Name: J.P. Hsu (email_not_shown)
Date: 11/16/11-04:54:10 PM Z


Hi, Dr. Mertig:
     Thank you very much. Yes, It works. JP

On 11/15/11 5:55 PM, Rolf Mertig wrote:
> (* Using FeynCalc 8.0.3 with Mathematica 8 *)
> (* the program below should print:
>
> ans = (4*C*(FV[k, mu]*FV[k, nu] - MT[mu, nu]*SP[k, k])*(6*m^2 - 6*A0[m^2] - SP[k, k] + 3*B0[SP[k, k], m^2, m^2]*(2*m^2 + SP[k, k])))/(9*SP[k, k])
>
> and
>
> res3 = (2*C*(-(FV[k, mu]*FV[k, nu]) + MT[mu, nu]*SP[k, k])*(2*(-2 + D)*A0[m^2] - B0[SP[k, k], m^2, m^2]*(4*m^2 + (-2 + D)*SP[k, k])))/((-1 + D)*SP[k, k])
>
>
> *)
>
> Needs["HighEnergyPhysics`FeynCalc`"]
> dm = GAD; ds = GS; mt = MT; fv = FV;
> (*Now write the numerator of the Feynman diagram.We define the \
> constant C=alpha/(4 pi)*)
> num = -C Tr[dm[mu].(ds[q] + m).dm[nu].(ds[q] + ds[k] + m)] ;
> (* By default FeynCalc to evaluates the integral in D dimensions *)
> (* By default it also takes the limit D -> 4 *)
> $LimitTo4 = True;
>
> (*Define the amplitude*)
> amp = num*FAD[{q, m}, {q + k, m}];
> (*Calculate the result*)
> res = (-I/Pi^2) OneLoop[q, amp] // PaVeReduce;
> ans = Simplify[res // FCE];
> Print["ans = ", ans // InputForm];
> (* this is more general (no limit is taken): *)
> res2 = OneLoopSimplify[amp, q] // Factor1
> $LimitTo4 = False;
> res3 = (-I/Pi^2) OneLoop[q, res2] // FCE // Simplify
> Print["res3 = ", res3 // InputForm]
>

-- 
HSU Jongping,
Chancellor Professor and Chair
Department of Physics
Univ. of Massachusetts Dartmouth,
North Dartmouth, MA 02747.  FAX (508)999-9115
  http://www.umassd.edu/engineering/phy/people/faculty/jhsu/



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