Changes between Version 1 and Version 2 of DeadCone
 Timestamp:
 04/06/12 16:33:02 (8 years ago)
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DeadCone
v1 v2 1 1 2 2 3 == QCD Radiation from heavy quarks: %$e^+ e^ \to Q \bar Q g$%==3 == QCD Radiation from heavy quarks: $e^+ e^ \to Q \bar Q g$ == 4 4 5 5 ==== 1. ==== … … 7 7 Compute the differential cross section for the case of massive final state using a program for symbolic calculations (such as Mathematica+FeynCalc) and compare your result with 8 8 9 %$ \frac{1}{\sigma^{LO}} \frac{d^2\sigma}{dx_1dx_2}= \frac{1}{\beta} C_F \frac{\alpha_S}{2\pi} \left[ \frac{2(x_1+x_21\rho/2)}{(1x_1)(1x_2)} \frac{\rho}{2} \left( \frac{1}{(1x_1)^2}+ \frac{1}{(1x_2)^2}\right) \left. + \frac{1}{1+\rho/2} \frac{(1x_1)^2+(1x_2)^2}{(1x_1)(1x_2)}\right]\, P, \right. $ %(1)9 %$ \frac{1}{\sigma^{LO}} \frac{d^2\sigma}{dx_1dx_2}= \frac{1}{\beta} C_F \frac{\alpha_S}{2\pi} \left[ \frac{2(x_1+x_21\rho/2)}{(1x_1)(1x_2)} \frac{\rho}{2} \left( \frac{1}{(1x_1)^2}+ \frac{1}{(1x_2)^2}\right) \left. + \frac{1}{1+\rho/2} \frac{(1x_1)^2+(1x_2)^2}{(1x_1)(1x_2)}\right]\, P, \right. $ (1) 10 10 11 11 where 12 12 13 %$ \rho=\frac{4 m^2}{s}\le 1\,,\qquad \beta=\sqrt{1\rho} $ %,13 %$ \rho=\frac{4 m^2}{s}\le 1\,,\qquad \beta=\sqrt{1\rho} $, 14 14 15 15 and 16 16 17 %$\sigma^{LO}= N_c (\sum_{f} Q_f^2) 4 \pi \alpha^2/(3s) $ %.17 %$\sigma^{LO}= N_c (\sum_{f} Q_f^2) 4 \pi \alpha^2/(3s) $. 18 18 ==== 2. ==== 19 19 … … 28 28 29 29 30 where %$z=2 E_g/\sqrt{s}$% is the energy fraction of the gluon and %$\theta$%the angle between the gluon and the quark. Plot the behaviour of the matrix element in the massless and massive cases and compare with the Fig.1 below. Explain this behaviour in terms of angular momentum conservation.30 where $z=2 E_g/\sqrt{s}$ is the energy fraction of the gluon and $\theta$ the angle between the gluon and the quark. Plot the behaviour of the matrix element in the massless and massive cases and compare with the Fig.1 below. Explain this behaviour in terms of angular momentum conservation. 31 31 32 32 ==== 4. ==== … … 50 50 51 51 52