Table Of ContentHyperbolic trajectories around Black Holes
7 E. A. Le´ona1, J. A. Nietoa2, E. R´ıos-L´opezb3
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0 aFacultad de Ciencias F´ısico-Matem´aticas de la Universidad Aut´onoma
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de Sinaloa, 80010, Culiac´an Sinaloa, M´exico
n
a bInstituto Nacional de Astrof´ısica, O´ptica y
J
Electr´onica (INAOE), 72840, Tonantzintla, Puebla, M´exico
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]
c
q
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r
g
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1 Abstract
v
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We analyse test particle’s trajectories around the geometry of a Schwarzschild
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3 black hole. In order to resemble sections of jets in the neighborhood of a black
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hole, we consider the conserved quantities corresponding to constraints im-
0
. posed on the trajectories of the test particles, namely conic and hyperboloidic
1
0 trajectories. As expected, the energy and angular momentum are closely re-
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lated to the solutions in the non-constrained case.
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:
v
i
X
r
a
Keywords: Schwarzschild geometry, Trajectories around Black holes, Con-
strained systems.
Pacs numbers: 04.20.Fy, 04.20.Jb, 04.70.Bw
January 2017
[email protected]
[email protected], [email protected]
[email protected]
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1. Introduction
The orbits and trajectories of objects around black holes have been extensively
studied in the literature [1]-[4]. However, there remain several interesting
scenarios where the theory can be tested and even those where one can search
for modifications of general relativity [5][6]. Eventually, this could lead to a
better understanding of dark energy and dark matter [7]-[9].
In this work we consider hyperbolic trajectories of test particles around
a Schwarzschild black hole. We start by reviewing the formalism used for
obtaining conserved quantities in this static geometry, mentioning how the
static limit can be viewed as a hamiltonian constrained system. Next, we
turn the attention to the case of constrained trajectories of test particles. The
motivation for this analysis is that the case of a constraint in the form of
hyperboloid can be contrasted with the case of particles in the surface of a jet
in the boundaries of a black hole, at least to some approximation.
2. Static geometry for black holes as hamiltonian system
We start by considering the static geometry described by the Schwarzschild
metric, given by the metric
ds2 = γdt2 +γ−1dr2 +r2(dθ2 +sin2θdφ2), (1)
−
where
R
S
γ = 1 . (2)
− r
Here, the quantity R = 2GM is the Schwarzschild radius. The notation in this
S c2
article is as follows: greek indices (µ, ν, ...) run from 0 to 3 and latin indices
(i, j, ...) run from 1 to 3, corresponding to purely spatial indices. Derivatives
respect to coordinate r will be denoted by primes, e. g. df = f′(r), while
dr
derivatives respect to τ, the proper time, shall be written by an over dot, as
t˙= dt. Further, from here on we consider units such that G = c = 1.
dτ
For obtaining the trajectories one can vary the function
1 r˙2
= m γt˙2 + +r2 θ˙2 +sin2θφ˙2 , (3)
L 2 − γ
(cid:20) (cid:21)
(cid:16) (cid:17)
where m denotes the test particle mass.
Let us check the simplest case, where we impose a particle to be moving
in the plane θ = π. Then (3) transforms into
2
1 r˙2
= m γt˙2 + +r2φ˙2 . (4)
L 2 − γ
(cid:18) (cid:19)
2
The Euler-Lagrange equation for t is
d
(mγt˙) = 0. (5)
dτ
For the coordinate r, the variation leads to
r¨ r˙γ˙ γ′t˙2 r˙2γ′
˙2
+ + rφ = 0. (6)
γ − γ2 2 2γ2 −
Since we have that γ˙ = dγ = dγ dr = γ′r˙, we can substitute this in (6) and
dτ dr dτ
multiply by γ, obtaining the simpler relation
r˙2γ′ γγ′t˙2
˙2
r¨ + γrφ = 0. (7)
− 2γ 2 −
In a similar way, the variation respect to φ yields
d
(mr2φ˙) = 0. (8)
dτ
The solutions for (5) and (8) are
a
t˙= (9)
γ
and
h
˙
φ = , (10)
r2
respectively, where a and h are constants. Substitution of both solutions into
(7) gives
r˙2γ′ a2γ′ h2γ
r¨ + = 0. (11)
− 2γ 2γ − r3
Now, in this case (θ = π/2) the Schwarzschild metric (1) can be expressed
as γt˙2 + r˙2 + r2φ˙2 = 1, and taking into account (9) and (10) this is the
− γ −
same as
h2γ
r˙2 = γ +a2 . (12)
− − r2
The combination of (11) with (12) allow us to cancel terms with a2, and
therefore we arrive to the relation
γ′ h2γ′ h2γ
r¨+ + = 0. (13)
2 2r2 − r3
3
Since γ = 1 2M, we have that γ′ = 2M. With this, and rearranging some
− r r2
terms in (13), we see that it is equivalent to
M h2 3M
r¨+ = 1 . (14)
r2 r3 − r
(cid:18) (cid:19)
Multiplying (14) by r˙ and integrating, we obtain
r˙2 M h2 Mh2
+ = E, (15)
2 − r 2r2 − r3
where E is a constant, that can identified with the energy of the system (per
mass unit), and which is related with a by E = a2−1. This allows to interpret
2
(15) in the form E = r˙2 +V, where V is the potential V = M + h2 Mh2.
2 − r 2r2 − r3
In Fig. 1 is plotted the potential V(r) for radial motion for different values of
h and also the Newtonian potential (dashed line), where can be seen that for
largevalues of r the relativistic potential is similar to the Newtonian potential.
This function has extrema in r± = h h √h2 12M2 . If h2 > 12M2, this
2M ± −
value indicates an (stable) outer circular orbit, as well as an (unstable) inner
(cid:0) (cid:1)
circular orbit for test particles.
0.10
h/M = 5
h/M = 4.6
h/M = 4.3
h/M = 4
h/M = 3.2
Newtonian
0.05
V
0.00
0.05
0 10 20 30 40 50 60 70 80
r/M
Figure 1: The relativistic and Newtonian potentials for radial motion for different values
of h/M.
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Furthermore, the equations can be recasted to a suitable form by mean of
the change of variable u = 1. This implies du = 1 dr, and we also use the
r dφ −r2dφ
relation r˙ = φ˙ dr = h dr, where in the last equality we have taken into account
dφ r2dφ
(10). By substituting all these in (15), we have:
h2 du 2 h2u2
Mu+ Mh2u3 = E. (16)
2 dφ − 2 −
(cid:18) (cid:19)
Differentiating respect to φ and rearranging terms, we see that (16) yields
the simplified version
d2u M
+u = 3Mu2 + . (17)
dφ2 h2
Equations (10) and (17) determine the trajectories of tests particles mov-
ing in the gravitational field imposed by Schwarzschild metric. Of course,
another approach is to solve directly the geodesic equation d2xµ +Γµ dxαdxβ.
dτ2 αβ dτ dτ
We can compare the mentioned equations with the corresponding relations in
newtonian theory, where we have dφ = h and d2u +u = M, respectively. The
dt r2 dφ2 h2
well-known relativistic correction is the term 3Mu2 in (17), that permits to
calculate the perihelion shift in Mercury, for instance [10].
3. Hyperbolic trajectories.
As observations indicate, the jets propelled by black holes consist of a colli-
mated beam of relativistic particles [11][12]. This suggests to constrain the
movement of particles to an adequate geometry that emulates sections of jets.
In this direction, we add a constriction to the Lagrangian (3) in such a way
that tests particles are forced to move along a hyperboloid (see Fig. 2). As
limit case, we first review the case of movement in a cone. The revolution axis
is z and the opening cone angle θ should be small in order to emulate the men-
tioned beam collimation. Both cases are included in the surfaces generated by
the relation Ax2+By2 Pz2 Q = 0. Here, we are assuming that A, B and P
− −
are positive constants, while the conic and hyperboloid geometry are imposed
by Q = 0 and Q > 0, respectively. Also, azimuthal symmetry is guaranteed
by A = B = 1. Then, without loss of generality we have as constraint:
x2 +y2 Pz2 Q = 0. (18)
− −
◦
Note that varying P between 0 and 1 allows to vary θ from 0 to 45 .
By mean of a Lagrange multiplier, we add the constraint (18) to the La-
grangian (3), in the form
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1 r˙2 N
= m γt˙2 + +r2 θ˙2 +sin2θφ˙2 + x2 +y2 Pz2 Q , (19)
L 2 − γ 2 − −
(cid:20) (cid:21)
(cid:16) (cid:17)
(cid:0) (cid:1)
where N is a Lagrange multiplier.
Z
X
Figure 2: The proposed constraint in Eq. (18) forces to the particles to move into an
adequategeometryemulating sectionsofrelativistic jets, specifically alongofa hyperboloid
as can be seen here for Q > 0 and a small value of P (Q = 0 is for the conic case). Also,
the maximum approach of the particles to the singularity is given when r =√Q .
Letsfirst consider thecaseofacone(Q = 0). Theideaistofollowsimilarsteps
to those shown in section 2. Using this expression we find that the Lagrangian
(19) can be rewritten as
1 r˙2 N
= m γt˙2 + +r2sin2θφ˙2 + [r2(sin2θ P cos2θ)]. (20)
L 2 − γ 2 −
(cid:20) (cid:21)
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Since in this expression we are assuming that the angle θ is constant, we shall
consider only variations of (20) with respect t, r, φ and N. From the Euler-
Lagrange equation for t, we obtain d (mγt˙) = 0 and therefore we find that
dτ
a
t˙= , (21)
γ
where a is a constant. After simplification, the variation of (21) with respect
to r yields
r˙2γ′ γγ′t˙2 N
r¨ + γr sin2θφ˙2 + (sin2θ P cos2θ) = 0. (22)
− 2γ 2 − m −
(cid:20) (cid:21)
The variation respect to φ leads to d (mr2sin2θφ˙) = 0, and thus we obtain
dτ
b
˙
φ = , (23)
r2sin2θ
where b is another constant.
For the Lagrange multiplier N, it results
r2(sin2θ P cos2θ) = 0, (24)
−
which is consistent with the constraint x2+y2 Pz2 = 0, obtained from (18)
−
when Q = 0. Note that from this we get
P = tan2θ. (25)
By using (21), (23) and (24) the equation (22) becomes
r˙2γ′ a2γ′ b2γ
r¨ + = 0. (26)
− 2γ 2γ − r3sin2θ
Note that if we make the association h2 b2 , (26) takes exactly the
→ sin2θ
same form as (11). But now, the Schwarzschild metric implies γt˙2 + r˙2 +
− γ
r2sin2θφ˙2 = 1, and by (21) and (23) this expression can be written as
−
b2γ
r˙2 = γ +a2 . (27)
− − r2sin2θ
Insertion of (27) into (26) leads to
γ′ b2γ′ b2γ
r¨+ + = 0. (28)
2 2r2sin2θ − r3sin2θ
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By following similar steps to the previous section, taking into account γ =
1 2M and γ′ = 2M, after a straightforward computation we see that (27)
− r r2
yields to the integration constant
r˙2 M b2 b2M
+ = E. (29)
2 − r 2r2sin2θ − r3sin2θ
The trajectories with angular momentum zero (b = 0) correspond to radial
paths, and then (29) converts to
r˙2 M
= E. (30)
2 − r
From (21) and (30) we have the components of the four velocity vector uα,
uα = a, γ +a2,0,0 (recall that E = a2−1 and γ = 1 2M), that clearly
γ ± − 2 − r
satisfi(cid:16)es thepnormalization(cid:17)uαuα = 1. In the component u1 = r˙ the positive
−
sign indicates particles going outwards while the negative sign corresponds to
free falling particles [13].
From (30) we have that r˙ = 2(Er+M), with solution
± r
q
Er(Er+M) M ln E(Er +M)+E√r = E√2E(τ τ˜). (31)
− ± −
p hp i
Since from (21) we have that dr = drdτ = 1 2M 2(Er+M), in
dt dτ dt ± − r (2E+1)r
Schwarzschild coordinates the solution is q
(cid:0) (cid:1)
√2E +1 Er(Er +M)+(4E 1)M ln E(Er +M)+E√r
− −
n h io
p p
4ME√2Etanh−1 (2E+1)r = E√2E(t t˜).
2(Er+M) ± −
hq i (32)
The particular case where the velocity of the particle tends to zero in r
→
corresponds to E = 0 in (30), and correspondingly r˙ = 2M. Integration
∞ r
yields q
2
3
3
r = 2M (τ τˆ) . (33)
s 2 −
(cid:20) (cid:21)
This result can be recasted in terms of Schwarzschild coordinates (t,r),
since from (21) we have dr = dr dτ = 2M 1 2M , with solution
dt dτ dt r − r
q
(cid:0) (cid:1)
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r 1 r 3/2 r
t = tˆ+4M + tanh−1 . (34)
2M 3 2M − 2M
(cid:20)r r (cid:21)
(cid:16) (cid:17)
In (31-34) the quantities τ˜, t˜, τˆ and tˆare integration constants.
These resultscorrespondtotheconicgeometry[Q = 0in(18)]. Thiscanbe
seen as a limiting case for a model with a point source for the collimated beam
modelling a jet. It can be also interpreted as the limit case for a hyperbolic
constraint. Therefore, we focus the attention in the case Q > 0 in (18). Thus,
the more general Lagrangian now becomes
1 r˙2 N
L = m γt˙2 + +r2 θ˙2 +sin2θφ˙2 + [r2(sin2θ P cos2θ) Q], (35)
2 − γ 2 − −
(cid:20) (cid:21)
(cid:16) (cid:17)
˙
where now θ = 0. The variations respect to t and φ give again (21) and (23),
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while the variations respect to r and θ results in
r˙2γ′ γγ′t˙2 N
r¨ + γr (θ˙2 +sin2θφ˙2)+ (sin2θ P cos2θ) = 0, (36)
− 2γ 2 − m −
(cid:20) (cid:21)
and
d
(mr2θ˙) mr2sinθcosθφ˙2 (1+P)Nr2sinθcosθ = 0, (37)
dτ − −
respectively.
It is useful to express the constraint in (18) as
Q
r2 = . (38)
sin2θ P cos2θ
−
Multiplying Eq. (37) by mr2θ˙ and inserting (23) and (38) in its second
˙
and third term, we get rid of the dependence on r and φ. The result is
˙ ˙
d cosθθ sinθcosθθ
mr2θ˙ (mr2θ˙) m2b2 mN(1+P)Q2 = 0. (39)
dτ − sin3θ − [sin2θ P cos2θ]2
−
This can be rewritten as
d m2b2 mNQ2
(mr2θ˙)2 + + = 0. (40)
dτ sin2θ sin2θ P cos2θ
(cid:20) − (cid:21)
We use again (23) and (38) to eliminate b and Q, and after integration (40)
gets converted to
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N l2
(θ˙2 +sin2θφ˙2)+ (sin2θ P cos2θ) = , (41)
m − m2r4
where l is constant. Now, this expression appears in square brackets in Eq.
(36), and then we have:
r˙2γ′ γγ′t˙2 l2γ
r¨ + = 0. (42)
− 2γ 2 − m2r3
In a convenient way, we define h = l and use [from (21)] t˙2 = a2 in the
m γ2
last relation, obtaining
r˙2γ′ a2γ′ h2γ
r¨ + = 0. (43)
− 2γ 2γ − r3
Again, we have the same relation as (11), although in this case integration
is less trivial. From the definition of solid angle, Ω˙2 = θ˙2 +sin2θφ˙2, and then
(41) can be rewritten as
h2 N
Ω˙2 = (sin2θ P cos2θ). (44)
r4 − m −
Also, the Schwarzschild metric implies the relation
r˙2 a2 r˙2
γt˙2 + +r2(θ˙2 +sin2θφ˙2) = + +r2Ω˙2 = 1, (45)
− γ −γ2 γ −
where we have used t˙2 = a2. From the last equality of the equation, isolating
γ2
r˙2 and using (23) and (44), results
h2 Nr2
r˙2 = a2 γ 1+ (sin2θ P cos2θ) . (46)
− r2 − m −
(cid:20) (cid:21)
Then we substitute this in (43) to obtain
γ′ h2 Nr2 h2γ
r¨+ 1+ (sin2θ P cos2θ) = 0. (47)
2 r2 − m − − r3
(cid:20) (cid:21)
Using sin2θ P cos2θ = Q[cf. Eq. (38)], γ = 1 2M and γ′ = 2M, (47)
− r2 − r r2
takes the form
M NQ h2 3M
r¨+ 1 = (1 ). (48)
r2 − m r3 − r
(cid:18) (cid:19)
Multiplying by r˙, this can be integrated to obtain the energy:
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