**Next message:**Vladyslav Shtabovenko: "Re: Polarization sums with dummy indices"**Previous message:**Kyrylo Bondarenko: "Re: Polarization sums with dummy indices"**Maybe in reply to:**Ben: "Polarization sums with dummy indices"**Next in thread:**Vladyslav Shtabovenko: "Re: Polarization sums with dummy indices"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Mail actions:**[ respond to this message ] [ mail a new topic ]

Hi,

first of all, you'll obtain the same result if you do this calculation

with pen and paper.

*>From the contraction of the two epsilon tensors you get
*

2 g^{c,f} g^{d,e} - 2 g^{c,e} g^{d,f}

Now in the first term two polarization vectors get contracted with each

other:

eps(w,la)^d eps*(w,la)^*e g^{d,e} = -1

which is the standard normalization. Hence, in the first term there are

no polarization vectors left.

Now, from the physical point of view, this "discrepancy" is not

surprising, as the polarization sum is by definition a gauge dependent

quantity, i.e. it is not physical. On the other hand, the matrix element

squared, where this sum enters, is a physical quantity and there it is

guaranteed that the gauge dependent terms involving the auxiliary vector

n^mu will cancel.

So I would suggest that you consider the full matrix element from which

you get your bla expression and check there if this difference between

bla and Contract[bla] changes the physical result, which it shouldn't.

P.S. By the way, one can see nicely see how the gauge dependent terms

cancel out in the gg->gg example (QCDGGToGGTree.m) included in the

development version of FeynCalc.

If you replace

ClearAll[pre];

Table[Print[" calculating product of the amplitudes ", i, " and ",

j," (CC), time = ", Timing[pre[i,j]=re[i,j]//polsums[#,k1,k2,

1/2]&//polsums[#,k2,k1,1/2]&//polsums[#,k3,k4,1]&//polsums[#,k4,k3,1]&//Simplify][[1]]];pre[i,j],{i,4},{j,i}];

with

ClearAll[pre];

Table[Print[" calculating product of the amplitudes ", i, " and ",

j," (CC), time = ", Timing[pre[i,j]=re[i,j]//polsums[#,k1,4k2+k3,

1/2]&//polsums[#,k2,6k1+5k4,1/2]&//polsums[#,k3,29k4,1]&//polsums[#,k4,5k3,1]&//Simplify][[1]]];pre[i,j],{i,4},{j,i}];

where the auxiliary vectors are now quite "ugly", then the evaluation

takes slightly longer, but the result (after applying TrickMandelstam)

remains the same, because the gauge dependent terms *must* cancel out in

all physical quantities.

Cheers,

Vladyslav

Am 28.02.2015 um 12:07 schrieb Kyrylo Bondarenko:

*> Another thing about DoPolarizationSums. Let
*

*> bla = LC[a, b, c, d] FV[w, c] PolarizationVector[w, d] LC[a, b, e, f] FV[w, e] Conjugate[PolarizationVector[w, f]]
*

*> tmp = Contract[bla]
*

*>
*

*> Then DoPolarizationSums[bla, w, n] gives
*

*> 6w^2 (as I expected)
*

*> but DoPolarizationSums[tmp, w, n] gives
*

*> 4w^2 + 2(w^2 - n^2w^2/(nw)^2)
*

*> which is very different answer. For example, for n=w the last answer is 4w^2.
*

*>
*

**Next message:**Vladyslav Shtabovenko: "Re: Polarization sums with dummy indices"**Previous message:**Kyrylo Bondarenko: "Re: Polarization sums with dummy indices"**Maybe in reply to:**Ben: "Polarization sums with dummy indices"**Next in thread:**Vladyslav Shtabovenko: "Re: Polarization sums with dummy indices"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Mail actions:**[ respond to this message ] [ mail a new topic ]

*
This archive was generated by hypermail 2b29
: 02/22/18-07:20:01 AM Z CET
*