Changes between Version 3 and Version 4 of DeadCone


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Timestamp:
04/19/12 08:54:51 (7 years ago)
Author:
md987
Comment:

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  • DeadCone

    v3 v4  
    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