# Changes between Version 3 and Version 4 of DeadCone

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Timestamp:
04/19/12 08:54:51 (8 years ago)
Comment:

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Unmodified
 v3 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 %$\frac{1}{\sigma^{LO}} \frac{d^2\sigma}{dx_1dx_2}= \frac{1}{\beta} C_F \frac{\alpha_S}{2\pi} \left[ \frac{2(x_1+x_2-1-\rho/2)}{(1-x_1)(1-x_2)} -\frac{\rho}{2} \left( \frac{1}{(1-x_1)^2}+ \frac{1}{(1-x_2)^2}\right) \left. + \frac{1}{1+\rho/2} \frac{(1-x_1)^2+(1-x_2)^2}{(1-x_1)(1-x_2)}\right]\, P, \right.$ (1) $\frac{1}{\sigma^{LO}} \frac{d^2\sigma}{dx_1dx_2}= \frac{1}{\beta} C_F \frac{\alpha_S}{2\pi} \left[ \frac{2(x_1+x_2-1-\rho/2)}{(1-x_1)(1-x_2)} -\frac{\rho}{2} \left( \frac{1}{(1-x_1)^2}+ \frac{1}{(1-x_2)^2}\right) \left. + \frac{1}{1+\rho/2} \frac{(1-x_1)^2+(1-x_2)^2}{(1-x_1)(1-x_2)}\right]\, P, \right.$ (1) where %$\rho=\frac{4 m^2}{s}\le 1\,,\qquad \beta=\sqrt{1-\rho}$, $\rho=\frac{4 m^2}{s}\le 1\,,\qquad \beta=\sqrt{1-\rho}$, and %$\sigma^{LO}= N_c (\sum_{f} Q_f^2) 4 \pi \alpha^2/(3s)$. $\sigma^{LO}= N_c (\sum_{f} Q_f^2) 4 \pi \alpha^2/(3s)$. ==== 2. ====