Subroutine ffhe3 (q2mev,rp,rn,rpsp,rnsp,nfcn,ache,amhe,ach) c *** see RH Landau, Program lpott, 1981, mods by m sagen c m paez eliminated use of fmmm- McMillan 3-nucleon ff c ffhe3 calculates the four form factors needed to describe he3with c (fsd is only calc as a cpmparison, it should be small and is c the apptox, or effectove, sd contib to the magnetic ff c the ff*s are calc by using c 1 fche3- expt of 7ccarthy et al for all q2,have fitted c 2 ff*s calc by m McMillan,for fcn=0 these fit h3 data c of collard and for larger q2( gt 6 f-2) they rep a c reasonable extrapltn which has a correct zero as he3 c 3 fmhe3 use the form of 7ccarthy et al but vary this since the exh c chage contib is included in e scatter, but presmbly not in our pi c scatt (nb we use mu he3 = un) c these ff include s,s* + d states exactly in f1c,f2c, c the d state is only put in approximatly in f1m,f2m as Gibson has c shown this to be small (ignore too the dd terms too) c nfcn=1 , include neutron form factor in escat data, =0 , Don*t c data seems best interpreted with no neutron ?charge? form factor c possibly this is due to polarization or meson exch efects && c c q2mev2 is in mev**2 not Fermi-2 Implicit Real*8(a-h,o-z) data up,un,uhe3/2.79278,-1.91315,-1.91315/ q2f2 = q2mev/(197.329**2) qf = Sqrt(q2f2) c nucleon ff fn=0. fp=1. If(nfcn.eq.0) Call ffpn (q2f2,fp,fn) If ((nfcn.ne.1).and.(nfcn.ne.0)) Write (6,40) nfcn c he3 and h3 charge ff fche3 = ffche3(qf) c use mmm wf*s for h3,this fits collard (if fn=0) and is reasnb,e c sp= s* prob, pd= d state prob, these a can be varied psp = 0.02 pd = 0.09 c Call fmmm (q2f2,f1c,f2c,psp,pd) c fch3 = (2.*fn+fp)*f1c+(fp-fn)*f2c c he3 mag ff fmhe3 = ffmhe3(qf) GoTo 30 entry fhe3gs(q2mev,rp,rn,rpsp,rnsp,nfcn,ache,amhe,ach) q2f2 = q2mev/(197.32**2) fn = 0. fp = 1. c Gaussian form factors fche3 = 0. rhl = ache*ache*q2f2 If (rhl.gt.150.) GoTo 10 fche3 = exp(-rhl) 10 fch3 = 0. rhl = ach*ach*q2f2 If (rhl.ge.150.) GoTo 20 fch3 = exp(-rhl) 20 fmhe3 = 0. rhl = amhe*amhe*q2f2 If (rhl.ge.150.) GoTo 30 fmhe3 = exp(-rhl) 30 Continue c c direct calc of ff for fcn=0 rp=fche3/fp rn=fch3/fp rhl=un/(up+2.*un)/fp rhl1=(fch3-4.*fche3)/3. rpsp=rhl*(fmhe3+rhl1)/2. rnsp=rhl*(2.*fmhe3-rhl1*(up/un)) Return c 40 format (55h ********** neutron form factor switch is strange,nfcn= 1,i3,11h **********,/) End