Hi Jongping,
essentially (see attachment),
Subscript[F,
0](0,0,p^2,0,0,p^2,0,p^2,p^2,0,p^2,p^2,p^2,p^2,p^2,0,0,0,0,0,0)==((D-10)
(D-8) (D-5) (D-3) Subscript[B, 0](p^2,0,0))/(4 p^8)
Your use case is actually quite interesting. Essentially, one needs
something like an inverse of ToPaVe in order to convert scalar
Passarino-Veltman functions back into FAD-integrals, so that one can
apply IBP-reduction. I will think of an implementation for that.
Cheers,
Vladyslav
Am 21.12.2016 um 02:27 schrieb Jongping Hsu:
> Hi,Vladyslav,
> What is the relation between
> p^4 F_0(0,0,p^2,0,0,p^2,0,p^2,p^2,0,p^2,p^2,p^2,p^2,p^2,0,0,0,0,0,0)and
> B_0(p^2,0,0)?
> Thanks. JP
>
> HSU Jongping,
> Chancellor Professor
> Department of Physics
> Univ. of Massachusetts Dartmouth,
> North Dartmouth, MA 02747. FAX (508)999-9115
> http://www.umassd.edu/engineering/phy/people/facultyandstaff/jong-pinghsu/
> recent monograph: Space-Time Symmetry and Quantum Yang–Mills Gravity
> (https://sites.google.com/site/yangmillsgravity123/)
>
> ------------------------------------------------------------------------
> *From: *"Vladyslav Shtabovenko" <noreply@feyncalc.org>
> *To: *feyncalc@feyncalc.org
> *Sent: *Monday, December 19, 2016 10:05:46 AM
> *Subject: *Re: Possible bug in OneLoop
>
> 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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