Eur. Phys. J. C (2019) 79 :161 Page 5 of 24 161
A =−Et + Lφ + A
r
(r) + A
θ
(θ) (6)
Here A
r
(r) and A
θ
(θ) correspond to the parts of the action
dependent on radial coordinate r and angular coordinate θ
respectively. Given the above structure of the action asso-
ciated with a particle moving in the Kerr–Newman-NUT
spacetime, the Hamilton–Jacobi equation becomes separa-
ble. Thus one can obtain separate equations for A
r
(r) and
A
θ
(θ) respectively, having the following structures,
dA
θ
dθ
2
+
(EP − L)
2
sin
2
θ
+ m
2
(l + a cos θ)
2
= K (7)
dA
r
dr
2
+ m
2
r
2
−
1
{E(r
2
+ a
2
+l
2
) − aL}
2
=−K (8)
In the above expressions the quantity K acts as the sepa-
ration constant and m is the mass of the orbiting particle.
Among others l is the NUT charge, a is the rotation parame-
ter and P = a sin
2
θ −2l cos θ . Since the components of the
momentum four vector of the orbiting particle correspond
to, ∂A/∂ x
μ
, it is possible to rewrite Eqs. (7) and (8)in
a more explicit form. This involves writing the separation
constant K appearing in both Eqs. (7) and (8), such that
K = λ +(L −aE)
2
, with λ as the Carter constant. Thus we
obtain the following geodesic equations,
m
2
ρ
4
dr
dτ
2
={E(r
2
+ a
2
+l
2
) − aL}
2
− (λ+m
2
r
2
) −
(
L − aE
)
2
(9)
m
2
ρ
4
sin
2
θ
dθ
dτ
2
= λ sin
2
θ +
(
L − aE
)
2
sin
2
θ
−
[
EP−L
]
2
−m
2
sin
2
θ(l+a cos θ)
2
(10)
where τ represents the affine parameter for a timelike
geodesic. Similar to the above case with massive particles, for
photons with zero rest mass, the geodesic equations become,
ρ
4
dr
dν
2
={E(r
2
+ a
2
+l
2
) − aL}
2
− λ −
(
L − aE
)
2
(11)
ρ
4
sin
2
θ
dθ
dν
2
=λ sin
2
θ +
(
L − aE
)
2
sin
2
θ −
[
EP − L
]
2
(12)
where ν stands for affine parameter along the null geodesic.
Given the above two geodesic equations one can proceed
towards solving these equations, which will ultimately lead to
the trajectory of a particle moving in the Kerr–Newman-NUT
spacetime. With the complicated structure of the geodesic
equations as presented above, it is very difficult to solve
them in general circumstances. However in some specific
situations, e.g., in the equatorial plane (located at θ = π/2)
it is indeed possible to solve the above equations analyti-
cally, which we will discuss next. Having grasped the analyt-
ical structure of the trajectory associated with the equatorial
plane, we will consider the general scenario later on.
3.1 Orbits confined on a given plane
The addition of the NUT charge will introduce nontrivial
difficulties while obtaining the orbital dynamics for a mas-
sive or even a massless particle. This can be understood by
employing the angular equations given in Eqs. (10) and (12)
respectively. For any trajectory to be confined on a particu-
lar plane θ = θ
0
, we bound to have
˙
θ =
¨
θ = 0. While the
condition
˙
θ = 0 indicates a trajectory moving on a constant
plane, the additional second derivative ensures that the first
derivative remains null throughout its motion. Therefore both
of these conditions are essentials to determine any planner
orbit in the presence of a NUT charge and unlike the Kerr
black hole, this constraint would introduce stringent bound
on the particle trajectories in Kerr–Newman-NUT spacetime.
In the upcoming discussions, we shall explore any possible
scenarios in which the conditions for a planner orbit can be
satisfied in a certain ranges of parameters. Regarding that, we
first introduce the massless particles and following that, the
phenomenon involving massive particles will be addressed.
3.1.1 The massless particles
For a massless particle, the effective radial potential can be
written as
V
eff
(r) ={E(r
2
+ a
2
+l
2
) − aL}
2
− {λ +
(
L − aE
)
2
}.
(13)
Remember, the value of the Carter constant λ would only be
fixedfromtheangularequations,i.e,
˙
θ =
¨
θ = 0. Eventhough
for a general E and L, the angular equations are compli-
cated to solve and often subjected to numerical calculations,
we start with a simple exercise which corresponds to the
following relation between angular momentum and energy:
L = aE. In this case the impact parameter, defined as the
ratio of angular momentum and energy, turns out to be D
sp
≡
L/E = a. On other hand, the angular equations become
V
eff
(θ) = λ sin
2
θ −{aE sin
2
θ − 2El cos θ − aE}
2
= 0,
V
eff
(θ)
dθ
= 2sinθ cos θλ − 2{aE sin
2
θ − 2El cos θ − aE}
123
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