Scale dependence of the q and T parameters of the Tsallis distribution in the process of jet fragmentation

Abstract

The dependence of the q and T parameters of the Tsallis-distribution-shaped fragmentation function (FF) on the fragmentation scale (found to be equal to the jet mass) is calculated via the resummation of the branching process of jet fragmentation in the leading-log appriximation (LLA) in the $\phi ^3$ theory. Jet and hadron spectra in electron-positron ($e^+e^-$) annihilations with 2- and 3-jet final states are calculated using virtual leading partons. It is found that jets, produced earlier in the branching process, are more energetic, and the energy, angle and multiplicity distributions of hadrons stemming from them are broader. It is also found that replacing the LL resummation in the branching process by a single splitting provides good approximation for the jet energy distribution in 2-jet events. Furthermore, a micro-canonical statistical event generator is presented for the event-by-event calculation of hadron momenta in $e^+e^-$ annihilations.

Appendices

6 Appendix

A Phasespace of n massless particles

$$\begin{aligned} \varOmega _n(P)= & {} \prod _{i=1}^n \int \frac{d^{D-1}\textbf{p}_i}{p_i^0}\, \delta ^{D}\left( \sum _j p_j^\mu -P^\mu \right) \nonumber \\\sim & {} \int d^Ds\, e^{-is_\mu P^\mu } \varphi ^n(s) , \end{aligned}$$

(40)

where the Fourier-transform of the single-particle phasespace $\varphi (s)$ can be evaluated in the frame, in which, $s = (\sigma ,{\textbf{0}})$ (with $\sigma ^2 = s_0^2 - \textbf{s}^2$), and $p = (p,{\textbf{p}})$:

$$\begin{aligned} \varphi (\sigma )= & {} \int \frac{d^{D-1}\textbf{p}}{p^0}\, e^{is_\mu p^\mu } \nonumber \\\sim & {} \int dp\,p^{D-3} e^{i\sigma p} \sim \frac{1}{(i\sigma )^{D-2}} . \end{aligned}$$

(41)

We may evaluate the inverse Fourier-transform in Eq. (1) in the frame, where $P = (M_0,{\textbf{0}})$, thus,

$$\begin{aligned} \varOmega _n(P)\sim & {} \int d^{D-1}\textbf{s} \nonumber \\{} & {} \int ds_0\, \frac{e^{-is_0 M_0} }{ [(s_0+|{\textbf{s}}|)(s_0-|{\textbf{s}}|)] ^{n(D-2)/2} }. \end{aligned}$$

(42)

As we have poles at $s_0 = \pm |{\textbf{s}}|$, we use Cauchy’s formula $\oint \frac{dzf(z)}{(z-z_0)^n}\sim f^{(n-1)}(z_0)$, and arrive at terms of the form of

$$\begin{aligned} \varOmega _n(P)\sim & {} \sum _j A^{\pm }_j \int d^{D-1}\textbf{s} \, \left( \frac{\partial }{\partial s_0}\right) ^j \nonumber \\{} & {} e^{-is_0 M_0} \left( \frac{\partial }{\partial s_0}\right) ^{n(D-2)/2-1-j}\nonumber \\{} & {} \left. \frac{1}{ (s_0\pm |{\textbf{s}}|) ^{n(D-2)/2} } \right| _{s_0=\pm |{\textbf{s}}|} \nonumber \\{} & {} \sim \sum _j A^{\pm }_j\, M_0^j \int ds\, s^{D-1 - n(D-2)+j} \nonumber \\{} & {} e^{-is M_0} \sim M_0^{n(D-2)-D} . \end{aligned}$$

(43)

The actual values of the constant factors $A^{\pm }_j$ multiplying the terms coming from the poles at $s_0 = \pm |{\textbf{s}}|$, are of no importance from the point of view of the particle distributions.

B Calculation of the splitting function

Via introducing the renormalized field and coupling $g=Z_g g_r$ and $\phi = Z_3^{1/2} \phi _r$, along with $Z_3 = 1 + \delta Z_3$ and $Z_g = 1 + \delta Z_g$, we arrive at the renormalised Lagrangian

$$\begin{aligned} \mathcal {L}= & {} \frac{1}{2}(\partial _\mu \phi _r)^2 + (Z_3-1)\frac{1}{2}(\partial _\mu \phi _r)^2 \nonumber \\{} & {} + \frac{g_r}{3!}\phi _r^3 + (Z_g Z_3^{3/2}-1)\frac{g_r}{3!}\phi _r^3\nonumber \\ {}= & {} \frac{1}{2}\left( \partial _\mu \phi _r\right) ^2 + \delta Z_3\frac{1}{2}\left( \partial _\mu \phi _r\right) ^2\nonumber \\{} & {} + \frac{g_r}{3!}\phi _r^3 + \left( \delta Z_g + \frac{3}{2}\delta Z_3 \right) \frac{g_r}{3!}\phi _r^3. \end{aligned}$$

(44)

As $\delta Z_3$ and $\delta Z_g$ are of $\mathcal {O}(g^2)$, terms proportional to them come as perturbative corrections. This way, a propagator of momentum p is $i/(p^2+i\epsilon )$, a vertex is $-ig_r$, the counter terms are $ip^2\delta Z_3$ and $-ig_r(\delta Z_g+ \frac{3}{2}\delta Z_3)$, and a cut propagator is $2\pi \delta (p^2)$. Propagators and vertices on the right of the cuts are complex conjugates.

We may write $A(z,P^2) = \delta (1-z)A_1(z,P^2) + A_2(z,P^2)$ in Eq. (12). When calculating $A_1(z,P^2)$, we use the identity

$\prod \nolimits _i \dfrac{1}{A_i^{\alpha _i}} = \dfrac{\varGamma (\alpha )}{\prod \nolimits _i \varGamma (\alpha _i)} \prod \nolimits _i\int \nolimits _0^1 \dfrac{d\xi _i\,\xi _i^{\alpha _i-1} \delta (1-\alpha )}{(\sum \xi _i A_i)^\alpha }$, with $\alpha = \sum \alpha _i$, thus,

$$\begin{aligned}{} & {} A_1(z,P^2) = \int \limits _0^1 d\xi _1 \int \frac{d^Dq}{(2\pi )^D}\nonumber \\{} & {} \quad \left\{ \int \limits _0^{1{-}\xi _1} d\xi _2 \frac{2i\,n_c}{\left[ (1{-}\xi _1{-}\xi _2)q^2 {+} \xi _1(q{-}{\hat{k}})^2 {+} \xi _2(P{-}q)^2\right] ^3} \right. \;\nonumber \\{} & {} \quad + \left. \frac{i\,n_d}{m_0^2\left[ (1-\xi _1)q^2 + \xi _1(q-{\hat{k}})^2\right] ^2} \right. \nonumber \\{} & {} \quad \left. + \frac{i\,n_e}{m_0^2 \left[ (1-\xi _1)q^2 + \xi _1 (P-q-{\hat{k}})^2\right] ^2} \right\} \nonumber \\{} & {} \quad + \frac{{\bar{n}}_c}{g^2} \left( \delta Z_g + \frac{3}{2}\delta Z_3\right) - \frac{{\bar{n}}_d + {\bar{n}}_e}{g^2}\delta Z_3 . \end{aligned}$$

(45)

Substituting ${\tilde{q}}=q - \xi _1k-\xi _2P$ and $L_1 = -P^2\xi _2(1-\xi _2)$ in the first term, ${\tilde{q}} = q-\xi _1k$ in the second term and ${\tilde{q}} = q-\xi _1(P-k)$ in the third term, we get

$$\begin{aligned} g^2 A_1(z,P^2)= & {} i\,g^2 \int \limits _0^1 d\xi _1 \int \frac{d^D{\tilde{q}}}{(2\pi )^D} \nonumber \\{} & {} \left\{ \int \limits _0^{1-\xi _1} d\xi _2 \frac{2n_c}{({\tilde{q}}^2 - L_1)^3} \;+\; \frac{n_d + n_e}{m_0^2\, {\tilde{q}}^4} \right\} \nonumber \\{} & {} \quad +\; {\bar{n}}_c \left( \delta Z_g + \frac{3}{2}\delta Z_3\right) - ({\bar{n}}_d + {\bar{n}}_e)\delta Z_3 .\nonumber \\ \end{aligned}$$

(46)

Substituting ${\tilde{q}} = (iq_E,{\textbf{q}}_E)$ (Wick-rotation) gives

$$\begin{aligned} g^2A_1(z,P^2)= & {} g^2\frac{\kappa _D}{(2\pi )^D} \int \limits _0^1 d\xi _1 \int \limits _0^\infty dq_E\,q_E^{D-1} \nonumber \\{} & {} \left\{ \int \limits _0^{1-\xi _1} d\xi _2 \frac{2n_c}{( q^2_E + L_1)^3} \;-\; \frac{n_d + n_e}{m_0^2\, {\tilde{q}}_E^4}e^{-\epsilon q_E/m_0} \right\} \nonumber \\{} & {} + {\bar{n}}_c \left( \delta Z_g + \frac{3}{2}\delta Z_3\right) - ({\bar{n}}_d + {\bar{n}}_e)\delta Z_3 . \end{aligned}$$

(47)

Using $\int \limits _0^\infty \dfrac{dx\,x^{a-1}}{(x+1)^{a+b}} = \varGamma (a)\varGamma (b)/\varGamma (a+b)$, $\varGamma (\epsilon ) = 1/\epsilon -\gamma _E$ and setting $D=6-2\epsilon$ along with $g\rightarrow g\mu ^\epsilon$, we obtain

$$\begin{aligned} g^2A_1(z,P^2)= & {} \frac{g^2 n_c}{2 (4\pi )^3} \left( \frac{1}{\epsilon } - \ln \frac{-P^2}{\mu ^2} -\gamma _E + \ln 4\pi \right. \nonumber \\{} & {} \left. - 2\int \limits _0^1 d\xi _1 \int \limits _0^{1-\xi _1} d\xi _2 \ln \xi _2(1-\xi _2) \right) \;\nonumber \\ {}{} & {} - \frac{g^2 }{2 (4\pi )^3} (n_d + n_e)\frac{1}{\epsilon ^2} \nonumber \\{} & {} +\; {\bar{n}}_c \left( \delta Z_g + \frac{3}{2}\delta Z_3\right) - ({\bar{n}}_d + {\bar{n}}_e)\delta Z_3 .\nonumber \\ \end{aligned}$$

(48)

If we remove the divergences along with the $P^2$ and $\mu$ independent constants by the counter terms, we are left with

$$\begin{aligned} A_1(z,P^2) \;=\; -\frac{n_c }{2 (4\pi )^3} \ln \frac{P^2}{\mu ^2} \end{aligned}$$

(49)

For the calculation of the second line of $A(z,P^2)$ in Eq. (12), we parametrize momenta as $P=(M,0,{\textbf{0}})$, $k=(Mz/2,Mz/2,{\textbf{0}})$ and $q=\alpha P + \beta k + q_T = (\alpha M + \beta Mz/2,\beta Mz/2,{\textbf{q}}_T)$. This way, the integration measure becomes $d^Dq = (M^2z/2)d\alpha \, d\beta \, d^{D-2}{\textbf{q}}_T$. Besides, $2Pq = M^2(2\alpha + \beta z)$, $2 k q = M^2\alpha z$ and $q^2 = M^2\alpha (\alpha +\beta z)-{\textbf{q}}_T^2$. Furthermore, due to the $\delta \left[ (q-k)^2\right]$ term, $q^2 = 2k q = M^2\alpha z$. This way, the second line of Eq. (12) becomes

$$\begin{aligned} g^2A_2(z,P^2)= & {} g^2\frac{z \kappa _{D-2}}{4(2\pi )^{D+1}} \int d\alpha \int d\beta \nonumber \\{} & {} \int dq_T^2\,(q_T^2)^{D/2-2} \nonumber \\{} & {} \times \; (2\pi )\delta \left[ M^2\alpha (\alpha + \beta z - z)-q_T^2\right] (2\pi )\delta \nonumber \\{} & {} \left[ M^2(1+\alpha z - 2\alpha -\beta z)\right] \nonumber \\{} & {} \times \; \left\{ \frac{n_f}{\alpha ^2 z^2 } + \frac{n_g}{(1-z)^2} \right. \nonumber \\{} & {} \left. + \frac{n_h}{ \alpha z (1-z)} + \frac{n_i}{\alpha z (1-2\alpha +z-\beta z)} \right\} .\nonumber \\ \end{aligned}$$

(50)

Using the integral for $\beta$ to eliminate the second $\delta$ function gives

$$\begin{aligned} g^2 A_2(z,P^2)= & {} \frac{g^2\kappa _{D-2}}{4(2\pi )^{D-1} M^2} \int d\alpha \int dq_T^2\,(q_T^2)^{D/2-2}\delta \nonumber \\{} & {} \left[ M^2\alpha (1-\alpha )(1 - z)-q_T^2\right] \;\nonumber \\{} & {} \times \; \left\{ \frac{n_f}{\alpha ^2 z^2 } + \frac{n_g}{(1-z)^2} \right. \nonumber \\{} & {} \left. + \frac{n_h}{ \alpha z (1-z)} + \frac{n_i}{\alpha (1-\alpha )z^2} \right\} \nonumber \\= & {} \frac{g^2\kappa _{D-2} M^{D-6} (1-z)^{D/2-2}}{4(2\pi )^{D-1}} \nonumber \\{} & {} \int d\alpha \,[\alpha (1-\alpha )]^{D/2-2} \nonumber \\{} & {} \times \left\{ \frac{n_f}{\alpha ^2 z^2 } + \frac{n_g}{(1-z)^2} \right. \nonumber \\{} & {} \left. + \frac{n_h}{ \alpha z (1-z)} + \frac{n_i}{\alpha (1-\alpha )z^2} \right\} \end{aligned}$$

(51)

Using the identity $\int \limits _0^1 d\alpha \alpha ^{a-1}(1-\alpha )^{b-1} = \varGamma (a)\varGamma (b)/\varGamma (a+b)$ and $D=6-2\epsilon$ dimensions, where the coupling acquires dimension $g\rightarrow g\mu ^\epsilon$, the solid angle is $\kappa _D = 2\pi ^{D/2}/\varGamma (D/2)$ and $\varGamma (-\epsilon )=-1/\epsilon -\gamma _E$, we obtain

$$\begin{aligned}{} & {} g^2 A_2(z,P^2) = \frac{g^2 (1-z)}{(4\pi )^3} \left( \frac{4\pi \mu }{M^2(1-z)} \right) ^\epsilon \nonumber \\{} & {} \quad \left\{ \frac{n_f}{z^2}\left( -\frac{1}{\epsilon }-\gamma _E \right) + \frac{n_g}{6(1-z)^2} + \frac{n_h}{2z (1-z)} + \frac{n_i}{z^2} \right\} \nonumber \\{} & {} \quad = \frac{g^2}{(4\pi )^3} \left\{ -\left[ \frac{1}{\epsilon }+\gamma _E + \ln \left( \frac{4\pi \mu }{M^2(1-z)} \right) \right] \frac{n_f(1-z)}{z^2 }\; \right. \nonumber \\{} & {} \quad + \left. \frac{n_g}{6(1-z)} + \frac{n_h}{2z} + \frac{n_i(1-z)}{z^2} \right\} . \end{aligned}$$

(52)

We made use of $\varGamma (-1+\epsilon )=-1/\epsilon +\gamma _E-1$, $\varGamma (-\epsilon )=-1/\epsilon -\gamma _E$. Note that the $1/\epsilon$ term being the collinear divergence, cannot be eliminated via renormalisation, however, it drops out of the splitting function, which is

$$\begin{aligned} \varPi (z)= & {} \frac{\partial }{\partial \ln P^2}A(z,P^2) \nonumber \\= & {} \frac{n_f}{(4\pi )^3}\frac{1-z}{z^2} - \frac{n_c}{2 (4\pi )^3} \delta (1-z) . \end{aligned}$$

(53)