Dynamics of Nambu monopole in two Higgs doublet models -- Cosmological Monopole Collider --

We study the dynamics of the Nambu monopole in two Higgs doublet models, which is a magnetic monopole attached by two topological $Z$ strings ($Z$ flux tubes) from two opposite sides. The monopole is topologically stable solution of the equation of motions when the Higgs potential has two global $U(1)$ and $\mathbb{Z}_2$ symmetries. In this paper, we consider more general cases without the $\mathbb{Z}_2$ symmetry, and find that it is no longer a static solution but moves along the $Z$ string being pulled by the heavier string. After analytically constructing an asymptotic form of the monopole, we confirm such a motion using the numerical relaxation method. In addition, we analyze the real time dynamics of the monopole based on a point-like approximation. Consequently, if there were long string networks with the monopoles in the early universe, the monopole accelerates nearly to the speed of light emitting electromagnetic radiations as a synchrotron accelerator, and collides to an anti-monopole on the string. This collision event, which we call the cosmological monopole collider, can produce much heavier particles than those we can see today, e.g., at the Large Hadron Collider.


Introduction
Topological solitons in field theories play important roles in various fields of physics such as particle physics, condensed matter physics and cosmology.Particularly, magnetic monopoles are prominent examples among them, which were first discovered by 't Hooft and Polyakov [1,2] in the SO(3) Yang-Mills-Higgs model as a field theoretical realization of the Dirac hypothesis [3] providing an explanation for the electric charge quantization.While theoretically they play crucial roles to study non-perturbative aspects of (non-)supersymmetric field theories [4][5][6], experimentelly such monopoles have never been found in reality, except for condensed-matter analogues [7,8].For instance, such magnetic monopoles are predicted in all grand unified theories (GUTs) [9][10][11][12], and their search have been extensively conducted.Nevertheless, no GUT monopoles have been found so far.
On the other hand, a magnetic monopole configuration in the Standard Model (SM) was first considered by Nambu [13], which is called the Nambu monopole.Since the Nambu monopole is attached by a string and consequently is pulled by the tension of the string, it cannot be a stable soliton.Nevertheless the Nambu monopoles might be phenemenologically or cosmologically useful; they were suggested to produce primordial magnetic fields before their disappearance [14,15].The reason of the absence of stable monopoles in the SM is its trivial topology, that is, the vacuum manifold is S 3 having a trivial second homotopy group π 2 .In the same way, the vacuum manifold S 3 has trivial π 0 for domain walls and π 1 for cosmic strings.Non-topological electroweak Z strings (or magnetic Z-fluxes) [16][17][18][19][20][21] have been studied extensively, but they were shown to be unstable in the realistic parameter region [22,23].The Nambu monopoles are the end points of these electroweak Z strings [13].
Incidentally, while the SM was established by the discoverly of the 125 GeV Higgs boson (h) at the Large Hadron Collider (LHC), there remain several problems that are unanswered by the SM.Two Higgs doublet models (2HDMs), in which one more Higgs doublet is added to the SM, are one of the most popular extensions of the SM with a potential to solve unsolved problems of the SM (for reviews, see, e.g., Refs.[24,25]).In addition to the 125 GeV Higgs boson (h), it has four additional scalar degrees of freedom: the charged Higgs bosons (H ± ), the CP-even Higgs boson (H) and the CP-odd Higgs boson (A).These additional scalars could be directly produced at the LHC, though there is no signal so far today, therefore placing lower bounds on masses of those additional scalar bosons.Those lower bounds highly depend on parameter choices of the 2HDM.For more detailed phenomenological studies, see, e.g., Refs.[26][27][28][29][30] and references therein.One of the most remarkable aspects of 2HDMs distinguishable from the SM may be that it has a much richer vacuum structure than the SM, thereby allowing a variety of topologically stable solitons, in addition to non-topological solitons [31][32][33][34][35][36][37][38][39] analogous to the SM; domain walls [40][41][42][43], membranes [44,45], and cosmic strings such as topological Z strings [42,43,46,47] (see also Ref. [48]).However, magnetic monopoles were not examined because of the trivial second homotopy group π 2 of the 2HDM, as in the SM.
In the previous paper [49], the present authors studied the Nambu monopole in the 2HDM, which is a magnetic monopole attached by two topological Z strings from two opposite sides.The monopole is topologically stable and a regular solution of the equation of motion (EOM) when the Higgs potential has two global symmetries; One is a global U (1) symmetry that ensures the stability of the topological Z strings.The other is a discrete symmetry Z 2 exchanging the topological Z strings.The string tensions pulling the monopole are balanced due to the Z 2 symmetry, and thus the monopole does not move unlike the Nambu monopole in the SM and can be regarded as a topologically stable Z 2 kink on one string.
If the symmetries were realized in nature, the monopoles are expected to be abundant in the early and present universe.However, the models with these two symmetries are phenomenologically disfavored.The U (1) symmetry should be explicitly broken to give the mass to the CP-odd Higgs boson and the Z 2 symmetry is broken by the Yukawa coulings with the SM fermions.Therefore, these two symmetries may not be realized in the Higgs potential, at least for phenomenologically viable 2HDMs, and the monopoles may not be exactly stable.In such a case, it is still important to investigate whether (un)stable monoples can exist, and if so, how (un)stable they are.If they are sufficiently stable even for the non-symmetric case, they can be abundant and might be observed in the current monopole searches.In addition, they could be useful to impose constraints on the parameter space of 2HDMs from cosmological arguments such as the cosmological monopole problem.
In this paper, we investigate the dynamics of the Nambu monopole for the nonsymmetric case.Especially, we consider the case without the Z 2 symmetry, in which the tensions of the Z strings are not degenerated.As a result, the monopole is no longer static, but moves being pulled by the heavier string.We can confirm this picture based on the numerical relaxation.Furthermore, the monopole continues to accelerate emitting electromagnetic radiations.If the monopoles and networks of the Z strings are produced in the early universe, a monopole on a long string accelerates sufficiently and reaches nearly to the speed of light.This can be regarded as an accelerating charged particle in a synchrotron accelerator with a radius of the horizon scale.After the acceleration, the monopole eventually collides to an anti-monopole on the string and would produce various high energy radiations and particles.The typical kinetic energy of the accelerated monopole is ∼ (cos 2β) 1/4 10 8 TeV with tan β being the ratio of the vacuum expectation values of the two Higgs doublets.This is a quite high-energy event in the early universe, which we call as the Cosmological Monopole Collider.Consequently, it is unlikely that the monopoles without the Z 2 symmetry remain abundant in the present universe.Instead, some remnants of the collision events could be detected by astrophysical and cosmological observations.This paper is organized as follows.Sec. 2 is devoted to introduce the model.We will explain the U (1) symmetry and the Z 2 symmetry and show mass spectrum in the model.In Sec. 3, the electroweak strings will be considered.We will first review the topological Z strings, and then analyze string tensions of the general electroweak strings.In addition, we will investigate asymptotic properties of the Z strings.In Sec. 4, we will consider the Nambu monopole in the 2HDM without the Z 2 symmetry.We will construct an asymptotic form of the monopole and observe that the magnetic flux spherically spreads from the monopole independently of tan β.After that, we will give a cylindrical ansatz describing the monopole and perform the numerical relaxation.We will show results of the relaxation for several parameter choices, in which the monopole slowly moves being pulled by the heavier string.In Sec. 5 we will analyze the real dynamics of the Nambu monopole based on an approximation where we regard the monopole as a point particle.Consequently, we find that it accelerates immediately nearly to the speed of light by the CMC.Finally we will give a discussion and conclusion in Sec. 6.In Appenix A, we will provide a derivation for expressions of the gauge fields in the presence of the monopole configuration of the Higgs field.

The Lagrangian and Higgs potential
We introduce two SU (2) doublets, Φ 1 and Φ 2 , both with the hypercharge Y = 1.The Lagrangian which describes the electroweak and Higgs sectors is written as Here, Y µν and W a µν describe field strength tensors of the hypercharge and weak gauge interactions, respectively, with µ (ν) and a being Lorentz and weak iso-spin indices, respectively.D µ represents the covariant derivative acting on the Higgs fields, and the index i runs i = 1, 2. The most generic quartic potential V (Φ 1 , Φ 2 ) for the two Higgs doublets is given by where we have imposed a (softly-broken) Z 2 symmetry, Φ 1 → +Φ 1 , Φ 2 → −Φ 2 , in order to suppress Higgs-mediated flavor-changing neutral current processes.In this paper, we assume that both the Higgs fields develop real vacuum expectation values (VEVs) as Φ 1 = (0, v 1 ) T , Φ 2 = (0, v 2 ) T . 1 Then the electroweak scale, v EW ( 246 GeV), can be expressed by these VEVs as ).For later use, we rewrite the Higgs fields in a two-by-two matrix form [50], H, defined by The matrix field H transforms under the electroweak SU (2) L × U (1) Y symmetry as where the group element acting from the left belongs to SU (2) L and the other element acting from the right beongs to U (1) Y .Therefore the covariant derivative on H can be expressed as The VEV of H is expressed by a diagonal matrix H = diag(v 1 , v 2 ), and the Higgs potential can be written by using H as follows: where |H|2 ≡ H † H and the relations between the parameters in Eq. (2.6) and in Eq. (2.2) are given by ) ) (2.9) In the rest of this paper we will mostly restrict ourselves to the limited parameter space When this condition holds, the Lagrangian is invariant under a global U (1) a transformation, which is defined by a rotation of the relative phase of the two doublets: After H gets the VEV, this U (1) a symmetry is spontaneously broken and the corresponding Nambu-Goldstone (NG) boson appears, which is nothing but the CP odd Higgs boson (A).The spontaneously broken U (1) a symmetry gives rise to non-trivial topological excitations in which we are interested in this work.
Because an experimental lower bound on the mass of A is typically O(100) GeV (which highly depends on how the doublets couple to the SM fermions), such a massless A is phenomenologically disfavored.Therefore, in realistic cases, we should break the U (1) a symmetry explicitly by switching on m 3 and α 5 , giving a mass to A. Nevertheless, we set m 3 = α 5 = 0 throughout this paper because otherwise an analysis of the dynamics of the monopole becomes very complicated as we see in Sec. 6.

Custodial symmetry and (Z 2 ) C symmetry
The custodial transformation acting on the matrix H [50,51] is defined as the following global SU (2) transformation: H → U HU † , U ∈ SU (2) C . 2 In addition, the SU (2) W gauge field transforms as an adjoint representation simultaneously.The potential V (H) given in Eq. (2.6) is invariant under this transformation, which we call as the custodial symmetry, when m 2 = α 3 = α 4 = 0. Note that the gauge kinetic term of H cannot be invariant under this transformation because of the presence of the U (1) Y gauge field.Thus the custodial symmetry is not exact symmetry of the theory but is explicitly broken by the gauge interaction.
The bosonic Lagrangian L in Eq. (2.1) still has a symmetry under a Z 2 transformation defined by which we call as the (Z 2 ) C symmetry.Since iσ 1 ∈ SU (2) C , (Z 2 ) C acts on H and W µ as a subgroup of the SU (2) C transformation, but not for Y µ .Imposing the (Z 2 ) C symmetry on L in Eq. (2.1) reads (Z 2 ) C condition : m 2 = α 4 = 0. (2.12) Note that, as we will manifestly show below, the (Z 2 ) C is not spontaneously broken in the vacuum because of H ∝ 1 2 (tan β ≡ v 2 /v 1 = 1).Since the transformation of H can be written as Φ 1 → iσ 2 Φ * 2 and Φ 2 → iσ 2 Φ * 1 , it can be regarded as a combination of exchanging the two doublets and the CP transformation in the Higgs sector.
Similarly to the U (1) a symmetry, the (Z 2 ) C symmetry may not be realized in Lagrangian for realistic models because it is broken by the Yukawa couplings between the doublets and SM fermions.For instance, in the type-I 2HDM, the Yukawa couplings are given by which is clearly not invariant under the (Z 2 ) C transformation.As a result, m 2 and α 4 should be generated by radiative corrections from the SM fermion loops even when the (Z 2 ) C condition in Eq. (2.12) is imposed at the tree level.We studied the Nambu monopole in the 2HDM under both the U (1) a and (Z 2 ) C conditions in our previous study [49].In this paper, we relax the latter and study the Nambu monopoles in more realistic cases.Namely, we will investigate the Nambu monopoles in the cases without the (Z 2 ) C condition by setting m 2 = 0, α 4 = 0, α 3 = 0 in the Higgs potential.

Higgs mass spectrum
For the U (1) a symmetric Higgs potential with the condition (2.10), the Higgs VEVs are given by and tan β is defined as Now, it is clear that tan β = 1 holds if we impose the (Z 2 ) C condition.
In the matrix notation, fluctuations around the VEVs can be parametrized as

17)
with σ A = (1, σ a ) (a = 1, 2, 3).Here π 0 is the NG boson for the U (1) a symmetry, which is called as the CP-odd Higgs boson A in the literature.On the other hand, π a 's are would-be NG bosons for SU (2) W × U (1) Y and eaten by the gauge bosons.
When the potential has the (Z 2 ) C symmetry, yielding tan β = 1, (χ 1 ± iχ 2 )/ √ 2 are called as the charged Higgs bosons H ± , respectively.On the other hand, χ 0 and χ 3 are CP-even neutral components.Note that they are not mixed because they are (Z 2 ) C even and odd, respectively.The lighter one is identified with the SM Higgs boson h while the other is called as the heavier CP-even neutral Higgs boson H. Their masses are given by The lighter one among χ 0 and χ 3 is identified with the SM Higgs boson h while the other is called as the heavier CP-even neutral Higgs boson H.
On the other hand, it is more complicated for the potential without the (Z 2 ) C symmetry.The charged Higgs bosons are given by more complicated linear combinations of χ 1 and χ 2 .In addition, χ 3 and χ 0 are mixed with the mixing angle α, which is defined by where with α (±) 134 ≡ α 1 + α 3 ± α 4 .After the rotation by α, χ 0 and χ 3 are transformed into the mass eigen states H and h.We regard the lighter one as the SM Higgs again.The mass eigen values for the Higgs bosons are given by where The mass of the gauge bosons are given by with the standard definitions cos

Electroweak strings
In Refs.[42,43,46,47], it is pointed out that, unlike in the SM case, 2HDMs allow topologically stable strings to exist thanks to the global U (1) a symmetry.First, consider topological strings with the Z flux (topological Z strings).There are two types of topological Z strings corresponding to which one of the two Higgs doublets is to be wound.To see that, let us take W ± µ = A µ = 0.

Z strings
The solution called the (1, 0)-string on the z axis is given by where ρ ≡ x 2 + y 2 and ϕ is the rotation angle around the z-axis.The boundary conditions imposed on the profile functions are f (1,0) . On the other hand, the solution called the (0, 1)-string is given by and The boundary conditions for f (0,1) , h (0,1) and w (0,1) are the same as the (1, 0)-string.
Looking at the asymptotic forms, it is clear that both the (1, 0)-and (0, 1)-strings have winding number 1/2 for the global U (1) a symmetry, and thus they are topological vortex strings of the global type.Similarly to standard global vortices, their tensions (masses per unit length) logarithmically diverge.It can be seen from the kinetic term of the Higgs field: for ρ → ∞.Note that β dependence of Z i in Eqs.(3.2) and (3.4) were determined in such a way that the logarithmic divergences are minimized.For tan β = 1, it becomes a quarter of that for a global U (1) a integer vortex because of the half winding number for U (1) a [42].
On the other hand, they also have a winding number 1/2 inside the gauge orbit U (1) Z ∈ SU (2) W ×U (1) Y , which lead to the Z fluxes flowing inside them.The amounts of the fluxes of (1, 0)-and (0, 1)-string are along the z-axis, respectively.For tan β = 1, they are half of that of a non-topological Z string in the SM because of the half winding number.Note that the Z fluxes for the (1, 0)and (0, 1)-strings are different for generic β, unlikely the logarithmic divergent energy.The Z flux is squeezed into a flux tube.It decays exponentially fast as a usual Abrikosov-Nielsen-Olesen vortex [52,53] in the Abelian-Higgs model, in contrast to the 1/ρ tail given in Eq. (3.5).In other words, contributions to the energy from the non-Abelian parts do not diverge.Therefore, the difference of the tensions of the (1, 0)-and (0, 1)-strings appears only in finite portion of the tensions, which are due to the non-symmetric Z fluxes and the Higgs potential energy for tan β = 1.
It is instructive to see the Z strings from the special point m 2 = α 3 = α 4 = 0.As we explained above, the Higgs potential has the custodial SU (2) C symmetry there.Furthermore, if sin θ W = 0, the custodial symmetry is an exact symmetry of the Lagrangian.However, the presence of a topological string solution spontaneously breaks it down to a U (1) C subgroup in the vortex core, giving S 2 ( SU (2) C /U (1) C ) orientational moduli to the vortex, as studied in Refs.[42,43].Each point on the S 2 moduli space corresponds to a physically different string solution with a different magnetic flux, having a common winding number 1/2 for the global U (1) a .We parametrize the S 2 moduli space by two parameters 0 ≤ ζ ≤ π, 0 ≤ χ < 2π, where ζ and χ correspond to the zenith and azimuth angles, respectively.We identify the (1,0)-string, Eq. (3.4), as the one associated with the north pole of the S 2 moduli space, ζ = π.On the other hand, the (0,1)-string, Eq. (3.2), corresponds to the south pole, ζ = 0.A string solution on a generic point of the S 2 moduli space can be obtained by acting an SU (2) C transformation on the (1, 0)-string.
However, sin θ W = 0 in nature, and it explicitly breaks the custodial symmetry even when m 2 = α 3 = α 4 = 0.As a consequence, almost all the points of the S 2 moduli space are energetically lowered, except for the equator ζ = π/2.As studied in Refs.[42,43], the two Z strings, (1, 0)-string and (0, 1)-string, are the most stable with degeneracy among the topological strings.On the other hand, the strings corresponding to the equatorial points of S 2 , which contain a W flux and are called as W strings, are the most unstable.The effect of α 3 = 0 also breaks the custodial symmetry and lifts or lowers the tensions on the moduli space depending on the sign of α 3 .Note that it does not break the (Z 2 ) C symmetry because the (Z 2 ) C condition is still satisfied, so that it keeps the degeneracy of the two Z strings as we will see in the next subsection.However, m 2 = 0 and α 4 = 0 break the degeneracy. 3They make the energetic structure of the "moduli space" quite complicated.In the next subsection, we investigate the effects of m 2 = 0, α 3 = 0 and α 4 = 0 on the string tensions.

Tensions for the electroweak strings
Let us investigate the effects of m 2 = 0, α 3 = 0 and α 4 = 0 on the string tensions.In Ref. [42], it was numerically studied for various parameter regions.In this subsection, we analytically study it both for the case with and without the (Z 2 ) C symmetry.For simplicity, we consider θ W = 0. 4We begin with the string for m 2 2 = α 3 = α 4 = 0.Because of the custodial symmetry, the string solution has the moduli parameters which are the coordinates ζ (0 ≤ ζ ≤ π) and χ (0 ≤ χ ≤ 2π) of the moduli space S 2 .Since the azimuth angle χ remains as the moduli even when we turn on m 2 , α 3 , and α 4 , we fix it as χ = 0. Then the generic solution for ζ is given by where we have ntroduced Here f 0 (ρ), h 0 (ρ) and w 0 (ρ) are determined by the EOMs for m 2 2 , α 3 , α 4 = 0.The (0, 1)-and (1, 0)-string correspond to ζ = 0 and π, respectively.The string tensions are degenerated for all ζ as long as the condition m 2 2 = α 3 = α 4 = 0 is kept.Next, we turn on the (Z 2 ) C breaking parameters, i.e., m 2 2 , α 3 , and α 4 , and estimate their effects on the string tensions.We use a perturbation with respect to the parameters assuming they are sufficiently small.Thus, the tension of the perturbed string is approximated by substituting the unperturbed string solution {H(ζ), W i (ζ)} into the energy functional.Then, we express the tension as follows: where T 0 is the ζ-independent part and ∆T is the ζ-dependent one, which is caused by the breaking of the custodial symmetry.After some algebra, ∆T (ζ) is obtained as where the ellipses stand for higher order corrections and the coefficients are given by ) For later use, we derive a sufficient condition that the W -strings (ζ = π/2) are heavier than the both Z strings Since A 2 = α 3 × (positive constant), it implies that α 3 should be negative with large |α 3 |.
For practical purposes, let us roughly estimate the condition Eq. (3.15): and by approximating the profile functions by h 0 (ρ) = 1 and f 0 (ρ) = tanh(ρv), we obtain an upper bound on α 3 : Note that this bound is not rigorous and valid only up to the leading order of m 2 , α 3 , α 4 , but provides a guide for parameter choices in numerical computations.

Asymptotics of Z strings
We here investigate the asymptotic forms of the Z strings at large distances.Such an investigation is important to understand the dynamics and the stability of the string network in cosmology.For general local vortices, e.g., the ANO vortex in the Abelian-Higgs model or superconductors, an asymptotic form is given by an exponentially damping tale whose typical size is the mass scale of the model.The stability of lattice structures of the ANO vortex (called as an Abrikosov lattice) is determined by a ratio between sizes of tales of the scalar field and gauge field, which is equal to the ratio of the scalar and gauge couplings.
On the other hand, for global vortices (e.g., axion strings), the asymptotic form is given by a power-law tale because of the massless NG boson (axion particle).This means that global vortices are much fatter than local ones and that they have logarithmically divergent tensions.In the present case for the 2HDM, there are various mass scales in the mass spectrum including a massless CP-odd Higgs A as shown in Sec. 2, so that the asymptotic form of the electroweak strings are quite non-trivial.This situation is quite similar to a non-Abelian vortex in dense QCD [54][55][56][57][58], see Ref. [59] as a review.Here we follow the analysis of the asymptotic forms in Refs.[56,59].
Let us consider the (1, 0)-string in the (Z 2 ) C symmetric case: By introducing new functions, the expression (3.1) can be rewritten as where Here, F and G are profile functions in the mass basis.The former corresponds to the custodial singlet component χ 0 in Eq. (2.17) and the latter is the σ 3 component of the (split) custodial triplet, χ 3 .We study the asymptotic forms of F , G and w (1,0) at large distances compared to the inverses of the mass scales.In this region, they are almost in the vacuum, so that it is convenient to expand them around the vacuum as The linearized EOMs for δF (ρ), δG(ρ) and δw(ρ) are given by where ∆ ρ ≡ 1 ρ ∂ ρ (ρ∂ ρ ).Eq. (3.25) can be rewritten as with δ w ≡ δw/(m Z ρ).
Let us solve Eq. (3.23).The equation with the right hand side being zero has a solution proportional to K 1/2 (m χ 0 ρ).Here K 1/2 is one of the modified Bessel functions of the second class, which solves Dealing with the right hand side iteratively, we obtain the asymptotic form of F as where q F is an integration constant and we have used a fact that K n (r) with 0 ≤ r ≤ 1 behaves as π 2r e −r for r 1.The first term in Eq. (3.28) is sufficiently small for ρ 1/m χ 0 .Therefore, δF (ρ) behaves as a power function 1/ρ 2 for large ρ, which leads to the logarithmic divergence in the string tension as discussed above.This power-law tale is caused by the right hand side in Eq. (3.23), which is a source term generated by the massless particle π 0 (CP-odd Higgs A), and a common feature for global vortices.
Let us next solve Eqs.(3.24) and (3.26) using the iteration.By setting the right hand sides in the equations to zero, we obtain where q G and q Z are integration constants.In realistic 2HDMs, the additional CP-even neutral Higgs H, as well as the SM Higgs h, is typically heavier than m Z , so that we take m χ 3 > m Z .In this case, δG(ρ) in Eq. (3.29) is negligible and ignoring the right hand side in Eq. (3.26) is good approximation.Thus, the leading expression Eq. (3.30) is justified up to the sub-leading order of the iteration.On the other hand, Eq. (3.29) should receive the sub-leading iteration by substituting Eq. (3.30) into Eq.(3.29), and we obtain Therefore, δG and δw have the same exponential tales e −m Z ρ .This result is a quite similar to a non-Abelian vortex in dense QCD [56,59], instead of the ordinary ANO vortex in which each field has an exponential tale with each own mass scale.
The coefficient q Z is determined only by a numerical computation solving the EOMs.Following the previous study of the (1, 0) string in Ref. [42], we solve the EOMs for various range of m χ 3 /m Z while fixing m χ 0 = m h = 125 (GeV).We fit the solutions by the asymptotic forms Eq. (3.30) and (3.31) to obtain q Z , which are summarized in Tab. 1 We have investigated the asymptotic form of (1, 0)-string for the (Z 2 ) C symmetric case.The string has two tale structures; one is the power-law tale of F associating with the custodial singlet component, and the other is the exponential tale in G and w (1,0) with the size of 1/m Z .Therefore, the string has the logarithmic divergence in the tension while the Z flux tube exponentially decays as stated above.The asymptotic forms of the (0, 1)string is the same as the above results thanks to the (Z 2 ) C symmetry.In addition, those of the W -strings can be obtained by replacing m Z with m W in the above analysis.On the other hand, when the potential does not have the (Z 2 ) C symmetry, the situation could be more complicated.However, the Z strings still have the two structures that we explained above.A quantitative discussion of the Z strings without the (Z 2 ) C symmetry requires a further analysis, which will be done elsewhere.
δG) 2.60787 1.82783 1.65704 1.59833 q Z (δw) 1.79159 1.61170 1.55712 1.53367 Table 1: The obtained values of q Z by fitting the solution of the (1, 0) string.q Z (δG) and q Z (δw) denote the one using the asymptotic forms of δG (Eq.(3.31)) and δw (Eq.(3.30)), respectively.There is a good agreement between them for larger m χ 3 but not for smaller m χ 3 .This is because the approximate expressions (3.30) and (3.31) are not valid for smaller m χ 3 .

Nambu monopoles
In this section, we study the Nambu monopole, which is a 't Hooft-Polyakov type magnetic monopole attached by the Z strings in the 2HDM.The static stable monopole was obtained under the restriction (Z 2 ) C symmetry in Ref. [49].In contrast, here, we will investigate the Nambu monopoles without the (Z 2 ) C symmetry.

A point-monopole approximation
Firstly, let us observe the monopole-string complex at large distance infinitely far from it.Namely, we analytically deal with the Z strings and Nambu monopole as infinitely thin and small objects.An actual regular form will be constructed by a numerical relaxation method in the next subsection.
In the thin string limit, we replace, for instance in Eqs.(3.1) and (3.2), f (1,0) (ρ) by Then, the singular Z-fluxes of the (1, 0)-and (0, 1)-strings are given by The Nambu monopole will play a role of a junction at which the two Z-fluxes are connected.In Ref. [49], we constructed the string-monopole complex at the tan β = 1 limit which is quite special in a sense that the (1, 0)-and (0, 1)-strings can be transformed to each other by the (Z 2 ) C symmetry.With the aid of (Z 2 ) C , we constructed the Nambu monopole at tan β = 1 by acting the "local" SU (2) C transformation that depends on the zenith angle θ in the real space.We transformed the (1, 0)-string which is put on the z-axis by the local SU (2 We have U (π) = iσ 1 , which is nothing but the (Z 2 ) C .Thus, we obtained a configuration made of (1, 0)-and (0, 1)-strings on the positive and negative sides of the z-axes (θ = 0, π), respectively.
However, this construction cannot be used in the present case because (Z 2 ) C is no longer a symmetry of Lagrangian and does not relate the two strings.Instead, we provide a more general way to construct an asymptotic form of the Nambu monopole.Let us start with SU (2) W adjoint composite fields n a 1 and n a 2 (a = 1, 2, 3) normalized to unity,5 defined by Note that they are analogues to a normalized adjoint scalar field for the 't Hooft-Polyakov monopole [1,2] because they determine the unbroken subgroup of the SU (2) W × U (1) Y gauge group.In the following, we take n a 1 = n a 2 because otherwise there is no unbroken subgroup and the photon becomes massive.Since the n i -fields are not well-defined at points in which |Φ i | 2 = 0, we introduce the following adjoint field which are well-defined even inside of the strings: whose norm is unity.Now, for a configuration with n a depending on x 1,2,3 , the field strength for the electromagnetism and Z-boson are naturally defined by respectively.Suppose there is a magnetic monopole at the origin surrounded by the vacuum of the 2HDM.Since the electromagnetic U (1) EM is unbroken there, the magnetic flux should spherically symmetrically spread from the monopole, as a usual Dirac or 't Hooft-Polyakov monopole.Therefore, it is natural to impose a spherical symmetry on n a , as described in Refs.[13,49]. 6Hence, we consider a configuration satisfying the so-called hedgehog structure: The topological number of n a is unity similarly to the case of the 't Hooft-Polyakov monopole: 1 4π r→∞ d S • n = 1.The corresponding configuration of the original Higgs fields is given by where we have used Φ † i Φ i = v 2 i except for the origin, and φ i 's are arbitrary real functions but we have to choose φ i in such a way that Φ i becomes single valued.Note that Eq. (4.8) is quite similar to the configuration discovered by Nambu [13] in the SM, except for the overall phase factor: which describes a point-like magnetic monopole at the origin attached by an infinitely thin (non-topological) Z string on θ = π.Since the 2HDM monopole is attached by two topological Z strings on the opposite sides (θ = 0 and π), we take φ i 's as which ensures the single valued-ness of Φ i .This can be manifestly seen by the two-by-two matrix notation as Eq. (4.11) describes the (1, 0)-string ((0, 1)-string) on θ = 0 (θ = π) up to the SU (2) W gauge transformation. 7This is a generalization of the one constructed in Ref. [49] for tan β = 1.
If we take φ 1 = φ 2 = ϕ instead of Eq. (4.10), we have another monopole-string configuration quite similar to the Nambu monopole in the SM.There are two Z strings on the negative side of the z axis (θ = π), but they are (0, 1)-and (0, −1)-strings.Thus, the configuration is not topologically protected at all, and we do not discuss it here.
As a next step, we consider the gauge fields W µ and Y µ in the presence of H mon. [Eq.(4.11)].They are determined to minimize the kinetic energy of the Higgs doublets, (4.12) After some lengthy calculations (see Appendix A), we obtain Note that the minimization condition for the Higgs fields can determine only the gauge fields corresponding to the broken generators.Regarding Y i as an arbitrary function, we have where we have used identities is present, the configuration has a line singularity δ(x)δ(y) on the z-axis, which is inconsistent with the fact that U (1) EM is unbroken.
Therefore, we must choose Y µ to cancel such unphysical singular structures.An appropriate choice is given by Plugging these into Eqs.(4.14) and (4.15), we get the final forms of the physical field strengths where we have used the identity From Eq. (4.21), it is clear that there is a magnetic flux from the origin in a spherical hedgehog form.The total amount of the magnetic flux Φ EM can be calculated by integrating the flux density Interestingly, the electromagnetic flux is independent of the ratio of the two Higgs VEVs, v 1 and v 2 .This is understandable because the electromagnetic U (1) remains unbroken, and the photon does not couple to the Higgs VEVs.
In addition, from Eq. (4.20), the Z-fluxes only exist on the z-axis as flowing on the positive and negative sides of the z-axes, respectively, from the origin.These amounts of the Z-fluxes agree with ones of the Z strings given in Eq. (3.6).Therefore, the total amount of the Z-fluxes flowing from the monopole at the origin is independent of tan β as with It is worthwhile to demonstrate the topological current of U (1) a in the configuration.The flux, corresponding to the winding of the U (1) a phase of the Higgs field, is defined by Substituting Eq. (4.11) to this, we have Importantly, A i is topologically conserved, ∂ i A i = 0, and independent of z.This indicates that not only the string parts but also the monopole itself has the topological charge of U (1) a .Before closing this subsection, let us give a summary picture of the monopole in the 2HDM.Let us denote the magnetic and Z fluxes by F EM and F Z , respectively for simplicity.Similarly, we denote the hypercharge-and n-magnetic fluxes by F Y and F n = n • F W , respectively.Now consider a large sphere Σ centered at the monopole, and let S and N be infinitesimally small regions at the south and north poles, respectively.Eq. (4.20) tells Similarly, from Eq. (4.21) we have Combining these with and Fig. 1 shows the schematic pictures of the magnetic fluxes.The magnetic monopole is the source for the Z and magnetic fluxes.On the other hand, all the hypercharge-magnetic fluxes entering inside the strings go out through the sphere.This is consistent with the fact that the hypercharge-magnetic field is divergenceless, so that they cannot be terminated.Note also that the net non-Abelian magnetic flux F n Σ = − 4π g equals to the one of the conventional 't Hooft-Polyakov monopole.

Ansatz for regular solutions
The configuration constructed above is singular because it is just the asymptotic form at infinitely large distance.In the next subsection, we will obtain regular solutions by numerically solving the equations of motion.This subsection is devoted to prepartion for it.Namely, we provide an appropriate ansatz for the profile functions of all the fields.The

⌃ < l a t e x i t s h a 1 _ b a s e 6 4 = " Y F n v D H q i r d I l s 6 i b r u 7 f S y W X A B A = " >
o s 4 m 5 D w 9 j l O E S J t n b I 4 Z c j E 1 m i s 0 m r f Z w 1 a 9 2 o 6 n l q l U z T q N i k j i L E X 9 s C 6 7 J k 9 s l f 2 8 W u t l l e j 5 6 V J s 9 L X c q s 8 c 7 a Q f f 9 X p d P s 4 u h L 9 a d n F x V s e F 4 F e b c 8 p n c L t a 9 v n F x 0 s   ansatz should be consistent with the asymptotic form obtained in the previous subsection, and moreover a numerically low-cost ansatz is preferable.Our starting point is rewriting the asymptotic gauge fields given in Eqs.(4.18) and (4.19) in terms of the cylindrical coordinates as

a t e x i t s h a 1 _ b a s e 6 4 = " M Y k k p D o I 6 m T u f k d v x G w k 9 e t G G l s = " >
Y ρ = 0, (4.40) This configuration has a cylindrical symmetry in the sense that it is invariant under the simultaneous rotations of the 2-dimensional real space (x, y) and the internal space (σ 1 , σ 2 ).
In addition, it is also invariant under the simultaneous parity transformations: Let us introduce profile functions {u 1 , u 2 , u 3 , u 4 , b} to smear the singularities keeping the cylindrical symmetry as Y ρ = 0, (4.46) Similarly, we make the ansatz for the Higgs fields by smearing the singular one given in Eq. (4.11) as These profile functions should behave at ρ → ∞ as where the Dirichlet conditions in the first line come from finiteness of the energy, 8 and the Neumann conditions in the second line follow from smoothness of the Higgs field.The last line is due to the parity symmetry under ρ → −ρ, σ 1(2) → −σ 1 (2) .The energy density is indeed written down as follows, with where we have used u = ∂ z u, u = ∂ ρ u, and g 2 Z = g 2 + g 2 .We want to obtain the profile functions by solving the equations of motion δE/δX = 0, where X stands for the profile functions X ∈ {f 1,2 , h 1,2 , u 1,2,3,4 , b}.In general, however, even numerically it is not easy to solve such complicated differential equations.So, instead of directly solving them, here we make use of the relaxation method.We introduce a fictitious time τ called as a flow time besides the real time t, and evolve them by the following differential equations (flow equations): starting from some appropriate functions satisfying the boundary conditions given above as an initial configuration at τ = 0.If the τ -evolution converges, namely ∂ τ X → 0 as τ evolves, the convergent profile functions are nothing but the static solution of the original equations of motion.Before going to solve the flow equations, however, we should note that kinetic terms for u 1 and u 4 in Eq. ( 4 which transforms the profile functions as where ω = ω(ρ, z) is an arbitrary real function of ρ and z.By taking ω that satisfies ω + ω = u1 − u 4 , we can choose a gauge : u1 − u 4 = 0. Equivalently, we can simply add the following gauge fixing term Then, we find the normal quadratic terms for u 1 and u 4 in this gauge as In summary, the improved flow equations to be solved are In the previously study [49] by the present authors we experienced the similar numerical computation but we did not make use of any symmetries to reduce numerical cost.The relaxation scheme in Ref. [49] dealt with 20 fields (12 gauge fields W a i , Y i and 8 scalar fields in H) which are dependent of the 3 spatial coordinates and the flow time τ .Compared with Ref. [49], the new relaxation scheme only includes 9 profile functions, and furthermore they are only dependent of the 2 spatial coordinates (ρ and z) and flow time τ .

Results of relaxation
We show several results of solving the flow equations.We will take the parameters in Lagrangian so that the W strings are heavier than the Z strings because we want to study the Nambu monopole attached by the two Z strings as is shown in Fig. 1.Otherwise, the two Z strings are unstable and would rapidly decay to the W strings, resulting in a single homogeneous (up to the U (1) EM modulus) W string without a monopole.A rough condition for the W strings to be heavier than the Z strings was obtained in Sec. 3 [See Eq. (3.17)].
Throughout this subsection, we fix experimentally observed three parameters θ W , m h , v EW as sin The other parameters such as the masses of the additional Higgs bosons and tan β are changed for several cases.

tan β = 1 case
Firstly, let us consider the case that the (Z 2 ) C is exact, where the monopole is topologically stable because the tensions of the (1, 0)-and (0, 1)-strings are balanced.This case was previously studied by the present authors in Ref. [49] by constructing the stable solution based on the 3D simulation without any ansatz.Therefore, we can check the consistency of our ansatz Eqs.(4.42)-(4.47)and (4.48) by comparing the result with that in Ref. [49].We take the following physical parameters in the 2HDM as9 and impose the alignment limit: cos(β − α) = 0.This is equivalent to choosing the parameters in the potential (2.6) as In this choice, the Higgs potential has the custodial SU (2) C symmetry, and, therefore, the (Z 2 ) C condition (2.12) is satisfied.The energy of the W strings are slightly heavier than those of the Z strings because of θ W = 0 as studied in Ref. [42].The first line in Fig. 2 shows the result at the flow time τ = 30; the energy density, magnetic flux and Z-flux from left to right.The monopole attached by the two Z strings on the opposite sides does not move and corresponds to a stable and static solution of the EOMs.The blue dots in Fig. 3 show the evolution of the energy for increasing τ .We can observe that it exponentially converges and that the variation of the energy density per flow time is O(10 −6 ) for τ ∼ 30, so that we regard it as the end of the relaxation.
These results agree well with those of Ref. [49].Thus, the present ansatz Eqs.(4.48)-(4.47)are consistent and correct.We emphasize that the ansatz is more economical and convenient in the sense of the numerical cost compared to the full 3D computation.

tan β = 1 case
Let us next consider the cases with tan β = 1 where the (Z 2 ) C is explicitly broken.In this case, the monopole cannot be static, since the monopole is attached by the two Z strings whose tensions are not equal each other.As a result, there are no static monopole solutions of the genuine EOMs, and correspondingly the energy does never converge in the relaxation.
As benchmark values, we take the following physical parameters: and impose the alignment limit: cos(β − α) = 0.Here θ W , m h , v EW are the same as Eq.(4.74).These are equivalent to choosing the parameters in the potential as These choices satisfy the condition Eq. (3.17).We should note that the (Z 2 ) C symmetry is explicitly broken because α 4 = 0 violates Eq. (2.12).We can observe that the monopole is pulled by the string as follows.There are two stages of the τ -evolution.The early stage is the period that the energy exponentially decreases since the artificial initial configuration rapidly releases the energy.After the first stage, the τ -evolution goes into the second stage in which the each piece of the configuration, namely the monopole, or (1, 0)-or (0, 1)-strings, is not modified very much.Instead, the monopole gradually shifts toward the heavier Z string, which results in the linear decreasing of the energy in the relaxation.We can observe the two-stage τ -evolution in several ways.Fig. 4 shows the evolution of the amplitude of the magnetic flux, | B| 2 .For 0 < τ 25, the size and the density of the amplitude change significantly.After τ 25, they do not change, but slowly move down along the z-axis.From this result, it is confirmed that the monopole slowly moves and is pulled by the heavier Z string unlike the previous case.The orange dots in Fig. 3 show the evolution of the energy for this case.Unlike the previous ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ • tanβ=1 ▲ tanβ=1.1  case of tan β = 1, the evolution does not converge.The first stage (0 < τ 25) is quite similar to the one of tan β = 1 but the second stage (τ 25) is peculiar to tan β = 1.The energy continues to decrease by ∆z × (T (1,0) − T (0,1) ) with ∆z being a distance the monopole moves.As a result, the time dependence of the energy is linear in τ for the large flow time.From the above two observations, we regard τ ∼ 25 as the end of the relaxation for the monopole.
The second line in Fig. 2 shows the energy density, magnetic flux and Z-flux at τ = 30 from left to right.Note that the amount of the Z fluxes of the two strings are slightly different since their ratio is given by tan 2 β = 1.21.As a result, their energy densities are also different.In addition, the monopole is not vertically symmetric, which is clear in Fig. 4.  The two-stage evolution in τ is qualitatively the same as the case of tan β = 1.1, so that we do not repeat the detailed explanations for the time evolution.We only show the plots of Z flux, the magnetic flux and the energy density in the third and fourth lines in Fig. 2, respectively.As expected, we can see that the amounts of the Z fluxes of the two strings are significantly different.Their ratio is tan 2 β = 1.69 and 9, respectively.The energy densities are also different between the two strings, however, the power-law tails coming from the winding of U (1) a phase are almost the same.Furthermore, the shape of the magnetic flux is significantly distorted around the center, but it is not so at large distances.This is clearly shown in Fig. 5. Finally, we comment on the topological current of U (1) a , A i , which corresponds to the winding of the U (1) a phase of the Higgs field.The current is defined by Eq. (4.27).
Importantly, A i is topologically conserved, and independent of z at large distances, see Eq. (4.29).Fig. 6 shows the U (1) a current for the tan β = 3 case (Eq.(4.82)).The density and width are different between the upper and lower sides in the vicinity of the string cores, but the total flux integrated over a cross section z = const.is always conserved.This indicates that not only the string parts but also the monopole itself has the topological charge for U (1) a .

Monopole dynamics and radiation
We here discuss the real-time dynamics of the Nambu monopole without the (Z 2 ) C symmetry.As shown in Sec. 4, the monopole moves along the strings being pulled by the heavier string.It is difficult to analyze such a dynamics solving the EOMs of the gauge fields and the Higgs fields, so that we deal with the monopole as a point-like object with a magnetic charge and solve the classical mechanics.As a result, the monopole is immediately accelerated to a velocity of order 1 (the speed of light) and emits electromagnetic radiation depending on the radius of curvature of the strings.If the monopoles and strings existed in the early universe, the accelerated monopole collides to an anti-monopole with a kinetic energy of order (cos 2β) 1/4 10 8 TeV, whose remnants could be observed by the CMB anisotropy, primordial gravitational waves or the 21cm observation.

Acceleration of Nambu monopole
The dynamics of a point-like monopole is described by the following energy conservation law: where K ≡ γM is the kinetic energy of the monopole with static mass M and γ is the rapidity: γ ≡ (1 − u(t) 2 ) −1/2 with u(t) being the velocity.P rad is the energy loss per unit time by the electromagnetic radiation from the monopole.To calculate this, we just need to replace electric and magnetic variables in the well-known formula of the radiation from an accelerated electric charge.Then, we find where Ω is the solid angle from the monopole point, n is a unit vector pointing from the position of the monopole to a point x, and q M is the magnetic charge of the monopole: q M = 4π sin θ W /g. On the other hand, P string in Eq. (5.1) is the energy gain per unit time from the string, where ∆T is the difference between the tensions of (1, 0) and (0, 1) strings.Since the difference of the string tensions mainly comes from the Z-flux squeezed inside the core of the strings, we estimate it as When the motion of the monopole is non-relativistic (u 1), the effect of the radiation P rad is negligible and we have which means that u soon increases to of order 1 with a time scale M/∆T .After that, we cannot ignore the relativistic breaking effect of the radiation.Let us discuss the cases that the Z-stirings are straight and curved.First, when the Z strings are on a straight line, u and d u/dt are parallel.Then, we have Combining Eqs.(5.6), (5.3) and (5.1), it is found that the monopole continues to accelerate emitting the radiation since du/dt = 0 is not a solution.The velocity approaches to the speed of light with infinite time.
The situation changes when the Z strings bend.Suppose the monopole runs along the string whose local curvature is R, see Fig. 7.Even though the monopole speed reaches a maximum constant value, the monopole is still accelerated toward the center of the circle, R < l a t e x i t s h a 1 _ b a s e 6 x 4 e t J 8 1 J I W h q m S j 5 P s 4 g v I U U + A K 3 j j C s = < / l a t e x i t > < l a t e x i t s h a 1 _ b a s e 6 4 = " g T   (5.7) Then, we can roughly solve P rad ∼ P string as where we have used du/dt = u 2 /R and u max ∼ 1.When u ∼ 1, the angular distribution has a strong peak at θ = 0 similarly to the usual synchrotron radiation.In addition, a typical frequency of a power spectrum of the synchrotron radiation is given by (5.9) Note that ω becomes larger for the larger radius R of the curvature because the maximum velocity u max becomes larger, producing the larger blue-shift effect.

Cosmological monopole collider
Let us consider such an acceleration of the monopole in the early universe.We assume that the monopole and both the Z strings are produced during the electroweak phase transition [60,61] even when the (Z 2 ) C symmetry is explicitly broken, which is justified if

Discussion and conclusion
In this paper, we have investigated the dynamics of the Nambu monopole in the 2HDM.Firstly we have studied the electroweak strings and found that asymptotic behaviours of profile functions of gauge and scalar fields decay with the mass of Higgs fields, unlike vortices in superconductors.We have derived the condition (3.17) that the Z strings are stabler than the W strings, equivalently that the Nambu monopoles are stable on the Z string.Next we have given an asymptotic form of the monopole in the case without the (Z 2 ) C symmetry based on the point-like approximation for the monopole.The two Z strings attached to the monopole have different Z fluxes with its ratio being tan 2 β because of the absence of (Z 2 ) C .On the other hand, the magnetic flux spreading from the monopole does not depend on tan β and is spherically symmetric at large distances.
Based on the asymptotic form, we have presented a cylindrical ansatz describing the regular monopole configuration.It is much more convenient than the full 3D one that we used in the previous work [49].After that, we have shown results of the relaxation method using the ansatz for several parameter choices.In the (Z 2 ) C symmetric case, tan β = 1, the monopole does not move and is static solution of the EOM, which is consistent with the previous work [49].On the other hand, the monopole moves on the string being pulled by the heavier one for the non (Z 2 ) C symmetric case, tan β = 1.See Figs. 2 and 4. In the last section, we have analyzed the real time dynamics of the monopole based on the point-like approximation.The monopole accelerates by the string emitting electromagnetic radiations like a synchrotron accelerator.We have considered the CMC, a high-energy phenomenon in the early universe that an accelerated monopole collides to an anti-monopole with kinetic energy ∼ (cos 2β) 1/4 1011 GeV.Let us comment on the CMC and its phenomenological implication.Interestingly, the center-of-mass energy of the collision event at the CMC is around 10 11 GeV for (cos 2β) 1/4 ∼ 1, which is much higher than those that our collider experiments can reach today.By the collision, heavy particles with masses of the order 10 11 GeV can be produced as in an ordinary synchrotron collider.Such remnants could remain as fluctuations of matter distributions in the present universe and be observed by the CMB anisotropy, primordial gravitational waves and the 21cm observations.Therefore, the CMC can be a tool built in nature to probe high energy physics beyond the SM such as inflation models and GUTs.This situation is quite similar to the so-called cosmological collider [63][64][65][66][67]. 11  In this paper, the U (1) a symmetry is imposed in the Higgs potential, so that the stability of the Z strings is topologically protected.As we stated in Sec. 2, however, it should be explicitly broken by switching on m 3 , α 5 to make the CP-odd Higgs boson massive.The effect of m 3 , α 5 = 0 is discussed by some of the present authors in Refs.[42,43].Similarly to the axion string and domain wall associated with the U (1) PQ symmetry, which is explicitly broken by the axial anomaly, the Z strings and the monopole are attached by one or two domain walls depending on the values of m 3 , α 5 .Therefore, our observation that the monopole moves along the string network would be more complicated, that is, the monopole is pulled by the string, and both of them are pulled by the wall.While our study is justified when the tension of the wall is small compared to ∆T , in general, we have to consider two directions of the acceleration of the monopole, and the CMC is no longer a simple synchrotron accelerator.A further quantitative study is needed to estimate the energy of the CMC, which is left for future work.
We discuss the abundance of the monopoles in the present universe, which is important to see whether the monopoles dominate the energy density of the universe (cosmological monopole problem) or not.As we stated above, the monopole accelerates by the CMC with a typical time scale ∼ M/∆T .After the acceleration, the monopole and anti-monopole collide and annihilate immediately, and hence the monopoles would not remain abundant unless ∆T is unnaturally small, i,e, fine tuned to be a small value. 12Consequently, it is unlikely that the monopoles without the Z 2 symmetry dominate the energy density of the universe.
We here comment on a relation of the Nambu monopole and the sphaleron in 2HDMs.As we have studied above, the monopole is pulled by the heavier Z string.If we twist the monopole relatively to the anti-monopole, there could arise a repulsive force between them.Thus, if the repulsive force and the tension of the string are balanced, the configuration would be a static and unstable solution of the EOM, which is a new type of the sphaleron in the 2HDM.This is infinitely long and has an infrared-divergent energy while the ordinary sphaleron in 2HDMs studied in Refs.[37,38,68,69] is compact and has a finite energy.It is interesting to consider if the new sphaleron can contribute to the baryon asymmetry in the universe.

. 14 )
We omit A 0 since it is independent of ζ.Note that cos ζ and cos 2 ζ are odd and even under the (Z 2 ) C transformation (ζ → −ζ), respectively.The coefficient A 1 of cos ζ only depends on m 2 and α 4 , whereas the coefficient A 2 of cos 2 ζ only depends on α 3 .This is consistent with the fact that α 3 does not break (Z 2 ) C because the (Z 2 ) C condition (2.12) is independent of α 3 .Namely, the cos 2 ζ term raises or lowers the tensions of both the Z strings (ζ = 0, π) keeping the degeneracy of them while the cos ζ term breaks the degeneracy.The both terms do not change the tensions of the W -strings (ζ ∼ π/2).
d H R j / w g E 9 n d R l t c d 5 f d V T D p D w R d 8 9 C p I C L 6 G V 3 6 A x 3 8 B 0 V H g y 4 d e l 0 X o q R 6 h 5 l 5 5 p n 3 e e e Z G c l Q F c t m r O s R R k b H x i e 8 k 7 6 p 6 R l / I D g 7 l 7 P 0 d H R j / w I E 9 n d R l t c d 5 f d V T D p D w R d 8 9 C p I C L 6 G V 3 6 A x 3 8 B 0 V H g y 4 d e l 0 X o q R 6 h 5 l 5 5 p n 3 e e e Z G c l Q F c t m r O s R R k b H x i e 8 k 7 6 p 6 R l / I D g 7 l 7 P 0

Figure 1 :
Figure 1: The schematic pictures of the magnetic monopole in the 2HDM for tan β = 1.The (1, 0)-string passes through the north pole while the (0, 1)-string does through the south pole.

Figure 2 :
Figure 2: Plots for the Nambu monopole for the (Z 2 ) C symmetric (tan β = 1) and non-symmetric (tan β = 1) cases.In all plots, v EW /2 is normalized to unity.(a): Energy density.The color represents its value.(b): Magnetic flux.(c): Z flux.The direction of the arrows indicates that of the fluxes.Also, the color and size of the arrows indicate the flux densities.

Figure 3 :
Figure 3: Plot of the evoluion of the energy for the flow time τ for the (Z 2 ) C symmetric case (tan β = 1.0) and the non-symmetric case (tan β = 1.1).We evaluated the energy by integrating the energy density over 0 < ρ < 10, −15 < z < 20 for tan β = 1.1 and 0 < ρ < 10, −15 < z < 15 for tan β = 1.0, respectively.We adopt the unit v EW /2 = 123 GeV → 1.The left panel shows the evolution for 0 < τ < 30 (tan β = 1.0 and 1.1 is the blue circle and yellow triangle, respectively).The right one shows the interpolated energy plot for 2 < τ < 30 (tan β = 1.0 and 1.1 is the blue solid line and yellow dashed line, respectively).Here we have shifted the latter line to compare the shape with the former one.

Figure 4 :
Figure 4: Snapshots of | B| 2 at τ = 10, 15, 20, 25, 30.The plots are cut at y = 0. Two horizontal dotted lines are shown to indicate the position of the monopole at τ = 25.It slowly moves down being pulled by the downside Z string.

Figure 5 :
Figure 5: Cut of the amplitude of the magnetic flux at y = 0 for tan β = 3.

Figure 6 :
Figure 6: Left panel: Plot of the U (1) a topological current A i .The direction, color and size of the arrows are the same as those for the magnetic flux.Right panel: Plot of A 3 versus ρ at z = 10, 0, −10.In both the panels, we take the parameter choice (4.82) (tan β = 3 case).
1 1 p d t 8 b Q S 4 d m e a T l Z i 1 4 F s m 9 / 6 t q 0u z g 6 E v 1 p 2 c H d W y 7 X l X y b r r M 8 B b K S N 8 + u R j k d r L x 7 j q 7 Z X 3 y f 8 N 6 7 I l u o L f f l L s M z 1 4 j Q B 8 g / n z u c V D Y T I j J R D K T j K V S 3 l f 4 s Y o 1 b N B 7 b y G F P a S R p 3 O P c Y 5 L X P n 6 Q l i I C C u j V M H n a Z bx L Y T Y J x j w i 9 8 = < / l a t e x i t > ũ < l a t e x i t s h a 1 _ b a s e 6 4 = " f e h t 4 B r w p 5 N x x m e A t 5 p G + f X A y y O 5 l o d 5 3 d s j 7 5 v 2 E 9 9 k Q 3 0 N p v 8 l 2 a Z 6 7 h o w + I / 3 z u c Z D f j M U T s U Q 6 E U n u u l / h x S r W s E H v v Y U k 9 p F C j s 4 9 x j k u c e X p C 0 E h J K y M U g W P q 1 n G t x A i n y e K i + g = < / l a t e x i t > ✓ < l a t e x i t s h a 1 _ b a s e 6 4 = " u M p R n o N O 8 9 F 5 l 4 T D l a w 6 R H X X o k e 6 p T e + / 1 m o G N X w v D Z 6 1 r l b Y h f j Z 1 P b b v 6 o q z x 7 K n 6 o / P X s o Y j n w K t m 7 H T D + L f S u v n 7 S 6 m y v b K W a 8 3 R N L + z / i p 7 p g W 9 g 1 l / 1 m 0 2 x d Y k Y f 0 D m + 3 P / B L u L 6 U w 2 n d 3 M J n O r 4 V d E M Y 0 5 L P B 7 L y G H d W w g z + e W c Y 4 W L i J t J a 5 M K l P d V C U S a i b w J Z T Z D 7 V B i 0 I = < / l a t e x i t > dũ/dt < l a t e x i t s h a 1 _ b a s e 6 4 = " o V 8 6 4 8 I E o o b L c p K b w X n 4 F 5 P l r u b 0 b P h z L v / 2 r M m g W 2 P t U / e l Z o I x F z 6 t G 3 m 2 P a d 1 C b e v r + 8 c v + a V c v D n N L t g T + T 9 n D + y W b m D W X 9 X L L M + d I E Q f k P z + 3 D / B + n w i m U q k s q l Y e t n / i i D G M Y V Z e u 8 F p L G K D A p 0 r o k j n O I s 8 C y N S h P S Z D t V C v i a E X w J a e Y D n j 6 N e w = = < / l a t e x i t > (0,1) string (1,0) string !< l a t e x i t s h a 1 _ b a s e 6 4 = " P S T f o m

8 b g q b 5 6 SFigure 7 : 2 M 4π du dt 2 dΩ ( 1 −
Figure7: Accelerating Nambu monopole on the Z strings.We approximate the string as a part of a circle with a radius R of curvature.The monopole generates the synchrotron radiation with a typical frequency ω.
are absent here.Due to this, the flow equations for u 1 and u 4 are not genuine diffusion equations, which are sometimes problematic because it is an obstacle to convergence.To resolve this, we can use the gauge transformation, W i → U W i + ig −1 ∂ i U † and H → U H, which does not change forms of the ansatz in Eqs.(4.42), (4.43), (4.44), and (4.48).Such U is given by