Hi,
could you provide a 2-loop example where it takes too much time?
For my own purposes I used to employ FVSeries
exp = -(CA^3*
FAD[{p1, I*mE}, {p2, I*mE}, {p1 - q, I*mE}, {p1 - q,
I*mE}]*(FVD[p1, Lor1] + FVD[p1 - q, Lor1])*(FVD[p1, Lor2] +
FVD[p1 - q, Lor2])*(2*omega[21] - 5*psi[19])*SD[a, b]*
SMP["g_s"]^6*(SPD[p2, p2] + SPD[p1 - q, p1 - q]))
ClearAll[FVSeries];
Options[FVSeries] = {List -> False, Dimension -> D};
FVSeries[expr_, {q_, q0_, n_}, OptionsPattern[]] :=
Block[{fvs, ex, res, dim}, dim = OptionValue[Dimension];
ex = FCI[expr];
res = Join[{EpsEvaluate[(ex /. q -> q0)]},
Table[fvs =
Table[FCI[
Pair[Momentum[q, dim], LorentzIndex[Unique[], dim]]], {j, 1,
i}]; ((1/i!) FourDivergence[ex, Sequence @@ fvs,
Contract -> False]) // ReplaceAll[#, q -> q0] & //
Contract[# Times @@ fvs] &, {i, 1, n}]];
If[! OptionValue[List], res = Total@res];
res]
The following requires around 12 second on my machine with FeynCalc 9.3
exp = FVSeries[exp, {q, 0, 6}]; // AbsoluteTiming
I guess one could also make it a bit faster, but I'm not sure how long
time does it require in your 2-loop calculation.
Cheers,
Vladyslav
Am 06.06.2017 um 16:30 schrieb Philipp:
> Hi,
>
> I wanted to expand the following expression in terms of the external momentum q or (q/mE)
>
> -(CA^3*FAD[{p1, I*mE}, {p2, I*mE}, {p1 - q, I*mE}, {p1 - q, I*mE}]*(FVD[p1, Lor1] + FVD[p1 - q, Lor1])*(FVD[p1, Lor2] + FVD[p1 - q, Lor2])*(2*omega[21] - 5*psi[19])*SD[a, b]*SMP["g_s"]^6*(SPD[p2, p2] + SPD[p1 - q, p1 - q]))
>
> Up to now I have done this as a series by using FourDivergence[] including first and second order derivatives. Is there a faster way to do that, as at two loop level this becomes computationally expansive very rapidly.
>
> Cheers,
> Philipp
>
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