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First, we briefly review the holographic model used in the paper. Following Ref. [35], this background metric at finite temperature is
$ {\rm d} {\boldsymbol{s}}^{2}={\rm{e}}^{s r^{2}} \frac{R^{2}}{r^{2}}\left(f(r){\rm d} t^{2}+{\rm d} \vec{x}^{\,2}+ f^{-1}(r) {\rm d} r^{2}\right)+{\rm{e}}^{-s r^{2}} g_{a b}^{(5)} {\rm d} \omega^{a} {\rm d} \omega^{b}. $
(1) This model parameterized by s is a one-parameter deformation of Euclidean
$ AdS_5 $ space with a constant radius R, and a five-dimensional compact space (sphere)$ {\boldsymbol{X}} $ , whose coordinates are$ \omega^a $ and$ f(r) $ , is a blackening factor. The Nambu-Goto action of a string is expressed as$ S=\frac{1}{2 \pi \alpha^{\prime}} \int_{0}^{1} {\rm d} \sigma \int_{0}^{T} {\rm d} \tau \sqrt{\gamma}, $
(2) where γ is an induced metric on the string world-sheet (with a Euclidean signature). For
$ AdS_5 $ space,$ f(r) = 1- {r^4}/{r_h^4} $ with the boundary conditions$ f(0) = 1 $ and$ f(r_h) = 0 $ .$ r_h $ is the position of the black hole (brane). The Hawking temperature identified with the temperature of a dual gauge theory can be defined as$ T=({1}/{4 \pi})\left|\partial_{r} f\right|_{r=r_{h}} $ . The motivations for this metric have been clarified in Ref. [35]. However, such a deformed$ AdS_5 $ metric leads to linear Regge-like spectra for mesons [39, 40] and the Cornell potential of a quark-antiquark [4]. The deformed metric satisfies the thermodynamics of lattice [41].Subsequently, we introduce the baryon vertex. According to the AdS/CFT correspondence, this is a five brane [42]. At leading order in
$ \alpha' $ , the brane action is$ S_{\rm vert} = {\cal{T}}_5 \times \int {\rm d}^6 \xi \sqrt{\gamma^{(6)}} $ , where$ {\cal{T}}_5 $ is the brane tension, and$ \xi^{i} $ are the world-volume coordinates. Because the brane is wrapped around the internal space, it appears point-like in$ AdS_5 $ . We choose a static gauge$ \xi^0 = t $ and$ \xi^a = \theta^a $ with$ \theta^a $ coordinates on$ {\boldsymbol{X}} $ . Thus, the action is$ S_{\rm{vert}}=\tau_{v} \int {\rm d} t \frac{{\rm{e}}^{-2 s r^{2}}}{r} \sqrt{f(r)}, $
(3) where
$ \tau_{v} $ is a dimensionless parameter defined by$ \tau_{v}= {\cal{T}}_{5} R {\rm{vol}}({\boldsymbol{X}}) $ , and$ {\rm{vol}}({\boldsymbol{X}}) $ is a volume of$ {\boldsymbol{X}} $ .Finally, we consider that the light quark at string endpoints is a tachyon field, which couples to the world-sheet boundary via
$ S_{q}=\int {\rm d} \tau e \;{T} $ ; this is the usual sigma-model action for a string propagating in the tachyon background [43]. The integral is over a world-sheet boundary parameterized by τ, and e is a boundary metric. We consider a constant background$ {T}(x,r) = {T}_0 $ and worldsheets whose boundaries are lines in the t direction. Thus, the action can be written as$ S_{q}={m} \int {\rm d} t \frac{{\rm{e}}^{\frac{s}{2} r^{2}}}{r} \sqrt{f(r)}, $
(4) where
$ {\rm{m}}=R \;{T}_{0} $ . This is the action of a point particle of mass$ T_0 $ at rest. We choose the model parameters as follows:$ {\boldsymbol{g}}=\dfrac{R^{2}}{2 \pi \alpha^{\prime}}, \;{k}=\dfrac{\tau_{v}}{3 {\boldsymbol{g}}} $ , and$ {n}=\dfrac{m}{\boldsymbol{g}} $ . -
The configuration of QQq is plotted in Fig. 1. The total action is the sum of the Nambu-Goto actions plus the actions for the vertex and background scalar.
Figure 1. (color online) Static string configuration at a small separation distance of a heavy-quark pair. The heavy quarks Q are placed on the x-axis, while the light quark q and baryon vertex V are on the r-axis at
$r = r_q$ and$r = r_v$ , respectively. The quarks and baryon vertex are connected by the blue string. The force exerted on the vertex and light quark are depicted by the black arrows.$r_h$ is the position of the black-hole horizon.$r_w$ is the position of a soft wall in the confined phase.$ S=\sum\limits_{i=1}^{3} S_{{\rm{NG}}}^{(i)}+S_{\rm{vert}}+S_{q}. $
(5) If we choose the static gauge
$ \xi^{1}=t $ and$ \xi^{2}=r $ , the boundary conditions for$ x(r) $ become$ x^{(1)}(0)=-\frac{L}{2} , \quad x^{(2)}(0)=\frac{L}{2}, \quad x^{(i)}\left(r_{v}\right)=x^{(3)}\left(r_{q}\right)=0. $
(6) The action can now be written as
$ \begin{aligned}[b] S=&{\boldsymbol{g}} T \Bigg(2 \int_{0}^{r_{v}} \frac{{\rm d} r}{r^{2}} {\rm{e}}^{s r^{2}} \sqrt{1+f(r)\left(\partial_{r} x\right)^{2}}\\& +\int_{r_{v}}^{r_{q}} \frac{{\rm d} r}{r^{2}} {\rm{e}}^{{s} r^{2}} \sqrt{1+f(r)\left(\partial_{r} x\right)^{2}}\\&+3 {k} \frac{{\rm{e}}^{-2 {s} r_{v}^{2}}}{r_{v}}\sqrt{f(r_v)} +{n} \frac{{\rm{e}}^{\frac{1}{2} {s} r_{q}^{2}}}{r_{q}}\sqrt{f(r_q)}\Bigg), \end{aligned} $
(7) where
$ \partial_{r} x={\partial x}/{\partial r} $ and$ T=\int_{0}^{T} {\rm d} t $ . We consider the first term in (7), which corresponds to string (1) and (2) in Fig. 1. The equation of motion (EoM) for$ x(r) $ can be obtained from the Euler-Lagrange equation. Thus, we have$ {\cal{I}}=\frac{w(r) f(r) \partial_{r} x}{\sqrt{1+f(r)\left(\partial_{r} x\right)^{2}}}, \quad w(r)=\frac{{\rm{e}}^{{\rm{s}} r^{2}}}{r^{2}}. $
(8) $ {\cal{I}} $ is a constant. At$ r_v $ , we have$ \left.\partial_{r} x\right|_{r=r_{v}}=\cot \alpha $ with$ \alpha>0 $ and$ {\cal{I}}=\frac{w(r_v) f(r_v) {\rm cot} \alpha}{\sqrt{1+f(r)\left({\rm cot} \alpha\right)^{2}}}. $
(9) $ \partial_{r} x $ can be solved as$ \partial_{r} x = \sqrt{\frac{\omega\left(r_{v}\right)^{2} f\left(r_{v}\right)^{2}}{\left(f\left(r_{\nu}\right)+\tan ^{2} \alpha\right) \omega(r)^{2} f(r)^{2}-f(r) w\left(r_{v}\right)^{2} f(r_v)^{2}}}. $
(10) Using (10), the integral over [
$ 0,r_v $ ] of dr is$ L=2 \int_{0}^{r_{v}} \frac{{\rm d}x}{{\rm d}r} {\rm d}r. $
(11) L is a function of
$ r_v $ , α, and$ r_h $ (or equally, temperature). The energy of string (1) can be found from the first term of (7).$ E_R=\frac{S}{T}={\boldsymbol{g}} \int_{0}^{r_{v}}\frac{{\rm d} r}{r^{2}} {\rm{e}}^{{\bf{s}} r^{2}} \sqrt{1+f(r)\left(\partial_{r} x\right)^{2}}. $
(12) Subtracting the divergent term
$ {\boldsymbol{g}} \int_{0}^{\infty}{\rm d}r\dfrac{1}{r^2} $ , we have the regularized energy:$ E_{1}=\frac{S}{T}={\boldsymbol{g}} \int_{0}^{r_{v}}\left(\frac{1}{r^{2}} {\rm{e}}^{{\bf{s}} r^{2}} \sqrt{1+f(r)\left(\partial_{r} x\right)^{2}}-\frac{1}{r^2}\right){\rm d} r - \frac{{\boldsymbol{g}}}{r_v} + c. $
(13) Here, c is a normalization constant. Because string (2) produces the same results, we move to string (3), whose action is given by the second term in (7). This string is a straight string stretched between the vertex and light. The energy can be calculated as
$ E_{2}={\boldsymbol{g}} \int_{r_{v}}^{r_{q}} \frac{{\rm d} r}{r^{2}} {\rm{e}}^{{\rm{s}} r^{2}}. $
(14) Now, we can present the energy of the configuration. From (7), it follows that
$ \begin{aligned}[b] E_{Q Qq} =& {\boldsymbol{g}} \Bigg(2\int_{0}^{r_{v}}\left(\frac{1}{r^{2}} {\rm{e}}^{{\boldsymbol{s}} r^{2}} \sqrt{1+f(r)\left(\partial_{r} x\right)^{2}}-\frac{1}{r^2}\right){\rm d} r\\& - \frac{2}{r_v} + \int_{r_{v}}^{r_{q}} \frac{{\rm d} r}{r^{2}} {\rm{e}}^{{s} r^{2}}+ {n} \frac{{\rm{e}}^{\frac{1}{2} s r_q^2}}{r_q}\sqrt{f(r_q)}\\& +3 {k} \frac{{\rm{e}}^{-2 s r_v^2}}{r_v}\sqrt{f(r_q)}\Bigg) +2 c. \end{aligned} $
(15) The energy is also a function of
$ r_v $ , α, and$ r_h $ . There are two steps to be compeleted: The first is to determine the position of the light quark, which is a function of temperature, and the second is to determine α. To achieve this goal, the net forces exerted on the light quark and vertex vanish. We first write the force balance equation of the light quark as$ {\boldsymbol{f}}_{q}+{\boldsymbol{e}}_{3}^{\prime}=0. $
(16) By varying the action with respect to
$ r_q $ , it is found that${\boldsymbol{f}}_{q}=\left(0,-{\boldsymbol{g}} n \partial_{r_{q}} \left(\big({{\rm{e}}^{\frac{1}{2} {sr}_{q}^{2}}}/{r_{q}}\big)\sqrt{f(r_q)}\right)\right)$ and$ {\boldsymbol{e}}_{3}^{\prime}={\boldsymbol{g}} w\left(r_{q}\right)(0,-1) $ . Hence, the force balance equation becomes$ 2{\rm e}^{({r_q^{2} s})/{2}}\sqrt{f(r_q)}+2n\left(-1+r_q^{2} s\right) f(r_q)+n r_q f^{\prime}(r_q)= 0. $
(17) Note that
$ r_q $ (the position of the light quark) is only a function of$ r_h $ . At fixed temperature,$ r_q $ can be fixed via the above equation. The force balance equation on the vertex is$ {\boldsymbol{e}}_{1}+{\boldsymbol{e}}_{2}+{\boldsymbol{e}}_{3}+{\boldsymbol{f}}_{v}=0. $
(18) $ {\boldsymbol{e}}_{i} $ is the string tension, which can be calculated in the same manner as in Ref. [37]. As a result, the force on the vertex is$ {\boldsymbol{f}}_{v}=\left(0,-3 {\boldsymbol{g}} k \partial_{r_{v}} \left(\big({{\rm{e}}^{-2 s r_{v}^{2}}}/{r_{v}}\big)\sqrt{f(r_v)}\right)\right) $ , and the string tensions are$\begin{aligned}[b]&{\boldsymbol{e}}_{1}={\boldsymbol{g}} w\left(r_{v}\right)\left(-\frac{f(r_{v})}{\sqrt{{\rm tan}^2 \alpha + f(r_{v})}},-\frac{1}{\sqrt{f(r_{v}) {\rm cot}^2 \alpha + 1}}\right), \\&{\boldsymbol{e}}_{2} ={\boldsymbol{g}} w\left(r_{v}\right)\left(\frac{f(r_{v})}{\sqrt{{\rm tan}^2 \alpha + f(r_{v})}},-\frac{1}{\sqrt{f(r_{v}) {\rm cot}^2 \alpha + 1}}\right),\\& {\boldsymbol{e}}_{3}={\boldsymbol{g}} w\left(r_{v}\right)(0,1).\end{aligned}$
The non-trivial component of the force balance equation is
$\begin{aligned}[b]& 2 {\rm e}^{3 s r_v^{2}}\left(1-\frac{2}{1+{\rm cot}\alpha^2 f(r_v)}\right)+6\left(k+4 k s r_v^{2} \right)\sqrt{f(r_{v})}\\-&\frac{3 k r_v f^{\prime}(r_v)}{\sqrt{f(r_v)}}=0. \end{aligned}$
(19) This equation provides the relationship between
$ r_v $ and α when the temperature is fixed. With (15) and (17), we can numerically solve the energy at small L. -
The total action in the configuration plotted in Fig. 2 is expressed as
Figure 2. (color online) Static string configuration at an intermediate separation distance of a heavy-quark pair. The heavy quarks Q are placed on the x-axis, while the light quark q and baryon vertex V are at the same point
$r_v$ on the r-axis. The forces exerted on the point are depicted by the black arrows.$r_h$ is the position of the black-hole horizon.$r_w$ is the position of a soft wall in the confined phase.$ S=\sum\limits_{i=1}^{2} S_{{\rm{NG}}}^{(i)}+S_{\rm{vert}}+S_{q}. $
(20) We still choose the same static gauge as before. The boundary conditions then take the form
$ x^{(1)}(0)=-\frac{L}{2}, \quad x^{(2)}(0)=\frac{L}{2} , \quad x^{(i)}\left(r_{v}\right)=0. $
(21) In this configuration, the expressions (11) and (13) still hold. Naturally, we can express the total energy of the string without the contribution from string (3) as follows:
$ \begin{aligned}[b] E_{Q Qq}=&{\boldsymbol{g}} (2\int_{0}^{r_{v}}\left(\frac{1}{r^{2}} {\rm{e}}^{{\boldsymbol{s}} r^{2}} \sqrt{1+f(r)\left(\partial_{r} x\right)^{2}}-\frac{1}{r^2}\right){\rm d} r - \frac{2}{r_v}\\ & + {n} \frac{{\rm{e}}^{\frac{1}{2} s r_v^2}}{r_v} \sqrt{f(r_v)} +3 {k} \frac{{\rm{e}}^{-2 s r_v^2}}{r_v}\sqrt{f(r_v)})+2 c. \end{aligned} $
(22) The force balance equation at the point
$ r = r_v $ is$ {\boldsymbol{e}}_{1}+{\boldsymbol{e}}_{2}+{\boldsymbol{f}}_{v}+{\boldsymbol{f}}_{q}=0. $
(23) Each force is given by
$ {\boldsymbol{f}}_{q}=\left(0,-n {\boldsymbol{g}} \partial_{r_{q}}\left(\frac{{\rm{e}}^{\frac{1}{2} s r_{q}^{2}}}{r_{q}} \sqrt{f\left(r_{q}\right)}\right)\right), $
$ {\boldsymbol{f}}_{v}=\left(0,-3 {\boldsymbol{g}} k \partial_{r_{v}}\left(\frac{{\rm{e}}^{-2 s r_{v}^{2}}}{r_{v}} \sqrt{f\left(r_{v}\right)}\right)\right), $
$ {\boldsymbol{e}}_{1}={\boldsymbol{g}} w\left(r_{v}\right)\left(-\frac{f(r_{v})}{\sqrt{{\rm tan}^2 \alpha + f(r_{v})}},-\frac{1}{\sqrt{f(r_{v}) {\rm cot}^2 \alpha + 1}}\right), $
$ {\boldsymbol{e}}_{2}={\boldsymbol{g}} w\left(r_{v}\right)\left(\frac{f(r_{v})}{\sqrt{{\rm tan}^2 \alpha + f(r_{v})}},-\frac{1}{\sqrt{f(r_{v}) {\rm cot}^2 \alpha + 1}}\right), $
where
$ r_{q} = r_{v} $ . The force balance equation leads to$ \begin{aligned}[b] &2(-{\rm e}^{({5 r_v^2 s})/{2}} n (-1 + r_v^2 s)+3 k (1+4r_v^2 s))\sqrt{f(r_v)}\\-&\frac{4{\rm e}^{3r_v^2 s}}{\sqrt{1+{\rm cot}\alpha^2 f(r_v)}}-\frac{(3k+{\rm e}^{({5r_v^2 s})/{2}})r_v f'(r_v)}{\sqrt{f(r_v)}}=0. \end{aligned} $
(24) -
The total action of the configuration plotted in Fig. 3 is the same as Eq. (20). However, it is convenient to choose another static gauge
$ \xi^{1}=t $ and$ \xi^{2}=x $ here. Then, the boundary conditions areFigure 3. (color online) Static string configuration at a large separation distance of a heavy-quark pair. The heavy quarks Q are placed on the x-axis, while the light quark q and baryon vertex V are at the same point
$r_v$ on the r-axis. The forces exerted on the point are depicted by the black arrows. There is a turning point located at$(x_0, r_0)$ for string (1).$r_w$ is the position of a soft wall in the confined phase.$r_h$ is the position of the black-hole horizon.$ r^{(1)}(-L / 2)=r^{(2)}(L / 2)=0, \quad r^{(i)}(0)=r_{v} . $
(25) The total action becomes
$ \begin{aligned}[b] S=&{\boldsymbol{g}} T\Bigg(\int_{-L / 2}^{0} {\rm d} x w(r) \sqrt{f(r)+\left(\partial_{x} r\right)^{2}}\\&+\int_{0}^{L / 2} {\rm d} x w(r) \sqrt{f(r)+\left(\partial_{x} r\right)^{2}}\\ & +3 {k} \frac{{\rm{e}}^{-2 s r_{v}^{2}}}{r_{v}}\sqrt{f(r_v)}+{n} \frac{{\rm{e}}^{\frac{1}{2} {\rm{s}} r_{v}^{2}}}{r_{v}}\sqrt{f(r_v)}\Bigg). \end{aligned} $
(26) We consider string (1), whose action is given by the first term in (26). The first integral has the form
$ {\cal{I}}=\frac{w(r)f(r)}{\sqrt{f(r)+\left(\partial_{x} r\right)^{2}}}. $
(27) At points
$ r_0 $ and$ r_v $ , we have$ \frac{w(r)f(r)}{\sqrt{f(r)+\left(\partial_{x} r\right)^{2}}}=w(r_0)\sqrt{f(r_0)}, $
(28) $ \frac{w(r_v)f(r_v)}{\sqrt{f(r_v)+ {\rm tan}\alpha^2}}=w(r_0)\sqrt{f(r_0)}. $
(29) The relationship between
$ r_0 $ ,$ r_v $ , and α can be obtained from Eq. (29). Using Eq. (28), the separation distance and energy can be subsequently obtained. As before,$ \partial_x r $ can be solved as$ \partial_x r =\sqrt{\frac{w(r)^2f(r)^2f(r_0) - f(r)w(r_0)^2 f(r_0)^2}{w(r_0)^2 f(r_0)^2}}. $
(30) The separation distance consists of two parts:
$ \begin{aligned}[b] L =& 2(L_1 + L_2) = 2\left(\int_0^{r_0} \frac{1}{r^{\prime}} {\rm d}r + \int_{r_v}^{r_0} \frac{1}{r^{\prime}} {\rm d}r\right)\\ =&2\left(\int_0^{r_0} \sqrt{\frac{w(r_0)^2 f(r_0)^2}{w(r)^2f(r)^2f(r_0) - f(r)w(r_0)^2f(r_0)^2}} {\rm d}r \right.\\ & + \left.\int_{r_v}^{r_0} \sqrt{\frac{w(r_0)^2 f(r_0)^2}{w(r)^2f(r)^2f(r_0) - f(r)w(r_0)^2f(r_0)^2}} {\rm d}r\right). \end{aligned} $
(31) The energy of string (1) can be obtained by summing two parts.
$ \begin{aligned}[b] E_R =& E_{R_1} + E_{R_2}\\ =&{\boldsymbol{g}} \int_0^{r_0} w(r)\sqrt{1 + f(r) x^{\prime 2}} {\rm d}r + {\boldsymbol{g}} \int_{r_v}^{r_0} w(r)\sqrt{1 + f(r) x^{\prime 2}} {\rm d}r \\ =&{\boldsymbol{g}} \int_0^{r_0} w(r) \sqrt{\frac{w(r)^2 f(r)^2 f(r_0)}{w(r)^2f(r)^2f(r_0) - f(r) w(r_0)^2 f(r_0)^2}} {\rm d}r \\ & + {\boldsymbol{g}} \int_{r_v}^{r_0} w(r) \sqrt{\frac{w(r)^2 f(r)^2 f(r_0)}{w(r)^2f(r)^2f(r_0) - f(r) w(r_0)^2 f(r_0)^2}} {\rm d}r . \end{aligned} $
(32) After subtracting the divergent term, the renormalized energy of string (1) is
$ \begin{aligned}[b] E =&{\boldsymbol{g}} \int_{r_v}^{r_0} \Bigg(w(r) \sqrt{\frac{w(r)^2 f(r)^2 f(r_0)}{w(r)^2f(r)^2f(r_0) - f(r) w(r_0)^2 f(r_0)^2}}\Bigg) {\rm d}r \\&+{\boldsymbol{g}} \int_0^{r_0} \Bigg( w(r) \sqrt{\frac{w(r)^2 f(r)^2 f(r_0)}{w(r)^2f(r)^2f(r_0) - f(r) w(r_0)^2 f(r_0)^2}}\\ &- \frac{1}{r^2}\Bigg) {\rm d}r - \frac{1}{r_0}+ 2c. \end{aligned} $
(33) The calculation of string (2) has the same procedure as before and gives the same expressions for L and E. Then, the total energy of the configuration can be written as
$ \begin{aligned}[b] E_{QQq} =& \Bigg{(} 2\int_{r_v}^{r_0} w(r) \sqrt{\frac{w(r)^2 f(r)^2 f(r_0)}{w(r)^2f(r)^2f(r_0) - f(r) w(r_0)^2 f(r_0)^2}} {\rm d}r\\&+ 2\int_0^{r_0} \Bigg(w(r) \sqrt{\frac{w(r)^2 f(r)^2 f(r_0)}{w(r)^2f(r)^2f(r_0) - f(r) w(r_0)^2 f(r_0)^2}} \\& - \frac{1}{r^2}\Bigg) {\rm d}r - \frac{1}{r_0}+ 3 {k} \frac{{\rm{e}}^{-2 \;{\rm{s}} r_{v}^{2}}}{r_{v}}+{n} \frac{{\rm{e}}^{\frac{1}{2} {\boldsymbol{s}} r_{v}^{2}}}{r_{v}}\Bigg){\boldsymbol{g}} + 2c. \end{aligned} $
(34) The force balance equation is the same as Eq. (23). Each force is similarly given in the previous section. With Eq. (29) and (24), we can numerically solve L and
$ E_{QQq} $ .Except in the symmetric case, the light quark can be in a position far from the r-axis. The non-symmetric configuration has been discussed in Ref. [33], which, as noted in Ref. [33], is not energetically favorable. In this paper, we mainly focus on the symmetric configuration.
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In the confined phase, the quarks are confined in the hadrons. Can QQq exist at extremely large distances? The answer is no. The light quarks and anti-quark will be excited from vacuum at large distances. We call this string breaking and consider the following decay mode:
$ Q Q q \rightarrow Q q q+Q \bar{q}. $
(35) Qqq consists of three fundamental strings: a vertex and two light quarks. Thus, the total action of Qqq is
$ S=\sum_{i=1}^{3} S_{{\rm{NG}}}^{(i)}+S_{\rm{vert}}+2S_{\rm{q}} $ .$ Q\bar{q} $ consists of a fundamental string and a light quark. Thus, the total action of Qqq is$ S= S_{{\rm{NG}}}+S_{q} $ . The total actions of$ Qqq $ and$ Q\bar{q} $ are$ \begin{aligned}[b] S_{ {Qqq} }=& {\boldsymbol{g}} \Bigg(2 \int_{r_v}^{r_q} \frac{{\rm e}^{s r^2}}{r^2} {\rm d}r + \int_{0}^{r_v} \frac{{\rm e}^{s r^2}}{r^2}+ 3k \frac{{\rm e}^{-2 s r_v^2} \sqrt{f(r_v)}}{r_v}\\&+2 n \frac{{\rm e}^{\frac{1}{2}s r_q^2}}{r_q} \sqrt{f(r_q)}\Bigg), \\ {S_ {Q\bar{q}} }=& {\boldsymbol{g}} \int_{0}^{r_q}\frac{{{\rm e}^{s r^2}}}{r^2} + n {\boldsymbol{g}} \frac{{\rm e}^{\frac{1}{2}s r_q^2}}{r_q} \sqrt{f(r_q)} . \end{aligned} $
(36) Varying the action with respect to
$ r_q $ gives Eq. (17), and varying the action with respect to$ r_v $ gives$ \begin{aligned}[b] &1 + 3k {\rm e}^{-3 s r_v^3} \sqrt{f(r_v)} + 12 k s {\rm e}^{-3 s r_v^3} r_v^2\sqrt{f(r_v)}\\ -& \frac{3 k {\rm e}^{-3 s r_v^2} r_v f'(r_v)}{2\sqrt{f(r_v)}} = 0. \end{aligned} $
(37) We can obtain
$ r_v=0.410 \;{\rm{GeV}}^{-1} $ or$ r_v=0.453\;{\rm{ GeV}}^{-1} $ . Because the difference in energy is extremely small for the two states, we take$ r_v=0.410\;{\rm{GeV}}^{-1} $ for simplicity. The configuration for$ Q q q+Q \bar{q} $ is shown in Fig. 20. The renormalized total energy is$ \begin{aligned}[b] E_{ {Qqq}}+E_{ {Q\bar{q}}} =& {\boldsymbol{g}} \Bigg(2 \int_{r_v}^{r_q} \frac{{\rm e}^{s r^2}}{r^2} {\rm d}r + \int_{0}^{r_q}\left(\frac{{{\rm e}^{s r^2}}}{r^2} - \frac{1}{r^2}\right) - \frac{1}{r_q} \\ &+\int_{0}^{r_v} \left(\frac{{\rm e}^{s r^2}}{r^2}-\frac{1}{r^2}\right)- \frac{1}{r_v}+ 3k \frac{{\rm e}^{-2 s r_v^2} \sqrt{f(r_v)}}{r_v}\\& + 3 n \frac{{\rm e}^{\frac{1}{2}s r_q^2}}{r_q} \sqrt{f(r_q)}\Bigg) + 2c. \end{aligned} $
(38) For fixed
$ r_q=1.146 \;{\rm{GeV}}^{-1} $ and$ r_v=0.410\;{\rm{ GeV}}^{-1} $ , we have$ E_{ {Qqq}}+E_{ {Q\bar{q}}} = 3.006\;{\rm{ GeV}} $ . Figure 21 is a schematic diagram of string breaking. To determine the string breaking distance, we plot the energy$ E_{QQq} $ and$ E_{Q\bar{q}}+ E_{Qqq} $ as a function of$ r_v $ at$ T = 0.1\;{\rm{GeV}} $ in Fig. 22. The cross point shown in the figure enables us to determine$ L_{QQq} $ at fixed temperature. We find the distance of string breaking to be$ L_{QQq} = 1.27\; {\rm{fm}} $ .Figure 21. (color online) Schematic diagram of string breaking from the third configuration of QQq to
$Q q q+Q \bar{q}$ .Figure 22. (color online) Green line is the energy
$E_{Qqq}+E_{Q\bar{q}}$ . The black line represents$E_{QQq}$ at small distances, the blue line represents$E_{QQq}$ at intermediate distances, and the red line represents$E_{QQq}$ at large distances. The unit of E is GeV, and the unit of$r_v$ is$\rm GeV^{-1}$ . -
The energy of
$ {Q\bar{Q}} $ has been extensively studied in many holographic models. In this section, we focus on comparing the energy of${Q\bar{Q}} $ with QQq in the confined and deconfined phases. First, we only show the results of the separation distance and energy of$ {Q\bar{Q}} $ $ L_{Q\bar{Q}} = 2 \int_{0}^{r_0} \left(\frac{g_2(r)}{g_1(r)}\left(\frac{g_2(r)}{g_2(r_0)}-1\right)\right)^{-{1}/{2}}, $
(39) $ E_{ {Q\bar{Q}}} =2 {\boldsymbol{g}} \left(\int_{0}^{z_0} \sqrt{\frac{g_2(r) g_1(r)}{g_2(r)-g_2(r_0)}}-\frac{1}{r^2}\right) - \frac{2 {\boldsymbol{g}}}{r_0} + 2c, $
(40) where
$ g_1(r) = \dfrac{{\rm e}^{2 s r^2}}{r^4},\; g_2(r) = \dfrac{{\rm e}^{2 s r^2}}{r^4} f(r) $ , and$ z_0 $ is the turning point of the U-shape string. A detailed calculation can be found in our previous paper [19, 24, 25]. We consider the decay mode$ {Q \bar{Q}\rightarrow Q\bar{q} + \bar{Q}q}. $ It is clear that$ E_{ {Q\bar{q}}} = {\boldsymbol{g}} \left(\int_{0}^{r_q} \frac{1}{r^2}({\rm e}^{s r^2} - 1)\right) -\frac{{\boldsymbol{g}}}{r_q} + {\boldsymbol{g}} n \frac{{\rm e}^{\frac{s}{2}r_q^2}}{rq}\sqrt{f(r_q)} + c. $
(41) Thus, we can calculate
$ E_{ {Q\bar{q}}}+E_{ {\bar{Q}q}} = 2.39 \;{\rm{GeV}} $ at$ T = 0.1\;{\rm{GeV}} $ . In the confined phase, such as when$ T = 0.1\;{\rm{GeV}} $ , we present the energy of$ E_{ {Q\bar{q}}}+E_{ {\bar{Q}q}} $ and$ E_{ {Q\bar{Q}}} $ in Fig. 23. Through a comparison with Fig. 22, we find the distance of string breaking$ L=1.25\; {\rm{fm}} $ is close to that of QQq.Figure 23. (color online) Blue dashed line is the energy of
$E_{ {Q\bar{q}}}+E_{ {\bar{Q}q}}$ , and the black dashed line is the energy of$E_{ {Q\bar{Q}}}$ . The temperature is$T = 0.1\; \rm{GeV}$ . The unit of E is GeV, and the unit of L is fm.In the deconfinement phase, we compare the energies of
$ E_{ {Q\bar{Q}}} $ and$ E_{ {QQq}} $ at$ T = 0.148\;{\rm{GeV}} $ . Fig. 24 shows that the screening distance of$ E_{ {QQq}} $ ($ L = 0.45\;\rm{fm} $ ) is significantly smaller than that of$ E_{ {Q\bar{Q}}} $ ($ L = 0.94\;\rm{fm} $ ) at the same temperature$ T = 1.48 \;{\rm{GeV}} $ . This indicates that$ {Q\bar{Q}} $ is more stable than QQq in the deconfined phase.
Studying the potential of QQq at finite temperature in a holographic model
- Received Date: 2021-12-28
- Available Online: 2022-07-15
Abstract: Using gauge/gravity duality, we investigate the string breaking and dissolution of two heavy quarks coupled to a light quark at finite temperature. It is found that three configurations of QQq exist with the increase in separation distance for heavy quarks in the confined phase. Furthermore, string breaking occurs at the distance