CALL PYSTOP(1)
ENDIF
+ DO 2 I = 1, 4000
+ DO 1 J = 1, 5
+ V(I,J) = 0.
+ 1 ENDDO
+ 2 ENDDO
C...Initial values for some counters.
MSTU(1)=0
MSTU(2)=0
C... Evolve massive quark creation separately.
MCRQQ=0
IF (MQMASS.NE.0) THEN
- PT2CR=(RMQ2+VINT(18))*(RML**(TPM/(TPL*PYR(0)**(-TML/WN)-TPM)))
- & -VINT(18)
+ IF (WN .EQ. 0.) THEN
+ ARG = -1.
+ ELSE
+ ARG = TPM/(TPL*PYR(0)**(-TML/WN)-TPM)
+ ENDIF
+ PT2CR=(RMQ2+VINT(18))*(RML**ARG)-VINT(18)
C...If massive quark also on opposite side, ensure sufficient remaining
C...phase space also for creation of that quark
TMINQQ = TMIN
& + (P(ICR(IS),3)+ P(ICR(IS+1),3))**2
PZTS = P(ICR(IS),1)*TX+P(ICR(IS),2)*TY+P(ICR(IS),3)*TZ
& + P(ICR(IS+1),1)*TX+P(ICR(IS+1),2)*TY+P(ICR(IS+1),3)*TZ
- PTTS = SQRT(PPS2 - PZTS**2)
+ PTTS = SQRT(MAX(0.D0,PPS2 - PZTS**2))
C... Mass of string piece in units of mpi (at least 1)
RMPI2 = 0.135D0
RM2STR = MAX(RMPI2,EES**2 - PPS2)
ENDIF
RETURN
+C -------------------------------------------------------------------------
+C Extracted from a private e-mail exchange with Torbjorn Sjostrand
+C
+C No, Lambda(1520) is not included and not foreseen.
+C So if you want it in Pythia, it would have to be a hack.
+C What you could do is:
+C 1) In PYKFDI, just before the RETURN above label 140, you could check if
+C a Lambda, Sigma0 or Sigma*0 has been produced, and with some small
+C probability switch such a particle to the Lambda(1520) code. That is,
+C if KF = 3122, 3212, or 3214 and a random number below some number, switch
+C to KF = 3124. (And correspondingly for anticparticles.)
+C 2) Use the PYUPDA routine (see manual) to include particle and decay data
+C for the Lambda(1520).
+C -------------------------------------------------------------------------
+
+ IF (IABS(KF).EQ.3122) THEN
+C Converting a fraction (0.20) of Lambda0 to Lambda(1520) + c.c.
+C This fraction is based on the experimental measurement at ISR
+C Bobbink 83, NP B217,11 (1983)
+C The region 0.5 < XF < 1.0 has been extrapolated to XF=0
+ IF(PYR(0).LE.0.20) KF=ISIGN(3124,KF)
+ ENDIF
+
+ IF(IABS(KF).EQ.3212) THEN
+C Converting a fraction (0.20) of Sigma0 to Lambda(1520) + c.c.
+C We suppose the same fraction as for Lambda0
+ IF(PYR(0).LE.0.20) KF=ISIGN(3124,KF)
+ ENDIF
+
+ IF (IABS(KF).EQ.3214) THEN
+C Converting a fraction (0.30) of Sigma0(1385) to Lambda(1520) + c.c.
+C This is conservative extimate supposing that the ratio
+C scales as (M_Sigma1385/M_Lambda0)^2 ~ 1.5
+ IF(PYR(0).LE.0.30) KF=ISIGN(3124,KF)
+ ENDIF
+ RETURN
+
C...Use tabulated probabilities to select new flavour and hadron.
140 IF(KTAB2.EQ.0.AND.MSTJ(12).LE.0) THEN
KT3L=1
> XMIN,XMAX,Q2MIN,Q2MAX)
IF (MSTP(194) .EQ. 0) THEN
CALL SETLHAPARM("EKS98")
- ELSE
+ ELSE IF (MSTP(194) .EQ. 9) THEN
+ CALL SETLHAPARM("EPS09LO")
+ ELSE IF (MSTP(194) .EQ. 19) THEN
+ CALL SETLHAPARM("EPS09NLO")
+ ELSE IF (MSTP(194) .EQ. 8) THEN
CALL SETLHAPARM("EPS08")
+ ELSE
+ CALL SETLHAPARM("EPS09LO")
ENDIF
ELSE
write(6,*) "-> pdfset"