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Two methods of solving QCD evolution equation Aleksander Kusina, Magdalena Sławińska
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2 Multiple gluon emission from a parton participating in a hard scattering process. The parton with hadron’s momentum fraction x 0 emits gluons. After each emission its momentum decreases: x 0 >x 1 >... > x n-1 > x n The evolution is described by momentum distribution function of partons D(x, t). t denotes a scale of a process. t = lnQ
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3 Evolution presented in a (t, x) diagram. The change of momentum distribution function D(x, t) is presented on a diagram by lines incoming and outgoing from a (t, x) cell.
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4 Evolution equation for gluons From many possible processes we consider only those involving one type of partons (gluons). The evolution equation is then one-dimentional: where z denotes gluon fractional momenta kernel P(z, t) stands for branching probability density
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5 We use regularised kernel: where P represents outflow of momentum and P – inflow of momentum. We discuss simplified case of stationary P. Proper normalisation of D, namely: requires: leading to:
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6 Monte Carlo Method t0t0 t1t1... t n -1 tntn x0x0 x1x1 x n -1 xnxn t max We generate values of momenta and ”time” according to proper probability distribution for each point in the diagram. (x 0, t 0 )->(x 1, t 1 )->...->(x n-1, t n-1 ) We obtain an evolution of a single gluon. Each dot represents a single gluon emission. Repeating the process many times we obtain a distribution of the momentum x.
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7 Monte Carlo Method Iterative solution We introduce the following formfactor:
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8 By using substitution we transform the evolution equation to the integral form: and obtain the iterative solution:
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9 to obtain the markovian form of the iterative equation we define transition probability: Which is properly normalized to unity Applying this probability to the iterative solution we obtain the markovian form: Markovianisation of the equation
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10 Now we introduce the exact form of the kernel so that we can explicitly write the probability of markovian steps The transmission probability factorizes into two parts where
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11 Once more the final form of the evolution equation The Monte Carlo algorithm: 1. Generate pairs (t i, z i ) from distributions p(t) and p(z) 2. Calculate T i = t 1 + t 2 +... + t i,x i = z 1 z 2... z i 3. In each step check if T i > t max (t max – evolution time) 4. If T i > t max, take the pair (T i -1, x i -1 ) as a point of distribution function D(x, t max ) and EXIT 5. Repeat the procedure: GO TO POINT 1 MC algorithm
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12 Results Starting with delta – distribution, now we demonstrate, how the gluon momenta distribution changes during evolution t=2t=5
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13 t=10t=15 t=50t=25
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14 From the histograms we see the character of the evolution – momenta of gluons are softening and the distribution resembles delta function at x=0. Now we investigate how the evolution depends on coupling constant s :
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15 s =0.3 s =1
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16 Semi- analytical Method The model Problems: ● How to interpret probability P(z) ? ● Discrete calculations Solutions: ● Many particles in the system their distribution according to P(z) distribution ● Calculations performed on a grid ● evolution steps of size t ● momenta fractions N bins of width x ● k th bin represents momentum fraction (k + ½) x
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17 Since time steps and fractional momenta are descreet, so must be the equation The interpretation of P(z) within this model: In each evolution step particles move - from k to k-1, k-2,..., 0 - from N-1, N – 2,..., k + 1 to k where
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18 s =0.3
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19 This is to emphasise that both calculation methods and computational algorithms differ very much. In MC the history of a single particle is generated according to probability distributions and its final momentum is remembered. These operations are repeated for 10 8 events (histories) so that a full momenta distribution is obtained. In semi- analytical approach, a momenta distribution function is calculated by considering all 10 4 emiter particles. At each scale a number of particles changing position from (t, i) to (t+1, k) is calculated. All particles are then redistributed and a new momenta distribution is obtained. To compare the methods we divided corresponding histograms. Comparison of the methods
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20 As we can see from division of final distribution functions, both methods give the same distribution within 2%! T = 4T = 10 T = 18
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21 References: [1] R. Ellis, W. Stirling and B. Webber, QCD and Collider Physics (Cambridge University Press, 1996)
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