Changes between Version 3 and Version 4 of DeadCone
 Timestamp:
 04/19/12 08:54:51 (8 years ago)
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DeadCone
v3 v4 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