# Remarks on Static Three-Quark Potentials, String Breaking and Gauge/String
Duality

Oleg Andreev  L.D. Landau Institute for Theoretical Physics, Kosygina 2,
119334 Moscow, Russia  V.A. Steklov Mathematical Institute, Gubkina 8, 119991,
Moscow, Russia  Arnold Sommerfeld Center for Theoretical Physics, LMU-München,
Theresienstrasse 37, 80333 München, Germany

###### Abstract

Making use of the gauge/string duality, it is possible to study some aspects
of the string breaking phenomenon in the three quark system. Our results point
out that the string breaking distance is not universal and depends on quark
geometry. The estimates of the ratio of the string breaking distance in the
three quark system to that in the quark-antiquark system would range
approximately from $\frac{2}{3}$ to $1$. In addition, it is shown that there
are special geometries which allow more than one breaking distance.

††preprint: LMU-ASC 47/20

##  I Introduction

Since the $J/\Psi$ discovery in 1974, there remains a serious puzzle
concerning triply heavy baryons bj . On the theoretical side, the challenge is
to explain the structure and properties of such baryons, and thus reach the
level of knowledge similar to that of charmonium and bottomonium. It might be
expected that, based on the success with the quarkonium spectroscopy, the
potential (quark) models would be useful in doing so richard . At that stage,
one would expect to gain important insights into understanding how baryons are
formed from quarks and gluons.

A static three-quark potential is one of the most important inputs to the
potential models and also a key to understanding the phenomenon of quark
confinement in baryons. The potential as a non-perturbative object has been
studied in the context of lattice gauge theory bali .111For more recent
developments, see 3Q-lattice . It is determined from the expectation value of
a baryonic Wilson loop. Such a loop is defined in a gauge-invariant manner as
$W_{3\text{\tiny
Q}}=\frac{1}{3!}\epsilon_{abc}\epsilon_{a^{\prime}b^{\prime}c^{\prime}}U_{1}^{aa^{\prime}}U_{2}^{bb^{\prime}}U_{3}^{cc^{\prime}}$,
with $U_{i}$ the path-ordered exponents along the paths shown in Figure 1. In
the limit of large $T$, the expectation value of $W_{3\text{\tiny Q}}$ can

Figure 1: A baryonic Wilson loop in $SU(3)$ gauge theory. A three-quark state
is generated at $t=0$ and then annihilated at $t=T$.

be written in the form

$\langle W_{3\text{\tiny
Q}}\rangle=\sum_{n=0}^{\infty}w_{n}\text{e}^{-V_{3\text{\tiny Q}}^{(n)}T}\,.$
(1.1)

Here $V_{3\text{\tiny Q}}^{(0)}$ is the ground state energy, also called the
three-quark potential, and the remaining $V_{3\text{\tiny Q}}^{(i)}$’s are
excited state energies. These are called hybrid three-quark potentials.

The crucial fact underlying the lattice calculations is that these were
performed in $SU(3)$ pure gauge theory. Because of this, the three-quark
potential increases as quarks are pulled apart. By contrast, it is expected
that in the presence of light quarks the potential flattens out. In string
models of hadrons such a phenomenon is interpreted as string breaking stringP
. It is worth noting that in the $Q\bar{Q}$ system string breaking was
established by numerical simulations bali . In particular, the estimates of a
scale (string breaking distance) characterizing this phenomenon were recently
obtained for different light quark masses bulava .

The strong decay of a heavy meson into a pair of heavy-light mesons

$Q\bar{Q}\rightarrow Q\bar{q}+\bar{Q}q\,$ (1.2)

can be interpreted as breaking a single string by light quark-antiquark pair
creation. In the case of light quarks with equal masses, one way to describe
the ground state energy of the $Q\bar{Q}$ system is to consider a two-state
system with a model Hamiltonian drum

${\cal H}=\begin{pmatrix}E_{\text{\tiny Q}\bar{\text{\tiny Q}}}&g\\\
g&2E_{\text{\tiny Q}\bar{\text{\tiny q}}}\\\ \end{pmatrix}\,.$ (1.3)

Here $E_{\text{\tiny Q}\bar{\text{\tiny Q}}}$ is the energy of two separated
heavy quark sources which are connected by a string and $2E_{\text{\tiny
Q}\bar{\text{\tiny q}}}$ is the energy of a non-interacting pair of heavy-
light mesons. The off-diagonal element $g$ describes the mixing between these
two states. The eigenvalues of this model Hamiltonian give the energy levels
of the $Q\bar{Q}$ system. In this case a characteristic scale, called the
string breaking distance, is naturally defined by drum ; bulava

$E_{\text{\tiny Q}\bar{\text{\tiny Q}}}(\boldsymbol{\ell}_{\text{\tiny
Q}\bar{\text{\tiny Q}}})=2E_{\text{\tiny Q}\bar{\text{\tiny q}}}\,.$ (1.4)

In the string models of hadrons one thinks of a baryon as a system of three
quarks connected by a Y-shaped string configuration stringP . Such a picture
implies the existence of several decay channels

$\begin{split}&\,QQq+Q\bar{q}\\\ \nearrow&\downarrow\\\
\,QQQ\,\,\rightarrow&\,\,Qqq+2Q\bar{q}\,\\\ \searrow&\downarrow\,\\\
&\,\,\,qqq+3Q\bar{q}\ \,,\end{split}\\!\\!\\!{\left\downarrow\vbox
to19.91684pt{}\right.\kern-1.2pt}$ (1.5)

defined by a number of broken strings or, equivalently, by a number of light
quark-antiquark pairs. To formalize the analogy with the mesonic case, one
could consider a model Hamiltonian

${\cal H}=\begin{pmatrix}\,E_{3\text{\tiny Q}}&g_{1}&g_{2}&g_{3}\\\
g_{1}&E_{\text{\tiny QQq}}+E_{\text{\tiny Q}\bar{\text{\tiny
q}}}&g_{12}&g_{13}\\\ g_{2}&g_{12}&E_{\text{\tiny Qqq}}+2E_{\text{\tiny
Q}\bar{\text{\tiny q}}}&g_{23}\\\ g_{3}&g_{13}&g_{23}&E_{3q}+3E_{\text{\tiny
Q}\bar{\text{\tiny q}}}\,\,\,\end{pmatrix}$ (1.6)

for a four-state system. The $E$’s represent energies of quark sources
connected by strings. Here we assume that hadrons are non-interacting, hence
the total energies are the sums of the energies of the individual hadrons. The
off-diagonal elements describe the mixing between these four states.

The two important differences from the $Q\bar{Q}$ system are (i) that it is
not clear how to define the string breaking distance by equating the diagonal
elements of the model Hamiltonian. Indeed, the relative position of the heavy
quarks, placed at vertices of a triangle, is defined by three parameters, and
the equations can only say how the minimal energy state evolves in the
parameter space. Transitions between such different minimal energy states
occur at critical parameter values. In general, given the critical values, one
can not determine the critical string distance without referring to any
particular string model. Though the exceptions exist for some symmetric cases,
as we will see in Secs.III and V; (ii) by equating the diagonal elements one
can get at least five scales. If so, it is not a priori that the ground state
energy depends only on one of them.

The basic task of this paper is to further advance the use of effective string
models in QCD. Here we continue our study of the static three-quark potentials
a-3q , elaborating on the phenomenon of string breaking. So far this issue has
not been discussed in the literature, neither in the context of lattice gauge
theory nor in the context of the gauge/string duality, particularly in
AdS/CFT. The rest of the paper is organized as follows. For orientation, we
begin in Sec.II by setting the framework and recalling some preliminary
results. Then, in Sec.III, we look at two special geometric configurations of
heavy quarks. An important feature of those is symmetry which allows one to
actually determine the string breaking distance from the critical values of
parameters without referring to any specific string model. We continue in
Sec.IV with two type of geometries. These interpolate between a diquark limit,
in which two heavy quarks are very close and the remaining one is far away
from these two, and the geometries of Sec.III. In Sec.V we provide more
details about the special geometries of Sec.III. Finally, we make concluding
remarks in Sec.VI. Appendix A contains our notation and definitions. To make
the paper more self-contained, we include many necessary results and technical
details in Appendices B-D.

##  II The model

In the gauge/string duality, the expression for the expectation value of a
baryonic Wilson loop can be put into a semiclassical form

$\langle\,W_{3\text{\tiny Q}}\,\rangle=\sum_{n}w_{n}\text{e}^{-S_{n}}\,,$
(2.1)

where $S_{n}$, whose relative weight is $w_{n}$, is expressed in terms of an
energy of the string configuration $E_{n}$ and a time interval $T$ by
$S_{n}=E_{n}T$. The $E_{n}$’s are the diagonal elements of the model
Hamiltonian (1.6). That is the key point which already allowed us to compare
the results obtained for the $Q\bar{Q}$ system in a-strb to those of lattice
gauge theory.

For three colors, string configurations are built in analogy with tree
diagrams of $\varphi^{3}$ field theory stringP . The dictionary is as follows.
A propagator now means a string which is represented by a solid line. A vertex
means a string junction, nowadays called the baryon vertex. The baryon vertex
always has three strings attached to it. Strings also end on quarks which play
a role of sources. An important ingredient in describing string configurations
on $\text{AdS}_{5}$-like geometries is a gravitational force. This force acts
on all objects. In particular, it bends strings which are now represented as
curves. Heavy quark sources are set on the boundary of five-dimensional space
and light quark sources in its interior.

To further illustrate these ideas, we employ a particular effective string
model. Though it is not exactly dual to QCD, our reasons for pursuing this
model are: (i) Because it would be good to gain some insight into any problems
for which there are no predictions from phenomenology and the lattice, but
which can be solved with the effective string model already at our disposal.
(ii) Because the results provided by this model on the quark-antiquark and
three-quark potentials are consistent with the lattice calculations and QCD
phenomenology a-3q ; az1 . (iii) Because analytic formulas are obtained by
solving the model. (iv) Because the aim of our work is to make predictions
which may then be tested by means of other non-perturbative methods, e.g.,
numerical simulations.

We will consider only the simplest class of 5-dimensional geometries which is
an extension of the geometry used for successful modeling the heavy quark
potentials in pure gauge theory. The extension is due to a scalar field which
is responsible for light quarks at string endpoints and, as a consequence, for
string breaking a-strb .222This is an attempt to describe light sea quarks
along the lines of son , but in the framework of the worldsheet formalism.
Since such a scalar signals an instability of a fundamental string, it seems
natural to call it a tachyon. If so, it could correspond to an open string
tachyon because of its action is a boundary term. One difficulty with this
interpretation is that the scalar field has a real mass son . So, this tachyon
is not the usual tachyon around the string perturbative vacuum in flat space.
We introduce a single field, because in what follows we will be interested
only in the case of two light quarks of equal mass. This background is of the
form

$ds^{2}=\text{e}^{\mathsf{s}r^{2}}\frac{R^{2}}{r^{2}}\Bigl{(}dt^{2}+d\vec{x}^{2}+dr^{2}\Bigr{)}\,,\qquad{\text{T}}={\text{T}}(r)\,,$
(2.2)

where T is the scalar field. In the absence of it, the above metric reduces to
a one-parameter deformation, with a deformation parameter $\mathsf{s}$, of
that for the Euclidean $\text{AdS}_{5}$ space of radius $R$. The geometry has
two important features:(i) aside from the boundary at $r=0$, there is a soft
wall at $z=1/\sqrt{\mathsf{s}}$ which prevents strings from going deep into
the bulk and (ii) a corresponding gravitational force has only a radial
component.

As noted above, we need three basic ingredients. The first is the fundamental
string governed by the Nambu-Goto action

$S_{\text{\tiny NG}}=\frac{1}{2\pi\alpha^{\prime}}\int
d^{2}\xi\,\sqrt{\gamma^{(2)}}\,.$ (2.3)

Here $\gamma$ is an induced metric, $\alpha^{\prime}$ is a string parameter,
and $\xi^{i}$ are worldsheet coordinates.

The second ingredient is the baryon vertex. In the context of AdS/CFT
correspondence it is a five brane witten . At leading order in
$\alpha^{\prime}$ its dynamics is described by the action
$S_{\text{vert}}={\cal T}_{5}\int d^{6}\xi\sqrt{\gamma^{(6)}}$, with ${\cal
T}_{5}$ a brane tension and $\xi^{i}$ world-volume coordinates. Since the
brane is wrapped on a five-dimensional internal space $\mathbf{X}$, the vertex
looks point-like in Euclidean space (2.2). Following a-3q , we pick a static
gauge $\xi^{0}=t$ and $\xi^{a}=\theta^{a}$, with $\theta^{a}$ coordinates on
$\mathbf{X}$. The action is then

$S_{\text{vert}}=\tau_{v}\int dt\,\frac{\text{e}^{-2\mathsf{s}r^{2}}}{r}\,,$
(2.4)

where $\tau_{v}$ is a dimensionless parameter defined by $\tau_{v}={\cal
T}_{5}R\,\text{vol}(\mathbf{X})$ and $\text{vol}(\mathbf{X})$ is a volume of
$\mathbf{X}$.

The last ingredient which takes account of light quarks at string endpoints is
the scalar field. Since we wish to mimic the $u$ and $d$ quarks of QCD which
have equal masses, we add to the worldsheet action a boundary term
$S_{\text{q}}=\int d\tau e\,\text{T}$. Such a term is usual for strings
propagating in an open string tachyon background. The integral is over a
worldsheet boundary parameterized by $\tau$ and $e$ is a boundary metric. As
in a-strb , we consider only the case of a constant background $\text{T}_{0}$
and worldsheets whose boundaries are lines in the $t$ direction. In that case,
the action written in the static gauge is

$S_{\text{q}}=\mathsf{m}\int
dt\frac{\text{e}^{\frac{\mathsf{s}}{2}r^{2}}}{r}\,,$ (2.5)

where $\mathsf{m}=R{\text{T}}_{0}$. This is obviously the action of a particle
of mass ${\text{T}}_{0}$ at rest.

Finally, let us note that in practice it is not so convenient to use the
constant factors in the above actions as the model parameters. Instead, we
choose the model parameters as follows:
$\mathsf{g}=\frac{R^{2}}{2\pi\alpha^{\prime}}$,
$\mathsf{k}=\frac{\tau_{v}}{3\mathsf{g}}$, and
$\mathsf{n}=\frac{\mathsf{m}}{\mathsf{g}}$.

##  III A first look at two special cases

To gain an initial intuition about string breaking in the three-quark system,
we consider two simple geometries. The first is a geometry with the heavy
quarks are at the vertices of an equilateral triangle and the second with the
heavy quarks are on a straight line with equal separation. The basic reason
for considering these particular geometries is that the string breaking
distance can be naturally defined in terms of the quark positions. In this
section, we restrict to the decay mode $QQQ\rightarrow QQq+Q\bar{q}$. We
return to this question in Sec.V, after describing the string configuration
for $qqq$.

###  III.1 Equilateral triangle geometry

We begin with the equilateral triangle geometry. As usual, we place the heavy
quarks at the boundary points of the five-dimensional space and consider
static string configurations in which each quark is a string endpoint. Strings
join at baryon vertices, or are terminated by light quarks in the interior.

####  III.1.1 String configurations

Since we are interested in a decay mode which is due to one pair of light
quarks, we need to consider only two string configurations. These are shown in
Figure 2.333The configuration shown in the right panel is appropriate only for
large $\ell$. A full set of the configurations for the $QQq$ system is as
shown in Figure 19. For notation, see Appendix A.

Figure 2: String configurations for the $QQQ$ system with a number of light
quarks less or equal to $2$. The heavy quarks are at the vertices of the
equilateral triangle of side $\ell$. Left: A connected configuration. Right: A
disconnected configuration.

First, let us consider the connected configuration. For the background
geometry (2.2), the energy as a function of quark separation, can be written
in parametric form a-3q

$\ell=\sqrt{\frac{3}{\mathsf{s}}}\,{\cal L}^{-}(\lambda,v)\,,\qquad
E_{3\text{\tiny Q}}=3\mathsf{g}\sqrt{\mathsf{s}}\biggl{(}{\cal
E}^{-}(\lambda,v)+\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}\biggr{)}+3c\,,$
(3.1)

where $v$ is a parameter and $c$ is a normalization constant (see Appendix B).
The functions ${\cal L}^{-}$, ${\cal E}^{-}$ are defined in Appendix A.
$\lambda$ is a function of $v$ such that

$\lambda(v)=-\text{ProductLog}\Bigl{[}-v\text{e}^{-v}\bigl{(}1-\mathsf{k}^{2}(1+4v)^{2}\text{e}^{-6v}\bigr{)}^{-\frac{1}{2}}\Bigr{]}\,.$
(3.2)

Here ProductLog denotes the ProductLog function wolf . The parameter $v$ runs
from $0$ to $v_{{\vartriangle}}$, where $v_{{\vartriangle}}$ is a solution to
$\lambda(v)=1$ (see also Eq.(4.14) below).

For future reference, note that the large $\ell$ behavior of $E_{3\text{\tiny
Q}}$ is

$E_{3\text{\tiny
Q}}=\sqrt{3}\sigma\ell-3\mathsf{g}\sqrt{\mathsf{s}}I_{3\text{\tiny
Q}}+3c+o(1)\,,$ (3.3)

with

$I_{3\text{\tiny Q}}={\cal
I}(v_{{\vartriangle}})-\mathsf{k}\frac{\text{e}^{-2v_{{\vartriangle}}}}{\sqrt{v_{{\vartriangle}}}}\,,\qquad\sigma=\text{e}\mathsf{g}\mathsf{s}\,.$
(3.4)

Here $\sigma$ is the string tension and ${\cal I}$ is defined by Eq.(A.6). As
one can easily recognize, such a behavior is governed by the $Y$-law.444In the
context of lattice gauge theory the $Y$-law was discussed in bali ; 3Q-lattice
.

To get further, consider the disconnected configuration. In the gauge/string
duality it is interpreted as a non-interacting pair of hadrons. So, the energy
of the configuration is the sum of $E_{\text{\tiny QQq}}$ and $E_{\text{\tiny
Q}\bar{\text{\tiny q}}}$, whose explicit expressions are given in Appendix D.
For our purposes, we will need to know about its large $\ell$ behavior. Using
the formulas (D.1) and (D.13), we get

$E_{\text{\tiny QQq}}+E_{\text{\tiny Q}\bar{\text{\tiny
q}}}=\sigma\ell+\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}-2I_{\text{\tiny
QQq}}\Bigr{)}+3c+o(1)\,.$ (3.5)

Here the function ${\cal Q}$ is defined by (A.5), $q$ is a solution to
equation (D.2), and $I_{\text{\tiny QQq}}$ is given by (D.14). Thus, we see
again that the asymptotic behavior for large $\ell$ is described by the
$Y$-law.

####  III.1.2 String breaking

Given the expressions for the energies, we can find a characteristic scale for
this decay mode by solving the equation $E_{3\text{\tiny
Q}}(\ell)=E_{\text{\tiny QQq}}(\ell)+E_{\text{\tiny Q}\bar{\text{\tiny q}}}$.
In general this is a complicated problem, best done numerically, but what
saves the day is that $\ell_{c}$ is an infrared scale. If so, then the
equation is linear in very good approximation. To see that this is the case,
we plot the energies in Figure 3. We use the parameter set $L$ defined below.
Indeed, both functions behave de facto

Figure 3: $E_{3\text{\tiny Q}}$ and $E_{\text{\tiny QQq}}+E_{\text{\tiny
Q}\bar{\text{\tiny q}}}$ vs $\ell$ in the case of the equilateral triangle
geometry for $\mathsf{k}=-0.102$ and $c=0.623\,\text{GeV}$.

as linear functions for $\ell\gtrsim 0.5\,\text{fm}$. Given this, the critical
value of $\ell$ is555Here and below, the superscript $(1)$ indicates that the
scale is set by equating $E_{3\text{\tiny Q}}$ to $E_{\text{\tiny
QQq}}+E_{\text{\tiny Q}\bar{\text{\tiny q}}}$, where $1$ stands for one pair
of $q\bar{q}$.

$\ell_{c}^{(1)}=\frac{1+\sqrt{3}}{2\text{e}\sqrt{\mathsf{s}}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+3I_{3\text{\tiny
Q}}-2I_{\text{\tiny QQq}}\Bigr{)}\,,$ (3.6)

as it follows from the $Y$-laws (3.3) and (3.5). It is finite and independent
of $c$, as expected.

There is a nontrivial issue concerning the question how to define the notion
of a string length in five dimensional models which is consistent with QCD in
four dimensions. In four dimensions, the string description is valid at large
quark separations, and as a result, the energy of the $QQQ$ system is
described by the $Y$-law. In this case, the string length is naturally defined
as a distance between its endpoints, one of which is attached to a heavy quark
and the other to a string junction. All the strings in question are straight
and the junction is at the Fermat point of a triangle $Q_{1}Q_{2}Q_{3}$. The
situation becomes less transparent when dealing with five dimensional string
models motivated by the gauge/string duality. The problem here is that a
baryon vertex is in the interior, while heavy quarks are on the boundary of a
five-dimensional curved space. To make a contact with QCD, it is natural to
define the string length as a distance between a string endpoint attached to a
heavy quark and a projection of the baryon vertex onto the boundary, i.e. by
$|Q_{i}Y|$ (see Figure 2). The good reason for this is that in those models
the $Y$-law is written precisely in terms of such defined string length a-3q0
.

In the present case, the definition of the string length is dictated by
symmetry. The point $Y$ is always at the triangle center which coincides with
its Fermat point, and hence $|Q_{i}Y|=\ell/\sqrt{3}$. Thus,
$\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=|Q_{i}Y|_{c}=\ell_{c}^{(1)}/\sqrt{3}$. Explicitly,

$\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=\frac{3+\sqrt{3}}{6\text{e}\sqrt{\mathsf{s}}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+3I_{3\text{\tiny
Q}}-2I_{\text{\tiny QQq}}\Bigr{)}\,.$ (3.7)

This is the formula we will need to make some preliminary estimates of the
string breaking distance in subsection C.

###  III.2 Symmetric collinear geometry

Our second special example is that of the collinear geometry. Without loss of
generality, we may place heavy quarks on the $x$-axis at points
$(-\ell,0,\ell)$. As before, these quarks are string endpoints. The other
endpoints are terminated at light quarks and vertices in the interior to form
color singlets.

####  III.2.1 String configurations

We begin by considering the configuration of Figure 4. It has been previously

Figure 4: A connected collinear configuration. It is symmetric under
reflection through the $r$-axis. So, $Q_{2}$ is the projection of $V$ onto the
$x$-axis.

described in a-3q , where the explicit formula for the energy was derived.
When written in terms of the functions defined in Appendix A, this formula is

$\ell=\frac{1}{\sqrt{\mathsf{s}}}\,{\cal L}^{-}(\lambda,v)\,,\qquad
E_{3\text{\tiny Q}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}2{\cal
E}^{-}(\lambda,v)+3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}+{\cal
Q}(v)\Bigr{)}+3c\,.$ (3.8)

Here $\lambda$ is a function of $v$ which is given by

$\lambda(v)=-\text{ProductLog}\biggl{(}-\frac{2}{\sqrt{3}}v\,\text{e}^{-v}\Bigl{(}1+2\mathsf{k}(1+4v)\text{e}^{-3v}-3\mathsf{k}^{2}(1+4v)^{2}\text{e}^{-6v}\Bigr{)}^{-\tfrac{1}{2}}\biggr{)}\,.$
(3.9)

The parameter $v$ is varying from $0$ to $\bar{v}$, where $\bar{v}$ is a
solution to

$\frac{4}{3}v^{2}\text{e}^{2(1-v)}=1+2\mathsf{k}(1+4v)\text{e}^{-3v}-3\mathsf{k}^{2}(1+4v)^{2}\text{e}^{-6v}\,$
(3.10)

or, equivalently, to $\lambda(v)=1$.

Just as in the above example, the large $\ell$ behavior is governed by the
$Y$-law which in the present case takes the form

$E_{3\text{\tiny
Q}}(\ell)=2\sigma\ell-2\mathsf{g}\sqrt{\mathsf{s}}\bar{I}_{3\text{\tiny
Q}}+3c+o(1)\,,\quad\text{with}\quad\bar{I}_{3\text{\tiny
Q}}=\mathcal{I}(\bar{v})-\frac{1}{2}\mathcal{Q}(\bar{v})-\frac{3}{2}\mathsf{k}\frac{\text{e}^{-2\bar{v}}}{\sqrt{\bar{v}}}\,.$
(3.11)

Note that for this geometry the Fermat point is coincident with $Q_{2}$.

Now let us consider the disconnected configurations shown in Figure 5. Here,
unlike the first example, there are

Figure 5: Typical disconnected collinear configurations. In the configurations
shown here, $\ell$ is large enough.

two possible configurations. The configuration shown in the left panel
corresponds to the case when one of the side strings breaks down, whereas the
configuration shown in the right panel to the case when the middle string
breaks down. Among of them, the first provides a dominant contribution to the
baryonic Wilson loop as its energy is smaller. The simple reason for this is
that its $QQq$ subsystem has a smaller length, $\ell$ rather than $2\ell$.
Since we assume that hadrons do not interact with each other, the dominant
configuration has the same energy as the disconnected configuration shown in
Figure 2. Hence its large $\ell$ behavior is as given by Eq.(3.5).

####  III.2.2 String breaking

We are now in position to estimate the string breaking distance for this
geometry. As before, we do so by solving the equation $E_{3\text{\tiny
Q}}(\ell)=E_{\text{\tiny QQq}}(\ell)+E_{\text{\tiny Q}\bar{\text{\tiny q}}}$.
This problem is simple because the linear approximation is applicable for
large $\ell$, as seen from the plot in Figure 6. Given a critical value
$\ell_{c}^{(1)}$, the string breaking distance is
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}=\ell_{c}^{(1)}$ as dictated

Figure 6: $E_{3\text{\tiny Q}}$ and $E_{\text{\tiny QQq}}+E_{\text{\tiny
Q}\bar{\text{\tiny q}}}$ vs $\ell$ for the symmetric collinear geometry,
plotted here for the parameter set $L$ with $\mathsf{k}=-0.102$ and
$c=0.623\,\text{GeV}$.

by symmetry. A short calculation, using the formulas (3.11) and (D.13), shows
that

$\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=\frac{1}{\text{e}\sqrt{\mathsf{s}}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+2\bar{I}_{3\text{\tiny
Q}}-2I_{\text{\tiny QQq}}\Bigr{)}\,,$ (3.12)

which is the analog of Eq.(3.7).

###  III.3 Some estimates

Let us make some estimates. In doing so, we use the two parameter sets
suggested in a-strb . The first is mainly a result of fitting the lattice QCD
data to the string model we are considering. So, we call it $L$ for brevity.
The value of $\mathsf{s}$ is fixed from the slope of the Regge trajectory of
$\rho(n)$ mesons in the soft wall model with the geometry (2.2). This gives
$\mathsf{s}=0.450\,\text{GeV}^{2}$ a-q2 . Then, using (3.4), we obtain
$\mathsf{g}=0.176$ by fitting the value of the string tension $\sigma$ to its
value in bulava . The parameter $\mathsf{n}$ is adjusted to reproduce the
lattice result for the string breaking distance in the $Q\bar{Q}$ system. With
$\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}=1.22\,\text{fm}$
bulava , this gives $\mathsf{n}=3.057$. In a-3q , the value of $\mathsf{k}$ is
adjusted to fit the three-quark potential to the lattice data for pure $SU(3)$
gauge theory. So far there is no such data available for QCD with two
dynamical quarks. We will use two different parameter values:
$\mathsf{k}=-0.102$ and $\mathsf{k}=-\frac{1}{4}\text{e}^{\frac{1}{4}}$. These
parameter values are of special interest. The first comes from matching to the
Lipkin rule $E_{\text{\tiny QQ}}=\frac{1}{2}E_{\text{\tiny Q}\bar{\text{\tiny
Q}}}$ lipkin , but at the same time it is rather close to the value of
$-0.083$ used in a-3q to describe the lattice data for the static three-quark
potentials in $SU(3)$ pure gauge theory 3Q-lattice . The reasons for the
second value are due to the particular model we are using here.666We discuss
them in Sec.V.

In bulava , the calculations were done at unphysical pion mass
$m_{\pi}=280\,\text{MeV}$. Because of this and in view of possible
applications to phenomenology, we now consider the second parameter set
denoted by $P$. In that case, the values of $\mathsf{s}$ and $\mathsf{g}$ are
extracted from the quarkonium spectrum obtained by using the heavy quark
potential derived from the model we are considering az1 . This is self-
consistent, and gives $\mathsf{s}=0.15\,\text{GeV}^{2}$ and $\mathsf{g}=0.44$
PC .777At first glance, the difference between these parameter sets is
significant. However, what is relevant for making the estimates of string
breaking distances is the value of the string tension (3.4). With the $L$ set,
it is about $0.215\,\text{GeV}^{2}$, while with $P$ about
$0.285\,\text{GeV}^{2}$. So, the values are different but not greatly so. Then
we determine $\mathsf{n}$ from the condition $E_{\text{\tiny
Qqq}}-E_{\text{\tiny Q}\bar{\text{\tiny
q}}}=M_{\Lambda_{c}^{+}}-M_{D^{0}}\approx 420\,\text{MeV}$. It results in
$\mathsf{n}=1.589$. We use $\mathsf{k}=-0.102$ and
$\mathsf{k}=-\tfrac{1}{4}\text{e}^{\frac{1}{4}}$, as before.

Having set the parameters, one can use Eqs.(3.7) and (3.12) to estimate the
string breaking distance $\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$. The
results are summarized in Table 1.

Geometry |  $\mathsf{s}\,(\text{GeV})$ |  $\mathsf{g}$ |  $\mathsf{n}$ |  $\mathsf{k}$ |  $\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}\,(\text{fm})$ |  $\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}\,(\text{fm})$ |  $\frac{\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}}{\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}}$
---|---|---|---|---|---|---|---
triangle | 0.450 | 0.176 | 3.057 | -0.102 | 0.910 | 1.220 | 0.746
triangle | 0.450 | 0.176 | 3.057 |  -$\frac{1}{4}\text{e}^{\frac{1}{4}}$ | 0.911 | 1.220 | 0.747
triangle | 0.150 | 0.440 | 1.589 | -0.102 | 0.810 | 1.074 | 0.754
triangle | 0.150 | 0.440 | 1.589 |  -$\frac{1}{4}\text{e}^{\frac{1}{4}}$ | 0.810 | 1.074 | 0.754
collinear | 0.450 | 0.176 | 3.057 | -0.102 | 1.100 | 1.220 | 0.902
collinear | 0.450 | 0.176 | 3.057 |  -$\frac{1}{4}\text{e}^{\frac{1}{4}}$ | 1.180 | 1.220 | 0.967
collinear | 0.150 | 0.440 | 1.589 | -0.102 | 0.934 | 1.074 | 0.870
collinear | 0.150 | 0.440 | 1.589 |  -$\frac{1}{4}\text{e}^{\frac{1}{4}}$ | 1.070 | 1.074 | 0.996
Table 1:  Results for the string breaking distance
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$ obtained from the $L$ and $P$
parameter sets at two values of $\mathsf{k}$.

We conclude this subsection by making a few remarks.

(i) We have used the different parameter sets to assess the robustness of the
conclusion that, in contrast to the string tension, the string breaking
distance is not universal. In the $QQQ$ system it may be smaller than that in
the $Q\bar{Q}$ system and, moreover, it may depend on geometry, i.e. on the
positions of the three heavy quarks in space.

(ii) We included the results of a-strb for the string breaking distance in the
$Q\bar{Q}$ system because it is of physical interest to compare both systems.
The explicit formula reads

$\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny
Q}}}=\frac{2}{\text{e}\sqrt{\mathsf{s}}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+I_{0}\Bigr{)}\,,$
(3.13)

where the constant $I_{0}$ is defined in Appendix A. Note that it allows one
to get rid of the explicit dependence of $\mathsf{s}$ and $\mathsf{g}$ in the
ratio $\frac{\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}}{\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}}$. So, in
contrast to the triangle geometry, a stronger $\mathsf{k}$-dependence is seen
for the collinear geometry.

(iii) It is interesting to make an estimate of the quark mass in our model. In
son , it was assumed that the quark mass is simply related to the parameter
${\text{T}}_{0}$ in Eq.(2.5), namely $m={\text{T}}_{0}$. However,
${\text{T}}_{0}$ does not enter explicitly in the formulas for the string
breaking distance because it is combined with $R$ to form a dimensionless
product. The unknown parameter $R$ makes a direct estimate impossible. Yet
what saves the day in the case with the $L$ parameter set is that in bulava
the sum of the quark masses ($m_{u/d}+m_{s}$) is approximately equal to its
physical value.888We thank F. Knechtli for pointing this out. This allows us
to estimate the value of $R$ and hence the quark masses. The details are as
follows. First, we extend the model by adding the strange quark. Then, using
(3.13), we get $\mathsf{n}_{s}=3.239$ by fitting the value of the string
breaking distance associated with a pair of strange quarks to
$1.29\,\text{fm}$ from bulava . Finally from the expression
$\mathsf{n}=\frac{R}{\mathsf{g}}m$ specialized to the $u/d$ and $s$ quarks, we
immediately find
$m_{u/d}=\frac{m_{u/d}+m_{s}}{\mathsf{n}+\mathsf{n}_{s}}\mathsf{n}$. With
$m_{u/d}+m_{s}=96\,\text{MeV}$, this gives $m_{u/d}=46.6\,\text{MeV}$. What is
an appropriate value of $m_{u/d}$ for the $P$ set? Suppose that the ratio
$\frac{R}{\mathsf{g}}$ is universal. In that case, the mass will be
$m_{u/d}=23.5\,\text{MeV}$. This is of course a crude estimate, but it is in
agreement with the fact that the string breaking distance for the $P$ set is
smaller than that for the $L$ set. One may wonder about the interpretation of
$m$. The answer could be that it is a running quark mass in the IR regime at
the energy scale of order $\Lambda_{\text{\tiny QCD}}$. The reason for this is
that the worldsheet action (2.5) was introduced to describe the IR phenomenon
of string breaking. Of course, further work is needed to clarify this issue.

(iv) This is not the whole story because of the other modes in (1.5). These
modes are discussed in Sec.V, but with the same conclusion.

##  IV More general geometries

Now we will describe two types of geometries that allow us to see how
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$ decreases from the value
$\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}$ to the values we
found in the previous Section. The basic idea is to start from a special case
in which two heavy quarks are very close and one is far away from these two,
or in other words from the diquark limit. In this limit the equation
$E_{3\text{\tiny Q}}=E_{\text{\tiny QQq}}+E_{\text{\tiny Q}\bar{\text{\tiny
q}}}$ can be easily solved. From heavy quark-diquark symmetry wise , it
follows that $E_{3\text{\tiny Q}}=E_{\text{\tiny Q}\bar{\text{\tiny
Q}}}+E_{\text{\tiny QQ}}$ and $E_{\text{\tiny QQq}}=E_{\text{\tiny
q}\bar{\text{\tiny Q}}}$+$E_{\text{\tiny QQ}}$. After substituting those in
the equation, we find $E_{\text{\tiny Q}\bar{\text{\tiny Q}}}=E_{\text{\tiny
Q}\bar{\text{\tiny q}}}+E_{\text{\tiny q}\bar{\text{\tiny Q}}}$ whose solution
is $\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}$. This implies that
in the diquark limit $\boldsymbol{\ell}_{3\text{\tiny
Q}}=\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}$. Then we depart
from this limit by more and more separating the neighbor quarks until we reach
one of the geometries of Section III. Importantly, the decay mode
$QQQ\rightarrow QQq+Q\bar{q}$ is dominant as long as the separation is not
large enough. In other words, the two remaining strings are short and,
therefore, do not break down. So, one can identify
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$ with the string breaking distance
seen from the ground state energy because this scale is the only possible
measure for string breaking.

###  IV.1 Isosceles triangle geometry

First consider a geometry in which the heavy quarks are at the vertices of an
isosceles triangle. Without loss of generality, we can place the triangle in
the $xy$-plane as shown in Figure 7. The base angle $\gamma$ runs from
$\frac{\pi}{3}$ to $\frac{\pi}{2}$. The

Figure 7: An isosceles triangle of base length $\ell$ and height $h$.

former corresponds to the equilateral triangle geometry we discussed in the
previous section, and the latter to the diquark limit, but with small enough
$\ell$. For convenience, we list the formulas

$\ell=2\cos\beta\,|Q_{1}Y|\,,\qquad
h=\sin\beta|Q_{1}Y|+|Q_{3}Y|\,,\qquad\tan\gamma=\tan\beta+\frac{1}{\cos\beta}\frac{|Q_{3}Y|}{|Q_{1}Y|}\,,$
(4.1)

all of which are simple, but useful.

####  IV.1.1 Connected string configurations

We are interested in the situation when a string connected to $Q_{3}$ breaks
down. For this to be the case, the string should be long enough. In practice,
this means that the string is of type shown in Figure 15 on the right. If so,
then there are only two possible string configurations. These are illustrated
in Figure 8.

Figure 8: Connected string configurations in which $Q_{3}$ is away from
$Q_{1}$ and $Q_{2}$. These are symmetric under reflection in the $yr$-plane.
Left: The case $\alpha_{1}\geq 0$. Right: The case $\alpha_{1}\leq 0$. Here
$\alpha_{1}$ is a tangent angle at $V$ for the string stretched between
$Q_{1}$ and $V$.

Such configurations have not been studied in the literature, except for the
diquark limit a-3q . So, it is convenient to begin with this limiting case and
then go further by increasing the base length $\ell$. If so, then the
corresponding configuration is that shown in the Figure on the left.

Given the configuration, we can use the general formulas (B.9) and (B.14) to
write the distances between the points $Q_{i}$ and $Y$ as

$|Q_{1}Y|=|Q_{2}Y|=\frac{1}{\sqrt{\mathsf{s}}}{\cal
L}^{+}(\alpha_{1},v)\,,\qquad|Q_{3}Y|=\frac{1}{\sqrt{\mathsf{s}}}{\cal
L}^{-}(\lambda_{3},v)\,.$ (4.2)

By symmetry, the shape of strings connected to $Q_{1}$ and $Q_{2}$ is the
same. This implies that $\alpha_{1}=\alpha_{2}$.

The string energies can be read off from the expressions (B.12) and (B.16).
Combining those with the gravitational energy of the vertex, we find the total
energy of the configuration

$E_{3\text{\tiny Q}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}2{\cal
E}^{+}(\alpha_{1},v)+{\cal
E}^{-}(\lambda_{3},v)+3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}\,\Bigr{)}\,+3c\,.$
(4.3)

With the help of the relation (B.17) and gluing conditions (C.2), the angles
$\alpha_{1}$, $\alpha_{3}$, and $\beta$ are explicitly written as

$\sin\alpha_{1}=\frac{1}{2}\Bigl{(}3\mathsf{k}(1+4v)\text{e}^{-3v}+\bigl{[}1-\cos^{2}\alpha_{3}\bigr{]}^{\frac{1}{2}}\Bigr{)}\,,\qquad\cos\alpha_{3}=\frac{v}{\lambda_{3}}\text{e}^{\lambda_{3}-v}\,,\qquad\sin\beta=\frac{1}{2}\cos\alpha_{3}\bigl{[}1-\sin^{2}\alpha_{1}\bigr{]}^{-\frac{1}{2}}\,.$
(4.4)

So the result for $E_{3\text{\tiny Q}}(\ell,h)$ can be conveniently expressed
in terms of two parameters $(v,\lambda_{3})$.

Now we wish to look at the behavior of $E_{3\text{\tiny Q}}$ in the diquark
limit: small $\ell$ and large $|Q_{3}Y|$. This case was discussed in relation
to the static quark-quark potential in a-3q . A starting point is that one has
to analyze the parametric equations in a small region near the point $(0,1)$.
We do not need all the results of the analysis. What is important for us to
know is the behavior of $\ell$. It behaves as

$\ell=\sqrt{\frac{v}{\mathsf{s}}}\bigl{(}\ell_{0}+o(1)\,\bigr{)}\,.$ (4.5)

Here
$l_{0}=\frac{1}{2}\xi^{-\frac{1}{2}}B\bigl{(}\xi^{2};\tfrac{3}{4},\tfrac{1}{2}\bigr{)}$,
$\xi=\frac{\sqrt{3}}{2}(1-2\mathsf{k}-3\mathsf{k}^{2})^{\frac{1}{2}}$, and
$B(z;a,b)$ is the incomplete beta function.

As we saw in Sec.III, to analyze string breaking, it helps to know that this
phenomenon occurs in IR limit in which the linear approximation (B.23) turns
out to be quite accurate. Since we assume that the string ended on $Q_{3}$
breaks down, we should take the limit $\lambda_{3}\rightarrow 1$. In that
limit, Eq.(4.3) for the energy becomes

$E_{3\text{\tiny Q}}=\sigma|Q_{3}Y|-\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}{\cal
I}(v)-2{\cal
E}^{+}(\alpha_{1},v)-3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}\,\Bigr{)}+3c+o(1)\,,$
(4.6)

and similarly Eqs.(4.4) for the auxiliary angles become

$\sin\alpha_{1}=\frac{1}{2}\biggl{(}3\mathsf{k}(1+4v)\text{e}^{-3v}+\Bigl{[}1-v^{2}\text{e}^{2(1-v)}\Bigr{]}^{\frac{1}{2}}\biggr{)}\,,\qquad\cos\alpha_{3}=v\text{e}^{1-v}\,,\qquad\sin\beta=\frac{1}{2}v\text{e}^{1-v}\Bigl{[}1-\sin^{2}\alpha_{1}\Bigr{]}^{-\frac{1}{2}}\,.$
(4.7)

The parameter $v$ varies from $0$ to $v_{{\blacktriangle}}$, where
$v_{{\blacktriangle}}$ is a solution to equation

$3\mathsf{k}(1+4v)\text{e}^{-3v}+\Bigl{[}1-v^{2}\text{e}^{2(1-v)}\Bigr{]}^{\frac{1}{2}}=0\,.$
(4.8)

The solution $v_{{\blacktriangle}}$ has a clear meaning. The angle
$\alpha_{1}$ precisely vanishes at $v=v_{{\blacktriangle}}$. As a result, for
larger values of $v$ the string configuration becomes as shown in Figure 8 on
the right. What is, however, important for our purposes is that $\ell(v)$ is
an increasing function of $v$. Thus the transition between these two types of
configurations gives rise to larger values of $\ell$.

Similarly, we may also analyze the remaining configuration. From (B.14) it
follows that

$|Q_{1}Y|=|Q_{2}Y|=\frac{1}{\sqrt{\mathsf{s}}}{\cal
L}^{-}(\lambda_{1},v)\,,\qquad|Q_{3}Y|=\frac{1}{\sqrt{\mathsf{s}}}{\cal
L}^{-}(\lambda_{3},v)\,.$ (4.9)

Because of the reflection symmetry, $\lambda_{2}$ is equal to $\lambda_{1}$.

The expression for the energy is simply obtained by replacing ${\cal E}^{+}$
with ${\cal E}^{-}$ in (4.3). So,

$E_{3\text{\tiny Q}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}\sum_{i=1}^{3}{\cal
E}^{-}(\lambda_{i},v)+3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}\,\Bigr{)}+3c\,.$
(4.10)

Combining the formulas (B.17) and (B.18) with the gluing conditions (C.2)
gives

$\lambda_{1}=-\text{ProductLog}\biggl{(}-v\,\text{e}^{-v}\biggl{[}1-\frac{1}{4}\biggl{(}3\mathsf{k}(1+4v)\text{e}^{-3v}+\Bigl{[}1-\frac{v^{2}}{\lambda_{3}^{2}}\text{e}^{2(\lambda_{3}-v)}\Bigr{]}^{\frac{1}{2}}\biggr{)}^{2}\,\biggr{]}^{-\frac{1}{2}}\,\biggr{)}\,,\qquad\sin\beta=\frac{\lambda_{1}}{2\lambda_{3}}\text{e}^{\lambda_{3}-\lambda_{1}}\,.$
(4.11)

All of this allows one to write $E_{3\text{\tiny Q}}(\ell,h)$ in parametric
form with the same parameters $v$ and $\lambda_{3}$ as before.

Since we are interested in the situation when the string connected to $Q_{3}$
is long enough, we have to consider the behavior near $\lambda_{3}=1$. If so,
then the expression for the energy takes the form

$E_{3\text{\tiny Q}}=\sigma|Q_{3}Y|-\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}{\cal
I}(v)-2{\cal
E}^{-}(\lambda_{1},v)-3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}\,\Bigr{)}+3c+o(1)\,,$
(4.12)

whereas the expressions for $\lambda_{1}$ and $\beta$ take the form

$\lambda_{1}=-\text{ProductLog}\biggl{(}-v\,\text{e}^{-v}\biggl{[}1-\frac{1}{4}\biggl{(}3\mathsf{k}(1+4v)\text{e}^{-3v}+\Bigl{[}1-v^{2}\text{e}^{2(1-v)}\Bigr{]}^{\frac{1}{2}}\biggr{)}^{2}\,\biggr{]}^{-\frac{1}{2}}\,\biggr{)}\,,\qquad\sin\beta=\frac{1}{2}\lambda_{1}\text{e}^{1-\lambda_{1}}\,.$
(4.13)

Here the parameter $v$ runs from $v_{{\blacktriangle}}$ to
$v_{{\vartriangle}}$, where $v_{{\vartriangle}}$ is a solution to the equation
$\lambda_{1}(v)=1$. We denote this solution by $v_{{\vartriangle}}$ because it
is the same as that in the case of the equilateral triangle geometry. One way
to see this is to rewrite the equation $\lambda_{1}(v)=1$ as

$\mathsf{k}(1+4v)\text{e}^{-3v}+\Bigl{[}1-v^{2}\,\text{e}^{2(1-v)}\Bigr{]}^{\frac{1}{2}}=0\,.$
(4.14)

In the limit $v\rightarrow v_{{\vartriangle}}$, the angle $\beta$ tends to
$\frac{\pi}{6}$, and hence $Y$ tends to the Fermat point of the
triangle.999$\beta(v)$ is an increasing function of $v$ which takes values on
the interval $[0,\pi/6]$. Thus, for a general $v$ the point $Y$ is different
from the Fermat point. Moreover, as $|Q_{1}Y|/|Q_{3}Y|\rightarrow 1$ the
isosceles triangle approaches the equilateral one. This is the IR limit of the
$QQQ$ system such that all the quarks are away from each other. The energy is
then given by

$E_{3\text{\tiny
Q}}=\sigma|Q_{3}Y|+\frac{2}{\sqrt{3}}\sigma\ell-3\mathsf{g}\sqrt{\mathsf{s}}\,I_{3\text{\tiny
Q}}+3c+o(1)\,$ (4.15)

that is nothing but the Y-law (3.3) expressed in terms of $|Q_{3}Y|$ and
$\ell$.

Thus, we started from the diquark limit at $v=0$ and finally reached the
equilateral triangle geometry at $v=v_{{\vartriangle}}$ by separating the
neighbor quarks $Q_{1}$ and $Q_{2}$.

####  IV.1.2 Disconnected configuration and string breaking

For the decay mode under consideration the corresponding disconnected
configuration is that of Figure 2 but with $Q_{3}$ placed at $y=h$. Under the
assumption of non-interacting hadrons, its energy is the same sum of
$E_{\text{\tiny QQq}}$ and $E_{\text{\tiny Q}\bar{\text{\tiny q}}}$.

Now, we want to find a critical value of $|Q_{3}Y|$ by solving the equation
$E_{3\text{\tiny Q}}=E_{\text{\tiny QQq}}+E_{\text{\tiny Q}\bar{\text{\tiny
q}}}$. That gives the string breaking distance

$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}=|Q_{3}Y|_{c}\,$ (4.16)

meaning that for fixed $\ell$, the disconnected configuration dominates in the
expectation value of the Wilson loop for $|Q_{3}Y|>|Q_{3}Y|_{c}$. Since both
functions $E_{3\text{\tiny Q}}$ and $E_{\text{\tiny QQq}}$ are defined
piecewise so that each subinterval has a size depending on a value of
$\mathsf{k}$, for definiteness we restrict to
$\mathsf{k}=-0.102$.101010Results for other values of $\mathsf{k}$ are
similar. We also choose the parameter set $L$. A straightforward but somewhat
tedious calculation, using the linear approximations (4.6) and (4.12), shows
that the equation $E_{3\text{\tiny Q}}=E_{\text{\tiny QQq}}+E_{\text{\tiny
Q}\bar{\text{\tiny q}}}$ is solved by

$\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=\begin{cases}\boldsymbol{\ell}_{0}+\frac{2}{\text{e}\sqrt{\mathsf{s}}}\bigl{(}{\cal
E}^{+}(\alpha^{\prime},v^{\prime})-{\cal
E}^{+}(\alpha_{1},v)\bigr{)}\quad&\text{if}\quad 0\leq\ell\leq
0.248\,\text{fm}\,,\\\
\boldsymbol{\ell}_{1}+\frac{2}{\text{e}\sqrt{\mathsf{s}}}\bigl{(}{\cal
E}^{+}(\alpha^{\prime},v^{\prime})-{\cal
E}^{+}(\alpha_{1},v)\bigr{)}\quad&\text{if}\quad 0.248\leq\ell\leq
0.779\,\text{fm}\,,\\\
\boldsymbol{\ell}_{1}+\frac{2}{\text{e}\sqrt{\mathsf{s}}}\bigl{(}{\cal
E}^{+}(\alpha^{\prime},v^{\prime})-{\cal
E}^{-}(\lambda,v)\bigr{)}\quad&\text{if}\quad 0.779\leq\ell\leq
1.110\,\text{fm}\,,\\\
\boldsymbol{\ell}_{1}+\frac{2}{\text{e}\sqrt{\mathsf{s}}}\bigl{(}{\cal
E}^{-}(\lambda^{\prime},v^{\prime})-{\cal
E}^{-}(\lambda,v)\bigr{)}\quad&\text{if}\quad 1.110\leq\ell\leq
1.577\,\text{fm}\,,\end{cases}$ (4.17)

with

$\begin{split}\boldsymbol{\ell}_{0}=\frac{1}{\text{e}\sqrt{\mathsf{s}}}\Bigl{(}2{\cal
Q}(q)-{\cal
Q}(v^{\prime})+2\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+3\mathsf{k}\Bigl{(}\frac{\text{e}^{-2v^{\prime}}}{\sqrt{v^{\prime}}}-\frac{\text{e}^{-2v}}{\sqrt{v}}\Bigr{)}+{\cal
I}(v)\Bigr{)}\,,\\\
\boldsymbol{\ell}_{1}=\frac{1}{\text{e}\sqrt{\mathsf{s}}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\Bigl{(}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+\frac{\text{e}^{\frac{1}{2}v^{\prime}}}{\sqrt{v^{\prime}}}\Bigr{)}+3\mathsf{k}\Bigl{(}\frac{\text{e}^{-2v^{\prime}}}{\sqrt{v^{\prime}}}-\frac{\text{e}^{-2v}}{\sqrt{v}}\Bigr{)}+{\cal
I}(v)\Bigr{)}\,.\end{split}$ (4.18)

Here $v$ is the parameter for the $QQQ$ configuration. The function
$\alpha_{1}(v)$ is defined by (4.7) and $\lambda(v)$ by (4.13). The parameter
$v^{\prime}$ refers to the $QQq$ configuration. The corresponding functions
$\alpha^{\prime}(v^{\prime})$ and $\lambda^{\prime}(v^{\prime})$ are defined
by (D.6), (D.9), and (D.12). The base length is expressed by the two functions
$\ell=\ell(v)$ and $\ell=\ell(v^{\prime})$ which are defined in the previous
subsection and Appendix D. Because of this, the parameters are not
independent. Note that the value $\ell=1.577\,\text{fm}$ corresponds to
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}=0.910\,\text{fm}$ obtained for the
equilateral triangle geometry in Sec.III.

Given the parametric equations $\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}(v)$ and $\ell=\ell(v)$, we
can eliminate the parameter and find $\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}(\ell)$ numerically. The
result is presented in Figure 9 on the left. We see that
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$ does decrease with increasing the
base

Figure 9: Left: $\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$ as a function of
$\ell$. Right: $h_{c}^{(1)}$ as a function of $\ell$. The dashed lines are the
asymptotic behaviors obtained from (4.20) and (4.21), and dotted line from
(4.22).

length of the triangle.

There is certainly more that we could say about this geometry. The point is
that for small enough $\ell$ the decay mode we are considering is dominant.
This means that the ground state is determined in terms of two states: $QQQ$
and $QQq+Q\bar{q}$, and the scale $\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$
is the only scale in question. One way to roughly estimate the upper bound for
$\ell$ is to accept that the $QQq$ system becomes unstable for
$\ell\gtrsim\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}$ a-QQq . If
so, then we expect that only one string breaks down if the separation between
$Q_{1}$ and $Q_{2}$ (length of the triangle’s base) is less than
$1.22\,\text{fm}$. For larger separations, the other decay modes become
relevant that makes the whole picture of string breaking quite complicated.

In order to make comparison with other approaches easier, we introduce a
critical height $h_{c}^{(1)}=|Q_{3}Y|_{c}+\frac{1}{2}\tan\beta\,\ell$.
Similarly to $\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$, we can also
eliminate $v$ and find $h_{c}^{(1)}=h_{c}^{(1)}(\ell)$ numerically. The result
of this is shown in Figure 9 on the right. In contrast to
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$, the critical height behaves in a
more complicated way. It is nearly $\boldsymbol{\ell}_{\text{\tiny
Q}\bar{\text{\tiny Q}}}$ for $\ell$ below $0.2\,\text{fm}$ and then, after a
small decrease, starts to linearly increase with $\ell$ for $\ell$ above
$0.3\,\text{fm}$.

Now consider the diquark limit so that $\ell\rightarrow 0$. It is useful to
write the first expression in (4.17) as

$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}=\boldsymbol{\ell}_{\text{\tiny
Q}\bar{\text{\tiny Q}}}+\frac{1}{\text{e}\sqrt{\mathsf{s}}}\Bigl{[}2{\cal
E}^{+}(\alpha^{\prime}_{1},v^{\prime})-2{\cal
E}^{+}(\alpha_{1},v)+3\mathsf{k}\Bigl{(}\frac{\text{e}^{-2v^{\prime}}}{\sqrt{v^{\prime}}}-\frac{\text{e}^{-2v}}{\sqrt{v}}\Bigr{)}-{\cal
Q}(v^{\prime})-\frac{1}{\sqrt{v}}+\int_{0}^{\sqrt{v}}\frac{du}{u^{2}}\bigl{(}\text{e}^{u^{2}}[1-u^{4}\text{e}^{2(1-u^{2})}]^{\frac{1}{2}}-1\bigr{)}\Bigr{]}\,,$
(4.19)

with $\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}$ given by (3.13).
The terms inside the square brackets vanish as $\ell\rightarrow 0$. This
immediately follows from the fact that this limit corresponds to $v\rightarrow
0$ and $v^{\prime}\rightarrow 0$. If so, then in leading order $v=v^{\prime}$
and
$\cos\alpha_{1}=\cos\alpha^{\prime}_{1}=\frac{\sqrt{3}}{2}(1-2\mathsf{k}-3\mathsf{k}^{2})^{\frac{1}{2}}$.
So, one obtains $\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}$, as expected
from quark-diquark symmetry.

The behavior for large enough $\ell$ is

$\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=\bigl{(}1-\tfrac{2}{\sqrt{3}}\bigr{)}\ell+\frac{1}{\text{e}\sqrt{\mathsf{s}}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+3I_{3\text{\tiny
Q}}-2I_{\text{\tiny QQq}}\Bigr{)}\,,$ (4.20)

as it follows from (4.15) and (D.13). For $\gamma=\frac{\pi}{3}$, where
$\ell=\sqrt{3}\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$, it reduces to the
expression (3.7) obtained for the equilateral triangle geometry. With the help
of this expression, one finds the asymptotic behavior of the critical height

$h_{c}^{(1)}=\bigl{(}1-\tfrac{\sqrt{3}}{2}\bigr{)}\ell+\frac{1}{\text{e}\sqrt{\mathsf{s}}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+3I_{3\text{\tiny
Q}}-2I_{\text{\tiny QQq}}\Bigr{)}\,.$ (4.21)

We conclude this discussion with one more remark. If one assumes that the
string breaking distance is universal, i.e.
$\boldsymbol{\ell}_{3Q}^{(1)}=\boldsymbol{\ell}_{\text{\tiny
Q}\bar{\text{\tiny Q}}}$, then the behavior of the critical heigh for large
$\ell$ is

$h_{c}^{(1)}=\tfrac{1}{2\sqrt{3}}\ell+\boldsymbol{\ell}_{\text{\tiny
Q}\bar{\text{\tiny Q}}}\,.$ (4.22)

This suggests that $h_{c}^{(1)}$ increases more rapidly for large values of
$\ell$ (see Figure 9). Hopefully, it will be possible eventually to determine
$h_{c}^{(1)}(\ell)$ reliably by computer simulations.

###  IV.2 Collinear geometry

Now let us consider the second type of geometry. It can be obtained from the
diquark limit in which $Q_{1}$ sits on top of $Q_{2}$ by moving $Q_{1}$ along
the axis connecting $Q_{2}$ and $Q_{3}$ in the direction away from $Q_{3}$. As
a result, we get the collinear geometry.

####  IV.2.1 Connected string configurations

For this case, the connected configurations were studied in a-3q . These are
shown in Figure 10.

Figure 10: Typical collinear configurations if $L\geq\ell$. Left: The case
$\alpha_{1}\geq 0$. Right: The case $\alpha_{1}\leq 0$.

An important observation which is inferred from moving $Q_{1}$ is that at some
point the string connected to it changes the shape from that of Figure 15 on
the left to that on the right. In the meantime the others keep their shape
intact. This change corresponds to a flip of the sign of $\alpha_{1}$.

We begin with configuration I describing the diquark limit. Using the formulas
of Appendix B for the case $\alpha\geq 0$, we immediately deduce that

$\ell=\frac{1}{\sqrt{\mathsf{s}}}\bigl{(}{\cal L}^{+}(\alpha_{1},v)+{\cal
L}^{+}(\alpha_{2},v)\bigr{)}\,,\quad
L=\frac{1}{\sqrt{\mathsf{s}}}\bigl{(}{\cal L}^{-}(\lambda_{3},v)-{\cal
L}^{+}(\alpha_{2},v)\bigr{)}\,,$ (4.23)

and the energy of the configuration is

$E_{3\text{\tiny Q}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}\sum_{i=1}^{2}{\cal
E}^{+}(\alpha_{i},v)\,+{\cal
E}^{-}(\lambda_{3},v)+3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}\Bigr{)}+3c\,.$
(4.24)

In addition, the angles $\alpha_{1}$ and $\alpha_{2}$ can be defined
implicitly as functions of $v$ and $\lambda_{3}$ by the force balance
equations (C.4). We have

$\cos\alpha_{1}=\cos\alpha_{2}+\frac{v}{\lambda_{3}}\text{e}^{\lambda_{3}-v}\,,\quad\Bigl{[}1-\Bigl{(}\cos\alpha_{2}+\frac{v}{\lambda_{3}}\text{e}^{\lambda_{3}-v}\Bigr{)}^{2}\Bigr{]}^{\frac{1}{2}}+\Bigl{[}1-\cos^{2}{}\alpha_{2}\Bigr{]}^{\frac{1}{2}}-\Bigl{[}1-\frac{v^{2}}{\lambda_{3}^{2}}\text{e}^{2(\lambda_{3}-v)}]\Bigr{]}^{\frac{1}{2}}=3\mathsf{k}(1+4v)\text{e}^{-3v}\,.$
(4.25)

Thus, the energy of the configuration is given in parametric form by the
equations $E_{3\text{\tiny Q}}=E_{3\text{\tiny Q}}(v,\lambda_{3})$,
$\ell=\ell(v,\lambda_{3})$, and $L=L(v,\lambda_{3})$. The parameters take
values on the interval $[0,1]$ and obey the inequality $v\leq\lambda_{3}$.

The behavior of this configuration for small $\ell$ was studied in a-3q . But
for our purposes, we don’t need all the results of this study. Only the
leading term in the expansion of $\ell(v)$ near $v=0$ matters. It is given by
Eq.(4.5), exactly as in the case of the triangle geometry.

On the other hand, we do need to know what happens if the string ended at
$Q_{3}$ is long enough. Taking the limit $\lambda_{3}\rightarrow 1$ in (4.24)
and (4.25), we get

$E_{3\text{\tiny
Q}}=\sigma|Q_{3}Y|+\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}\sum_{i=1}^{2}{\cal
E}^{+}(\alpha_{i},v)+3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}-{\cal
I}(v)\Bigr{)}+3c\,$ (4.26)

and

$\cos\alpha_{1}=\cos\alpha_{2}+v\text{e}^{1-v}\,,\quad\Bigl{[}1-\Bigl{(}\cos\alpha_{2}+v\text{e}^{1-v}\Bigr{)}^{2}\Bigr{]}^{\frac{1}{2}}+\Bigl{[}1-\cos^{2}{}\alpha_{2}\Bigr{]}^{\frac{1}{2}}-\Bigl{[}1-v^{2}\text{e}^{2(1-v)}]\Bigr{]}^{\frac{1}{2}}=3\mathsf{k}(1+4v)\text{e}^{-3v}\,.$
(4.27)

The angles $\alpha_{1}$ and $\alpha_{2}$ depend only on $v$ which varies from
$0$ to $\tilde{v}$, where $\tilde{v}$ is a solution to

$\Bigl{[}1-\bigl{(}1-v\text{e}^{1-v}\bigr{)}^{2}\Bigr{]}^{\frac{1}{2}}-\Bigl{[}1-v^{2}\text{e}^{2(1-v)}]\Bigr{]}^{\frac{1}{2}}=3\mathsf{k}(1+4v)\text{e}^{-3v}\,.$
(4.28)

This parameter value corresponds to the transition between the configurations
of Figure 10 because $\alpha_{1}(\tilde{v})=0$.

It is straightforward to obtain the corresponding expressions for
configuration II. The only modifications of the above formulas are due to the
string which changes its shape. Using the formulas of Appendix B for the case
$\alpha\leq 0$, we find

$\ell=\frac{1}{\sqrt{\mathsf{s}}}\bigl{(}{\cal L}^{-}(\lambda_{1},v)+{\cal
L}^{+}(\alpha_{2},v)\bigr{)}\,,\quad
L=\frac{1}{\sqrt{\mathsf{s}}}\bigl{(}{\cal L}^{-}(\lambda_{3},v)-{\cal
L}^{+}(\alpha_{2},v)\bigr{)}\,,$ (4.29)

and

$E_{3\text{\tiny Q}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}\sum_{i=1,3}{\cal
E}^{-}(\lambda_{i},v)+{\cal
E}^{+}(\alpha_{2},v)+3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}\Bigr{)}+3c\,.$
(4.30)

With the help of (B.17), the force balance equations (C.4) can be written as

$\begin{split}\cos\alpha_{2}=&v\text{e}^{-v}\Bigl{(}\frac{\text{e}^{\lambda_{1}}}{\lambda_{1}}-\frac{\text{e}^{\lambda_{3}}}{\lambda_{3}}\Bigr{)}\,,\\\
\Bigl{[}1-v^{2}\text{e}^{-2v}\Bigl{(}\frac{\text{e}^{\lambda_{1}}}{\lambda_{1}}-\frac{\text{e}^{\lambda_{3}}}{\lambda_{3}}\Bigr{)}^{2}\Bigr{]}^{\frac{1}{2}}-\Bigl{[}1-\frac{v^{2}}{\lambda_{1}^{2}}&\text{e}^{2(\lambda_{1}-v)}\Bigr{]}^{\frac{1}{2}}-\Bigl{[}1-\frac{v^{2}}{\lambda_{3}^{2}}\text{e}^{2(\lambda_{3}-v)}]\Bigr{]}^{\frac{1}{2}}=3\mathsf{k}(1+4v)\text{e}^{-3v}\,.\end{split}$
(4.31)

In general, we are unable to explicitly express one parameter as a function of
two others in the last equation. In practice it is convenient to choose $v$
and $\lambda_{3}$ as independent parameters and then solve this equation for
$\lambda_{1}$ numerically.

In the limit $\lambda_{3}\rightarrow 1$, the expression for the energy takes
the form

$E_{3\text{\tiny Q}}=\sigma|Q_{3}Y|+\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}{\cal
E}^{-}(\lambda_{1},v)+{\cal
E}^{+}(\alpha_{2},v)+3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}-{\cal
I}(v)\Bigr{)}+3c\,,$ (4.32)

and the equations (4.31) take the form

$\begin{split}\cos\alpha_{2}=&v\text{e}^{1-v}\Bigl{(}\frac{\text{e}^{\lambda_{1}-1}}{\lambda_{1}}-1\Bigr{)}\,,\\\
\Bigl{[}1-v^{2}\text{e}^{2(1-v)}\Bigl{(}\frac{\text{e}^{\lambda_{1}-1}}{\lambda_{1}}-1\Bigr{)}^{2}\Bigr{]}^{\frac{1}{2}}-\Bigl{[}1-\frac{v^{2}}{\lambda_{1}^{2}}&\text{e}^{2(\lambda_{1}-v)}\Bigr{]}^{\frac{1}{2}}-\Bigl{[}1-v^{2}\text{e}^{2(1-v)}]\Bigr{]}^{\frac{1}{2}}=3\mathsf{k}(1+4v)\text{e}^{-3v}\,.\end{split}$
(4.33)

The last equation now determines $\lambda_{1}$ as a function of $v$. The
parameter $v$ runs from $\tilde{v}$ to $\bar{v}$. The upper bound satisfies
$\lambda_{1}(\bar{v})=1$. It is the same as that for the symmetric collinear
configuration studied in Sec.II. Thus, starting from the diquark limit, we
reach the symmetric configuration by separating the heavy quarks $Q_{1}$ and
$Q_{2}$.

####  IV.2.2 Disconnected configuration and string breaking

As before, the relevant disconnected configuration is of type shown in Figure
5 on the left. There is a little difference between those configurations. The
quark $Q_{3}$ is now at $x=L$, with $L\geq\ell$. However, this has no effect
on the total energy of the configuration under the assumption of non-
interacting hadrons. So both configurations have the same energy
$E_{\text{\tiny QQq}}+E_{\text{\tiny Q}\bar{\text{\tiny q}}}$.

Having understood the string configurations, we can now find the critical
value of $|Q_{3}Y|$ that defines the string breaking distance
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}=|Q_{3}Y|_{c}$ by solving the
equation $E_{3\text{\tiny Q}}=E_{\text{\tiny QQq}}+E_{\text{\tiny
Q}\bar{\text{\tiny q}}}$. So, at fixed $\ell$ the disconnected configuration
dominates in the expectation value of the Wilson loop for
$|Q_{3}Y|>|Q_{3}Y|_{c}$. As in the previous subsection, we take the parameter
set $L$ and $\mathsf{k}=-0.102$. Using the linear approximations (4.26) and
(4.32) makes the things easy. So, we get

$\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=\begin{cases}\boldsymbol{\ell}_{0}+\frac{1}{\text{e}\sqrt{\mathsf{s}}}\bigl{(}{2\cal
E}^{+}(\alpha^{\prime},v^{\prime})-{\cal E}^{+}(\alpha_{1},v)-{\cal
E}^{+}(\alpha_{2},v)\bigr{)}\quad&\text{if}\quad 0\leq\ell\leq
0.087\,\text{fm}\,,\\\
\boldsymbol{\ell}_{0}+\frac{1}{\text{e}\sqrt{\mathsf{s}}}\bigl{(}{2\cal
E}^{+}(\alpha^{\prime},v^{\prime})-{\cal E}^{-}(\lambda_{1},v)-{\cal
E}^{+}(\alpha_{2},v)\bigr{)}\quad&\text{if}\quad 0.087\leq\ell\leq
0.248\,\text{fm}\,,\\\
\boldsymbol{\ell}_{1}+\frac{1}{\text{e}\sqrt{\mathsf{s}}}\bigl{(}{2\cal
E}^{+}(\alpha^{\prime},v^{\prime})-{\cal E}^{-}(\lambda_{1},v)-{\cal
E}^{+}(\alpha_{2},v)\bigr{)}\quad&\text{if}\quad 0.248\leq\ell\leq
1.110\,\text{fm}\,,\\\
\boldsymbol{\ell}_{1}+\frac{1}{\text{e}\sqrt{\mathsf{s}}}\bigl{(}{2\cal
E}^{-}(\lambda^{\prime},v^{\prime})-{\cal E}^{-}(\lambda_{1},v)-{\cal
E}^{+}(\alpha_{2},v)\bigr{)}\quad&\text{if}\quad 1.110\leq\ell\lesssim
1.110+\epsilon\,\text{fm}\,,\end{cases}$ (4.34)

where $\boldsymbol{\ell}_{0}$ and $\boldsymbol{\ell}_{1}$ are the same two
functions as in (4.17). The parameters $v$ and $v^{\prime}$ refer respectively
to the connected configurations for $QQQ$ and $QQq$. These are not independent
because of the relation $\ell=\ell(v)=\ell^{\prime}(v^{\prime})$. The
functions $\alpha_{1}(v)$, $\alpha_{2}(v)$, and $\lambda_{1}(v)$ are defined
implicitly by (4.27) and (4.33). As before in (4.17),
$\alpha^{\prime}(v^{\prime})$ and $\lambda^{\prime}(v^{\prime})$ are defined
by (D.6), (D.9), and (D.12). The last interval in (4.34) is really small, with
$\epsilon\approx 0.0003$. The string breaking distance tends to that of the
symmetric configuration as $\ell$ approaches the upper bound.

One can numerically eliminate the remaining parameter from the parametric
equations $\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}=\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}(v)$ and $\ell=\ell(v)$,
and find a corresponding function $\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}(\ell)$. The result is shown in Figure 11 on the left. We see that
the string

Figure 11: $\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$ and $L_{c}^{(1)}$ as
functions of $\ell$. The dashed horizontal lines represent the value
$1.110\,\text{fm}$ and dotted line the value $1.22\,\text{fm}$. The former
corresponds to the value we found in Sec.III and the latter to the value of
$\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}$ from bulava .

breaking distance coincides with $\boldsymbol{\ell}_{\text{\tiny
Q}\bar{\text{\tiny Q}}}$ at $\ell=0$ and then decreases with increasing the
separation between the neighbor quarks. A notable feature is that for
separations larger than $0.5\,\text{fm}$ it becomes almost the same as that
for the symmetric collinear geometry.

Before proceeding further, we pause here to say that for small $\ell$ the
decay mode $QQQ\rightarrow QQq+Q\bar{q}$ is dominant. So, as before for the
isosceles triangle geometry, the ground state is determined in terms of two
states and $\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$ is the only scale. The
present case shows that the estimate of the upper bound for $\ell$ via the
instability of $QQq$ is not so accurate. It gives $1.22\,\text{fm}$, but
$\ell$ is limited to $1.100\,\text{fm}$ which corresponds to the symmetric
case. Thus, another assumption is needed to determine the upper bound more
accurately.

Now we introduce a critical length $L_{c}^{(1)}=|Q_{3}Y|_{c}-|Q_{2}Y|$ that
can be useful in looking for possible ways to make comparisons with other non-
perturbative approaches. The meaning of $L_{c}^{(1)}$ is that at a given value
of $\ell$, the disconnected configuration dominates for lengths longer than
$L_{c}^{(1)}$. Like above, we can eliminate $v$ and find
$L_{c}^{(1)}=L_{c}^{(1)}(\ell)$. The result is presented in Figure 11. We see
that so defined $L_{c}^{(1)}$ behaves quite similarly to
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$.

The small $\ell$ behavior of $\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$ is
determined through the same steps as before. First, we bring the first
expression in (4.34) into the following form

$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}=\boldsymbol{\ell}_{\text{\tiny
Q}\bar{\text{\tiny Q}}}+\frac{1}{\text{e}\sqrt{\mathsf{s}}}\Bigl{[}2{\cal
E}^{+}(\alpha^{\prime},v^{\prime})-\sum_{i=1}^{2}{\cal
E}^{+}(\alpha_{i},v)+3\mathsf{k}\Bigl{(}\frac{\text{e}^{-2v^{\prime}}}{\sqrt{v^{\prime}}}-\frac{\text{e}^{-2v}}{\sqrt{v}}\Bigr{)}-{\cal
Q}(v^{\prime})-\frac{1}{\sqrt{v}}+\int_{0}^{\sqrt{v}}\frac{du}{u^{2}}\bigl{(}\text{e}^{u^{2}}[1-u^{4}\text{e}^{2(1-u^{2})}]^{\frac{1}{2}}-1\bigr{)}\Bigr{]}\,.$
(4.35)

Then, from the equation $\ell(v)=\ell^{\prime}(v^{\prime})$ we obtain that in
leading order $v=v^{\prime}$ and $\alpha_{1}=\alpha_{2}=\alpha^{\prime}$. This
implies that the terms inside the square brackets cancel each other out as $v$
and $v^{\prime}$ go to zero. Thus, we get the desired result
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}=\boldsymbol{\ell}_{\text{\tiny
Q}\bar{\text{\tiny Q}}}$.

To see what happens for large $\ell$, we consider the double limit
$\lambda_{1}\rightarrow 1$ and $\lambda^{\prime}\rightarrow 1$ such that
$\frac{\ln(1-\lambda_{1})}{\ln(1-\lambda^{\prime})}\rightarrow 1$. The
requirement implies that $\frac{\ell}{L}\rightarrow 1$.111111This follows from
the asymptotic expansion (B.19). Thus this limit gives rise to the symmetric
case. A simple calculation shows that the last expression in (4.34) does
reduce to that of (3.12). The latter is shown in the Figure by the dashed
horizontal lines.

Finally, let us mention that if one assumes that the string breaking distance
is universal, then the critical length is simply
$L_{c}^{(1)}=\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}$. In other
words, so defined critical length is a constant, which is independent of
$\ell$. It is shown in the Figure by the dotted line.

##  V More detail on the special cases

For completeness, we include the analysis of the remaining decay modes in
(1.5) which respectively contain two and three pairs of light quarks each. To
actually compute the energies of non-interacting hadrons within the
gauge/string duality formalism, one would have to consider disconnected string
configurations such as those sketched in Figure 12. There is an important
subtlety that arises when one tries to

Figure 12: Static string configurations for non-interacting hadrons. The
ordering does not matter. Left: A heavy-light baryon $Qqq$ and two heavy-light
mesons $Q\bar{q}$. Right: A light baryon $qqq$ and three heavy-light mesons.

construct a static configuration for a heavy-light baryon $Qqq$. The analysis
of Eq.(D.4) shows that this equation has a solution in the interval $[0,1]$ if
and only if
$-\frac{\text{e}^{3}}{15}\leq\mathsf{k}\leq-\frac{1}{4}\text{e}^{\frac{1}{4}}$
a-QQq . Clearly, the phenomenologically motivated value $\mathsf{k}=-0.102$ is
out of the interval. This is one of the limitations of using the model of
Sec.II. We take the upper bound as it is most close to $\mathsf{k}=-0.102$ and
leads to the exact solution $v_{\shortparallel}=\frac{1}{12}$ which makes the
calculations simpler.

First, let us consider the configuration sketched in the left panel. It
describes the products of the baryon decay into $Qqq+2Q\bar{q}$. Under the
assumption of non-interacting hadrons, the energy of the configuration is the
sum of energies of individual hadrons. These energies can be read off from
Eqs.(D.1) and (D.3). So, we have

$E_{\text{\tiny Qqq}}+2E_{\text{\tiny Q}\bar{\text{\tiny
q}}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}4{\cal Q}(q)-{\cal
Q}(v_{\shortparallel})+4\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+3\mathsf{k}\frac{\text{e}^{-2v_{\shortparallel}}}{\sqrt{v_{\shortparallel}}}\,\Bigr{)}+3c\,,$
(5.1)

where $q$ and $v_{\shortparallel}$ are the solutions of (D.2) and (D.4),
respectively.

The right panel represents a string configuration for the products of the
baryon decay into $qqq+3Q\bar{q}$. The novelty here is a light baryon that has
not been discussed in the literature. In the static limit, it looks like the
light quarks sit on top of the baryon vertex. The total action is therefore
given by

$S=\frac{T}{r_{v}}\Bigl{(}\tau_{v}\text{e}^{-2\mathsf{s}r_{v}^{2}}+3\mathsf{m}\text{e}^{\frac{\mathsf{s}}{2}r_{v}^{2}}\Bigr{)}\,.$
(5.2)

By varying it with respect to $r_{v}$, we get

$\mathsf{n}(1-v)+\mathsf{k}(1+4v)\text{e}^{-\frac{5}{2}v}=0\,.$ (5.3)

This equation says that the gravitational force acting on the vertex is
equilibrated by that acting on the quarks. As a result, the baryon is at rest.
The rest energy is then

$E_{3\text{\tiny q}}=3\mathsf{g}\sqrt{\frac{\mathsf{s}}{v_{3\text{\tiny
q}}}}\bigl{(}\mathsf{k}\text{e}^{-2v_{3\text{\tiny
q}}}+\mathsf{n}\text{e}^{\frac{1}{2}v_{3\text{\tiny q}}}\Bigr{)}\,,$ (5.4)

where $v_{3\text{\tiny q}}$ is a solution of equation (5.3) on the interval
$[0,1]$. Combining this with the expression (D.1), we find the energy of the
configuration

$E_{3\text{\tiny q}}+3E_{\text{\tiny Q}\bar{\text{\tiny
q}}}=3\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+\mathsf{k}\frac{\text{e}^{-2v_{3\text{\tiny
q}}}}{\sqrt{v_{3\text{\tiny
q}}}}+\mathsf{n}\frac{\text{e}^{\frac{1}{2}v_{3\text{\tiny
q}}}}{\sqrt{v_{3\text{\tiny q}}}}\,\Bigr{)}+3c\,.$ (5.5)

###  V.1 Equilateral triangle geometry

With all the diagonal elements of the model Hamiltonian at our disposal, we
can see a more complete picture of string breaking. We consider first the
equilateral triangle geometry. To this end, in Figure 13 we plot the diagonal
elements as a function of $\ell$.

Figure 13: Various $E$ vs $\ell$ plots for the equilateral triangle geometry,
shown in the left panel for $L$ and in the right panel for $P$. Here
$\mathsf{k}=-\frac{1}{4}\text{e}^{\frac{1}{4}}$ and $c=0.623\,\text{GeV}$.

The most interesting observation is that the three-quark potential (ground
state energy) is, in fact, determined in terms of only two diagonal elements:
$E_{3\text{\tiny Q}}$ and $E_{\text{\tiny Qqq}}+2E_{\text{\tiny
Q}\bar{\text{\tiny q}}}$. Thus, the string breaking distance should be defined
from the equation $E_{3\text{\tiny Q}}=E_{\text{\tiny Qqq}}+2E_{\text{\tiny
Q}\bar{\text{\tiny q}}}$, i.e. the $QQQ\rightarrow Qqq+2Q\bar{q}$ mode is
dominant. If so, then with the help of (3.3) and (5.1), we find that

$\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(2)}=\frac{1}{3\text{e}\sqrt{\mathsf{s}}}\Bigl{(}4{\cal Q}(q)-{\cal
Q}(v_{\shortparallel})+4\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+3\mathsf{k}\frac{\text{e}^{-2v_{\shortparallel}}}{\sqrt{v_{\shortparallel}}}+3I_{3\text{\tiny
Q}}\Bigr{)}\,.$ (5.6)

Here the superscript (2) indicates that the string breaking distance refers to
the decay mode $QQQ\rightarrow Qqq+2Q\bar{q}$.

Quite apart from the above observation, it is to be noticed that the decay
$Qqq\rightarrow qqq+Q\bar{q}$ is energetically forbidden because of
$E_{qqq}+E_{\text{\tiny Q}\bar{\text{\tiny q}}}>E_{Qqq}$. In the string model
we are using, this means that the string stretched along the radial direction
(see the right panel of Figure 18) does not break down.

Let us make a simple estimate of $\boldsymbol{\ell}_{3\text{\tiny Q}}^{(2)}$.
At $\mathsf{k}=-\frac{1}{4}\text{e}^{\frac{1}{4}}$, for $L$ and $P$ we get
correspondingly

$\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(2)}=0.804\,\text{fm}\,,\qquad\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(2)}=0.700\,\text{fm}\,.$ (5.7)

These values are smaller than those obtained for
$\boldsymbol{\ell}_{3\text{\tiny Q}}^{(1)}$ in Sec.III, as also seen from the
plots. In addition, we get121212Curiously, both values are close to
$\frac{2}{3}$.

$\frac{\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(2)}}{\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny
Q}}}}=0.659\,,\,\,\qquad\frac{\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(2)}}{\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}}=0.652\,.$
(5.8)

Thus, for the equilateral triangle geometry, the string theory prediction is
actually that the three quark potential is determined in terms of the energies
$E_{3\text{\tiny Q}}$ and $E_{\text{\tiny Qqq}}+2E_{\text{\tiny
Q}\bar{\text{\tiny q}}}$ with a single scale which characterizes the string
breaking effect.

###  V.2 Symmetric collinear geometry

Similarly, we can analyze the symmetric collinear geometry. Under the
assumption of non-interacting hadrons, the energies of the two remaining
disconnected configurations are given by Eqs.(5.1) and (5.5). In Figure 14 we
plot the diagonal elements of ${\cal H}$ as a function of $\ell$. The
noticeable difference from the previous case is that

Figure 14: Various $E$ vs $\ell$ plots for the symmetric collinear geometry.
The notation here is the same as in Figure 13.

the ground state energy is determined in terms of three diagonal elements:
$E_{3\text{\tiny Q}}$, $E_{\text{\tiny QQq}}+E_{\text{\tiny Q}\bar{\text{\tiny
q}}}$, and $E_{\text{\tiny Qqq}}+2E_{\text{\tiny Q}\bar{\text{\tiny
q}}}$.131313Because the difference between $\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}$ and $\boldsymbol{\ell}_{QQq}$ is of order $0.003\,\text{fm}$ for
$P$, it is not visible in the right panel. This fact suggests that a scale set
by $E_{\text{\tiny QQq}}(\ell)+E_{\text{\tiny Q}\bar{\text{\tiny
q}}}=E_{\text{\tiny Qqq}}+2E_{\text{\tiny Q}\bar{\text{\tiny q}}}$ is
relevant. In fact, this scale was recently discussed in the context of the
$QQq$ system a-QQq . It is given by

$\boldsymbol{\ell}_{\text{\tiny
QQq}}=\frac{3}{\text{e}\sqrt{\mathsf{s}}}\biggl{(}{\cal Q}(q)-\frac{1}{3}{\cal
Q}(v_{\shortparallel})+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+\mathsf{k}\frac{\text{e}^{-2v_{\shortparallel}}}{\sqrt{v_{\shortparallel}}}+\frac{2}{3}I_{\text{\tiny
QQq}}\biggr{)}\,.$ (5.9)

The estimates give $\boldsymbol{\ell}_{\text{\tiny QQq}}=1.257\,\text{fm}$ for
the parameter set $L$ and $\boldsymbol{\ell}_{\text{\tiny
QQq}}=1.073\,\text{fm}$ for P. These values are very close to those for
$\boldsymbol{\ell}_{\text{\tiny Q}\bar{\text{\tiny Q}}}$. All of this suggests
that the $QQQ$ system undergoes a cascade of decays. At small $\ell$ it looks
like a triply heavy baryon. Then, at $\ell=\boldsymbol{\ell}_{3\text{\tiny
Q}}^{(1)}$ this baryon decays into a double-heavy baryon and a heavy light
meson. Finally, at $\ell=\boldsymbol{\ell}_{\text{\tiny QQq}}$ the double
heavy baryon decays into a pair of heavy-light baryon and meson. Thus, for the
symmetric collinear geometry, the string model predicts that the potential is
determined in terms of the energies $E_{3\text{\tiny Q}}$, $E_{\text{\tiny
QQq}}+E_{\text{\tiny Q}\bar{\text{\tiny q}}}$, and $E_{\text{\tiny
Qqq}}+2E_{\text{\tiny Q}\bar{\text{\tiny q}}}$ with two scales which
characterize the string breaking effects.

Finally, let us note that small enough deformations of these special
geometries will give similar pictures of string breaking.

##  VI Concluding remarks

(i) In this paper, we have initiated a study of the phenomenon of string
breaking in the three quark system. Our findings could be an early indication
that this phenomenon is much more complex than one could naively expect based
on the universality of the string breaking distance. The main funding is that
the string breaking distance is not universal. Moreover, there are special
geometries, such as the symmetric collinear geometry and its small
deformations, which allow more than one characteristic scale.

(ii) Going back to the question of universality, we see from Eq. (B.23) that
in the IR limit the energy $E_{3\text{\tiny Q}}$ is141414This is valid if the
largest angle of the triangle $Q_{1}Q_{2}Q_{3}$ is at most $2\pi/3$.

$E_{3\text{\tiny
Q}}=\sigma\sum_{i=1}^{3}|Q_{i}Y|\,-3\mathsf{g}\sqrt{\mathsf{s}}\,{\cal
I}(v_{{\vartriangle}})+3c+o(1)\,,$ (6.1)

with $v_{{\vartriangle}}$ a solution of Eq.(4.14). In this limit, $Y$
coincides with the Fermat point of the triangle a-3q0 . The coefficient
$\sigma$ is the string tension which is universal in the sense that it is the
same for all the known examples of heavy quark systems. The two remaining
constant terms combine to $C_{\text{IR}}=3c-3\mathsf{g}\sqrt{\mathsf{s}}{\cal
I}(v_{{\vartriangle}})$. Using the fact that in UV limit (at small separations
between the quark sources) the constant term in the expansion of
$E_{3\text{\tiny Q}}$ is simply $C_{\text{UV}}=3c$, we get

$C_{\text{UV}}-C_{\text{IR}}=3\mathsf{g}\sqrt{\mathsf{s}}{\cal
I}(v_{{\vartriangle}})\,.$ (6.2)

This difference being positive is also scheme and geometry independent. The
latter means that it is universal and not limited to the particular case of a
triangle if its largest angle is at most $\frac{2}{3}\pi$. This statement is
an extension of that in a-3q , which was formulated for the equilateral
triangle.

(iii) Although from the ten-dimensional perspective the five-dimensional
string model used as an example throughout this paper is oversimplified, it
allows us to get the results on the $QQQ$ system analytically and uses only a
few model parameters. There are, however, some limitations and caveats for
this model, as for any effective model. Let us mention a couple of them.
First, the light quarks are incorporated along the lines of the hadro-
quarkonium picture so that $r_{q}$ denotes an averaged position of a light
quark or a center of a corresponding quark cloud. The model certainly is far
from accounting for all the features of $u$ and $d$ quarks. Second, it is not
clear how to compute the off-diagonal elements of the model Hamiltonians and
thus reach the level of understanding the quark-antiquark system in lattice
gauge theory.

###### Acknowledgements.

We would like to thank M. Catillo, P. de Forcrand, M. K. Marinković, and P.
Weisz for discussions and reading the manuscript. We also thank the Arnold
Sommerfeld Center for Theoretical Physics for hospitality. This research is
supported by Russian Science Foundation grant 20-12-00200 in association with
Steklov Mathematical Institute.

##  Appendix A Notation and definitions

In all Figures throughout the paper, heavy and light quarks (antiquarks) are
denoted by $Q$ and $q\,(\bar{q})$, and a baryon vertex and its projection onto
the boundary of five-dimensional space by $V$ and $Y$. A square indicates that
a light quark sits on top of a vertex. We assume that all strings are in the
ground state. So, these strings are represented by curves without cusps,
loops, etc. When not otherwise noted, we usually set light quarks (antiquarks)
at $r=r_{q}$ and a vertex at $r=r_{v}$. For convenience, we introduce two
dimensionless variables $q=\mathsf{s}r_{q}^{2}$ and $v=\mathsf{s}r_{v}^{2}$.
They take values on the interval $[0,1]$ and show how far from the soft-wall
these objects are.

To make formulas more compact, we introduce a set of basic functions. The
motivation for that comes from the analysis of a static string in Appendix B.

The non-negative function ${\cal L}^{+}$ is defined by the integral

${\cal
L}^{+}(\alpha,x)=\cos\alpha\sqrt{x}\int^{1}_{0}du\,u^{2}\,\text{e}^{x(1-u^{2})}\Bigl{[}1-\cos^{2}{}\alpha\,u^{4}\text{e}^{2x(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}\,,\qquad
0\leq\alpha\leq\frac{\pi}{2}\,,\qquad 0\leq x\leq 1\,.$ (A.1)

This function vanishes if $\alpha=\frac{\pi}{2}$ or $x=0$. It diverges at
$(0,1)$.

The other non-negative function ${\cal L}^{-}$ is defined as follows

${\cal
L}^{-}(y,x)=\sqrt{y}\biggl{(}\,\int^{1}_{0}du\,u^{2}\,\text{e}^{y(1-u^{2})}\Bigl{[}1-u^{4}\,\text{e}^{2y(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}+\int^{1}_{\sqrt{\frac{x}{y}}}du\,u^{2}\,\text{e}^{y(1-u^{2})}\Bigl{[}1-u^{4}\,\text{e}^{2y(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}\,\biggr{)}\,,\quad
0\leq x\leq y\leq 1\,.$ (A.2)

It vanishes at the origin and diverges at $y=1$. At $y=x$, ${\cal L}^{-}$
reduces to ${\cal L}^{+}$ with $\alpha=0$.

It is also useful to define two other functions. The first is given by

${\cal
E}^{+}(\alpha,x)=\frac{1}{\sqrt{x}}\int^{1}_{0}\,\frac{du}{u^{2}}\,\biggl{(}\text{e}^{xu^{2}}\Bigl{[}1-\cos^{2}{}\alpha\,u^{4}\text{e}^{2x(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}-1-u^{2}\biggr{)}\,,\qquad
0\leq\alpha\leq\frac{\pi}{2}\,,\qquad 0\leq x\leq 1\,$ (A.3)

and the second by

${\cal
E}^{-}(y,x)=\frac{1}{\sqrt{y}}\biggl{(}\int^{1}_{0}\,\frac{du}{u^{2}}\,\Bigl{(}\text{e}^{yu^{2}}\Bigl{[}1-u^{4}\,\text{e}^{2y(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}-1-u^{2}\Bigr{)}+\int^{1}_{\sqrt{\frac{x}{y}}}\,\frac{du}{u^{2}}\,\text{e}^{yu^{2}}\Bigl{[}1-u^{4}\,\text{e}^{2y(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}\biggr{)}\,,\,\,\,0\leq
x\leq y\leq 1\,.$ (A.4)

The function ${\cal E}^{+}$ diverges at $x=0$ and $(0,1)$, whereas the
function ${\cal E}^{-}$ at $(0,0)$ and $y=1$. Just like for the ${\cal L}$’s,
${\cal E}^{-}$ reduces to ${\cal E}^{+}$ at $y=x$.

A special case of ${\cal E}^{+}$ is obtained by taking $\alpha=\frac{\pi}{2}$.
In this case, the integral can be performed explicitly, yielding

${\cal E}^{+}(\tfrac{\pi}{2},x)={\cal
Q}(x)=\sqrt{\pi}\text{erfi}(\sqrt{x})-\frac{\text{e}^{x}}{\sqrt{x}}\,.$ (A.5)

Here $\text{erfi}(x)$ is the imaginary error function.

Finally, we define a function

${\cal
I}(x)=I_{0}-\int_{\sqrt{x}}^{1}\frac{du}{u^{2}}\text{e}^{u^{2}}\Bigl{[}1-u^{4}\text{e}^{2(1-u^{2})}\Bigr{]}^{\frac{1}{2}}\,\,,\qquad
I_{0}=\int_{0}^{1}\frac{du}{u^{2}}\Bigl{(}1+u^{2}-\text{e}^{u^{2}}\Bigl{[}1-u^{4}\text{e}^{2(1-u^{2})}\Bigr{]}^{\frac{1}{2}}\Bigr{)}\,,\qquad
0\leq x\leq 1\,,$ (A.6)

which appears in the limit of sufficiently elongated strings. For the constant
$I_{0}$, a simple numerical calculation gives $I_{0}\approx 0.751$.

##  Appendix B A static string with fixed endpoints

In this Appendix we give a summary of the results on a static Nambu-Goto
string in the curved geometry (2.2). These results provide the grounds for
building string configurations of Secs. III and IV. Most of the material can
be found in a-3q whose conventions we generally follow.151515The only new
result presented here is a calculation of the constant term in the expansion
of the energy of long strings. So, we can be relatively brief.

To proceed, choose the static gauge $\xi^{1}=t$, giving for the Nambu-Got
action

$S_{\text{\tiny
NG}}=\frac{T}{2\pi\alpha^{\prime}}\int_{0}^{1}d\xi^{2}\,\sqrt{\gamma^{(2)}}\,,$
(B.1)

with $T=\int_{0}^{T}dt$. Consider now a string stretched between two points
$Q$ and $V$ in the $xr$-plane, as shown in Figure 15. This implies the
boundary conditions

$x(0)=0\,,\quad x(1)=x_{v}\,,\quad r(0)=0\,,\quad r(1)=r_{v}\,.$ (B.2)

Figure 15: A static string stretched between two points, one of which lies on
the boundary of space. The two forces exerted at point V are depicted by the
arrows. $\alpha$ is a tangent angle. The string does not cross the soft-wall
located at $r=1/\sqrt{\mathsf{s}}$. Left: The case $\alpha\geq 0$. Right: The
case $\alpha\leq 0$. Here $O$ is a turning point.

In this case, the Nambu-Goto action takes the form

$S_{\text{\tiny
NG}}=\mathsf{g}T\int_{0}^{1}d\xi^{2}\,w(r)\sqrt{x^{\prime}{}^{2}+r^{\prime}{}^{2}}\,,\quad\text{with}\quad
w(r)=\frac{\text{e}^{\mathsf{s}r^{2}}}{r^{2}}\,.$ (B.3)

A prime stands for a derivative with respect to $\xi^{2}$.

Since in what follows we set infinitely heavy quarks at boundary points, we do
not vary the action with respect to the position of point $Q$. The variation
of the action is therefore

$\delta S_{\text{\tiny
NG}}=\mathsf{g}T\biggl{[}\int_{0}^{1}d\xi^{2}\,\biggl{(}\partial_{r}w\,\sqrt{x^{\prime}{}^{2}+r^{\prime}{}^{2}}-\Bigl{(}\frac{wr^{\prime}}{\sqrt{x^{\prime}{}^{2}+r^{\prime}{}^{2}}}\Bigr{)}^{\prime}\,\biggr{)}\delta
r-\Bigl{(}\frac{wx^{\prime}}{\sqrt{x^{\prime}{}^{2}+r^{\prime}{}^{2}}}\Bigr{)}^{\prime}\delta
x+w(r_{v})\cos\alpha\,\delta x_{v}+w(r_{v})\sin\alpha\,\delta
r_{v}\,\biggr{]}\,,$ (B.4)

where $\partial_{r}=\frac{\partial}{\partial r}$. From this formula one can
find the string tension at point $V$.161616For static configurations, the
energy is simply $E=S/T$, while the components of $\mathbf{e}$ are
$\mathbf{e}_{x}=-\delta E/\delta x_{v}$ and $\mathbf{e}_{r}=-\delta E/\delta
r_{v}$. It comes out to be

$\mathbf{e}=-\mathsf{g}w(r_{v})\bigl{(}\cos\alpha,\,\sin\alpha\bigr{)}\,.$
(B.5)

There is one very simple but useful observation to add at this point. The
magnitude of $\mathbf{e}$ equals $\|\mathbf{e}\|=\mathsf{g}w(r_{v})$. This
clarifies the role of the fifth dimension in describing strings with different
tensions amp .

In order to keep the string at rest, some external force must be acting on it.
We exert such a force at point $V$, as shown in Figure 15. The force balance
equation is then

$\mathbf{e}+\mathbf{f}=0\,.$ (B.6)

Thus, given the string tension, the external force is
$\mathbf{f}=-\mathbf{e}$.

For further analysis it is convenient to completely fix the gauge by choosing
$\xi^{2}=x$. In that case, solving the Euler-Lagrange equation for $x$ gives

$I=\frac{w(r)}{\sqrt{1+(\partial_{x}r)^{2}}}\,,$ (B.7)

with $I=const$. It is easy to see that $I$ is nothing else but the first
integral for the equation for $r$ malda . For future reference, we note that

$I=w(r_{v})\cos\alpha\,$ (B.8)

at the endpoint $V$.

###  B.1 The case $\alpha\geq 0$

We begin with the case of positive $\alpha$ sketched in the left panel of
Figure 15. Combining (B.7) and (B.8), we get a differential equation
$w(r_{v})\cos\alpha=w(r)/\sqrt{1+(\partial_{x}r)^{2}}$ which can be further
integrated over $x$ and $r$ to yield a string length along the $x$-axis

$|QY|=\frac{1}{\sqrt{\mathsf{s}}}{\cal L}^{+}(\alpha,v)\,,$ (B.9)

where ${\cal L}^{+}$ and $v$ are defined earlier in Appendix A.

To compute the string energy, we first use (B.7) to explicitly express the
integrand in terms of $r$ and then find that

$E_{R}=\frac{S_{R}}{T}=\mathsf{g}\sqrt{\frac{\mathsf{s}}{v}}\int^{1}_{\sqrt{\tfrac{\mathsf{s}}{v}}\epsilon}\,\frac{du}{u^{2}}\,\text{e}^{vu^{2}}\Bigl{[}1-\cos^{2}{}\alpha\,u^{4}\text{e}^{2v(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}\,.$
(B.10)

Here the ultraviolet divergence at $r=0$ is regularized by placing a lower
bound on $r$ so that $r\geq\epsilon$. This expression behaves for
$\epsilon\rightarrow 0$ as

$E_{R}=\frac{\mathsf{g}}{\epsilon}+E+O(\epsilon)\,.$ (B.11)

After subtracting the $\tfrac{1}{\epsilon}$ term and letting $\epsilon=0$, we
arrive at

$E=\mathsf{g}\sqrt{\mathsf{s}}\,{\cal E}^{+}(\alpha,v)+c\,.$ (B.12)

where the function ${\cal E}^{+}$ is defined by (A.3) and $c$ is a
normalization constant.

If we set $\alpha=\frac{\pi}{2}$ in (B.12), we get a special case:

$E=\mathsf{g}\sqrt{\mathsf{s}}{\cal Q}(v)+c\,,$ (B.13)

with the function ${\cal Q}$ defined in (A.5). This formula gives the energy
of a string stretched along the $r$-axis.

Thus, the string energy is given in parametric form by $E=E(v)$ and
$|QY|=|QY|(v)$. The parameter takes values on the interval $[0,1]$. The string
becomes infinitely long at $v=1$, when its endpoint $V$ touches the soft-wall.

###  B.2 The case $\alpha\leq 0$

It is straightforward to extend the above analysis to the case of negative
$\alpha$, sketched in the right panel of Figure 15. The only novelty, which
applies to all expressions below, is that there are two contributions: one
comes from the interval $[0,x_{0}]$ and the other from the interval
$[x_{0},x_{v}]$. This is so because one can think of a string with $\alpha<0$
as two strings with $\alpha=0$ glued at the turning point.

Taking $I=w(r_{0})$ and integrating the differential equation
$w(r_{0})=w(r)/\sqrt{1+(\partial_{x}r)^{2}}$ over $x$ and $r$, we get for a
string length along the $x$-axis

$|QY|=\frac{1}{\sqrt{\mathsf{s}}}{\cal L}^{-}(\lambda,v)\,,$ (B.14)

where ${\cal L}^{-}$ is the function defined in Appendix A and $\lambda$ is a
dimensionless parameter defined as $\lambda=\mathsf{s}r_{0}^{2}$. It shows how
close to the soft-wall the turning point is. Clearly, at $\lambda=v$ the above
expression reduces to that of (B.9) with $\alpha=0$. This value also gives the
lower bound $\frac{1}{\sqrt{\mathsf{s}}}{\cal L}^{-}(v,v)\leq|QY|$.

As before, one can compute the string energy by first expressing the integrand
in terms of $r$ and then imposing the short-distance cutoff on $r$. This gives

$E_{R}=\mathsf{g}\sqrt{\frac{\mathsf{s}}{\lambda}}\biggl{(}\int^{1}_{\sqrt{\tfrac{\mathsf{s}}{\lambda}}\epsilon}\,\frac{du}{u^{2}}\,\text{e}^{\lambda
u^{2}}\Bigl{[}1-u^{4}\,\text{e}^{2\lambda(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}+\int^{1}_{\sqrt{\frac{v}{\lambda}}}\,\frac{du}{u^{2}}\,\text{e}^{\lambda
u^{2}}\Bigl{[}1-u^{4}\,\text{e}^{2\lambda(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}\biggr{)}\,.$
(B.15)

Near $\epsilon=0$ it behaves like that in (B.11). So, subtracting the
$\frac{1}{\epsilon}$ and then letting $\epsilon=0$ yields

$E=\mathsf{g}\sqrt{\mathsf{s}}\,{\cal E}^{-}(\lambda,v)+c\,.$ (B.16)

Here ${\cal E}^{-}$ is defined by (A.4) and $c$ is the same normalization
constant as in (B.12). The last guarantees that the string energy is a
continuous function of $\alpha$.

For future reference, it will be useful to explicitly express $\lambda$ in
terms of $v$ and $\alpha$. This can be done by using the first integral (B.7).
Equating its values at $O$ and $V$, one finds

$\frac{\text{e}^{\lambda}}{\lambda}=\frac{\text{e}^{v}}{v}\cos\alpha\,.$
(B.17)

From this, $\lambda$ can be expressed in terms of $v$ and $\alpha$ as

$\lambda=-\text{ProductLog}(-v\,\text{e}^{-v}/\cos\alpha)\,.$ (B.18)

The ProductLog function is the principal solution for $w$ in
$z=w\,\text{e}^{w}$ wolf .

To make estimates of the string breaking distance, we will need to know the
behavior of $E$ for large $|QY|$, or in other words what happens when the
string approaches the soft-wall. In fact, the problem reduces to examining the
limit $\lambda\rightarrow 1$ while keeping $v$ fixed. A calculation along the
lines of az1 gives the leading terms

$|QY|=-\frac{1}{\sqrt{\mathsf{s}}}\ln(1-\lambda)+O(1)\,,\qquad
E=-\mathsf{g}\text{e}\sqrt{\mathsf{s}}\,\ln(1-\lambda)+O(1)\,.$ (B.19)

Combining these equations results in

$E=\sigma|QY|+O(1)\,.$ (B.20)

As expected, at leading order $E$ is linear in $|QY|$. The constant of
proportionality $\sigma$ is called the string tension, given explicitly in
Eq.(3.4). It is universal and independent of $v$.

To find the constant term in the asymptotic expansion of $E$ for large $|QY|$,
consider

$\begin{split}E-\sigma|QY|=\mathsf{g}&\sqrt{\frac{\mathsf{s}}{\lambda}}\biggl{(}\int_{0}^{1}\frac{du}{u^{2}}\Bigl{(}\text{e}^{\lambda
u^{2}}\Bigl{[}1-\lambda
u^{4}\text{e}^{1+\lambda(1-2u^{2})}\Bigr{]}\Bigl{[}1-u^{4}\text{e}^{2\lambda(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}-1-u^{2}\Bigr{)}\\\
+&\int_{\sqrt{\frac{v}{\lambda}}}^{1}\frac{du}{u^{2}}\text{e}^{\lambda
u^{2}}\Bigl{[}1-\lambda
u^{4}\text{e}^{1+\lambda(1-2u^{2})}\Bigr{]}\Bigl{[}1-u^{4}\text{e}^{2\lambda(1-u^{2})}\Bigr{]}^{-\frac{1}{2}}\biggr{)}\,+c\,.\end{split}$
(B.21)

Letting $\lambda=1$, we get

$E-\sigma|QY|=\mathsf{g}\sqrt{\mathsf{s}}\biggl{(}\int_{0}^{1}\frac{du}{u^{2}}\Bigl{(}\text{e}^{u^{2}}\Bigl{[}1-u^{4}\text{e}^{2(1-u^{2})}\Bigr{]}^{\frac{1}{2}}-1-u^{2}\Bigr{)}+\int_{\sqrt{v}}^{1}\frac{du}{u^{2}}\text{e}^{u^{2}}\Bigl{[}1-u^{4}\text{e}^{2(1-u^{2})}\Bigr{]}^{\frac{1}{2}}\biggr{)}\,+c+o(1)\,.$
(B.22)

To this order, the expansion of $E$ is therefore

$E=\sigma|QY|-\mathsf{g}\sqrt{\mathsf{s}}\,{\cal I}(v)+c+o(1)\,,$ (B.23)

where the function ${\cal I}$ is defined by Eq.(A.6).

It is instructive to see how good this linear approximation actually is. To
this end, in Figure 16 we plot both

Figure 16: $E$ vs $|QY|$ for $|QY|\geq{\cal L}^{-}(v,v)/\sqrt{s}$ and fixed
$v$. Here we use the $L$ set with $c=0.623\,\text{GeV}$. The dashed lines
represent the asymptotic expression (B.23).

functions. We see that the linear approximation is perfect for strings longer
that $0.6\,\text{fm}$. This fact enables one to find expressions for the
string breaking distance analytically.

##  Appendix C Gluing conditions

The string solutions we discussed in Appendix B are building blocks of multi-
string configurations. What is still missing are gluing conditions for such
blocks. Our goal here is to provide those conditions, with particular
attention to the configurations of Sect. III and IV. More detail can be found,
for example, in a-3q .

A general force balance equation at a baryon vertex, takes the form

$\mathbf{e}_{1}+\mathbf{e}_{2}+\mathbf{e}_{3}+\mathbf{f}_{v}=0\,.$ (C.1)

Here $\mathbf{e}$’s are the string tensions and $\mathbf{f}_{v}$ is a
gravitational force acting on the vertex. The presence of this force is the
main difference between string models in flat space and those in curved
spaces. In the model we are considering $\mathbf{f}_{v}$ has only one non-zero
component which points in the $r$-direction. A formula for this component can
be derived from the action $S_{\text{vert}}=TE_{\text{vert}}$. Explicitly,
$\mathbf{f}_{v}^{r}=-\delta E_{\text{vert}}/\delta
r=-\tau_{v}\partial_{r}\frac{\text{e}^{-2\mathsf{s}r^{2}}}{r}$, with $r$ the
coordinate of the vertex.

Now consider two important examples. The first example is illustrated in the
left panel of Figure 17.

Figure 17: Three strings meeting at a baryon vertex in the bulk. The
gravitational force $\mathbf{f}_{v}$ acting on the vertex is directed in the
downward vertical direction. Left: The non-planar case with heavy quarks are
at the vertices of an isosceles triangle. Right: The planar case with heavy
quarks are on the $x$-axis.

Using the formula (B.5) for the string tension, it can be shown that in the
present case the tensions are written in components as
$\mathbf{e}_{1}=-\mathsf{g}w(r_{v})(\cos\beta\cos\alpha_{1},\sin\beta\cos\alpha_{1},\sin\alpha_{1})$,
$\mathbf{e}_{2}=-\mathsf{g}w(r_{v})(-\cos\beta\cos\alpha_{1},\sin\beta\cos\alpha_{1},\sin\alpha_{1})$,
and
$\mathbf{e}_{3}=-\mathsf{g}w(r_{v})(0,-\cos\alpha_{3},\sin\alpha_{3})$.171717On
symmetry grounds, the angles $\alpha_{1}$ and $\alpha_{2}$ are equal. With
this, the resulting equations for the $y$ and $r$ components take the form

$\cos\alpha_{3}-2\sin\beta\cos\alpha_{1}=0\,,\qquad
2\sin\alpha_{1}+\sin\alpha_{3}-3\mathsf{k}(1+4v)\text{e}^{-3v}=0\,.$ (C.2)

The equation for the $x$ component is trivially satisfied because of symmetry.

As a special case, consider $\alpha_{3}=\alpha_{1}$ which corresponds to an
equilateral triangle. The force balance equation now has only one non-trivial
component that gives rise to

$\sin\alpha_{1}-\mathsf{k}(1+4v)\text{e}^{-3v}=0\,.$ (C.3)

The second example is illustrated in the right panel of the Figure. The
analysis proceeds in a similar manner as above. The string tensions are
written in components as
$\mathbf{e}_{1}=-\mathsf{g}w(r_{v})(\cos\alpha_{1},\sin\alpha_{1})$,
$\mathbf{e}_{2}=\mathsf{g}w(r_{v})(\cos\alpha_{2},-\sin\alpha_{2})$, and
$\mathbf{e}_{3}=\mathsf{g}w(r_{v})(\cos\alpha_{3},-\sin\alpha_{3})$. So, in
component form the force balance equation reads

$\cos\alpha_{1}-\cos\alpha_{2}-\cos\alpha_{3}=0\,,\qquad\sin\alpha_{1}+\sin\alpha_{2}+\sin\alpha_{3}=3\mathsf{k}(1+4v)\text{e}^{-3v}\,.$
(C.4)

##  Appendix D Some string configurations

In order to make this paper more self-contained, in this Appendix we briefly
review some basic results on the static string configurations with light
quarks. For more details on these results, see a-strb ; a-QQq .

###  D.1 The configurations for $Q\bar{q}$ and $Qqq$

First, consider the configuration of Figure 18 on the left. In the context of
the gauge/string duality it provides the description of a heavy-light meson in
the static limit.

Figure 18: Left: A static configuration for a heavy-light meson $Q\bar{q}$.
Right: A static configuration for a heavy-light baryon $Qqq$.

The configuration is governed by the action which is the sum of the Nambu-Goto
action $S_{\text{NG}}$ and the boundary term $S_{\text{q}}$,
$S=S_{\text{NG}}+S_{\text{q}}$.

A short calculation performed in a-strb shows that the energy $E$ is expressed
in terms of $q$ by

$E_{\text{\tiny Q}\bar{\text{\tiny
q}}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}{\cal
Q}(q)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}\,\Bigr{)}+c\,.$ (D.1)

Here $c$ is a normalization constant. The function ${\cal Q}$ is as defined in
Appendix A. $q$ is a solution to the equation

$\text{e}^{\frac{q}{2}}+\mathsf{n}(q-1)=0\,,$ (D.2)

which is obtained by varying the action with respect to $r_{q}$. A noteworthy
fact is that such a solution only exists for $q<1$. The physical meaning of
this equation is that it is nothing else but the force balance equation at the
string endpoint which says that the net force acting on the light antiquark is
zero.

Now let us discuss the remaining configuration on the right, which describes a
static heavy-light baryon. This configuration is governed by the following
action $S=\sum_{i=1}^{3}S_{\text{NG}}^{(i)}+2S_{\text{q}}+S_{\text{vert}}$,
which is the sum of the Nambu-Goto actions, boundary actions $S_{\text{q}}$,
and action $S_{\text{vert}}$.

The energy of this configuration can be written in the form a-strb

$E_{\text{\tiny Qqq}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}2{\cal Q}(q)-{\cal
Q}(v_{\shortparallel})+2\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+3\mathsf{k}\frac{\text{e}^{-2v_{\shortparallel}}}{\sqrt{v_{\shortparallel}}}\,\Bigr{)}+c\,,$
(D.3)

where $v_{\shortparallel}$ is a solution to

$1+3\mathsf{k}(1+4v)\text{e}^{-3v}=0\,.$ (D.4)

The above equation defines the position of the baryon vertex in the bulk,
whereas the equation (D.2) defines the position of the light quarks. At
vanishing baryon chemical potential light quarks and antiquarks are at the
same radial distance from the boundary. Note that the stable configuration
exists only for $q\geq v$.

###  D.2 The configurations for $QQq$

Finally, we consider the string configurations of Figure 19.

Figure 19: Three types of static string configurations that contribute to the
ground state of the $QQq$ system. $\alpha$ denotes a tangent angle of the left
string at point $V$.

Those configurations were proposed to model the $QQq$ system in a-QQq , whose
conventions we follow here. The important point is that the shape of the
configuration changes with the increase of heavy quark separation.

For small $\ell$, the corresponding configuration is labeled by (S). In this
case, the relation between the energy and heavy quark separation is written in
parametric form

$\ell^{\text{\tiny(S)}}=\frac{2}{\sqrt{\mathsf{s}}}{\cal
L}^{+}(\alpha,v)\,,\qquad E_{\text{\tiny
QQq}}^{\text{\tiny(S)}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}2{\cal
E}^{+}(\alpha,v)+\mathsf{n}\frac{\text{e}^{\frac{1}{2}q}}{\sqrt{q}}+3\mathsf{k}\frac{\text{e}^{-2v}}{\sqrt{v}}+{\cal
Q}(q)-{\cal Q}(v)\Bigr{)}+2c\,,$ (D.5)

with the parameter $v$ varying from $0$ to $q$. The value of $q$ is determined
from equation (D.2). The functions ${\cal L}^{+}$ and ${\cal E}^{+}$ are as
defined in Appendix A. $c$ is a normalization constant. The tangent angle
$\alpha$ can be expressed in terms of $v$ by using the force balance equation
at the vertex, with the result

$\sin\alpha=\frac{1}{2}\bigl{(}1+3\mathsf{k}(1+4v)\text{e}^{-3v}\bigr{)}\,.$
(D.6)

At this point, it is worth mentioning that in the limit $\ell\rightarrow 0$
the energy reduces to a sum of energies: $E_{\text{\tiny
QQq}}^{\text{\tiny(S)}}=E_{\text{\tiny q}\bar{\text{\tiny Q}}}+E_{\text{\tiny
QQ}}$, as expected from heavy quark-diquark symmetry. This limit corresponds
to small values of $v$, where the function $\ell(v)$ behaves exactly as
described by Eq.(4.5).

For intermediate values of $\ell$, the configuration is labeled by (M). It
looks like one of the strings is missing so that the position of the light
quark coincides with that of the vertex. So, one has $q=v$. The distance
$\ell$ is expressed in terms of $v$ and $\alpha$ by the same formula as
before, only for another parameter range, whereas the energy by

$E_{\text{\tiny
QQq}}^{\text{\tiny(M)}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}2{\cal
E}^{+}(\alpha,v)+\frac{1}{{\sqrt{v}}}\bigl{(}\mathsf{n}\text{e}^{\frac{1}{2}v}+3\mathsf{k}\text{e}^{-2v}\bigr{)}\Bigr{)}+2c\,.$
(D.7)

Now $v$ varies from $q$ to $v_{\text{\tiny M}}$, where $v_{\text{\tiny M}}$ is
a solution to

$\mathsf{n}(1-v)+3\mathsf{k}(1+4v)\text{e}^{-\frac{5}{2}v}=0\,.$ (D.8)

The force balance equation in this case gives

$\sin\alpha=\frac{1}{2}\bigl{(}\mathsf{n}(1-v)\text{e}^{-\frac{1}{2}v}+3\mathsf{k}(1+4v)\text{e}^{-3v}\bigr{)}\,.$
(D.9)

A noteworthy fact is that $\alpha(v_{\text{\tiny M}})=0$.

For large $\ell$, the proper configuration is that labeled by (L). In fact,
what happens in the transition from (M) to (L) is that the tangent angle
changes the sign from positive to negative. Keeping this in mind makes it much
easier to arrive at the relation between the energy and quark separation.
Replacing ${\cal L}^{+}$ and ${\cal E}^{+}$ by ${\cal L}^{-}$ and ${\cal
E}^{-}$, it becomes

$\ell^{\text{\tiny(L)}}=\frac{2}{\sqrt{\mathsf{s}}}{\cal
L}^{-}(\lambda,v)\,,\qquad E_{\text{\tiny
QQq}}^{\text{\tiny(L)}}=\mathsf{g}\sqrt{\mathsf{s}}\Bigl{(}2{\cal
E}^{-}(\lambda,v)+\frac{1}{\sqrt{v}}\bigl{(}\mathsf{n}\text{e}^{\frac{1}{2}v}+3\mathsf{k}\text{e}^{-2v}\bigr{)}\Bigr{)}+2c\,,$
(D.10)

with the parameter $v$ varying from $v_{\text{\tiny M}}$ to $v_{\text{\tiny
L}}$. The upper bound is found by solving the non-linear equation

$2\sqrt{1-v^{2}\text{e}^{2(1-v)}}+3\mathsf{k}(1+4v)\text{e}^{-3v}+\mathsf{n}(1-v)\text{e}^{-\frac{1}{2}v}=0\,$
(D.11)

on the interval $[0,1]$. A function $\lambda(v)$ is written explicitly as

$\lambda(v)=-\text{ProductLog}\Bigl{[}-v\text{e}^{-v}\Bigl{(}1-\frac{1}{4}\Bigl{(}3\mathsf{k}(1+4v)\text{e}^{-3v}+\mathsf{n}(1-v)\text{e}^{-\frac{1}{2}v}\Bigr{)}^{2}\,\Bigr{)}^{-\frac{1}{2}}\,\Bigl{]}\,,$
(D.12)

as it follows from Eq.(B.17) with the tangent angle found from the force
balance equation at $V$. Note that $\lambda(v_{\text{\tiny L}})=1$ that
corresponds to the limit of infinitely long strings.

One can summarize all this by saying that the energy as a function of the
heavy quark separation is given in parametrical form by the two piecewise
functions $E_{\text{\tiny QQq}}=E_{\text{\tiny QQq}}(v)$ and $\ell=\ell(v)$.

For future reference, it is worth noting that the asymptotic behavior of
$E_{\text{\tiny QQq}}(\ell)$ for large $\ell$ is

$E_{\text{\tiny QQq}}=\sigma\ell-2\mathsf{g}\sqrt{\mathsf{s}}I_{\text{\tiny
QQq}}+2c+o(1)\,,$ (D.13)

with

$I_{\text{\tiny QQq}}={\cal I}(v_{\text{\tiny
L}})-\frac{1}{2\sqrt{v_{\text{\tiny
L}}}}\Bigl{(}\mathsf{n}\text{e}^{\frac{1}{2}v_{\text{\tiny
L}}}+3\mathsf{k}\text{e}^{-2v_{\text{\tiny L}}}\Bigr{)}\,$ (D.14)

and the same string tension $\sigma$ as in (B.23).

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