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- C The triangle anomaly.
- Letter to Jackiw, Jan 20, 1991.
-
- C Anomaly 1. Evaluation of the coefficients C(i,j), i=1,2, in terms
- of C0 and B0.
- C Anomaly 2. The axial current triangle graphs. Vertex: i*G5*G(al).
- C Anomaly 3. The axial current triangle graphs. Vertex: i*G5*G(al).
- An expansion in terms of the external momenta is done
- (assuming them to be small with respect to the loop mass m).
- The result shows that the axial current is at least of third
- order in the momenta.
- C Anomaly 4. The pseudo scalar graphs. Vertex: 2*m*G5.
-
- *end
-
- C Anomaly 1. Evaluation of the coefficients C(i,j), i=1,2, in terms
- of C0 and B0.
-
- P ninput
-
- BLOCK BFUN{}
- Id,B22(x~,M~,m~)=(-0.5*Ax(m)+M**2*B0(x,M,m)
- -0.5*(x+m^2-M**2)*B1(x,M,m))/[1-N]
- Id,B21(x~,M~,m~)=-((0.5*N-1)*Ax(m)
- -0.5*N*(x+m^2-M**2)*B1(x,M,m)
- +M**2*B0(x,M,m) )/x/[1-N]
- Id,B1(x~,M~,m~)= (0.5*Ax(M)-0.5*Ax(m)
- -0.5*(x+m^2-M**2)*B0(x,M,m) )/x
- ENDBLOCK
-
-
- BLOCK CASE6{}
-
- C Case: all masses equal.
-
- Id,M3=m
- Al,M2=m
- Al,M1=m
- Al,B0(x~,M1~,M2~)=B0(x,m)
- Id,Ax(M1~)=Ax(m)
- Al,B0(x~,M1~,M2~)=B0(x,m)
- ENDBLOCK
-
- A Pi,N,N_,m,M,M1,M2,M3,x,y
- V p,k
- F Ax
-
- D XX(I)= pDp,-pDk,-pDk,kDk
-
- X X(I,J)= XX( 2*I-2+J )/Det
-
- X f1=M1^2-M2^2-kDk
- X f2=M2^2-M3^2-pDp-2*pDk
- *fix
-
- D R1(I)=(1/2*f1*c0+1/2*B0(qDq,M1,M3)-1/2*B0(pDp,M2,M3)),
- (1/2*f2*c0 + B0(kDk,M1,M2)/2 - B0(qDq,M1,M3)/2)
-
- Z c11=DS(K,1,2,(X(1,K)*R1(K)))
- Z c12=DS(K,1,2,(X(2,K)*R1(K)))
- Id,Multi,pDk^2=kDk*pDp-Det
- Keep c11,c12
- B i,Pi,Det,c0
- CASE6{}
- *next
-
- Z c24=i*Pi^2/4-1/2*M1^2*c0+1/4*(B0(pDp,M2,M3)-f1*c11-f2*c12)
-
- Id,Multi,pDk^2=kDk*pDp-Det
- CASE6{}
- B i,Pi,Det,c0
- Keep c11,c12,c24
- *next
-
- D R3(I)=(1/2*f1*c11+1/2*B1(qDq,M1,M3)+1/2*B0(pDp,M2,M3)-c24),
- (1/2*f2*c11+1/2*B1(kDk,M1,M2)-1/2*B1(qDq,M1,M3))
- D R4(I)=(1/2*f1*c12+1/2*B1(qDq,M1,M3)-1/2*B1(pDp,M2,M3)),
- (1/2*f2*c12-1/2*B1(qDq,M1,M3)-c24)
-
- Z c21=DS(K,1,2,(X(1,K)*R3(K)))
- Z c23=DS(K,1,2,(X(2,K)*R3(K)))
-
- Z C23p=DS(K,1,2,(X(1,K)*R4(K)))
- Z c22=DS(K,1,2,(X(2,K)*R4(K)))
-
- B i,Pi,Det,c0
- BFUN{}
- Id,Multi,pDk^2=kDk*pDp-Det
- Id,N*[1-N]^-1=-1+[1-N]^-1
- Id,Ax(m~)*[1-N]^-1= -1/3*Ax(m) + 2*i*Pi^2*m^2/9
- Al,B0(x~,M~,m~)*[1-N]^-1= -1/3*B0(x,M,m) - 2*i*Pi^2/9
- CASE6{}
- Keep c11,c12,c24,c21,c23,C23p,c22
- Nprint C23p
- *next
- P input
-
- C Check: must be zero.
-
- Z Diff23=c23-C23p
- *end
-
- C Anomaly 2. The axial current triangle graphs. Vertex: i*G5*G(al).
-
- C The triangle anomaly.
- Computing the axial current.
- The expressions for the C(i,j) have been computed separately,
- and are contained in the block CFU.
-
- P ninput
- BLOCK CFU{}
- P ninput
- Id,C(1,1,m) =
- + Det^-1*C0(m)
- * ( 1/2*pDp*pDk + 1/2*pDp*kDk )
-
- - C0(m)
-
- + B0(pDp,m)*Det^-1
- * ( - 1/2*pDp )
-
- + B0(kDk,m)*Det^-1
- * ( - 1/2*pDk )
-
- + B0(qDq,m)*Det^-1
- * ( 1/2*pDp + 1/2*pDk )
-
- Al,C(1,2,m) =
- + Det^-1*C0(m)
- * ( - 1/2*pDp*kDk - 1/2*pDk*kDk )
-
- + B0(pDp,m)*Det^-1
- * ( 1/2*pDk )
-
- + B0(kDk,m)*Det^-1
- * ( 1/2*kDk )
-
- + B0(qDq,m)*Det^-1
- * ( - 1/2*pDk - 1/2*kDk ) + 0.
-
-
- Al,C(2,4,m) =
- + 1/4*i*Pi^2
-
- + Det^-1*C0(m)
- * ( - 1/4*pDp*pDk*kDk - 1/8*pDp*kDk^2 - 1/8*pDp^2*kDk )
-
- + C0(m)
- * ( - 1/2*m^2 )
-
- + B0(pDp,m)*Det^-1
- * ( 1/8*pDp*pDk + 1/8*pDp*kDk )
-
- + B0(kDk,m)*Det^-1
- * ( 1/8*pDp*kDk + 1/8*pDk*kDk )
-
- + 1/4*B0(qDq,m)
-
- + B0(qDq,m)*Det^-1
- * ( - 1/8*pDp*pDk - 1/4*pDp*kDk - 1/8*pDk*kDk ) + 0.
-
-
- Al,C(2,1,m) =
- + i*Pi^2*Det^-1
- * ( - 1/4*pDp )
-
- + Det^-2*C0(m)
- * ( 3/4*pDp^2*pDk*kDk + 3/8*pDp^2*kDk^2 + 3/8*pDp^3*kDk )
-
- + Det^-1*C0(m)
- * ( 1/2*m^2*pDp - pDp*pDk - pDp*kDk - 1/4*pDp^2 )
-
- + C0(m)
-
- + B0(pDp,m)*Det^-2
- * ( - 3/8*pDp^2*pDk - 3/8*pDp^2*kDk )
-
- + B0(pDp,m)*Det^-1
- * ( pDp )
-
- + B0(kDk,m)*Det^-2
- * ( - 3/8*pDp*pDk*kDk - 3/8*pDp^2*kDk )
-
- + B0(kDk,m)*Det^-1
- * ( 1/4*pDp + 3/4*pDk )
-
- + B0(qDq,m)*Det^-2
- * ( 3/8*pDp*pDk*kDk + 3/8*pDp^2*pDk + 3/4*pDp^2*kDk )
-
- + B0(qDq,m)*Det^-1
- * ( - 5/4*pDp - 3/4*pDk )
-
- Al,C(2,3,m) =
- + i*Pi^2*Det^-1
- * ( 1/4*pDk )
-
- + Det^-2*C0(m)
- * ( - 3/8*pDp*pDk*kDk^2 - 3/8*pDp^2*pDk*kDk - 3/4*pDp^2*kDk^2 )
-
- + Det^-1*C0(m)
- * ( - 1/2*m^2*pDk + pDp*kDk + 1/2*pDk*kDk )
-
- + B0(pDp,m)*Det^-2
- * ( 3/8*pDp*pDk*kDk + 3/8*pDp^2*kDk )
-
- + B0(pDp,m)*Det^-1
- * ( - 1/8*pDp - 1/2*pDk )
-
- + B0(kDk,m)*Det^-2
- * ( 3/8*pDp*pDk*kDk + 3/8*pDp*kDk^2 )
-
- + B0(kDk,m)*Det^-1
- * ( - 5/8*kDk )
-
- + B0(qDq,m)*Det^-2
- * ( - 3/4*pDp*pDk*kDk - 3/8*pDp*kDk^2 - 3/8*pDp^2*kDk )
-
- + B0(qDq,m)*Det^-1
- * ( 1/8*pDp + 1/2*pDk + 5/8*kDk )
-
- Al,C(2,2,m) =
- + i*Pi^2*Det^-1
- * ( - 1/4*kDk )
-
- + Det^-2*C0(m)
- * ( 3/4*pDp*pDk*kDk^2 + 3/8*pDp*kDk^3 + 3/8*pDp^2*kDk^2 )
-
- + Det^-1*C0(m)
- * ( 1/2*m^2*kDk - 1/4*kDk^2 )
-
- + B0(pDp,m)*Det^-2
- * ( - 3/8*pDp*pDk*kDk - 3/8*pDp*kDk^2 )
-
- + B0(pDp,m)*Det^-1
- * ( - 1/4*pDk + 1/4*kDk )
-
- + B0(kDk,m)*Det^-2
- * ( - 3/8*pDp*kDk^2 - 3/8*pDk*kDk^2 )
-
- + B0(qDq,m)*Det^-2
- * ( 3/8*pDp*pDk*kDk + 3/4*pDp*kDk^2 + 3/8*pDk*kDk^2 )
-
- + B0(qDq,m)*Det^-1
- * ( 1/4*pDk - 1/4*kDk ) + 0.
- P input
- ENDBLOCK
- P input
-
- V p,r,q,k
- I al,mu,nu,L1,L2,L3
- A N,N_,Pi,m,Ax,Det
- F Fx,Cx,C,c,C0,Bx,B0,B1
-
- C Triangle graphs.
- (k,al) => (p,mu),(q,nu) with all momenta pointing inwards.
- There are two graphs, differing with respect to each other by
- reversal of the fermion direction (or by the interchange p <=> q
- and mu <=> nu).
-
- Z A(al,mu,nu) =
- - i*Cx(m)*G5(1,2)*G(2,3,al)*
- (i*G(3,4,k) + i*G(3,4,r) + m*Gi(3,4))
- *G(4,5,mu)*
- (i*G(5,6,r) + i*G(5,6,k) + i*G(5,6,p) + m*Gi(5,6))
- *G(6,7,nu)*
- (i*G(7,1,r) + m*Gi(7,1))*Ax
-
- - i*Cx(m)*G5(1,2)*G(2,3,al)*
- (- i*G(3,4,r) + m*Gi(3,4))
- *G(4,5,nu)*
- (- i*G(5,6,r) - i*G(5,6,k) - i*G(5,6,p) + m*Gi(5,6))
- *G(6,7,mu)*
- (- i*G(7,1,r) - i*G(7,1,k) + m*Gi(7,1))*Ax
-
- Id,Gammas,"C
- *yep
-
- C Try to work out terms with three r, to avoid the C(3,i).
- Such terms have more than one r inside the trace.
- Move them towards each other, to produce rDr
- Do this by moving them to the right.
-
- Id,G(1,"t,"4,G5,al,r,L1~,L2~,L3~,L4~) =
- - G(1,"t,"4,G5,al,L1,r,L2,L3,L4)
- + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4)
- Id,G(1,"t,"4,G5,al,L2~,r,L1~,L3~,L4~) =
- - G(1,"t,"4,G5,al,L2,L1,r,L3,L4)
- + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4)
- Id,G(1,"t,"4,G5,al,L2~,L3~,r,L1~,L4~) =
- - G(1,"t,"4,G5,al,L2,L3,L1,r,L4)
- + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4)
- Id,G(1,"t,"4,G5,al,L2~,L3~,L4~,r,L1~) =
- - G(1,"t,"4,G5,al,L2,L3,L4,L1,r)
- + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4)
- Id,Gammas,"C
- Id,G(1,"t,"4,G5,al,r,L1~,L2~,L3~,L4~) =
- - G(1,"t,"4,G5,al,L1,r,L2,L3,L4)
- + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4)
- Id,G(1,"t,"4,G5,al,L2~,r,L1~,L3~,L4~) =
- - G(1,"t,"4,G5,al,L2,L1,r,L3,L4)
- + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4)
- Id,G(1,"t,"4,G5,al,L2~,L3~,r,L1~,L4~) =
- - G(1,"t,"4,G5,al,L2,L3,L1,r,L4)
- + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4)
- Id,G(1,"t,"4,G5,al,L2~,L3~,L4~,r,L1~) =
- - G(1,"t,"4,G5,al,L2,L3,L4,L1,r)
- + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4)
- Id,Gammas,"C
- *yep
-
- C There may still be some traces with two r's left.
-
- Id,G(1,"t,"4,G5,al,r,L1~,L2~) =
- - G(1,"t,"4,G5,al,L1,r,L2)
- Id,Gammas,"C
- *yep
-
- C Work out the rDr terms, writing rDr = rDr+m^2 - m^2.
- In the resulting two point function one has
- 1/((r+k)^2+m^2)*((r+p+k)^2+m^2). Shift r => r-k.
-
- IF Cx(m)*rDr=Bx(m)*Shift - Cx(m)*m^2
- IF Shift=1
- Id,r(mu~)=r(mu)-k(mu)
- Al,Func,r(mu~)=r(mu)-k(mu)
- Al,Dotpr,r(mu~)=r(mu)-k(mu)
- ENDIF
- ENDIF
- Id,Gammas,"C
- *yep
-
- C Now do the integration over the loop momentum r.
-
- Id,All,r,N,Fx
-
- Id,Adiso,Cx(m)*Fx(L1~,L2~)=
- + k(L1)*k(L2)*C(2,1,m) + p(L1)*p(L2)*C(2,2,m)
- + k(L1)*p(L2)*C(2,3,m) + p(L1)*k(L2)*C(2,3,m)
- + D(L1,L2)*C(2,4,m)
- Id,Adiso,Cx(m)*Fx(L1~) = k(L1)*C(1,1,m) + p(L1)*C(1,2,m)
- Al,Adiso,Bx(m)*Fx(L1~) = p(L1)*B1(pDp,m)
- Id,Cx(m)= C0(m)
- Al,Bx(m)= B0(pDp,m)
-
- Id,Gammas
- *yep
-
- C Work out the C(i,j).
-
- CFU{}
-
- Id,B1(x~,m) = -0.5*B0(x,m)
-
- Id,p(al)*Epf(mu,nu,la~,ka~)=
- p(mu)*Epf(al,nu,la,ka)
- + p(nu)*Epf(mu,al,la,ka)
- + p(la)*Epf(mu,nu,al,ka)
- + p(ka)*Epf(mu,nu,la,al)
- Al,k(al)*Epf(mu,nu,la~,ka~)=
- k(mu)*Epf(al,nu,la,ka)
- + k(nu)*Epf(mu,al,la,ka)
- + k(la)*Epf(mu,nu,al,ka)
- + k(ka)*Epf(mu,nu,la,al)
-
- Id,Multi,pDk^2=kDk*pDp-Det
- Id,Det=kDk*pDp-pDk^2
-
- *yep
-
- C Write the result, i.e. the axial current, in the more standard form.
-
- Id,k(mu~)=-q(mu)-p(mu)
- Id,Func,k(mu~)=-q(mu)-p(mu)
- Id,Dotpr,k(mu~)=-q(mu)-p(mu)
- Id,Multi,pDq^2=qDq*pDp-Det
- P output
- *yep
-
- C Take any one of the following three options.
-
- Id,Ax=k(al) ! Taking the divergence of the axial current.
- C Id,Ax=p(mu) ! Must be zero: gauge invariance.
- C Id,Ax=q(nu) ! Must be zero: gauge invariance.
- Id,k(mu~)=-q(mu)-p(mu)
- Id,Func,k(mu~)=-q(mu)-p(mu)
- Id,Dotpr,k(mu~)=-q(mu)-p(mu)
- Id,Multi,pDq^2=qDq*pDp-Det
- *end
-
- C Anomaly 3. The axial current triangle graphs. Vertex: i*G5*G(al).
- An expansion in terms of the external momenta is done
- (assuming them to be small with respect to the loop mass m).
- The result shows that the axial current is at least of third
- order in the momenta.
-
- C The triangle anomaly.
- Computing the axial current to order 3 in the external momenta.
-
- V p,r,q,k
- I al,mu,nu,L1,L2,L3,L4
- A N,N_,Pi,m,Ax,ax,Det,Den
- F Ln,Fx,Cx,C,c,C0,Bx,B0,B1
- X DY(L1,L2,L3,L4)=D(L1,L2)*D(L3,L4)+D(L1,L3)*D(L2,L4)+D(L1,L4)*D(L2,L3)
- X DZ(L1,L2,L3,L4,L5,L6)=
- D(L1,L2)*DY(L3,L4,L5,L6)
- + D(L1,L3)*DY(L2,L4,L5,L6)
- + D(L1,L4)*DY(L3,L2,L5,L6)
- + D(L1,L5)*DY(L3,L4,L2,L6)
- + D(L1,L6)*DY(L3,L4,L5,L2)
-
- C Triangle graphs.
- (k,al) => (p,mu),(q,nu) with all momenta pointing inwards.
- There are two graphs, differing with respect to each other by
- reversal of the fermion direction (or by the interchange p <=> q
- and mu <=> nu).
-
- Z A(al,mu,nu) =
- - i*Cx(m)*G5(1,2)*G(2,3,al)*
- (i*G(3,4,k) + i*G(3,4,r) + m*Gi(3,4))
- *G(4,5,mu)*
- (i*G(5,6,r) + i*G(5,6,k) + i*G(5,6,p) + m*Gi(5,6))
- *G(6,7,nu)*
- (i*G(7,1,r) + m*Gi(7,1))*Ax
-
- - i*Cx(m)*G5(1,2)*G(2,3,al)*
- (- i*G(3,4,r) + m*Gi(3,4))
- *G(4,5,nu)*
- (- i*G(5,6,r) - i*G(5,6,k) - i*G(5,6,p) + m*Gi(5,6))
- *G(6,7,mu)*
- (- i*G(7,1,r) - i*G(7,1,k) + m*Gi(7,1))*Ax
-
- Id,Gammas,"C
-
- C Cx(m) is the three point function:
- 1 / (r^2+m^2)*((r+k)^2+m^2)*((r+k+p)^2+m^2)
- Expand the second and third denominator, use k+p = -q. Below Den stands
- for 1/(r^2+m^2).
-
- Id,Cx(m)=Den^3*Dev2*Dev3
- Id,Dev2= 1 - (2*kDr+kDk)*Den + (2*kDr+kDk)^2*Den^2 - (2*kDr+kDk)^3*Den^3
- Al,Dev3= 1 - (-2*qDr+qDq)*Den + (-2*qDr+qDq)^2*Den^2 - (-2*qDr+qDq)^3*Den^3
- Id,Multi,Den^7=0
- *yep
- Id,All,r,N,Fx
-
- C Keep up to and including third order.
-
- Id,Count,ax,"F,G,"F,Fx,k,1,q,1,p,1
- Id,Multi,ax^4=0
- Id,ax=1
-
- C Doing the momentum integrals.
-
- Id,Fx(L1~,L2~,L3~,L4~,L5~)=0
- Al,Fx(L1~,L2~,L3~)=0
- Al,Fx(L1~)=0
-
- Id,Fx(L1~,L2~,L3~,L4~,L5~,L6~)*Den^6 =
- DZ(L1,L2,L3,L4,L5,L6)*i*Pi^2/960/m^2
- Al,Fx(L1~,L2~,L3~,L4~,L5~,L6~)*Den^5 =
- DZ(L1,L2,L3,L4,L5,L6)*(i*DEL/192-i*Pi^2/192*Ln(m))
- Al,Fx(L1~,L2~,L3~,L4~,L5~,L6~)*Den^4 =
- DZ(L1,L2,L3,L4,L5,L6)*m^2*(-i*DEL/48+i*Pi^2/48*(-1+Ln(m)))
-
- Id,Fx(L1~,L2~,L3~,L4~)*Den^6 = i*Pi^2/480/m^4*DY(L1,L2,L3,L4)
- Al,Fx(L1~,L2~,L3~,L4~)*Den^5 = i*Pi^2/96/m^2*DY(L1,L2,L3,L4)
- Al,Fx(L1~,L2~,L3~,L4~)*Den^4 =
- DY(L1,L2,L3,L4)*(i*DEL/24 - i*Pi^2/24*Ln(m))
- Al,Fx(L1~,L2~,L3~,L4~)*Den^3 =
- DY(L1,L2,L3,L4)*m^2*(-i*DEL/8 + i*Pi^2/8*(-1 + Ln(m)))
-
- Id,Fx(L1~,L2~)*Den^5 = i*Pi^2/48/m^4*D(L1,L2)
- Al,Fx(L1~,L2~)*Den^4 = i*Pi^2/12/m^2*D(L1,L2)
- Al,Fx(L1~,L2~)*Den^3 = D(L1,L2)*(i*DEL/4 - i*Pi^2/4*Ln(m))
-
- Id,Den^5=i*Pi^2/12/m^6
- Al,Den^4=i*Pi^2/6/m^4
- Al,Den^3=i*Pi^2/2/m^2
-
- *yep
-
- C Take the trace.
-
- B DEL,i,Pi
- Id,Gammas
- *yep
-
- Id,Multi,m^-4=0
- Id,N=4+N_
- Id,N_*DEL=-2*Pi^2
- Id,N_=0
- Id,k(mu~)=-q(mu)-p(mu)
- Al,Func,k(mu~)=-q(mu)-p(mu)
- Al,Dotpr,k(mu~)=-q(mu)-p(mu)
- *yep
-
- Id,p(al)*Epf(mu,nu,la~,ka~)=
- p(mu)*Epf(al,nu,la,ka)
- + p(nu)*Epf(mu,al,la,ka)
- + p(la)*Epf(mu,nu,al,ka)
- + p(ka)*Epf(mu,nu,la,al)
- Al,q(al)*Epf(mu,nu,la~,ka~)=
- q(mu)*Epf(al,nu,la,ka)
- + q(nu)*Epf(mu,al,la,ka)
- + q(la)*Epf(mu,nu,al,ka)
- + q(ka)*Epf(mu,nu,la,al)
-
- P output
- *yep
-
- C Take any one of the following three options.
-
- C Id,Ax=k(al) ! Taking the divergence of the axial current.
- C Id,Ax=p(mu) ! Must be zero: gauge invariance.
- Id,Ax=q(nu) ! Must be zero: gauge invariance.
- Id,k(mu~)=-q(mu)-p(mu)
- Id,Func,k(mu~)=-q(mu)-p(mu)
- Id,Dotpr,k(mu~)=-q(mu)-p(mu)
- *end
-
- C Anomaly 4. The pseudo scalar graphs. Vertex: 2*m*G5.
-
- C The triangle anomaly.
- Computing the pseudo scalar triangle diagrams.
-
- V p,r,q,k
- I al,mu,nu,L1,L2,L3
- A N,N_,Pi,m,Ax,Det
- F Fx,Cx,C,c,C0,Bx,B0,B1
-
- C Triangle graphs.
- (k) => (p,mu),(q,nu) with all momenta pointing inwards.
- There are two graphs, differing with respect to each other by
- reversal of the fermion direction (or by the interchange p <=> q
- and mu <=> nu).
-
- Z A(al,mu,nu) =
- - Cx(m)*G5(1,3)*2*m*
- (i*G(3,4,k) + i*G(3,4,r) + m*Gi(3,4))
- *G(4,5,mu)*
- (i*G(5,6,r) + i*G(5,6,k) + i*G(5,6,p) + m*Gi(5,6))
- *G(6,7,nu)*
- (i*G(7,1,r) + m*Gi(7,1))*Ax
-
- - Cx(m)*G5(1,3)*2*m*
- (- i*G(3,4,r) + m*Gi(3,4))
- *G(4,5,nu)*
- (- i*G(5,6,r) - i*G(5,6,k) - i*G(5,6,p) + m*Gi(5,6))
- *G(6,7,mu)*
- (- i*G(7,1,r) - i*G(7,1,k) + m*Gi(7,1))*Ax
-
- Id,Gammas,"C
- *yep
-
- C Now do the integration over the loop momentum r.
-
- Id,All,r,N,Fx
-
- Id,Adiso,Cx(m)*Fx(L1~,L2~)=
- + k(L1)*k(L2)*C(2,1,m) + p(L1)*p(L2)*C(2,2,m)
- + k(L1)*p(L2)*C(2,3,m) + p(L1)*k(L2)*C(2,3,m)
- + D(L1,L2)*C(2,4,m)
- Id,Adiso,Cx(m)*Fx(L1~) = k(L1)*C(1,1,m) + p(L1)*C(1,2,m)
-
- Id,Cx(m)= C0(m)
-
- Id,Gammas
-
- C Write the result in the more standard form.
-
- Id,k(mu~)=-q(mu)-p(mu)
- Id,Func,k(mu~)=-q(mu)-p(mu)
- Id,Dotpr,k(mu~)=-q(mu)-p(mu)
- Id,Multi,pDq^2=qDq*pDp-Det
- *end
- ə