1)^{4}(as_{2}+1)^{2}\sqrt{\frac{\left(a^{2}+1\right)^{2}E_{2}^{2}\zeta^{2}}{(as_{2}-1)^{2}}};$
(87)

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