Changes between Version 1 and Version 2 of DeadCone


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Timestamp:
04/06/12 16:33:02 (7 years ago)
Author:
trac
Comment:

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  • DeadCone

    v1 v2  
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    3 == QCD Radiation from heavy quarks: %$e^+ e^- \to Q \bar Q g$% ==
     3== QCD Radiation from heavy quarks: $e^+ e^- \to Q \bar Q g$ ==
    44
    55==== 1. ====
     
    77Compute 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
    88
    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_2-1-\rho/2)}{(1-x_1)(1-x_2)} -\frac{\rho}{2} \left( \frac{1}{(1-x_1)^2}+ \frac{1}{(1-x_2)^2}\right) \left. + \frac{1}{1+\rho/2} \frac{(1-x_1)^2+(1-x_2)^2}{(1-x_1)(1-x_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_2-1-\rho/2)}{(1-x_1)(1-x_2)} -\frac{\rho}{2} \left( \frac{1}{(1-x_1)^2}+ \frac{1}{(1-x_2)^2}\right) \left. + \frac{1}{1+\rho/2} \frac{(1-x_1)^2+(1-x_2)^2}{(1-x_1)(1-x_2)}\right]\, P, \right. $ (1)
    1010
    1111where
    1212
    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} $,
    1414
    1515and
    1616
    17 %$\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) $.
    1818==== 2. ====
    1919
     
    2828
    2929
    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.
     30where $z=2 E_g/\sqrt{s}$ 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.
    3131
    3232==== 4. ====
     
    5050
    5151
     52