# General properties of the Penrose process with neutral particles in the
equatorial plane

O. B. Zaslavskii  Department of Physics and Technology, Kharkov V.N. Karazin
National University, 4 Svoboda Square, Kharkov 61022, Ukraine<EMAIL_ADDRESS>
###### Abstract

We consider the background of a rotating axially symmetric black hole. Let
particle 0 decay to two fragments 1 and 2 in the direction parallel to that of
particle 0. It is shown that if decay occurs inside the ergoregion, both
particles 1 and 2 move in the same direction as particle 0. For the scenario,
when decay happens in the turning point of all three particles, we find the
condition when angular momenta of both particles 1 and 2 have the same sign.
We elucidate the relation between the approach of Wald that imposes constraint
on maximum and minimum energies of fragments and our approach. In doing so, we
express the results in terms of characteristics of particle 0 and all particle
masses. The conditions of the maximum efficiency depending on the relation
between masses is discussed. We find an explicit expression for angular
momenta of particles 1 and 2. We discuss also particle decay for static black
holes, when the Penrose process is impossible. Because of the absence of the
ergoregion in the static case, scenarios of decay for static black holes can
significantly differ from those in the rotating background.

energy extraction, rotating black hole

###### pacs:

04.70.Bw, 97.60.Lf

##  I Introduction

The Penrose process (PP) was discovered more than 50 years ago pen , pen2  but
currently it is experiencing a new wave of interest. It concerns both
theoretical and astrophysically relevant issues. It is an essential ingredient
in the discussion of energy extraction from black holes in high energy
scenarios, in particular in a magnetic field dad1  \- rufkerr . Originally,
the PP was found for rotating black holes. Meanwhile, it was shown later that
a similar phenomenon exists also for static electrically charged black holes
ruf , den . Energy extraction for combination of rotation and electric charge
is considered in fh . Moreover, some analogs of PP exist even in the flat
space-times df , flat , naked . New types of PP were found and elaborated in a
recent decade. This includes the collisional PP shn  playing a crucial role in
high energy collisions between particles near black holes, wormholes and
singularities. It is an important ingredient for the Bañados-Silk-West (BSW)
effects and its analogues ban  \- pir3 . Another example is a confined Penrose
process that takes place for massive particles inside a cavity or reflecting
shells conf , myconf , being a counterpart of wave amplification and
superradiance st1 \- pani . A concise review with a useful list of references
can be found in the Introduction of win .

All this makes it quite necessary to have well-elaborated and model-
independent formulas describing the PP. Meanwhile, strange as it might seem,
in spite of an investigation of the PP in a wide context including even quite
sophisticated models j , there are some gaps in basics of the PP. In a
particular, a quite complicated method of investigation of the PP became
rather popular (see Sec. XV), although more simple and straightforward one
wald  went into the shadows.

The goal of the present paper is twofold. (i) We elaborate basic formalism
describing PP for a rotating axially symmetric quite generic stationary
background. (ii) We give classification of possible scenarios and describe
some generic features of processes in the ergosphere that remained unnoticed
up to now. In particular, we trace carefully which scenario is possible (or
impossible) if decay happens near the horizon. To this end, we list in detail
different scenarios, indicating explicitly where the ”plus” or ”minus” sign
should stand. As by product, we dispel some incorrect statements made in the
literature during past several years. Also, we stress that the properties of
decay in the static background is not just a particular case of a more general
case in rotating one. This comparison reveals some qualitative difference,
especially for decay near the horizon, where one should be careful.

The paper is organized as follows. In Sec. II we list the general form of a
metric and equations of particle motion. In Sec. III we give general relations
between characteristics of a parent particle 0 and particles 1 and 2 to which
it decays. In Sec. IV we suggest classification of possible scenarios of
decay. In Sec. V we discuss general features of a scenario in which particles
1 and 2 are ejected parallel to the direction of motion of particle 0. In Sec.
VI we discuss properties of corresponding scenarios inside and outside the
ergoregion. In Sec. VII we show that the equations governing motion of
particles in the scenarios of parallel ejection can be also obtained from the
requirement of maximization (minimization) of the energy of ejected particles.
In Sec. VIII we demonstrate equivalence between the derived formulas and Wald
bounds but instead of velocity in the center of mass frame we use the values
of particle masses, thus reformulating the aforementioned bounds in other
terms. In Sec. IX we consider a special kind of scenario when decay occurs in
the turning point for all three particles. For the aforementioned scenario, in
Sec. X we consider possible signs of angular momenta of particles 1 and 2 in
the ergoregion. In Sec. XI we formulate the threshold for the Penrose process
in terms of particle masses. In Sec. XII we discuss general features of the
efficiency of the decay process. In Sec. XIV we discuss peculiarities of the
decay for static black holes. In Sec. XV we compare our approach and another
one rather popular in literature. In Sec. XVI we summarize the results
obtained.

##  II Equations of motion

Let us consider the metric

$ds^{2}=-N^{2}dt^{2}+g_{\phi}(d\phi-\omega
dt)^{2}+\frac{dr^{2}}{A}+g_{\theta}d\theta^{2},$ (1)

where we used notations $g_{\phi}\equiv g_{\phi\phi}$ and $g_{\theta}\equiv
g_{\theta\theta}$ for brevity. We restrict ourselves by stationary and axially
symmetric metrics only, so we assume that the metric coefficients do not
depend on $\phi$ and $t$.

We assume the symmetry with respect to the equatorial plane
$\theta=\frac{\pi}{2}$, so that the metric coefficient depend algebraically on
$\sin^{2}\theta$. Then, there exist geodesics just in this plane, and we
restrict ourselves by particle motion within this plane. Then, equations of
motion for a free particle read

$m\dot{t}=\frac{X}{N^{2}}\text{,}$ (2) $p^{r}\equiv
m\dot{r}=\sigma\frac{\sqrt{A}}{N}P\text{, }P=\sqrt{X^{2}-\tilde{m}^{2}N^{2}},$
(3) $X=E-\omega L\text{,}$ (4) $\tilde{m}^{2}=m^{2}+\frac{L^{2}}{g_{\phi}},$
(5) $m\dot{\phi}=\frac{L}{g_{\phi}}+\frac{\omega X}{N^{2}}=\frac{\omega
E}{N^{2}}-\frac{Lg_{00}}{N^{2}g_{\phi}},$ (6)

where dot denotes differentiation with respect to the proper time. The
direction of motion is characterized by a quantity $\sigma$, where $\sigma=+1$
for motion in the outward direction and $\sigma=-1$ for the inward case. Here,
$E$ is the particle energy, $L$ being angular momentum, $m$ its mass. Eqs. (2)
and (6) are direction consequences of the conservation of the energy and
angular momentum, eq. (3) uses the normalization conditions for the four-
velocity $u_{\mu}u^{\mu}=-1$.

As a consequence of the above equations, the angular velocity

$\Omega\equiv\frac{d\phi}{dt}=\omega+\frac{LN^{2}}{g_{\phi}X}.$ (7)

The requirement $\dot{t}>0$ leads to the forward-in-time condition

$X\geq 0.$ (8)

For the four-velocity one can write in coordinates $(t,r,\phi)$

$u^{\mu}=(\frac{X}{mN^{2}}\text{,
}\sigma\frac{\sqrt{A}}{mN}P,\frac{L}{mg_{\phi}}+\frac{\omega
X}{mN^{2}})\text{.}$ (9)

We omit the component $u^{\theta}=0$.

##  III General setup

Let in the point $r=r_{0}$ a parent particle 0 decay to particles 1 and 2.
Hereafter, we call it ”scenario AP” (since this happens in an arbitrary
point). We assume the conservation laws in this point:

$E_{0}=E_{1}+E_{2}\text{,}$ (10) $L_{0}=L_{1}+L_{2}\text{,}$ (11)
$p_{0}^{r}=p_{1}^{r}+p_{2}^{r}\text{,}$ (12)

whence

$X_{0}=X_{1}+X_{2}\text{.}$ (13)

Then, using these equations and (3), (4), one can obtain

$X_{2}=\frac{1}{2\tilde{m}_{0}^{2}}\left(X_{0}\tilde{b}_{2}+P_{0}\delta\sqrt{\tilde{d}}\right),$
(14)

$X_{1}=\frac{1}{2\tilde{m}_{0}^{2}}\left(X_{0}\tilde{b}_{1}-P_{0}\delta\sqrt{\tilde{d}}\right),$
(15) $\tilde{b}_{1}=\tilde{m}_{0}^{2}+\tilde{m}_{1}^{2}-\tilde{m}_{2,}^{2},$
(16) $\tilde{b}_{2}=\tilde{m}_{0}^{2}+\tilde{m}_{2}^{2}-\tilde{m}_{1,}^{2},$
(17)

where $i=0,1,2$, $\delta=\pm 1$,

$\tilde{m}_{i}^{2}=m_{i}^{2}+\frac{L_{i}^{2}}{g_{\phi}}\text{,}$ (18)
$\tilde{m}_{0}\geq\tilde{m}_{1}+\tilde{m}_{2}\text{.}$ (19)
$\tilde{d}=\tilde{b}_{1}^{2}-4\tilde{m}_{0}^{2}\tilde{m}_{1}^{2}=\tilde{b}_{2}^{2}-4\tilde{m}_{0}^{2}\tilde{m}_{2}^{2},$
(20) $P_{2}=\left|\frac{P_{0}\tilde{b}_{2}+\delta
X_{0}\sqrt{\tilde{d}}}{2\tilde{m}_{0}^{2}}\right|,$ (21)

$P_{1}=\left|\frac{P_{0}\tilde{b}_{1}-\delta
X_{0}\sqrt{\tilde{d}}}{2\tilde{m}_{0}^{2}}\right|\text{.}$ (22)

Here, parameter $\delta$ appears formally in the course of solving quadratic
equations after eliminating square roots in (13). Its presence shows that, for
given characteristics of particle 0, there are two different (in general,
nonequivalent) scenarios with fixed masses of particles 1 and 2.

A useful equality follows from (16), (17):

$\tilde{b}_{1}+\tilde{b}_{2}=2\tilde{m}_{0}^{2}.$ (23)

The above expressions are similar to the results already obtained in eqs. (19)
— (30) of centr  for particle collisions. The difference is that now we are
dealing with particle decay but the equations retain the same structure. We
use now particle labels 1 and 2 instead of 3 and 4 respectively in centr .

In what follows, we will exploit useful equalities

$\tilde{b}_{1,2}=b_{1,2}+\frac{2L_{0}L_{1,2}}{g_{\phi}},$ (24)
$\tilde{d}=d+4b_{1}\frac{L_{0}L_{1}}{g_{\phi}}-\frac{4L_{1}^{2}}{g_{\phi}}m_{0}^{2}-4\frac{L_{0}^{2}}{g_{\phi}}m_{1}^{2}.$
(25)

Equivalently,

$\tilde{d}=d+4b_{2}\frac{L_{0}L_{2}}{g_{\phi}}-\frac{4L_{2}^{2}}{g_{\phi}}m_{0}^{2}-4\frac{L_{0}^{2}}{g_{\phi}}m_{2}^{2}.$
(26)

Here,

$b_{1}=m_{0}^{2}+m_{1}^{2}-m_{2,}^{2},$ (27)
$b_{2}=m_{0}^{2}+m_{2}^{2}-m_{1,}^{2},$ (28)
$d=b_{1}^{2}-4m_{0}^{2}m_{1}^{2}=b_{2}^{2}-4m_{0}^{2}m_{2}^{2}.$ (29)

##  IV Classification of scenarios

We are mainly interested in the situation, when particle 0 moves from larger
radii to smaller ones, so $\sigma_{0}=-1$. Let, say, the metric describe a
black hole. Then, it follows from (12) that at least one of daughter particles
moves towards a black hole. By definition, we assume that $\sigma_{1}=-1$. For
particle 2, we cannot fix $\sigma_{2}$ beforehand. The conservation law gives
us

$-P_{0}=-P_{1}+\sigma_{2}P_{2}\text{.}$ (30)

Using general expressions (21) and (22) as well as (30), we can enumerate all
possible scenarios.

It is convenient to describe a scenario as a four-component vector
$(\sigma_{2}$,$h_{2}$,$h_{1},\delta)$ in the parametric space. Here,

$h_{1,2}=sign(\tilde{b}_{1,2}N-2\tilde{m}_{1,2}X_{0}),$ (31)

this quantity arises when different signs inside the absolute values in (21),
(22) are taken into account.

Then, from the conservation laws we have the following types of scenarios:

$\text{I }(+,+,h_{1},-)\text{, II }(-,h_{2},-,+),\text{ III
}(-,-,h_{1},-)\text{. }$ (32)

They are equivalent to scenarios

$\text{1 }(+,+,+,-),\text{ 2 }(+,+,-,-),\text{3 }(-,-,-,+)\text{, 4
}(-,+,-,+)\text{, 5 }(-,-,-,-)\text{, 6 }(-,-,+,-).$ (33)

from centr . (But a reader should bear in mind that there is a typo in
scenario 6 in centr ).

More explicitly, if $\delta=-1,$

$P_{2}=\frac{(X_{0}\sqrt{\tilde{d}}-P_{0}\tilde{b}_{2})h_{2}}{2\tilde{m}_{0}^{2}}\text{,}$
(34)
$P_{1}=\frac{P_{0}\tilde{b}_{1}+X_{0}\sqrt{\tilde{d}}}{2\tilde{m}_{0}^{2}}.$
(35)

If $\delta=+1,$

$P_{2}=\frac{X_{0}\sqrt{\tilde{d}}+P_{0}\tilde{b}_{2}}{2\tilde{m}_{0}^{2}}\text{,}$
(36)
$P_{1}=\frac{X_{0}\sqrt{\tilde{d}}-P_{0}\tilde{b}_{1}}{2\tilde{m}_{0}^{2}}h_{1}.$
(37)

If $\sigma=-1$, $P_{1}+P_{2}=P_{0}$ and we must take $\delta=-1$, $h_{2}$
$=-1$ or $\delta=+1$, $h_{1}=-1$

If $\sigma=+1$, $P_{1}-P_{2}=P_{0}$ and we must take $\delta=-1$, $h_{2}=+1$

##  V Ejection along the direction of motion

The above equations are valid in a general case. They allow us to find
$E_{1},E_{2}$ and $P_{1},P_{2}$ in terms of quantities characterizing initial
particle 0, plus one more free parameter (say, $L_{2}\,$). This uncertainty
disappears, if we impose an additional constraint. Let us assume that in the
coordinate frame (1) particles 1 and 2 are ejected parallel to the direction
of motion of particle 0. This means that the tangent vector to the trajectory
keeps the same direction after decay. Therefore, the ratio $u^{r}/u^{\phi}$
remains the same. Correspondingly, for particle 1 that moves in the same
direction as particle 0 ($\sigma_{1}=\sigma_{0}=-1$) we have from (9)

$\frac{P_{1}}{P_{0}}=\frac{\frac{L_{1}}{g_{\phi}}+\frac{\omega
X_{1}}{N^{2}}}{\frac{L_{0}}{g_{\phi}}+\frac{\omega X_{0}}{N^{2}}},$ (38)

where subscript denotes a particle’s label. For a static metric, $\omega=0$,
whence

$\frac{P_{1}}{P_{0}}=\frac{L_{1}}{L_{0}}\text{,}$ (39)

and we return to the situation considered in rocket  for the Schwarzschild
metric. However, for $\omega\neq 0,$ a more general relation (38) should hold.
It follows from it that

$L_{1}=\frac{P_{1}}{P_{0}}(L_{0}+g_{\phi}\frac{\omega}{N^{2}})-\frac{\omega}{N^{2}}g_{\phi}X_{1}\text{.}$
(40)

Let us consider, say, scenario I. It follows from (22), (15) that

$L_{1}=\frac{L_{0}}{2\tilde{m}_{0}^{2}}\tilde{b}_{1}+\frac{\sqrt{\tilde{d}}}{2\tilde{m}_{0}^{2}P_{0}}(L_{0}X_{0}+g_{\phi}\omega\tilde{m}_{0}^{2})\text{,}$
(41)
$L_{2}=\frac{L_{0}}{2\tilde{m}_{0}^{2}}\tilde{b}_{2}-\frac{\sqrt{\tilde{d}}}{2\tilde{m}_{0}^{2}P_{0}}(L_{0}X_{0}+g_{\phi}\omega\tilde{m}_{0}^{2})\text{.}$
(42)

This is not the end of story since the quantities $\tilde{b}_{1,2}$ and
$\tilde{d}$ themselves contain $L_{1,2}$.

Our goal is, given $E_{0},L_{0}$, $m_{0},m_{1},m_{2,}$ to find $E_{1}$,
$E_{2}$, $L_{1}$, $L_{2}$. Although, formally, calculations are quite direct,
they are rather lengthy. And, what is surprising, the quantities in terms of
tilted ones can be expressed in terms of original characteristics without a
tilde in a rather simple form.

Omitting details, we list the results.

###  V.1 Scenario I

$L_{2}=\frac{L_{0}b_{2}}{2m_{0}^{2}}-\frac{\sqrt{d}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m^{2}}}(L_{0}E_{0}+g_{\phi}\omega
m_{0}^{2})\text{,}$ (43)
$L_{1}=\frac{L_{0}b_{1}}{2m_{0}^{2}}+\frac{\sqrt{d}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m^{2}}}(L_{0}E_{0}+g_{\phi}\omega
m_{0}^{2}).$ (44)
$E_{2}=E_{0}\frac{b_{2}}{2m_{0}^{2}}-\frac{\sqrt{d}\sqrt{m_{0}^{2}g_{00}+E_{0}^{2}}}{2m_{0}^{2}}$
(45)
$E_{1}=E_{0}\frac{b_{1}}{2m_{0}^{2}}+\frac{\sqrt{d}\sqrt{m_{0}^{2}g_{00}+E_{0}^{2}}}{2m_{0}^{2}}$
(46)
$P_{2}=\frac{P_{0}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m_{0}^{2}}}(E_{0}\sqrt{d}-b_{2}\sqrt{E_{0}^{2}+g_{00}m_{0}^{2}})\text{,}$
(47)
$P_{1}=\frac{P_{0}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m_{0}^{2}}}(b_{1}\sqrt{E_{0}^{2}+g_{00}m_{0}^{2}}+E_{0}\sqrt{d}).$
(48)

###  V.2 Scenario II

$L_{1}=\frac{L_{0}}{2m_{0}^{2}}b_{1}+\frac{\sqrt{d}}{2m_{0}^{2}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}}(L_{0}E_{0}+g_{\phi}\omega
m_{0}^{2})$ (49)

$L_{2}=\frac{L_{0}}{2m_{0}^{2}}b_{2}-\frac{\sqrt{d}}{2m_{0}^{2}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}}(L_{0}E_{0}+g_{\phi}\omega
m_{0}^{2})$ (50)

$E_{1}=\frac{E_{0}b_{1}}{2m_{0}^{2}}+\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}},$
(51)
$E_{2}=\frac{b_{2}E_{0}}{2m_{0}^{2}}-\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}.$
(52)
$P_{1}=\frac{P_{0}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m^{2}}}(b_{1}\sqrt{E_{0}^{2}+g_{00}m_{0}^{2}}+E_{0}\sqrt{d})\text{,}$
(53)
$P_{2}=\frac{P_{0}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m^{2}}}(b_{2}\sqrt{E_{0}^{2}+g_{00}m_{0}^{2}}-E_{0}\sqrt{d}).$
(54)

In this scenario, we can interchange labels 1 and 2 since each of the two
particles falls in a black hole.

Additionally, the condition

$m_{0}\geq m_{1}+m_{2}$ (55)

should be fulfilled.

In the nonrelativistic case, the mass is conserved, $m_{0}=m_{1}+m_{2}$.
However, for relativistic particles, inequality $m_{0}>m_{1}+m_{2}$ is
possible due to the contribution of kinetic energy. In particular, decay to
particles with negligible masses of particles 1 and 2 is allowed. It is worth
mentioning that eq. (55) is not an additional assumption but can be derived
from equations of motions - see eqs. (21) - (25) in centr .

##  VI Properties of scenarios

###  VI.1 Scenario I

Let us discuss scenario I firstly. We remind a reader that $P$ is an absolute
value, so it is necessary that $P\geq 0$. As $d\leq b_{1,2}$, it is clear that
inside the ergoregion, where $g_{00}>0$, inequality $P_{2}>0$ in (47) cannot
be satisfied. Therefore, scenario I is not realized there at all. Particle 2
is drifted by flow of space, so particle 2 is ejected inward in the frame
comoving with particle 0, $\dot{r}_{2}<0$ anyway, and only scenario II
remains. It does not mean, of course, that particle 2 cannot reach infinity.
Instead, it only means that after decay it moves towards the horizon. In
principle, it can bounce back after reflection from the potential barrier and
escape to infinity. This depends on concrete properties of the metric.

On the first glance, the impossibility of scenario I inside the ergoregion
looks paradoxical but it can be explained as follows. If two particles move
along the same line tangent to the trajectory (in the same or opposite
directions), they must have $d\phi/dr$ of the same sign. Let $dr_{1}<0$ and
$dr_{2}>0$. Then, $d\phi_{1}$ and $d\phi_{2}$ must have the opposite signs.
However, inside the ergoregion the sign of the angular velocity is fixed,
particles rotate in the same direction as a black hole due to strong frame-
dragging. Therefore, $d\phi_{1}$ and $d\phi_{2}$ have the same signs, so
scenario I in the ergoregion is impossible.

Outside the ergoregion $g_{00}=-\left|g_{00}\right|<0$, and scenario I is
possible, provided (47) is positive, so

$E_{0}<\frac{b_{2}\sqrt{\left|g_{00}\right|}}{2m_{2}}\text{.}$ (56)

However, the Penrose process is impossible as it should be outside the
ergoregion, since $0<E_{2}<E_{0}$, $0<E_{1}<E_{0}$. Decay occurs without
energy extraction from a black hole.

###  VI.2 Scenario II

This type of decay is possible outside the ergoregion ($g_{00}<0$), provided

$E_{0}>\frac{b_{2}\sqrt{\left|g_{00}\right|}}{2m_{2}}\text{.}$ (57)

Inside the ergoregion ($g_{00}>0$), this scenario is always possible. Can the
PP be possible as well? It is seen from (51), (52) that $E_{1}>0$. If we
require $E_{2}<0$, this entails

$E_{0}<\frac{\sqrt{d}}{2m_{2}}\sqrt{g_{00}}\text{.}$ (58)

However, in this scenario both particles (at least, immediately after decay)
move towards a black hole. In principle, particle 1 can bounce back from the
potential barrier and escape to infinity. Such a scenario is model-dependent
and is beyond the scope of the present paper.

###  VI.3 Inverse scenarios

In a similar way, we can consider the cases when particle 0 moves with
$\sigma_{0}=+1$ (in the outward direction). For example, it can arrive from
infinity, bounce back in the turning point and keep moving in the outward
direction. Then, the conservation of radial momentum gives us

$P_{0}=P_{1}+\sigma_{2}P_{2}$ (59)

If $\sigma_{2}=+1$, both particles move outward. In doing so, the formulas for
$E_{i}$, $P_{i}$ and $L_{i}$ ($i=1,2$) coincide with those for scenario II.
Particle 1 can, in principle, move directly to infinity. Particle 2 can have
$E_{2}<0$, provided (58) holds in the ergoregion. However, it is inevitable
that in this case particle 2 bounces back and moves further towards a black
hole in agreement with general properties of trajections with negative energy
gpneg .

Scenario with $\sigma_{2}=-1$ is similar to scenario I.

##  VII Maximization of energy of debris

We can look at the problem from a somewhat different viewpoint. For fixed
$L_{0}\,$, $E_{0}$ and given masses $m_{0}$, $m_{1}$, $m_{2}$ eqs. (14), (15)
leave 1 free parameter, say $L_{1}$. Once it is fixed, one obtains
$L_{2}=L_{0}-L_{1}$ and $E_{1}$, $E_{2}$ from aforementioned equations and the
definition (4). If $L_{1}$ is allowed to vary, we have energies as functions
of $L_{1}$: $E_{1}=E_{1}(L_{1})$ and $E_{2}=E_{2}(L_{1})$. Let us require that
the quantity $E_{1}$ reach the maximum with respect to $L_{1}$. Then,

$\frac{\partial E_{1}}{\partial L_{1}}=0.$ (60)

As $E_{2}=E_{0}-E_{1}$, for $E_{2}$ we will have minimum. Taking into account
general formulas (15), (14), one obtains

$-\delta\sqrt{\tilde{d}}(\omega\tilde{m}_{0}^{2}+\frac{L_{0}X_{0}}{g_{\phi}})=\frac{P_{0}}{g_{\phi}}(2L_{2}m_{0}^{2}-b_{2}L_{0}).$
(61)

After some manipulations, one can show easily that its solution gives us (43),
(44) or (49), (50).

Thus both approaches are equivalent.

##  VIII Comparison with the Wald approach

There is one more approach. For decay of particles, general inequalities were
derived by Wald wald :

$E_{\min}\leq E\leq E_{\max}\text{,}$ (62)

where

$E_{\max}=\gamma_{2}\frac{m_{2}}{m_{0}}(E_{0}+v_{2}\sqrt{E_{0}^{2}+m_{0}^{2}})=E_{2}\text{,}$
(63)
$E_{\min}=\gamma_{1}\frac{m_{1}}{m_{0}}(E_{0}-v_{1}\sqrt{E_{0}^{2}+m_{0}^{2}})=E_{1}\text{.}$
(64)

Here, $v_{2}$ is the velocity of particle 2 with respect to particle 0. In
other words, this is a relative velocity. Here,
$\gamma_{2}=\frac{1}{\sqrt{1-v_{2}^{2}}}$ is the corresponding Lorentz gamma
factor. Similar quantities and notations stand for particle 1.

Comparing (52), (51) and (63), (64) we see that both expressions coincide, if
we identify

$v_{2}=\frac{\sqrt{d}}{b_{2}}\text{,
}\gamma_{2}=\frac{b_{2}}{2m_{0}m_{2}}\text{.}$ (65)

In a similar way,

$v_{1}=\frac{\sqrt{d}}{b_{1}}\text{, }\gamma_{1}=\frac{b_{1}}{2m_{0}m_{1}}.$
(66)

This is not the end of story. We must prove that this is not formal
coincidence, but $v$ and $\gamma$ defined according to (65) do have the
kinematic meaning of a relative velocity and Lorentz gamma factor. To this
end, we can calculate the relative gamma-factor between particle 0 and 2:
$\gamma_{2}=-u_{\mu(0)}u^{\mu(2)}$.

Using the explicit expressions (9) for each of particles 0 and 2, one can show
that

$\gamma_{2}=\frac{X_{0}X_{2}-\sigma
P_{0}P_{2}}{N^{2}}-\frac{L_{2}L_{0}}{g_{\phi}}\text{,}$ (67)

where $\sigma=-1$ for scenario I and $\sigma=+1$ for scenario II.

Using (50) - (54), we obtain after some calculations that $\gamma_{2}$ (hence,
also $v_{2}$) does indeed coincide with (65). Thus two ways of calculations
(from equations of motion or by comparing two expressions ((52), (51) and
(63), (64)) give the same result. The similar statement holds for particle 1.
Thus $v_{1}$ and $v_{2}$ depend on the relationship between masses only.

##  IX Decay in turning point

There exists a special type of scenario, when decay occurs in the turning
point for all three particles, $P_{i}=0$ for $i=0,1,2$. For brevity, we will
call it scenario TP. The condition (40) loses its sense in this case and turns
into an empty identity. Therefore, we must consider the problem all over
again.

Then, according to (3), in the point of decay

$X_{i}=\tilde{m}_{i}N\text{,}$ (68)

$i=0,1,2$. It is clear from (13) that this requires

$\tilde{m}_{0}=\tilde{m}_{1}+\tilde{m}_{2}\text{.}$ (69)

Then, it follows from (3), (16), (17), (20) that
$\tilde{b}_{1}=2\tilde{m}_{1}\tilde{m}_{0}$,
$\tilde{b}_{2}=2\tilde{m}_{0}\tilde{m}_{2}$, $\tilde{d}=0$. Then, we have from
(68), (69) that

$X_{1}=\frac{X_{0}\tilde{m}_{1}}{\tilde{m}_{0}}\text{,
}X_{2}=\frac{X_{0}\tilde{m}_{2}}{\tilde{m}_{0}}\text{.}$ (70)

If the process occurs in the egorsphere, $g_{00}>0$ that is equivalent to

$N<\omega\sqrt{g_{\phi}}\text{.}$ (71)

Eq. (30) for particle 0 gives us the quadratic equation for $L_{0}$. In
general, it has two roots:

$L_{0}=E_{0}\omega\frac{g_{\phi}}{g_{00}}+\frac{N\sqrt{g_{\phi}}}{g_{00}}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}$
(72)

or

$L_{0}=E_{0}\omega\frac{g_{\phi}}{g_{00}}-\frac{N\sqrt{g_{\phi}}}{g_{00}}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}\text{.}$
(73)

Then,

$X_{0}=\tilde{m}_{0}N=-\frac{E_{0}N^{2}}{g_{00}}\pm\frac{\omega
N\sqrt{g_{\phi}}}{g_{00}}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}\text{.}$ (74)

However, in the ergoregion $g_{00}>0$, so only root (73) is compatible with
(8). Then, one should take the upper sign in (74). Outside the ergoregion,
both signs in (74) are suitable. In doing so, sign ”minus” in (74 corresponds
to (72).

The expressions for $L_{1},L_{2}$ follow from (69) and (11). Then, after some
calculations that include taking the square, one obtains

$L_{1}=\frac{b_{1}}{2m_{0}^{2}}L_{0}\pm\frac{\sqrt{g_{\phi}d}\tilde{m}_{0}}{2m_{0}^{2}},$
(75)
$L_{2}=\frac{b_{2}}{2m_{0}^{2}}L_{0}\mp\frac{\sqrt{g_{\phi}d}\tilde{m}_{0}}{2m_{0}^{2}},$
(76)
$\tilde{m}_{1}=\frac{b_{1}\sqrt{L_{0}^{2}+g_{\phi}m_{0}^{2}}\pm\sqrt{d}L_{0}}{2m_{0}^{2}\sqrt{g_{\phi}}}=\frac{b_{1}}{2m_{0}^{2}}\tilde{m}_{0}\pm\frac{\sqrt{d}L_{0}}{2m_{0}^{2}\sqrt{g_{\phi}}}$
(77)
$\tilde{m}_{2}=\frac{b_{2}}{2m_{0}^{2}}\tilde{m}_{0}\mp\frac{\sqrt{d}L_{0}}{2m_{0}^{2}\sqrt{g_{\phi}}}.$
(78)

It is seen that (69) is satisfied as it should be.

Using also the definition of $X$ and (72), (73), we have after some algebraic
manipulations that

$E_{1}=\frac{b_{1}}{2m_{0}^{2}}E_{0}\pm\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}\text{,}$
(79)
$E_{2}=\frac{b_{2}}{2m_{0}^{2}}E_{0}\mp\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}$
(80)

that agrees with (46), (45) or (52), (51).

It is essential that signs in (79), (80) coincide with those in (75), (76).
Now, particles 1 and 2 can be considered on equal footing. However, the subtle
point is that signs in (75),(76) should agree with signs for the energy.

Further, we can substitute $L_{0}$ and express $L_{1,2}$ in terms of energy
$E_{0}$:

$L_{2}=E_{0}\frac{b_{2}\omega g_{\phi}\mp
N\sqrt{d}\sqrt{g_{\phi}}}{2m_{0}^{2}g_{00}}+\frac{\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}}{2m_{0}^{2}g_{00}}(-b_{2}N\sqrt{g_{\phi}}\pm\sqrt{d}g_{\phi}\omega),$
(81) $L_{1}=E_{0}\frac{b_{1}\omega g_{\phi}\pm
N\sqrt{d}\sqrt{g_{\phi}}}{2m_{0}^{2}g_{00}}+\frac{\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}}{2m_{0}^{2}g_{00}}(-b_{1}N\sqrt{g_{\phi}}\mp\sqrt{d}g_{\phi}\omega).$
(82)

In the case under discussion, classification of scenarios that includes I and
II is not valid any longer since all $P_{i}=0$ by definition. If we are
interested in the question, whether or not particle (say, 2) escapes after
decay, one is led to condition $h_{2}>0$ where $h_{2}$ is given by (31).

##  X Penrose process and signs of angular momenta for decay in turning point

For decay in a generic point we discussed above some features of the process
connected with radial components of particle velocities. In scenario TP, by
definition, radial components of momenta $P_{i}=0$ for $i=0,1,2$. Meanwhile,
here it is of some interest to elucidate the properties of the scenario that
are connected with angular motion. This concerns both angular velocities and
momenta.

###  X.1 Angular velocities

It is instructive to find $\Omega_{0}.$ Using eq. (7) and (73) we obtain

$\Omega_{0}=\omega+\frac{\varepsilon_{0}N}{\sqrt{g_{\phi}}\sqrt{\varepsilon_{0}^{2}+g_{00}}}\text{,}$
(83)

where $\varepsilon=\frac{E}{m}.$

It follows from (7) and (70) that

$\Omega_{1,2}=\omega+\frac{LN}{\sqrt{g_{\phi}}\sqrt{L_{1,2}^{2}+m_{1,2}^{2}g_{\phi}}}.$
(84)

As is well known, the condition that the interval is time-like or light-like
gives for a particle with $r=const$

$\Omega_{-}\leq\Omega\leq\Omega_{+}\text{,}$ (85)

where

$\Omega_{+}=\omega+\frac{N}{\sqrt{g_{\phi}}}$ (86)

is the maximum possible value for a particle,

$\Omega_{-}=\omega-\frac{N}{\sqrt{g_{\phi}}}$ (87)

is the minimum one. Inside the ergoregion, $\Omega_{-}>0$ and this causes
orbital motion of a particle.

If $m_{i}=0$, (84) gives us

$\Omega_{1,2}=\omega+\frac{\sigma_{1,2}N}{\sqrt{g_{\phi}}}\text{,}$ (88)

where $\sigma_{i}=\pm 1$ depending on the sign of $L_{i}$. Is it possible to
have the same signs for both massless particles? Below, we show in a rather
general setting that the answer is ”no”.

If $m_{1}=m_{2}=0$, we can derive some more features of angular momenta. In
this case, $b_{1}=b_{2}=m_{0}^{2}=\sqrt{d}$. Then, it follows from (43), (44)
or (49), (50) that $signL_{1}L_{2}=signY$, where

$Y=L_{0}^{2}-\frac{1}{(E_{0}^{2}+m_{0}^{2}g_{00})}(L_{0}E_{0}+g_{\phi}\omega
m_{0}^{2})^{2}.$ (89)

For scenario TP under discussion, with (73) taken into account,

$Y=-m_{0}^{2}g_{\phi}<0\text{.}$ (90)

Thus $L_{1}L_{2}<0$. Correspondingly, for one of particles
$\Omega=\Omega_{+}$, for another one $\Omega=\Omega_{-}$. Both particles
appear with opposite angular momenta and realize the maximum and minimum
values of $\Omega$.

For massive particles, a general inequality (85) holds. What can be said about
the relative signs of angular momenta in this case? Let particle 0 decay
inside the ergoregion and the Penrose process occur. Then, for one of
particles (say, particle 1) $E_{1}<0$. We must have $L_{1}<0$ according to the
forward-in-time condition (8). Now, we are going to elucidate, whether it is
possible to have $L_{2}<0$ as well.

Let us assume that indeed $L_{2}<0$. Then,
$L_{0}=L_{1}+L_{2}=-\left|L_{0}\right|<0$ as well. According to (50), we have
the following general condition on $\left|L_{0}\right|$:

$\left|L_{0}\right|>\frac{\sqrt{d}g_{\phi}\omega
m_{0}^{2}}{\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}}(b_{2}+\sqrt{d}\frac{E_{0}}{\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}})^{-1}.$
(91)

In scenario TP, further restrictions can be obtained. Then, $L_{0}$ is not
independent quantity but is given by (73), so $L_{0}<0$ entails

$N\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}>E_{0}\omega\sqrt{g_{\phi}}.$ (92)

In combination with (71) this gives us

$N_{0}\equiv\omega\sqrt{g_{\phi}}\frac{E_{0}}{\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}}<N<\omega\sqrt{g_{\phi}}.$
(93)

Also, we obtain from (91) and (73) that

$N>N_{1}\equiv\frac{1}{\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}}(E_{0}\omega\sqrt{g_{\phi}}+\frac{g_{00}\sqrt{d}\omega\sqrt{g_{\phi}}m_{0}^{2}}{b_{2}\sqrt{E_{0}^{2}+m_{0}^{2}g_{00}}+E_{0}\sqrt{d}})$
(94)

Obviously, $N_{1}>N_{0}$, so (94) is a more tight condition that $N>N_{0}$.
Thus for $L_{2}<0$ we must have

$N_{1}<N<\omega\sqrt{g_{\phi}}\text{,}$ (95)

whence $N_{1}<\omega\sqrt{g_{\phi}}$. It is easy to check that this holds for
any $m_{2}\neq 0$. Then, only in the region (95) the angular momentum
$L_{2}<0$.

Meanwhile, if particle 2 is massless, $m_{2}=0$, this strip degenerates to the
point, so actually $L_{2}<0$ is impossible in accordance with what is written
above.

##  XI Threshold for the Penrose process

We are interested in the situation when $E_{2}>E_{0}$, so that the efficiency
$\eta\equiv\frac{E_{2}}{E_{0}}>1$. Let us consider scenario TM assuming that
particle 2 moves after decay in the outward direction. Then, it follows from
(80) where sign ”plus” is taken that

$b_{2}+\sqrt{d}\sqrt{1+\frac{g_{00}}{\varepsilon_{0}^{2}}}>2m_{0}^{2}\text{,}$
(96)

whence

$v_{2}>v_{2\min}=\frac{b_{1}}{b_{2}}\frac{1}{\sqrt{1+\frac{\left(g_{00}\right)_{H}}{\varepsilon^{2}}}},$
(97)

where we used (65) and (27), (28).

In a similar way, condition $E_{1}<0$ gives us

$v_{1}>v_{1\min}=\frac{1}{\sqrt{1+\frac{g_{00}}{\varepsilon_{0}^{2}}}}=\frac{\left(v_{2}\right)_{\min}b_{2}}{b_{1}}=\frac{1}{\sqrt{1+\frac{g_{00}}{\varepsilon_{0}^{2}}}}\text{.}$
(98)

Both conditions are equivalent to each other, as it should be, according to
(65), (66).

For the Kerr metric, taking $g_{00}=1$ on the extremal horizon in the
equatorial plane and choosing $\varepsilon_{0}=1$, we obtain
$v_{1\min}=\frac{1}{\sqrt{2}}$. For circular orbits around the near-extremal
horizon, we can put $\varepsilon_{0}=\frac{1}{\sqrt{3}}$, then
$v_{1\min}=\frac{1}{2}$. These results agree with 72 .

Eqs. (97) and (98) with (65), (66) taken into account can be rewritten as

$b_{1}>2m_{0}m_{1}\gamma_{1\min}\text{,}$ (99)
$m_{0}^{2}-2m_{0}m_{1}\gamma_{1\min}+m_{1}^{2}-m_{2}^{2}>0\text{,}$ (100)

where
$\gamma_{1\min}=\frac{1}{\sqrt{1-\left(v_{1}\right)_{\min}^{2}}}=\frac{\sqrt{\varepsilon_{0}^{2}+g_{00}}}{\sqrt{g_{00}}}$,
whence

$m_{0}>m_{+},\text{ }$ (101)

or

$m_{0}<m_{-}\text{,}$ (102)

where

$m_{+}=\frac{m_{1}\sqrt{\varepsilon_{0}^{2}+g_{00}}+\sqrt{m_{1}^{2}\varepsilon_{0}^{2}+m_{2}^{2}g_{00}}}{\sqrt{g_{00}}},$
(103)
$m_{-}=\frac{m_{1}\sqrt{\varepsilon_{0}^{2}+g_{00}}-\sqrt{m_{1}^{2}\varepsilon_{0}^{2}+m_{2}^{2}g_{00}}}{\sqrt{g_{00}}}=\frac{(m_{1}^{2}-m_{2}^{2})\sqrt{g_{00}}}{m_{1}\sqrt{\varepsilon_{0}^{2}+g_{00}}+\sqrt{m_{1}^{2}\varepsilon_{0}^{2}+m_{2}^{2}g_{00}}}\text{.}$
(104)

It is clear that $m_{+}>m_{1}+m_{2}$, so (101) is a more tight condition than
(55) due to the account for contribution of kinetic energy of particles 1 and
2. Also, it is easy to show that (55) is inconsistent with (102). Therefore,
we are left with (101) only.

If $m_{1}=m_{2}\equiv m$, we must have

$m_{0}>\frac{2m}{\sqrt{g_{00}}}\sqrt{\varepsilon_{0}^{2}+g_{00}}\text{.}$
(105)

###  XI.1 Circle orbits

Further information about the scenario under discussion can be found if we
consider not simply turning points but assume that the corresponding point
lies on the circle orbit. General estimates for ”dirty” (surrounded by matter)
extremal black holes according to eq. (66) of near  give us for near-horizon
circular orbits

$\varepsilon=\frac{\omega_{H}\sqrt{g_{\phi}}}{\sqrt{B_{1}^{2}g_{\phi}-1}},$
(106)

where $B_{1}$ is the coefficient in the near-horizon expansion

$\omega=\omega_{H}-B_{1}N+...$ (107)

For the extremal Kerr-Newman

$\varepsilon=\frac{a}{M\sqrt{4\frac{a^{2}}{M^{2}}-1}}\text{,}$ (108)
$B_{1}=\frac{2a}{a^{2}+M^{2}}\text{,}$ (109)
$\left(g_{00}\right)_{H}=\left(\omega^{2}g_{\phi}\right)_{H}=\frac{a^{2}}{M^{2}},$
(110)

where $M$ is the mass of a black hole, $a$ being its angular momentum. Then,

$v_{1}=\frac{1}{2}\frac{M}{a}\text{,
}v_{2}=\frac{1}{2}\frac{M}{a}\frac{b_{1}}{b_{2}}.$ (111)

In accordance with wald  and eq. A15 of 72  and (see also page 373 of chandra
), $v_{1}>\frac{1}{2}$.

##  XII Rotating black holes: efficiency in particular cases

Now, we are going to analyze the efficiency of the process depending on the
mass of particles participating in decay, especially we will discuss the near-
horizon limit. We assume that it is particle 2 that has positive energy and
escapes - either immediately (in scenario TP) or later, after reflection from
the potential barrier (in scenario AP). In doing so, its energy is described
by eq. (80) with sigh ”plus”.

Then, the efficiency

$\eta=\frac{E_{2}}{E_{0}}=\frac{b_{2}}{2m_{0}^{2}}+\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{1+\frac{g_{00}}{\varepsilon_{0}^{2}}}\text{,}$
(112)

$\varepsilon=\frac{E}{m}$.

It is remarkable that the metric enters the efficiency through the component
$g_{00}$ only. This simple circumstance has some important consequences. If
$g_{00}$ is a monotonically decreasing function of the radial coordinate (like
for the Kerr metric), the maximum of efficiency is reached when the turning
point is as close to the horizon as possible. Let us consider it in more
detail.

####  XII.0.1 Decay on the horizon

$\eta=\frac{E_{2}}{E_{0}}=\frac{b_{2}}{2m_{0}^{2}}+\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{1+\frac{\left(g_{\phi}\omega^{2}\right)_{H}}{\varepsilon^{2}}}.$
(113)

Hereafter, subscript ”H” means that a corresponding quantity is calculated on
the horizon.

For given masses $m_{0}$, $m_{1}$ and $m_{2}$ and $E_{0}\geq m_{0}$, the
maximum of (112) is reached for the minimum possible value $\varepsilon=1$, if
a particle 0 falls from infinity.

Let $m_{0}$ be fixed. Obviously,

$m_{0}\geq m_{1}+m_{2}.$ (114)

Let us consider different particular cases.

$m_{0}=m_{1}+m_{2}$.

Then, $d=0$,

$\eta=\frac{m_{2}}{m_{0}}<1\text{,}$ (115)

so that the energy extraction is impossible.

$m_{2}=0$.

Then, $d=b_{2}^{2}$,

$\eta=\frac{b_{2}}{2m_{0}^{2}}(1+\sqrt{1+\frac{g_{00}}{\varepsilon^{2}}}).$
(116)

If also $m_{1}=0$ (two photons)

$\eta=\frac{1+\sqrt{1+\frac{g_{00}}{\varepsilon^{2}}}}{2}.$ (117)

If particle 0 comes from infinity, the maximum is achieved, if decay occurs on
the horizon and $\varepsilon=1$, then

$\eta=\frac{1+\sqrt{1+g_{00}}}{2}.$ (118)

This eq. coincides with eq. (7) of j  if one puts there $g_{t\phi}=-\omega
g_{\phi}$ and takes into account that on the horizon $N=0$, so
$g_{00}=\omega^{2}g_{\phi}.$

On the extremal horizon of the Kerr metric $g_{00}=+1$ and shn

$\eta=\frac{1+\sqrt{2}}{2}\text{, }\eta-1=\frac{\sqrt{2}-1}{2}.$ (119)

For the extremal Kerr circle orbit, $\varepsilon_{0}=\frac{1}{\sqrt{3}}$,
$g_{00}=1$, for decay to two photons we have

$\eta=\frac{1+\sqrt{1+3g_{00}}}{2}.$ (120)

On the near-horizon orbit $g_{00}=1$, $\eta=\frac{3}{2}$.

Let $m_{1}=m_{2}=m$. Then, $b_{2}=b_{1}=m_{0}^{2}$,
$d=m_{0}^{2}(m_{0}^{2}-4m^{2})$. Putting also $\varepsilon=1$ and for the
horizon taking again $g_{00}=+1$, we obtain

$\eta=\frac{1}{2}+\frac{\sqrt{1-4\frac{m_{2}^{2}}{m_{0}^{2}}}\sqrt{2}}{2}.$
(121)

that coincides, for example, with eq. 3.30 of pani . Then, $\eta\geq 1$,
provided

$\text{ }\frac{m_{2}}{m_{0}}\leq\frac{1}{2\sqrt{2}}\approx 0.35.$ (122)

If $\varepsilon\gg 1$,

$\eta_{\max}\approx 1+\frac{1}{4}\frac{g_{00}}{\varepsilon^{2}}.$ (123)

If $\varepsilon\rightarrow 0$, so particle 0 starts its motion not from
infinity but from the point that is already close to the horizon from the very
beginning, then

$\eta_{\max}\approx\frac{\sqrt{g_{00}}}{2\varepsilon}.$ (124)

Now we pose the following question. For a given $m_{0}$, how to choose the
masses $m_{1}$, $m_{2}$ if we want to achieve the maximum efficiency?

It follows from (113) that

$\eta=\frac{1}{2}(1+x_{2}-x_{1})+\frac{1}{2}\sqrt{1-2x_{1}-2x_{2}+(x_{2}-x_{1})^{2}}B\text{,}$
(125)

where $x_{i}=\frac{m_{i}}{m_{0}}\leq 1$, $i=1,2$,

$B=\sqrt{1+\frac{\left(g_{\phi}\omega^{2}\right)_{H}}{\varepsilon_{0}^{2}}}.$
(126)

One can calculate

$2\left(\frac{\partial\eta}{\partial
x_{1}}\right)_{x_{2}}=-1+\frac{B(x_{1}-x_{2}-1)}{\sqrt{1-2x_{1}-2x_{2}+(x_{2}-x_{1})^{2}}}<0$
(127)

As each term here is negative, $\left(\frac{\partial\eta}{\partial
x_{1}}\right)_{x_{2}}<0$. The maximum is reached for the minimum possible
$x_{1}=0$. Then, (113) gives us

$\eta=\frac{m_{0}^{2}+m_{2}^{2}}{2m_{0}^{2}}+\frac{m_{0}^{2}-m_{2}^{2}}{2m_{0}^{2}}B.$
(128)

This should be supplemented with the escaping condition which is model-
dependent.

In the horizon limit $N\rightarrow 0$

$X_{2}\approx\frac{N\omega\sqrt{g_{\phi}}}{2m_{0}^{2}\left(\omega^{2}g_{\phi}\right)_{H}}[b_{2}\sqrt{E_{0}^{2}+m_{0}^{2}\left(\omega^{2}g_{\phi}\right)_{H}}\pm
E_{0}\sqrt{d}],$ (129) $L_{2}\approx\omega_{H}g_{\phi
H}(E_{0}\frac{b_{2}}{2m_{0}^{2}\omega_{H}^{2}g_{\phi_{H}}}\pm\frac{\sqrt{E_{0}^{2}+m_{0}^{2}\omega_{H}^{2}g_{\phi_{H}}}}{2m_{0}^{2}\omega_{H}^{2}g_{\phi_{H}}}\sqrt{d}).$
(130)

For both signs, eq. (8) is fulfilled.

$E_{2}\approx\frac{E_{0}}{2m_{0}^{2}}b_{2}+\frac{\sqrt{d}\sqrt{E_{0}^{2}+\omega^{2}g_{\phi
H}}}{2m_{0}^{2}}\text{,}$ (131)
$E_{1}\approx\frac{E_{0}b_{1}}{2m_{0}^{2}}-\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{E_{0}^{2}+\omega^{2}g_{\phi
H}}.$ (132)

If, in scenario TP, we want particle 2 to escape, we must impose condition
$h_{2}>0$ where $h_{2}$ is given by eq. (31). In a general case, in the
horizon limit $N\rightarrow 0$ this condition cannot be fulfilled. Formally,
there is an exception when

$m_{2}=0\text{, }L_{2}=0.$ (133)

However, the latter condition is inconsistent with (76), for any $m_{0}\neq
0$. Actually, this means that the process with escape is impossible in this
case.

Meanwhile, this can becomes possible if both particles are ejected not in the
turning point and not parallel to the trajectory of particle 0. Then, the
efficiency is less than its maximum value. But, in this case eq. (133) opens a
possibility for particle (say, 2) to escape even if decay occurs near the
horizon.

##  XIII Efficiency: short summary of properties for the rotating space-times

###  XIII.1 Scenario TP

Let decay happen in the turning point. Then, there are the following
properties. 1) If $g_{00}$ is monotonically increasing function of $r$, the
maximum of $\eta$ is achieved at the horizon. 2) If particle 0 comes from
infinity, the maximum of $\eta$ is reached for $\varepsilon=1$. 3) For a given
$m_{0}$, the maximum of $\eta$ is reached if an escaping particle is massless.
The properties 1) - 2) are mentioned in literature time to time for different
concrete models. Here, we established them in a model-independent way.
Property 3) seems to be new.

###  XIII.2 Scenario AP

For decay in the ergoregion, both particles 1 and 2 move immediately after
decay in the same direction as particle 0 (provided decay occurs not in the
turning point). The calculation of efficiency has meaning, only if particle 2
bounces back and eclipse to infinity or if particle 0 moves before decay in
the outward direction.

We would like to stress that our conclusions are not universal. If ejection is
performed not parallel to the direction of motion, they do not apply. In
particular, then particle 2 can escape if $m_{2}=0$, $L_{2}=0$. Thus, there
are cases beyond the scope of the present article (see, e.g. esc1  \- tz ).

##  XIV Static black holes and neutral particles

In this section we consider characteristics of particle decay in the static
case. As before, we assume that all particles are neutral. Correspondingly,
the Penrose process is impossible. Nonetheless, such a decay can be of
interest since it models, for example, the processes of ejecting fuel from a
relativistic rocket rocket  when the Penrose process is absent, so the
efficiency $\eta<1$. There is one more motivation for discussion of this
issue. There exists a version of the Penrose process for static black holes
with electrically charged particles den , ruf . Therefore, it is desirable to
have formulas for neutral ones for future comparison of results for the
electric Penrose process and verifying corresponding limiting transitions when
all charges tend to zero.

There is a crucial difference between the static and stationary cases in the
sense that, as we saw it, if the process occurs in the ergosphere (that is
absent for static black holes) some scenarios are forbidden. For static
metrics, they are in general allowed. Therefore, decay in the static
background should be considered not just as a particular case of the process
near rotating black holes but, rather, as a separate issue. Now, $\omega=0$,
$g_{00}=-N^{2}$. As before, we consider the most interesting case when debris
of decay are ejected along the trajectory.

###  XIV.1 Scenario I

$L_{2}=\frac{L_{0}}{2m_{0}^{2}}(b_{2}-\frac{\sqrt{d}}{\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}}E_{0})\text{,}$
(134)
$L_{1}=\frac{L_{0}}{2m_{0}^{2}}(b_{1}+\frac{\sqrt{d}}{\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}}E_{0}).$
(135)
$E_{2}=E_{0}\frac{b_{2}}{2m_{0}^{2}}-\frac{\sqrt{d}\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}}{2m_{0}^{2}}$
(136)
$E_{1}=E_{0}\frac{b_{1}}{2m_{0}^{2}}+\frac{\sqrt{d}\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}}{2m_{0}^{2}}$
(137)
$P_{2}=\frac{P_{0}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m_{0}^{2}}}(E_{0}\sqrt{d}-b_{2}\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}})\text{,}$
(138)
$P_{1}=\frac{P_{0}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m_{0}^{2}}}(b_{1}\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}+E_{0}\sqrt{d}).$
(139) $E_{0}<\frac{b_{2}}{2m_{2}}N\text{.}$ (140)

###  XIV.2 Scenario II

$L_{1}=\frac{L_{0}}{2m_{0}^{2}}(b_{1}+\frac{\sqrt{d}}{\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}}E_{0})\text{,}$
(141)
$L_{2}=\frac{L_{0}}{2m_{0}^{2}}(b_{2}-\frac{\sqrt{d}}{\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}}E_{0}).$
(142)

$E_{1}=\frac{E_{0}b_{1}}{2m_{0}^{2}}+\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}$
(143)
$E_{2}=\frac{b_{1}E_{0}}{2m_{0}^{2}}-\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}$
(144)
$P_{1}=\frac{P_{0}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m^{2}}}(b_{2}\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}+E_{0}\sqrt{d})\text{,}$
(145)
$P_{2}=\frac{P_{0}}{2m_{0}^{2}\sqrt{E_{0}^{2}+g_{00}m^{2}}}(b_{2}\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}-E_{0}\sqrt{d}).$
(146) $E_{0}>\frac{b_{2}}{2m_{2}}N\text{.}$ (147)

Here, conditions (140), (147) follow from requirements $P_{1,2}>0$. The
efficiency in AP, scenario I

$\eta_{I}=\frac{E_{2}}{E_{0}}=\frac{b_{2}}{2m_{0}^{2}}-\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{1-\frac{N^{2}}{\varepsilon^{2}}}\text{.}$
(148)

As $b_{2}-\sqrt{d}<2m_{0}^{2}$, the efficiency $\eta<1$, since for static
metrics the ergoregion for neutral particles is absent, and the Penrose
process is absent as well.

###  XIV.3 Scenario TP

it follows from (72), (73) with $\omega=0$ that

$L_{0}=\pm\frac{1}{N}\sqrt{g_{\phi}(E_{0}^{2}-m_{0}^{2}N^{2})}$ (149)

can have any sign.

For scenario TP the maximum efficiency

$\eta_{TP}=\frac{E_{2}}{E_{0}}=\frac{b_{2}}{2m_{0}^{2}}+\frac{\sqrt{d}}{2m_{0}^{2}}\sqrt{1-\frac{N^{2}}{\varepsilon^{2}}}\text{.}$
(150)

This can also be rewritten as

$\eta=\frac{b_{2}}{2m_{0}^{2}}+\frac{\sqrt{d}}{2m_{0}^{2}}\frac{\left|L_{0}\right|}{\sqrt{L_{0}^{2}+g_{\phi}m_{0}^{2}}}.$
(151)

As $b_{2}+\sqrt{d}<2m_{0}^{2}$, this efficiency $\eta<1$, as it should be.

Different velocities are related according to the local Lorentz transformation

$V_{2}=\frac{v_{2}-V_{0}}{1-v_{2}V_{0}}\text{,}$ (152)

where $v_{2}$ is the velocity of particle 2 in the frame comoving with
particle 0 and is given by (65). The velocities $V_{0,2}$ are measured in the
laboratory (static) frame. They can be found from the relation

$E=\frac{mN}{\sqrt{1-V^{2}}}\text{.}$ (153)

One can check by straightforward calculations that (152) is indeed valid.

###  XIV.4 Near-horizon limit

To elucidate, what can happen to decay near a black hole, let us consider the
near-horizon limit when $r\rightarrow r_{h}$, where $r_{h}$ is the horizon
radius.

Let particle 0 turn into particles 1 and 2. Scenario TP is impossible now
since near the horizon where is no turning point for any particle coming from
infinity. Indeed, according to (68) with $\omega=0$, we would have
$E_{0}=O(N)$. However, for a particle under discussion $E_{0}\geq m_{0}$. Let
us consider scenario of type AP for a decay in a point which is not the
turning one.

####  XIV.4.1 Scenario I

Particle 1 falls in a black hole, particle 2 moves away from the horizon. It
follows from (140) that for $m_{2}\neq 0$ the point of decay cannot approach
the horizon arbitrarily close to the horizon since the limit $N\rightarrow 0$
is inconsistent with $E_{0}\geq m_{0}$. However, if $m_{2}=0$ it becomes
possible. The same is true if $m_{2}=O(N)$ is extremely small. For simplicity,
we assume that $m_{2}=0$ exactly. Then, $b_{2}=\sqrt{d}=m_{0}^{2}-m_{1}^{2}$,
$b_{1}=m_{0}^{2}+m_{1}^{2}$,

$P_{2}=\frac{P_{0}(m_{0}^{2}-m_{1}^{2})}{2m_{0}^{2}}(\frac{E_{0}}{\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}}-1),$
(154)
$P_{1}=\frac{P_{0}}{2m_{0}^{2}}(b_{1}+\frac{E_{0}b_{2}}{\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}}),$
(155)
$E_{2}=\left(\frac{b_{2}}{2m_{0}^{2}}\right)(E_{0}-\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}),$
(156)
$\eta=\frac{E_{2}}{E_{0}}=\left(\frac{b_{2}}{2m_{0}^{2}}\right)(1-\sqrt{1-\frac{m_{0}^{2}}{E_{0}^{2}}N^{2}}),$
(157)
$E_{1}=\left(\frac{b_{1}}{2m_{0}^{2}}\right)(E_{0}+\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}),$
(158)
$L_{2}=b_{2}\frac{L_{0}}{2m_{0}^{2}}(1-\frac{E_{0}}{\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}})\text{,}$
(159)
$L_{1}=\frac{L_{0}}{2m_{0}^{2}}(b_{1}+\frac{b_{2}}{\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}}E_{0}).$
(160)

For decay near the horizon, $N\ll 1$ and

$E_{2}\approx\frac{m_{0}^{2}-m_{1}^{2}}{4E_{0}}N^{2}\text{,}$ (161)
$E_{1}\approx E_{0}-\frac{(m_{0}^{2}-m_{1}^{2})N^{2}}{4E_{0}}\text{,}$ (162)
$P_{2}\approx\frac{P_{0}(m_{0}^{2}-m_{1}^{2})N^{2}}{4E_{0}^{2}},$ (163)
$P_{1}\approx P_{0}(1-\frac{m_{0}^{2}-m_{1}^{2}}{4E_{0}^{2}}N^{2})\text{,}$
(164) $L_{2}\approx L_{0}\frac{(m_{0}^{2}-m_{1}^{2})N^{2}}{4E_{0}^{2}},$ (165)
$L_{1}\approx L_{0}-\frac{(m_{0}^{2}-m_{1}^{2})N^{2}}{4E_{0}^{2}}L_{0}.$ (166)

It is particle 2 that escapes but it has almost vanishing energy: the closer
point of decay to the horizon, the smaller the energy.

####  XIV.4.2 Scenario II

Both particles fall in a black hole. Then, it follows from (50) - (54) that
near the horizon

$L_{1}\approx\frac{L_{0}}{2m_{0}^{2}}(b_{1}-\sqrt{d}-\sqrt{d}\frac{N^{2}m_{0}^{2}}{2E_{0}^{2}})\text{,}$
(167)
$L_{2}\approx\frac{L_{0}}{2m_{0}^{2}}(b_{2}+\sqrt{d}+\sqrt{d}\frac{N^{2}m_{0}^{2}}{2E_{0}^{2}})\text{,}$
(168)
$P_{1}\approx\frac{P_{0}}{2m_{0}^{2}}(b_{2}+\sqrt{d}+\frac{\sqrt{d}N^{2}}{2E_{0}^{2}}\text{,}$
(169)
$P_{2}\approx\frac{P_{0}}{2m_{0}^{2}}(b_{2}-\sqrt{d}-\frac{N^{2}m_{0}^{2}\sqrt{d}}{2E_{0}^{2}})\text{,}$
(170)
$E_{2}\approx\frac{E_{0}}{2m_{0}^{2}}(b_{2}-\sqrt{d}+\sqrt{d}N^{2}\frac{m_{0}^{2}}{2E_{0}^{2}})\text{,}$
(171)
$E_{1}\approx\frac{E_{0}}{2m_{0}^{2}}(b_{1}+\sqrt{d}-\sqrt{d}N^{2}\frac{m_{0}^{2}}{2E_{0}^{2}}).$
(172)

Now, for $m_{2}\neq 0$ both particles have comparable energies.

It is worth noting that, according to (140), in scenario I only massless
particle can escape that makes this scenario possible as exception. This can
be explained as follows on the basis of eq. (152). In the center of mass frame
(that coincides with particle 0) particles should move after decay in opposite
directions. To return to the laboratory frame, one has to make a local Lorentz
boost. For a massive particle, near the horizon this boost grows indefinitely
and overcomes an initial velocity, so particle is drifted towards a black
hole. However, if a particle is massless, the velocity is equal to the speed
of light from the very beginning and the boost with large but finite value
cannot change the direction of propagation (if we compare both frames).

And, in scenario II, the situation is opposite: according to (147), only in
the massless case particle 2 cannot fall in a black hole. The reason is the
same. Particle 1 in scenario II falls in a black hole, so in the center of
mass frame particle 2 moves in the outward direction. If particle 2 is
massless, even big but finite boost cannot overcome this, and particle 2
cannot move inwardly, scenario II fails. Instead, we can take just particle 1
to be massless. Then, nothing wrong occurs and scenario II is satisfied.

An important reservation is in order. These reasonings are valid for static
black holes but fail in the ergosphere of a rotating black holes. There,
another factors come into play and make it impossible for particle 2 to move
outwardly - see above discussion before eq. (56).

It is worth stressing the difference between rotating and static black holes
in the context under discussion. For rotating black holes, scenario I is
forbidden and particle 2 can escape only after preliminary bounce from the
potential barrier. For static black holes, this scenario is possible even if
decay occurs on the horizon. But, in the latter case, one should take the sign
”minus” before a square root in the efficiency (157). Then, in the near-
horizon limit, $\eta\rightarrow 0$. From another hand, scenario TP in the
near-horizon limit is impossible for static black holes and neutral particles
since it is inconsistent with the condition $E_{0}\geq m_{0}$.

###  XIV.5 Comparison with examples from literature

The formalism developed above enables us to reproduce the results for
particular metrics.

####  XIV.5.1 Schwarzschild metric

In this case, our eqs. (134), (135) agree with eq. (110) of rocket . It
follows from (136) that for scenario I

$\eta_{Sch}=\frac{E_{2}}{E_{0}}=\frac{b_{2}}{2m_{0}^{2}}-\frac{\sqrt{d}\sqrt{1-\frac{N^{2}}{\varepsilon_{0}^{2}}}}{2m_{0}^{2}}\text{,}$
(173)

where $\varepsilon_{0}=\frac{E_{0}}{m_{0}}$.

There is serious disagreement between our formulas and ju . In that paper, the
authors considered the efficiency according to their eqs. (34), (35). If one
neglects there the electric charges, it follows from the aforementioned
equations that $E_{3}\approx E_{1}$ for decay that occurs on the horizon, so
the efficiency $\eta\approx 1$. But, according to our treatment above, this
consideration applies only to a particle that falls in a black hole.
Meanwhile, the efficiency of the process should be evaluated for a particle
that escapes. Such a particle has almost vanishing energy (161) of the order
$N^{2}$, so the efficiency $\eta\ll 1$.

####  XIV.5.2 Magnetic field

The Penrose process in the electromagnetic field is a separate issue beyond
the scope of the present article. We only discuss very briefly for
illustration purposes motion of neutral particles in the Ernst background
ernst ,

$N^{2}=(1-\frac{2M}{r})(1+\frac{B^{2}r^{2}}{4}\sin^{2}\theta)\text{.}$ (174)

Full-fledged picture of particle motion in this space-time is quite nontrivial
and includes chaotic behavior kar92 . However, for our illustrative purposes
it is quite sufficient to consider the case of motion within the plane
$\theta=\frac{\pi}{2}$, according to the subject of the present article. Fore
mor details about properties of particle motion (charged or neutral) in this
background see 78 .

Then, for motion in the plane $\theta=\frac{\pi}{2}$, in scenario I according
to (156) we have

$E_{2}=\left(\frac{b_{2}}{2m_{0}^{2}}\right)(E_{0}-\sqrt{E_{0}^{2}-m_{0}^{2}N^{2}}),$
(175)
$\eta=\frac{E_{2}}{E_{0}}=\frac{b_{2}}{2m_{0}^{2}}-\frac{\sqrt{d}}{2m_{0}^{2}\varepsilon}\sqrt{\varepsilon^{2}-(1-\frac{2M}{r_{0}})(1+\frac{B^{2}r_{0}^{2}}{4})\text{.}}$
(176)

Here, it is assumed that $\varepsilon=\frac{E_{0}}{m_{0}}>N(r_{0})$. It is
seen that for given $r_{0}$, $\eta>\eta_{Sch}$ where
$\eta_{Sch}\equiv\eta(B=0)$ corresponds to the Schwarzschild metric.

If $B\rightarrow 0$,

$\eta=\eta_{Sch}+\frac{\sqrt{d}}{2m_{0}^{2}\varepsilon}\frac{(1-\frac{2M}{r_{0}})\frac{B^{2}r_{0}^{2}}{2}}{\sqrt{\varepsilon^{2}-(1-\frac{2M}{r_{0}})}}+O(B^{4}).$
(177)

It is worth noting that the Penrose process in this background was considered
quite recently in sh . In our view, the corresponding result for efficiency
described by Eq. (31) in the aforementioned paper is incorrect. It gives
$\eta=0$ in the neutral case $q=0$, $Q=0$, $B=0$ instead of the Schwarzschild
value $\eta_{Sch}$.

##  XV  Alternative approach

In this section, we compare two seemingly different approaches for the
description of particle decay. In this section, we proceed along the lines of
Refs. dad1 , win , j , ju , sh , nz  and demonstrate that they are equivalent
to general formulas derived above.

One can write

$p^{t}=-\frac{E}{g_{tt}+\Omega g_{t\phi}}\text{,}$ (178)
$p^{\phi}=p^{t}\Omega.$ (179)

It is easy to check using (7) that

$-p_{t}=E,\text{ }$ (180)

as it should be.

For a circle orbit, the equation $g_{\mu\nu}p^{\mu}p^{\nu}=-m^{2}$ gives us

$\Omega=\frac{-g_{t\phi}(\varepsilon^{2}+g_{tt})+\sqrt{(\varepsilon^{2}+g_{tt})\varepsilon^{2}g_{\phi}N}}{(g_{t\phi}^{2}+g_{\phi}\varepsilon^{2})}.$
(181)

As on the circle orbit $\dot{r}=0\,$, this equation should coincide with (83)
where $\Omega$ in the turning point is given. After some algebraic
manipulation one can confirm that this is indeed the case.

Using the conservation laws, we can write

$p_{(0)}^{t}Y_{(0)}=p_{(1)}^{t}Y_{(1)}+p_{(2)}^{t}Y_{(2)}\text{,}$ (182)
$p_{(0)}^{t}\Omega_{(0)}=p_{(1)}^{t}\Omega_{(1)}+p_{(2)}^{t}\Omega_{(2)}\text{,}$
(183)

where

$Y\equiv g_{00}+\Omega g_{t\phi},$ (184)

one obtains for the efficiency

$\chi=\frac{E_{2}}{E_{0}}=\frac{P_{(2)}^{t}Y_{(0)}}{Y_{(2)}P_{(0)}^{t}}$ (185)

that is equivalent to

$\chi=\frac{\Omega_{0}-\Omega_{1}}{\Omega_{2}-\Omega_{1}}\frac{Y_{0}}{Y_{2}}$
(186)

in agreement with Refs. mentioned in the beginning of this Section.

One can check that (112) agrees completely with (185).

Thus both approaches give the same results. The approach based on formulas
(43) - (54), has the advantages in that (i) it operates directly with
characteristics which are given in the problem as initial conditions (energy
$E_{0}$ and angular momentum $L_{0}$), (ii) it gives not only energies of
particles 1 and 2 but also their angular momenta, (iii) is applicable not only
in the point of decay or on a circle orbit.

##  XVI Summary and conclusions

We considered the background of a rotating axially symmetric black hole.
Classification of scenarios of decay to two particles is suggested. We mainly
restricted ourselves by the case when both particles are ejected in the
parallel direction with respect to the motion of a parent particle. It is this
particular case which is of the maximum interest since it enables us to obtain
maximum efficiency. In particular, we showed that inside the ergoregion there
is severe restriction on some scenarios of this type. Thus, it is shown that
if decay occurs in a generic point (not a turning point for radial motion)
inside the ergoregion, both new particles 1 and 2 move after decay in the same
radial direction that coincides with direction of a parent particle 0.

We argued that description of the Penrose process can be taken directly from
the formulas derived by Wald many years ago. However, these formulas were
written in terms of the velocities of particles with respect to a parent
particle 0. Meanwhile, we transformed the corresponding formulas, giving them
in terms of particle masses directly. In particular, in this manner, we found
the expressions for the gamma factor of relative motion of particles before
and after decay. Scenarios with decay in the turning point are also considered
in a model-independent way. We also discussed one more approach for
description of decay, popular in recent years, and showed its equivalence to
ours (with advantage that our approach gives the answer explicitly in terms of
characteristics of particle 0 and all three particle masses).

Our results enabled us to reject some misleading formulas in previously
published works where physically incorrect results appeared due to confusion
between particles that fall in a black hole and those that escape. Especially,
this concerns the decay near the horizon, where it is important to know the
relation between the energy of escaping particle and direction of its motion.

We also considered in detail possible variants of motion in the angular
direction. In doing so, the angular momenta of particles 1 and 2 are found
explicitly. We traced in detail how the ”plus” and ”minus” signs are
interwoven in different scenarios.

We collected and enlarged the results, how efficiency depends on the scenario,
location of a point where decay occurred, and particle passes. Some part of
corresponding statements were already known but existed as some folklore
scattered over literature and discussed for different models. Now, they are
given more accurate substantiation, also in a model-independent way.

Of interest is to apply the developed approach to decay processes in the
external electromagnetic field and consider astrophysical applications.

##  XVII Acknowledgment

I am grateful to Yuri Pavlov for interest to this work and useful discussion.
I also thank Naresh Dadhich for useful correspondence and some references. I
thank for hospitality Institute of Theoretical Physics of Charles University
(Prague) and CENTRA (Lisbon). I acknowledge financial support from grant No.
GAČR 21/11268S of the Czech Science Foundation and Fundac̃ao para a Ciência e
a Tecnologia - FCT, Portugal, project No. UIDB/00099/2020.

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