Name: Nikita Belyaev (email_not_shown)
Date: 03/13/15-12:57:01 AM Z


Good night,

Some time ago I've noticed a problem with the imaginary parts of the traces. You can see the example below:

//----------------------------------------------------------------\\
Clear["Global`*"];
(*<<C:\Users\1\AppData\Roaming\Mathematica\Applications\FeynArts-3.9\FeynArts39.m*)
<<HighEnergyPhysics`FeynCalc`
(*{Date[],$Version,$FeynCalcVersion};*)
Needs["FeynCalcFormLink`"]
$LeviCivitaSign = -1;

ScalarProduct[p,p] = m^2;
ScalarProduct[p1,p1] = m^2;
ScalarProduct[p2,p2] = m^2;
ScalarProduct[k1,k1] = 0;
ScalarProduct[k2,k2] = 0;
ScalarProduct[q,q] = u^2;
ScalarProduct[q,s] = 0;

Line1:= 4*m*m*u*(GA[\[Beta]].GS[p2].GS[k1].GS[p1]-m*m*GA[\[Beta]].GS[k1]);
Line2:= (-GS[q].GS[s].GA[\[Beta]]+GS[s].GA[\[Beta]].(GS[q]-GS[p1]-GS[p2])+GA[\[Beta]].GS[q].GS[s]+(GS[q]-GS[p1]-GS[p2]).GA[\[Beta]].GS[s]).GS[k2].(1-GA[5]);

Tr1= DiracTrace[Line1];
Tr2= DiracTrace[Line2];

TrA1A2mod4=FeynCalcFormLink[4*Tr1.Tr2];

Print["\!\(\*FractionBox[\(A1A2mod4\), \(1024\)]\)"]
FullSimplify[TrA1A2mod4/1024]
//----------------------------------------------------------------\\

The result of such calculation is:
A1A2mod4/1024 = 1/8 m^2 u (k1\[CenterDot]s (4 (k2\[CenterDot]p1+k2\[CenterDot]p2) (m^2+p1\[CenterDot]p2)-I \[Epsilon]^(k2p1p2q))+k1\[CenterDot]q (-8 k2\[CenterDot]s (m^2+p1\[CenterDot]p2)-I \[Epsilon]^(k2p1p2s))+4 p1\[CenterDot]s (k1\[CenterDot]p1 k2\[CenterDot]p2-m^2 k1\[CenterDot]k2)-4 m^2 k1\[CenterDot]k2 p2\[CenterDot]s+I k2\[CenterDot]q \[Epsilon]^(k1p1p2s)+I k2\[CenterDot]s \[Epsilon]^(k1p1p2q)-I p1\[CenterDot]q \[Epsilon]^(k1k2p2s)-I p1\[CenterDot]s \[Epsilon]^(k1k2p2q)+I p2\[CenterDot]q \[Epsilon]^(k1k2p1s)+I p2\[CenterDot]s \[Epsilon]^(k1k2p1q)+8 k1\[CenterDot]p2 k2\[CenterDot]s p1\[CenterDot]q+8 k1\[CenterDot]p1 k2\[CenterDot]s p2\[CenterDot]q-4 k1\[CenterDot]p2 k2\[CenterDot]p2 p1\[CenterDot]s-4 k1\[CenterDot]k2 p1\[CenterDot]p2 p1\[CenterDot]s-4 k1\[CenterDot]p1 k2\[CenterDot]p1 p2\[CenterDot]s+4 k1\[CenterDot]p2 k2\[CenterDot]p1 p2\[CenterDot]s-4 k1\[CenterDot]k2 p1\[CenterDot]p2 p2\[CenterDot]s)

It contains imaginary parts proportional to symmetric tensors like (s_{a}q_{b}+s_{b}q_{a}) and so on. So the imaginary parts of these traces is non-zero.

But if we replace "Line2:= (-GS[q].GS[s]..." with "Line2:= (GS[s].GS[q]..." (GS[q].GS[s]+GS[s].GS[q]=0, because ScalarProduct[q,s] is also 0) we'll get the real result without any imaginary parts:
A1A2mod4/1024 = 1/2 m^2 u (-2 k1\[CenterDot]q k2\[CenterDot]s (m^2+p1\[CenterDot]p2)+k1\[CenterDot]s (k2\[CenterDot]p1+k2\[CenterDot]p2) (m^2+p1\[CenterDot]p2)+p1\[CenterDot]s (k1\[CenterDot]p1 k2\[CenterDot]p2-m^2 k1\[CenterDot]k2)-p2\[CenterDot]s (k1\[CenterDot]k2 (m^2+p1\[CenterDot]p2)+k2\[CenterDot]p1 (k1\[CenterDot]p1-k1\[CenterDot]p2))+2 k1\[CenterDot]p2 k2\[CenterDot]s p1\[CenterDot]q+2 k1\[CenterDot]p1 k2\[CenterDot]s p2\[CenterDot]q-k1\[CenterDot]p2 k2\[CenterDot]p2 p1\[CenterDot]s-k1\[CenterDot]k2 p1\[CenterDot]p2 p1\[CenterDot]s)

So couldn't you please explain why the result is sensitive to mathematically equal transformations and how I can solve this problem?

I've tried to calculate it in Mathematica 6 and 8 with various FeynCalc versions and the result was always the same.

Best Regards,
Nikita Belyaev



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