4 Advances in High Energy Physics
where
𝑎𝜇]
=(1/2)
𝜇]𝜅𝜆
𝑎
𝜅𝜆
.roughoutthispaper,wewill
consider only static congurations. e dierence between
monopole and dyon is whether
𝑎
0
is zero or nonzero,
respectively. For monopole, the Bianchi identity becomes
D
𝑖
𝑎
𝑖
=0.
(15)
Here
𝑎
𝑖
=(1/2)
𝑖𝑗𝑘
𝑗𝑘
and ,,=1,2,3are the spatial
indices. For dyon,
𝑎
0
=0, there are additional equations of
motion for “electric” part since the Gauss law is nontrivial,
D
𝑖
𝑏
𝑖
=−
𝑏𝑐𝑎
𝑐
D
0
𝑎
,
(16)
where
𝑎
𝑖
=
𝑎
0𝑖
.
We could write the energy-momentum tensor
𝜇]
by
varying the action with respect to the space-time metric. e
energy density is then given by
00
component,
00
=
1
2
D
0
𝑎
D
0
𝑎
+D
𝑖
𝑎
D
𝑖
𝑎
+
𝑎
𝑖
𝑎
𝑖
+
𝑎
𝑖
𝑎
𝑖
+
.
(17)
In [5], it is possible to obtain the exact solutions of the Euler-
Lagrange equations in the BPS limit, i.e., =0,butstill
maintaining the asymptotic boundary conditions of ,and
we dene a new parameter such that
00
=
1
2
D
0
𝑎
D
0
𝑎
+D
𝑖
𝑎
D
𝑖
𝑎
sin
2
+
𝑎
𝑖
𝑎
𝑖
+D
𝑖
𝑎
D
𝑖
𝑎
cos
2
+
𝑎
𝑖
𝑎
𝑖
=
1
2
D
0
𝑎
2
+D
𝑖
𝑎
sin
𝑎
𝑖
2
+D
𝑖
𝑎
cos
𝑎
𝑖
2
±
𝑎
𝑖
D
𝑖
𝑎
sin ±
𝑎
𝑖
D
𝑖
𝑎
cos .
(18)
e last two terms can be converted to total derivative
𝑎
𝑖
D
𝑖
𝑎
=
𝑖
𝑎
𝑖
𝑎
−
D
𝑖
𝑎
𝑖
𝑎
=
𝑖
𝑎
𝑖
𝑎
,
(19a)
𝑎
𝑖
D
𝑖
𝑎
=
𝑖
𝑎
𝑖
𝑎
−
D
𝑖
𝑎
𝑖
𝑎
=
𝑖
𝑎
𝑖
𝑎
,
(19b)
aer employing the Gauss law (16) and Bianchi identity (15).
ey are related to the “Abelian” electric and magnetic elds
identied in [1], respectively. Since the total energy is =
∫
3
00
, the total derivative terms can be identied as the
electric and magnetic charges accordingly
Q
𝐸
=
𝑖
𝑎
𝑖
𝑎
, (20a)
Q
𝐵
=
𝑖
𝑎
𝑖
𝑎
, (20b)
with
𝑖
denoting integration over the surface of a 2-sphere
at →∞. erefore the total energy is ≥±(Q
𝐸
sin +
Q
𝐵
cos )since the other terms are positive semidenite. e
total energy is saturated if the BPS equations are satised as
follows [24]:
D
0
𝑎
=0,
(21a)
D
𝑖
𝑎
sin =
𝑎
𝑖
,
(21b)
D
𝑖
𝑎
cos =
𝑎
𝑖
.
(21c)
Solutions to these equations are called BPS dyons; they are
particularly called BPS monopoles for =0.eenergyof
this BPS conguration is simply given by
𝐵𝑃𝑆
=±Q
𝐸
sin +Q
𝐵
cos .
(21d)
Adding the constant contained in sin and cos is some-
how a bit tricky. We will show later using BPS Lagrangian
method that this constant comes naturally as a consequence
of identifying two of the eective elds.
Employing the ’t Hoo-Polyakov, together with Julia-Zee,
ansatz[1,2,4]
𝑎
=
(
)
𝑎
,
(22a)
𝑎
0
=
(
)
𝑎
,
(22b)
𝑎
𝑖
=
1−
(
)
𝑎𝑖𝑗
𝑗
2
,
(22c)
where
𝑎
≡(,,)and
𝑖
≡(,,)as well, denotes the
Cartesian coordinate. Notice that the Levi-Civita symbol
𝑎𝑖𝑗
in ((22a), (22b), and (22c)) mixes the space-index and the
group-index. Substituting the ansatz ((22a), (22b), and (22c))
into Lagrangian (9) we can arrive at the following eective
Lagrangian:
L
s
=−
2
2
−
2
+
2
2
2
+
2
−
2
−
1
2
2
−1
2
2
−,
(23)
where ≡/; otherwise it means taking derivative over the
argument. As shown in the eective Lagrangian above there
is no dependency over angles coordinates and despite the
fact that the ansatz ((22a), (22b), and (22c)) depends on and
. us we may derive the Euler-Lagrange equations from the
eective Lagrangian (23) which are given by
−
1
2
2
+
2
2
2
=−
,
(24a)
−
2
2
+
2
2
2
=0,
(24b)
2
−1
2
+
2
2
−
2
−
=0.
(24c)
Later we will also consider the case for generalize Lagrangian
of (9) by adding scalar-dependent couplings to the kinetic
terms as follows [25]:
L
G
=−
1
4
𝑎
𝜇]
𝑎𝜇]
+
1
2
D
𝜇
𝑎
D
𝜇
𝑎
−
.
(25)