Parton Energy Loss in an Unstable Quark-Gluon Plasma
Margaret E. Carrington, Katarzyna Deja, Stanislaw Mrowczynski
PParton Energy Lossin an Unstable Quark-Gluon Plasma ∗ Margaret E. Carrington
Department of Physics, Brandon University,Brandon, Manitoba, R7A 6A9 Canada andKatarzyna Deja
National Center for Nuclear Research, 00-681 Warsaw, Poland andStanis(cid:32)law Mr´owczy´nski
Institute of Physics, Jan Kochanowski University,25-406 Kielce, Polandand National Center for Nuclear Research, 00-681 Warsaw, Poland (Received November 18, 2011)
The energy loss of a fast parton scattering elastically in a weakly cou-pled quark-gluon plasma is formulated as an initial value problem. Theapproach is designed to study an unstable plasma, but it reproduces thewell known result in the case of an equilibrium plasma. Contributions tothe energy loss due to unstable modes are shown to exponentially grow intime. An unstable two-stream system is considered as an example.
1. Introduction
When a highly energetic parton travels through the quark-gluon plasma(QGP), it losses its energy due to elastic interactions with plasma con-stituents. This is the so-called collisional energy loss which for the equi-librium QGP is well understood, see the review [1] and the handbook [2].The quark-gluon plasma produced in relativistic heavy-ion collisions, how-ever, reaches the state of local equilibrium only after a short but finite time ∗ Presented by K. Deja at the HIC-for-FAIR Workshop & XXVIII Max-Born Sympo-sium ‘Three Days on Quarkyonic Island’, Wroc(cid:32)law, Poland, May 18-21, 2011. (1) a r X i v : . [ h e p - ph ] N ov e-loss-Wroclaw printed on November 7, 2018 interval, and during this period the momentum distribution of plasma par-tons is anisotropic. Collisional energy loss has been computed for such ananisotropic QGP [3]. However, the plasma with anisotropic momentum dis-tribution is unstable due to the chromomagnetic modes (for a review see[4]), and the fact that unstable systems are explicitly time dependent asunstable modes exponentially grow in time has not been taken into accountin the study [3].Our aim is to formulate an approach where the energy loss is found as thesolution of an initial value problem. The parton is treated as an energeticclassical particle with SU(3) color charge. For the equilibrium plasma werecover the known results but for the unstable plasma the energy loss isshown to strongly depend on time. Our approach to the energy-loss problemis similar to the method used earlier to study the momentum broadening ˆ q ofa fast parton in anisotropic plasma [5]. Analogous methods have also beenused to study the spectra of chromodynamic fluctuations of an unstableplasma [6] which, in particular, are responsible for the collisions integrals oftransport equations [7].Throughout the paper we use the natural system of units with c = (cid:126) = k B = 1 and the signature of our metric tensor is (+ , − , − , − ).
2. General formula
We consider a classical parton which moves across a quark-gluon plasma.Its motion is described by the Wong equations [8] dx µ ( τ ) dτ = u µ ( τ ) , (1) dp µ ( τ ) dτ = gQ a ( τ ) F µνa (cid:0) x ( τ ) (cid:1) u ν ( τ ) , (2) dQ a ( τ ) dτ = − gf abc u µ ( τ ) A µb (cid:0) x ( τ ) (cid:1) Q c ( τ ) , (3)where τ , x µ ( τ ), u µ ( τ ) and p µ ( τ ) are, respectively, the parton’s proper time,its trajectory, four-velocity and four-momentum; F µνa and A µa denote thechromodynamic field strength tensor and four-potential along the parton’strajectory and Q a is the classical color charge of the parton; g is the couplingconstant and α s ≡ g / π is assumed to be small. We also assume that thepotential vanishes along the parton’s trajectory i.e. our gauge conditionis u µ ( τ ) A µa (cid:0) x ( τ ) (cid:1) = 0. Consequently, due to Eq. (3) the classical parton’scharge Q c ( τ ) is constant within the chosen gauge.The energy loss is given directly by Eq. (2) with µ = 0. Using the time t = γτ instead of the proper time τ and replacing the strength tensor F µνa -loss-Wroclaw printed on November 7, 2018 by the chromoelectric E a ( t, r ) and chromomagnetic B a ( t, r ) fields, Eq. (2)gives dE ( t ) dt = gQ a E a ( t, r ( t )) · v , (4)where v is the parton’s velocity. Since we consider a parton which is veryenergetic, v is assumed to be constant and v = 1. Introducing the currentgenerated by the parton j a ( t, r ) = gQ a v δ (3) ( r − v t ), Eq. (4) gives dE ( t ) dt = (cid:90) d r E a ( t, r ) · j a ( t, r ) . (5)Since we deal with an initial value problem, we apply to the field andcurrent not the usual Fourier transformation but the so-called one-sidedFourier transformation defined as f ( ω, k ) = (cid:90) ∞ dt (cid:90) d re i ( ωt − k · r ) f ( t, r ) , (6)with the inverse f ( t, r ) = (cid:90) ∞ + iσ −∞ + iσ dω π (cid:90) d k (2 π ) e − i ( ωt − k · r ) f ( ω, k ) , (7)where the real parameter σ > ω is taken along a straight line in the complex ω − plane, parallel to thereal axis, above all singularities of f ( ω, k ). Using Eqs. (6) and (7), Eq. (5)can be rewritten: dE ( t ) dt = gQ a (cid:90) ∞ + iσ −∞ + iσ dω π (cid:90) d k (2 π ) e − i ( ω − ¯ ω ) t E a ( ω, k ) · v , (8)where ¯ ω ≡ k · v .The next step is to compute the chromoelectric field E a . Applying theone-sided Fourier transformation to the linearized Yang-Mills equations weget the set of equations familiar from electrodynamics ik i ε ij ( ω, k ) E ja ( ω, k ) = ρ a ( ω, k ) , ik i B ia ( ω, k ) = 0 ,i(cid:15) ijk k j E ka ( ω, k ) = iωB ia ( ω, k ) + B i a ( k ) , (9) i(cid:15) ijk k j B ka ( ω, k ) = j ia ( ω, k ) − iωε ij ( ω, k ) E ja ( ω, k ) − D i a ( k ) , where ρ a is the color-charge density, the fields with the index 0 are ini-tial values; the chromoelectric induction D a is expressed as D ia ( ω, k ) = e-loss-Wroclaw printed on November 7, 2018 ε ij ( ω, k ) E ja ( ω, k ) with ε ij ( ω, k ) being chromodielectric tensor which carriesall information about the medium. For an anisotropic plasma it equals ε ij ( ω, k ) = δ ij + g ω (cid:90) d p (2 π ) v i ω − k · v + i + ∂f ( p ) ∂p k (cid:104)(cid:16) − k · v ω (cid:17) δ kj + k k v j ω (cid:105) , where f ( p ) is the momentum distribution of plasma constituents. The colorindices a, b are dropped as ε ( ω, k ) is a unit matrix in color space.Although equations (9) strongly resemble those of electrodynamics, thegluon contribution to the color charge density ρ a and color current j a , whichis a genuine nonAbelian effect, is fully incorporated in these equations.Using Eq. (9), the field E ia ( ω, k ) is found to be E ia ( ω, k ) = − i (Σ − ) ij ( ω, k ) (cid:104) ωj ja ( ω, k ) + (cid:15) jkl k k B l a ( k ) − ωD j a ( k ) (cid:105) , (10)where Σ ij ( ω, k ) ≡ − k δ ij + k i k j + ω ε ij ( ω, k ) . (11)Substituting the expression (10) into Eq. (8), we get the final formula dE ( t ) dt = gQ a v i (cid:90) ∞ + iσ −∞ + iσ dω πi (cid:90) d k (2 π ) e − i ( ω − ¯ ω ) t (Σ − ) ij ( ω, k ) (12) × (cid:104) iωgQ a v j ω − ¯ ω + (cid:15) jkl k k B l a ( k ) − ωD j a ( k ) (cid:105) . The integral over ω is controlled by the poles of the matrix Σ − ( ω, k ) whichdetermine the collective modes in the system. Equivalently, these modes arefound as solutions of the equation det[Σ( ω, k )] = 0. Equation (12) needs tobe treated differently for stable and for unstable systems.
3. Stable systems
When the plasma is stable, all modes are damped and the poles ofΣ − ( ω, k ) are located in the lower half-plane of complex ω . Consequently,the contributions to the energy loss corresponding to the poles of Σ − ( ω, k )exponentially decay in time. The only stationary contribution is given bythe pole ω = ¯ ω ≡ k · v which comes from the current j a ( ω, k ). Therefore,the terms in Eq. (12), which depend on the initial values of the fields, canbe neglected, and the energy loss of a fast parton in a stable plasma is dE ( t ) dt = − ig C R v i v j (cid:90) d k (2 π ) ¯ ω (Σ − ) ij (¯ ω, k ) , (13) -loss-Wroclaw printed on November 7, 2018 where the bar indicates that averaging over parton’s color state has beenperformed, and the factor C R is 4/3 for a quark and 3 for a gluon.When the plasma is isotropic, the dielectric tensor can be expressed in astandard way through its longitudinal (cid:0) ε L ( ω, k ) (cid:1) and transverse (cid:0) ε T ( ω, k ) (cid:1) components and the matrix Σ ij ( ω, k ) (11) can be easily inverted. The energyloss (13) then equals dEdt = − ig C R (cid:90) d k (2 π ) ¯ ω k (cid:20) ε L (¯ ω, k ) + k v − ¯ ω ¯ ω ε T (¯ ω, k ) − k (cid:21) , (14)which corresponds to the standard energy loss due to soft collisions [2].
4. Unstable systems
When the plasma is unstable, the matrix Σ − ( ω, k ) contains poles inthe upper half-plane of complex ω , and the contributions to the energy lossfrom these poles grow exponentially in time. After a sufficiently long timethe parton’s energy loss will be dominated by the fastest unstable mode.For an unstable plasma, the terms in Eq. (12), which depend on the initialvalues of fields D and B , cannot be neglected, as these terms are amplifiedby a factor exponentially growing in time. Using Eq. (9) the initial valuescan be computed as D i a ( k ) = − igQ a ¯ ω ε ij (¯ ω, k )(Σ − ) jk (¯ ω, k ) v k , (15) B i a ( k ) = (cid:90) ∞−∞ dω π ω (cid:0) k × E a ( ω, k ) (cid:1) i = − igQ a (cid:15) ijk k j (Σ − ) kl (¯ ω, k ) v l . (16)Substituting Eqs. (15) and (16) into Eq. (12) and averaging over the par-ton’s color as before we obtain: dE ( t ) dt = g C R v i v l (cid:90) ∞ + iσ −∞ + iσ dω π (cid:90) d k (2 π ) e − i ( ω − ¯ ω ) t (Σ − ) ij ( ω, k ) (17) × (cid:104) ωδ jl ω − ¯ ω − ( k j k k − k δ jk )(Σ − ) kl (¯ ω, k ) + ω ¯ ω ε jk (¯ ω, k )(Σ − ) kl (¯ ω, k ) (cid:105) . Equation (17) gives the energy loss of a fast parton flying across the unstableplasma. The key point is that the presence of an unstable mode will producea time dependent exponential growth of the form e Im ω ( k ) t .
5. Two-stream system
In order to calculate the energy loss, one must invert the matrix Σ ij ( ω, k )defined by Eq. (11) to substitute the resulting expression into Eq. (17). For e-loss-Wroclaw printed on November 7, 2018 a general anisotropic system this is a tedious calculation, and therefore weconsider for simplicity the example of the two-stream system which has un-stable longitudinal electric modes. We assume that the chromodynamic fieldis dominated by the longitudinal chromoelectric field and take B ( ω, k ) = 0and E ( ω, k ) = k (cid:0) k · E ( ω, k ) (cid:1) / k . Then, Σ ij ( ω, k ) is trivially inverted as(Σ − ) ij ( ω, k ) = 1 ω ε L ( ω, k ) k i k j k , ε L ( ω, k ) ≡ ε ij ( ω, k ) k i k j k , (18)and Eq. (17) simplifies to dE ( t ) dt = g C R (cid:90) ∞ + iσ −∞ + iσ dω π (cid:90) d k (2 π ) e − i ( ω − ¯ ω ) t ω ε L ( ω, k ) ¯ ω k (cid:104) ωω − ¯ ω + ¯ ωω (cid:105) . (19)Eq. (19) gives a non-zero energy loss in the vacuum limit when ε L → /ε L by 1 /ε L − ε L ( ω, k ). With the distribution function ofthe two-stream system in the form f ( p ) = (2 π ) n (cid:104) δ (3) ( p − q ) + δ (3) ( p + q ) (cid:105) , (20)where n is the effective parton density in a single stream, one finds [6] ε L ( ω, k ) = (cid:0) ω − ω + ( k ) (cid:1)(cid:0) ω + ω + ( k ) (cid:1)(cid:0) ω − ω − ( k ) (cid:1)(cid:0) ω + ω − ( k ) (cid:1)(cid:0) ω − ( k · u ) (cid:1) , (21)where u ≡ q /E q is the stream velocity, µ ≡ g n/ E q is a parameteranalogous to the Debye mass squared, and ± ω ± ( k ) are the four roots of thedispersion equation ε L ( ω, k ) = 0 which are ω ± ( k ) = 1 k (cid:104) k ( k · u ) + µ (cid:0) k − ( k · u ) (cid:1) (22) ± µ (cid:113)(cid:0) k − ( k · u ) (cid:1)(cid:0) k ( k · u ) + µ (cid:0) k − ( k · u ) (cid:1)(cid:1) (cid:105) . It is easy to see that 0 < ω + ( k ) ∈ R for any k . For k ( k · u ) ≥ µ (cid:0) k − ( k · u ) (cid:1) , the minus mode is also stable, 0 < ω − ( k ) ∈ R , but for k · u (cid:54) = 0and k ( k · u ) < µ (cid:0) k − ( k · u ) (cid:1) one finds that ω − ( k ) is imaginary whichis the well-known two-stream electric instability.Strictly speaking, the stream velocity u given by the distribution func-tion (20) equals the speed of light. However, the distribution (20) should -loss-Wroclaw printed on November 7, 2018 Μ t10203040506070 (cid:72) g Μ (cid:76) (cid:45) dEdt Fig. 1. The energy loss per unit length as a function of time for a parton flyingalong the direction of streams. The lowest (green) curve corresponds to k max = 20,the middle (red) one to k max = 50 and the most upper (blue) curve to k max = 100. be treated as an idealization of a two-bump distribution with bumps of fi-nite width. Then, the momenta of all partons are not exactly parallel orantiparallel and the velocity u , which enters Eqs. (21, 22), obeys u ≤ ω can be computed analytically as it deter-mined by the six poles of the integrand located at ω = ± ω + ( k ) , ± ω − ( k ) , ¯ ω and 0. The remaining integral over k must be done numerically.Performing the calculations we have redefined all dimensional quantitiesby multiplying them by the appropriate power of µ to obtain dimensionlessvariables. We have chosen the following values of the parameters: g =1, | v | = 1, | u | = 0 . C R = 3. To take the integral over k , cylindricalcoordinates with the axis z along the stream velocity u have been used.Since the integral appears to be divergent, it has been taken over a finitedomain such that − k max ≤ k L ≤ k max and 0 ≤ k T ≤ k max . In Fig. 1 weshow the parton’s energy loss per unit length as a function of time for aparton flying along the streams for k max = 20 , , k max : the magnitude of energy loss growsand the period of oscillations shrinks when k max increases. This ultravioletsensitivity of our results is not very surprising, as our approach is fullyclassical. In the case of equilibrium (stable) plasma, the energy loss dueto soft interactions diverges logarithmically with k max [1]. The divergence e-loss-Wroclaw printed on November 7, 2018 signals a necessity to combine the classical contribution to the energy lossat small wave vectors with the quantum contribution at higher ones. Aquantum approach to the parton energy loss in unstable plasma needs tobe developed.Although the two-stream system is not directly relevant to the quark-gluon plasma which is produced in relativistic heavy-ion collisions, let uscomment on the potential interest of our study. Choosing µ = 200 MeV,which is a rough estimate of Debye mass in the QGP from nuclear collisions,the units on the horizontal and vertical axis in Fig. 1 are, respectively, fmand 200 MeV/fm. Since the parton’s energy loss in the QGP observed inrelativistic heavy-ion collisions is of order 1 GeV/fm, the magnitude of theenergy loss discussed here is certainly of phenomenological interest.
6. Conclusions
We have developed a formalism where the energy loss of a fast partonin a plasma medium is found as the solution of initial value problem. Theformalism allows one to obtain the energy loss in unstable plasma wheresome modes exponentially grow in time. In the case of stable plasma, onereproduces correctly the standard energy-loss formula. As an example ofan unstable system we have studied a two-stream system. The energy lossper unit length is not constant, as in an equilibrium plasma, but it exhibitsstrong time dependence.
Acknowledgments
This work was partially supported by Polish Ministry of Science andHigher Education under grants N N202 204638 and 667/N-CERN/2010/0.REFERENCES [1] S. Peigne and A. V. Smilga, Phys. Usp. , 659 (2009).[2] M. Le Bellac, Thermal Field Theory (Cambridge University Press, Cambridge,2000).[3] P. Romatschke and M. Strickland, Phys. Rev. D , 125008 (2005).[4] St. Mr´owczy´nski, Acta Phys. Polon. B , 427 (2006).[5] A. Majumder, B. M¨uller and St. Mr´owczy´nski, Phys. Rev. D , 125020(2009).[6] St. Mr´owczy´nski, Phys. Rev. D , 105022 (2008).[7] St. Mr´owczy´nski and B. M¨uller, Phys. Rev. D , 065021 (2010).[8] S. K. Wong, Nuovo Cim. A65