Dear both,
with the current stable version the issue does not apper anymore.
This
S1 = OneLoop[k,
FVD[k, \[Mu]] SPD[k, p] SPD[
k] FAD[{k + p - q, mw}, {k + p + q, mw}, {k, my}]];
S2 = OneLoop[k,
FVD[k, \[Mu]] SPD[k, p] SPD[
k] FAD[{k + p - q, mw}, {k + p + q, mw}, {k, mu}]];
NewM = S1 - S2;
Simplify[PaVeReduce[NewM /. mu -> my]]
gives zero, as well as
ampy = SP[k] SP[k, p] FourVector[k,
a] FAD[{k + p - q, mw}, {k + p + q, mw}, {k, my}] // FCI //
ChangeDimension[#, D] &
ty = PaVeReduce[OneLoop[k, ampy]];
ampy1 = FDS[ampy];
ty1 = PaVeReduce[OneLoop[k, ampy1]];
ty - ty1
and
T1 = OneLoop[k,
FVD[k, a] SPD[k, p] SPD[
k] FAD[{k, my}, {k + p - q, mw}, {k + p + q, mw}],
DenominatorOrder -> False];
T2 = OneLoop[k,
FVD[k, a] SPD[k, p] SPD[
k] FAD[{k, my}, {k + p - q, mw}, {k + p + q, mw}],
DenominatorOrder -> True];
Simplify[PaVeReduce[T1 - T2]]
I will nevertheless add your examples to our testsuite.
Cheers,
Vladyslav
> I don't have an answer to your question. Howver, I agree
> that there is a bug, or at least a terrible inconsistency, in how
> OneLoop handles this kind of UV divergent integral. This concerns me
> because I've been using FeynCalc to manipulate some integrals whose
> sum is finite, but with individual terms that are UV divergent.
>
> In fact, one does not have to rename the mass variable to reveal a problem.
> The answer depends on the order of the propagators.
>
> ==========================
>
> SetOptions[OneLoop,Prefactor->1/(I Pi^2)];
>
> First we define, as you did:
>
> ampy = SP[k] SP[k,p] FourVector[k,a] FAD[{k+p-q,mw},{k+p+q,mw},{k,my}]//FCI;
>
> k^2 k.p k[a]
> ampy = ----------------------------------------------------------
> ((k + p - q)^2 - mw^2) ((k + p + q)^2 - mw^2) (k^2 - my^2)
>
> The one loop integral is:
>
> ty = PaVeReduce[OneLoop[k,ampy]];
>
> Now instead put it in standard order using FeynAmpDenominatorSimplify:
>
> ampy1 = FDS[ampy];
>
> k^2 k.p k[a]
> ampy1 = ----------------------------------------------------------
> (k^2 - my^2) ((k + p - q)^2 - mw^2) ((k + p + q)^2 - mw^2)
>
> ty1 = PaVeReduce[OneLoop[k,ampy1]];
>
> ty is not equal to ty1. The difference, after simplifying B0[0,m0,m1] is
>
> -(2 mw^2 + 4 my^2 - p^2 - q^2 - 2 p.q) (p[a] + q[a])
> ty-ty1 = ----------------------------------------------------
> 48
>
> If we now change the mass my -> mu, and so define ampu = ampy/.my->mu,
> and ampu1 = ampy1/.my->mu, then the analogous tu and tu1 ARE in fact
> equal. However, as you discovered, tu does not equal ty
> (after replacing mu->my), with the difference being
>
> (3 mw^2 - 2 q^2) q[a]
> tu-ty = ---------------------
> 6
>
> Going back to the ampy amplitude, if we apply ScalarProductCancel we find
> (equivalent to regrouping k^2 -> k^2-my^2 + my^2 and cancelling propagators):
>
> SPC[ampy]
>
> k.p k[a]
> = --------------------------------------------- +
> ((k + p - q)^2 - mw^2) ((k + p + q)^2 - mw^2)
>
> my^2 k.p k[a]
> ----------------------------------------------------------
> ((k + p + q)^2 - mw^2) (k^2 - my^2) ((k + p - q)^2 - mw^2)
>
> Here the one loop integral gives the same result as the standard ordered
> result ty1. Interestingly, evaluating the one loop integral of SPC[ampu]
> also gives the SAME result as ty1 (after replacing mu->my). That this
> should happen is clear from the form of the amplitude above, since the
> first term is independent of my^2, and the second is explicitly
> multiplied by my^2. So I suspect that this form gives the "correct"
> answer (or perhaps the "preferred" answer).
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