•Plots - neutral pions

Following 't Hooft and Veltman one easily obtains the following expression for C _ 0:

c0 = Integrate[1/(MandelstamU y) Log[ParticleMass[Pion, RenormalizationState[0]]^2/(MandelstamU y^2 + -MandelstamU y + ParticleMass[Pion, RenormalizationState[0]]^2)], {y, 0, 1}]

(Li _ 2((2 u^(1/2))/(u^(1/2) - (u - 4 (m _ π^(ó  ))^2)^(1/2))) + Li _ 2((2 u^(1/2))/(u^(1/2) + (u - 4 (m _ π^(ó  ))^2)^(1/2))))/u

Now, according to the rule (Bürgi): Li _ 2(y) + Li _ 2(-y/(1 - y)) = 1/2 ln^2(1 - y), we get finally for C _ 0:

c0 = 1/(2 s) Log[((s - 4 mPi^2)^(1/2) + s^(1/2))/((s - 4 mPi^2)^(1/2) - s^(1/2))]^2

log^2((s^(1/2) + (s - 4 mPi^2)^(1/2))/((s - 4 mPi^2)^(1/2) - s^(1/2)))/(2 s)

Apart from a factor 2i, the amplitude below is in agreement with Bijnens and Cornet  (bugfixed - multiplied with a factor m _ π^2):

MandelstamReduce[ampfinalpi0, OnMassShell -> True, Cancel -> MandelstamS, Masses -> {ParticleMass[Pion, RenormalizationState[0]], 0, ParticleMass[Pion, RenormalizationState[0]], 0}] /. {C0[__] -> c0, MandelstamU -> s, -p1 - p2 - p3 -> p4} // Simplify

((e^( ))^2 µ ( p _ 2 ) · µ^* ( p _ 4 ) (s - (m _ π^(ó  ))^2) (log^2 ((s^(1/2) + (s - 4 mPi^2)^(1/2))/((s - 4 mPi^2)^(1/2) - s^(1/2))) (m _ π^(ó  ))^2 + s))/(8 π^2 s (f _ π^(ó  ))^2)

We can now insert some values for the contants and plot the cross section:

valueRules = {ParticleMass[__] -> 137 10^6, DecayConstant[__] -> 134 10^6, mPi -> 137 10^6, mPi -> 137 10^6, fPi -> 134 10^6, α -> 1/137.036} ;

This is then the amplitude to be plotted:

ampfinalpi00 = MandelstamReduce[ampfinalpi0, OnMassShell -> True, Cancel -> MandelstamS, Masses -> {ParticleMass[Pion, RenormalizationState[0]], 0, ParticleMass[Pion, RenormalizationState[0]], 0}] /. {C0[__] -> c0, CouplingConstant[QED[1], RenormalizationState[0]]^2 -> α 4 π, Pair[Momentum[Polarization[__], ___], Momentum[Polarization[__], ___]] -> 1} /. MandelstamU :> MandelstamS

-(α ((m _ π^(ó  ))^2 - s) ((log^2 ((s^(1/2) + (s - 4 mPi^2)^(1/2))/((s - 4 mPi^2)^(1/2) - s^(1/2))) (m _ π^(ó  ))^2)/s + 1))/(2 π (f _ π^(ó  ))^2)

β[s_] = (s - 4 mPi^2)/s^(1/2) ;

sigma00[ss_, z_] := 1/2 (2 π z)/(64 π^2 ss) β[ss] Abs[2 i ampfinalpi00]^2 /. {MandelstamS :> ss, s :> ss} /. valueRules

plot = Plot[0.38939 10^24 sigma00[e^2, 1.0], {e, (0.274 10^9), (0.7 10^9)}] ;

[Graphics:../HTMLFiles/index_44.gif]

0.38939 10^24 sigma00[(0.7 10^9)^2, 1.0]

21.693210517201024`

For comparison, here is Bijnens and Cornet's amplitude:

amp00Bij[s_] := 4 i α 4 π mPi^2 1/(16 π^2 fPi^2) (1 - s/mPi^2) (1 + (s/mPi^2)^(-1) Log[((s/mPi^2 - 4)^(1/2) + s/mPi^2^(1/2))/((s/mPi^2 - 4)^(1/2) - s/mPi^2^(1/2))]^2) ;

sigma00Bij[s_, z_] := 1/2 (2 π z)/(64 π^2 s) β[s] Abs[amp00Bij[s]]^2 /. valueRules

plotBij = Plot[0.38939 10^24 sigma00Bij[e^2, 1.0], {e, (0.274 10^9), (0.7 10^9)}] ;

[Graphics:../HTMLFiles/index_50.gif]

0.38939 10^24 sigma00Bij[(0.7 10^9)^2, 1.0]

21.69321051720102`

And here is Holstein, Donoghue and Lin's cross section:

zPlus[s_] := (1 + (1 - 4 mPi^2/s)^(1/2))/2

zMinus[s_] := (1 - (1 - 4 mPi^2/s)^(1/2))/2

fHol[s_] := 2 (Abs[Log[zPlus[s]/zMinus[s]]]^2 - π^2) + mPi^2/s (Abs[Log[zPlus[s]/zMinus[s]]]^2 + π^2)^2

sigma00Hol[s_] := α^2/(256 π^3 fPi^4) (s - mPi^2)^2/s (1 - (4 mPi^2)/s)^(1/2) (1 + mPi^2/s fHol[s]) /. valueRules

plotHol = Plot[{0.38939 10^24 4 sigma00Hol[e^2] /. valueRules, 0.38939 10^24 sigma00Bij[e^2, 1.0]}, {e, (0.274 10^9), (0.7 10^9)}] ;

[Graphics:../HTMLFiles/index_58.gif]

All plots agree:

Show[plot, plotBij, plotHol] ;

[Graphics:../HTMLFiles/index_60.gif]

Converted by Mathematica  (July 10, 2003)