55 Part III Solution to Equation 1.9 Using Separability Chapter 18 rr Application: Gyromagnetic Ratios H After separation of variables (section 17.6) the “r” component of equation 1.9 can be rewritten as: dt    d j3/2  g m m F c g  f 0 (18.1) ds oo p p  rr dr r  dt    d j1/2  g m m f c g  F  0 (18.2) ds 00 p p  rr dr r  Using the above Dirac equation it is easiest to find the gyromagnetic ratios gy for the spin polarized F=0 case. Recall the usual calculation of rate of the change of spin S gives dS/dtmgyJ from the Heisenberg equations of motion. We note that 1/g rescales dr in rr  d J 3/2  g  f in equation 18.1. Thus to have the same rescaling of r in the second  rr dr r  term we must multiply the second term denominator (i.e.,r) and numerator (i.e., J+3/2) each by 1/g and set the numerator equal to 3/2+J(gy), where gy is now the gyromagnetic ratio. This rr makes our equation 18.1 compatible with the standard Dirac equation allowing us to substitute the gy into the standard dS/dtmgyJ to find the correction to dS/dt. Thus again: [1/g ]( 3/2 +J)=3/2+Jgy, Therefore for J= ½ we have: rr [1/g ]( 3/2+½)=3/2+½gy= 3/2+½(1+gy) (18.3) rr Then we solve equation 18.3 for gy and substitute it into the above dS/dt equation. S States: Recall  and  and S states from 4.12 and section 2.2. Noting in equation 4.12 the 1+ cancels we get the gyromagnetic ratio of the electron with g =1/(1+/(1+)) and =0 for rr electron. Thus solve equation 18.3 for g = (1+/(1+))=  (1+/(1+0))=  (1+.0005799/1). rr Thus from equation 18.3[1/ (1+.0005799)](3/2 + ½)= 3/2 + ½(1+gy). Solving for gy gives gyromagnetic rato of the electron gy=.00116 Going to higher energies (so 0 in equation 18.3) we get the gyromagnetic ratio of the muon. 2P states: Recall the 2P3/2 states from chapter 3. Note also that k can be positive or negative 3/2 since 4k=Z in our Lagrangian with a positive k meaning at least one charge is not canceled.. 00 Therefore 1/g =1k/r+ (using our Frobenius solution expansion near rr of eq.19.5 below rr H multiply through by (1+/4)((1++..) 1+.08=1+’ so a pion mass is then added to the protons) from the nature of Z . Therefore we have two cases at the boundary r=k oo 56 CASE 1 1/g =1+k/k+ charge 1 (core case) rr CASE 2 1/g =1- k/k+ charge 0 (use m from case 1) rr Note: The effect of a zero charge is to make metric component g (=1/g ) contribution zero in oo rr case 2. Thus the effect of nonzero charge is to increase the dimensionality. This provides the reason that Kaluza Klein theory (adding a 5th dimension) is so successful at injecting E&M into general relativity. But Kaluza Klein theory must be avoided at all costs because it adds unnecessary mathematical complexity. 2D is sufficient as we showed in Chapter 1. CASE 1: Plus +k, therefore is the proton + charge component. 1/g =1+k/k + = 2+  . Thus rr from equation 18.1, 18.2 2(1.5+.5)=1.5+.5(gy), gy=2.8 The gyromagnetic ratio of the proton (therefore that above r k stability was indeed proton stability as we concluded) mass=m dt/dsg =1/g =E=m p . oo oo p CASE 2: negative k, thus charge cancels, zero charge: 1/g =1-k/k +=  Therefore from equation 18.3 and case 1 1/g =1+k/k+: rr rr (1.5+.5)=1.5+.5(gy), gy=-1.9, the gyromagnetic ratio of the neutron with the other charged and neutral hyperon magnetic moments scaled using their masses by these values respectively. 57 Chapter 19 Frobenius series solution method applied to CASE1, CASE2 19.1 Series Solutions  Ansatz Near rr H Recall equations 18.1, 18.2: dt    d j3/2  g m m F c g  f 0 ds oo p p  rr dr r  dt    d j1/2  g m m f c g  F  0 ds 00 p p  rr dr r  Recall from the previous section g =1-k/r-(ε+Δε). Also recall our Dirac doublet (equation 4.17 oo and section 1.5 originating in 1.2e) must have a left handed zero mass component will be called case 1 and case 3 respectively below. Also we need the equivalent of the singlet equation 1.2a is our below case 2. Also in equation 2.6 at r=r the eigenvalue is ++1=2m for that principle H p quantum which then must be the same for the 2P state Here we write out the left handed Dirac 3/2 Doublet Eq.4.17 in the general representation of the Dirac matrices. Also recall from chapter 3 that the 2P state (and its sp2 hybrid) for this new electron Dirac equation gives a azimuthal 3/2 trifolium, 3 lobe shape and thus a /3 spherical harmonic wavelength so that for covalent bonding r’r /3 in  =1-r’/r. This /3 also is used in section 16.1 to calculate P wave scattering. H oo To use the f & F components of the equation 18.1, 18.2 Dirac equation we write the Dirac equation for free particle motion along the symmetry axis z (r=ratio of momentum to energy) to find the chirality of the components in the general representation.of equation 4.17. We then compare this z motion free particle Dirac equation eigenfunction structure with radial component structure to arrive at a sense of which components of the radial equation are left handed and which aren’t. This step is a little more complicated here because we are not using the chiral representation of the Dirac matrices, but the standard representation instead. In any case given that the electron is positive energy, then (as we see below) for the positron -E gives left handed f and F implying that this object must have a positive charge since this left handedness(doublet, 4.17) results from the fractalness (There is a corresponding argument for G and g). The proton indeed is positive charged. So:  dt    d j3/2 ds goo mpmpg c grr dr  r G 0 c2u1+cpu3-Epu1  dt    d j1/2 ds g00mpmpGc grr dr  r g 0 cpu1- c2u3-Epu3  dt    d j1/2 ds goo mpmpf c grr dr  r F 0 c2u2-cpu4-Epu2  dt    d j3/2 ds g00mpmpF c grr dr  r F 0 -cpu2 -c2u4-Epu4 58 where to get correspondence from these two Dirac equation structures we see that at +E: uR= 1  0  r 0         0  1  0 r =g, uL= =f; -E: No (vR= here), vL= =F, Note in general (with r0) here:         r 0 1 0                 0 r 0 1 ig uR     if  uL   = . So we have the solution that in the standard representation of the left handed     G vR         F vL  doublet is given by F and f only for –E of the electron (here a positron needed below for + proton hadron excited states) at the horizon. Dirac matrices F q, m0,E, forF So for the left handed doublet:  we have respectively (19.4)    f  q 0,m0,E, forf L Or more succinctly equation 1.9 in the Dirac doublet form implies: CASE 1 1/g =1+k/k+ F charge 1, m=1 (core case) rr CASE 2 1/g =1- k/k+ F charge 0, m from case 1) rr CASE 3 f charge 0, m=0 We solve these equations only near r≈k since that is where the stability is to be found (and also H fortunately were these equations are linear differential equations). Thus our first step is to expand √g about this radius and drop the higher order terms. rr The Frobenius series solution method can now be used to solve equations 18.1 and 18.2 at rr . H See for example Mathematical Methods of Physics, Arfken 3rd ed. Page 454. First we solve the f in equation 18.1, plug that into equation 18.2 and then have an equation in only F. There we substitute a series solution ansatz F= ∑a rn in the resulting combined equations. We can then n separate out the results into coefficients of respective rn and get recursion relations that will give us series that must be terminated at some N. Note the energy Eigenvalue ‘E’ will be in this series as dt/ds√g so we can then solve for the mass energy of these hadrons at specific J. We will need oo an indicial equation for the first term to start out this process. Also in this Frobenius solution method ‘n’ turns out to be a multiple of ½ and the series must start at n=-1. Finally to get the charge zero case the charged case must be done first and its constant masses used in the uncharged state calculations. The method appears to be working correctly. For example at the ∑ particle mass there is a square root that gives two charges instead of the usual 1. Also the L=1 solution has a mass of two protons associated with it, giving that (otherwise very mysterious) deuteron L=1 eigenstate. In any case all we are doing here is solving equation 1.9 for the left handed Dirac doublet case (our section 1.5 and equation 4.11 fundamental use of the Dirac equation 1.9) inside the horizon r and just working out the math details of what has to be done to accomplish this. The energy in H 12 is split between two 2P lobes, as with the deuteron, each lobe giving a proton 59 mass.[m= (tauon+muon)/2]. Thus we divide by two here: the proton mass is taken to be the unit mass. Not that r=k is inversely proportional to m .(i.e., m  1/k ) For m =.5 then k =2 for each p p H p H of these 2P lobes. In the energy square roots (below) then we can normalized out the .5 (because k squared is always reciprocal of m 2 in the energies) so the proton mass energy becomes unit 1 H p in all the rest of these calculations. 19.2 CASE 1 Excited States for F, m0, q Again case 1 is one of the equation 18.1 possibilities. Therefore let R=k -r, r<<R (for stability) H we can write in 18.1: R g 1/ 1k /R  (19.1) rr H Rk R H k r H  (19.2)   k rk  k r H H H k r H  (19.3)     k 2 r 1 H  r r2   1  ..  1/4 2k 8k2  H H = (19.4)   2 r r2 1  ..   4k 16k2   H H r 1 r 3r2 4k ((1- /4)/2) (1  ..)  H (19.5) 4k 32k2 2 H H Note that including the above 1/4 the compensating (1/4) in the next r term has the effect of a multiplying the derivative terms by 1/4. This rescales r to allow us to still say that the stable boundary is still at r . Thus we could use it to also rescale t in the first term of equations 18.1 H and 18.2 or note that (1+/4) (1+)=1+5/4 thus renormalizing 1+  to 1+4/3 =1+’ everywhere. Also the 3r2/32k 2 terms must be included. We drop these perturbative terms until H the end. Therefore substituting in equation 19.5 we find that equation 18.1 reads:  3   j  Em F c1r/k 4 d  2 f 0= (19.6) p  H 2dr k r  H       d 3  Em F c1r/k 4 1r/k (j )/k f 0; also: p  H 2dr H 2 H   1   j    d 2 Em f c g  F 0 p  rr dr r      60    d  1  Em f c1r/k 4 1r/k  j /k F 0 p  H 2dr H  2 H  Therefore c  d  1  f  c 1r/k 4 1r/k  j /k F substituting into Em  H 2dr H  2 H  p  3   j  Em F c1r/k 4 d  2 f 0 (19.7) p  H 2dr k r  H    We find solving for f and substituting back in:    d  Em F c1r/k 4 1r/k j1.5/k  p  H 2dr H H  c 1r/k 4 d 1r/k (j 1)/k F =Em F+ Em  H 2dr H 2 H  p p c2  d d2 1    1r/4k  1r/4k 2 1r/4k (j )/k2 F Em 2 H k 4 2dr H 2dr2 H 2 H    p H c2  d 1  + 13r/k 4j1.5 1r/k 2j1.5(j )/k2 F = Em  H 2k dr H 2 H    p H    c2   3 1 1   Emp Em  j 2(j 2)/kH2 (j 2)/ 2kH2 F +   p  c2   3 1 1  Em  22 2 j 2(j 2)/kH3 (j 2)/4kH3 rF+ p c2  1  3 1 dF     j   +   Emp 2 kH4 2  2kH  dr c2  1  3 3  dF     j  r + Emp 2 kH216 2  24kH2  dr c2  1 d2F     + Em 2  2 dr2 p c2  1  d2F    r Em 2 2 2k  dr2   p H Here r=2k is a regular singular point. Next substitute in F=  a rn with again half integer n H n n allowed as well: 61 N EmpEcm2  j 23(j 12)/kH2 (j 12)/ 2kH2 anrn + (19.8) M   p  N c2   3 1 1  Em  22 2 j 2(j 2)/kH3 (j 2)/4kH3 an1rn + (19.9) M p N c2  1  3 1      j  n1a rn + (19.10) M Emp 2kH4 2  2kH  n1 N c2  1  3 3      j  na rn  (19.11) M Emp 2kH216 2  24kH2  n N c2  1     n2n1a rn + (19.12) Em 2  2 n2 M p N c2  1     n1na rn 0 (19.13) Em 22 2k  n1 M p  H  Note from equation 19.12 that this series diverges. To terminate the series we now take 19.8 and 19.11 together and 19.10 and 19.13 together (since they have the same a ). For example n combining the equation 19.8 and 19.11 terms    c2   3 1 1   Emp Em  j 2(j 2)/kH2 (j 2)/ 2kH2 +   p  c2  1  3 3      j  n, Emp 2 kH216 2  24kH2  Replacing the normalization m m (1 ) (from section 2.1): p p     3 1 1  1  1  3 3  E2 m2p  j 2(j 2)/kH2 (j 2)/ 2kH2 - 2 kH216 2  j 24kH2 N =0 Therefore after rearranging: E  m2  1 j2 1.7071j1.10355j.5303.8269N , (19.14) p k2 H We have for a general Laurent series ansatz: ..a r1 a r1/2 a ro a r1/2 a r1 .. F 1 1/2 o 1/2 1 Note also that equations 19.8-19.13 imply that the coefficients of a given rn are independent. Thus adding together the coefficients of rn for equations 19.8–19.13 at a given n: 19.9(j-½)a +(19.8+19.11)a +(19.10+19.13)(n+1)a +19.12(n+2)(n+1)a =0 (19.15) n-1 n n+1 n+2 62 Method of Solving Equation 19.15 For the outside observer an F=0 finite boundary condition at infinity applies for flat vacuum value n=0, j=½ and for ro, r-½, r-1 and for complete vacuum for N=0, J=0. Here then the generalized Laurent series ..a r1 a r1/2 a ro a r1/2 a r1 .. F 1 1/2 o 1/2 1 reduces to ..+a r1 a r1/2 a ro  F. Thus either set 19.9(j-½)a =0 or 1 1/2 o n-1 (19.10+19.13)(n+1)a +19.12(n+1)(n+1)a =0 separately in eq.19.15 or set both equal to zero: n-1 n+2 J= ½, sets eq.19.9=0 1) N=-1, in equation 19.14 gives mass eigenvalue for  Exact solution for all possible a , sets none of them to zero. n 2) N=0, in equation 19.14 gives mass eigenvalue for nucleon. dro/dr=0 so all derivative of F terms are then zero and this solution applies inside as well. N=0 flat J=0 allowed flat vacuum gives  and with free e, j= ½ muon. 3) N= -½, in equation 19.14 gives mass eigenvalue of two  s since a plus and minus square root of r. These 19.9=0 cases have case 2 zero charge representations as well. N=-1, Principle QM number Also a =0 -2 1) J=0, in equation 19.14 gives mass eigenvalue for K 2) J=1, gives deuteron mass eigenvalue (bonding) given N=0,J=0 fills first (i.e., pion). Thereafter use nuclear shell model-Schrodinger equation many body techniques with these nonrelativistic lobes with this (bound state) force acting like a outer layer surface tension, finite height square well potential . Get a aufbau principle that then gives the D,F,G,..nuclear shell model states. Alternatively can fill that first S state in with free 1S (next ½ state to filled state) and we have j= 3/2 - filling in some (i.e., uds) of the 2P states (see Ch.3) and thereby also deriving from first principles 3/2 Gell Man’s 1963 eight fold way for hyperon eigenvalue classification (to finish that effort need case II zero charge and case III  as well). M is o p replaced by 2 in c hyperons, by 4 for b hyperons as indicated in fig. 16-1 for how to fill in the cbt 2P harmonic states given the requirement to use r2 then. Also, to include higher order r expansion term effects in equation 19.5 we must include those perturbative 1+/4 and 3r2/32k 2 contributions which gives a n(n-1)/6.4 added to the “n” term H component inside the radical of equation 19.14. In our new pde J=o through LS spin-orbit coupling so the three spin½s and the L=1 add to a minimum. 1-½-½+½ =½ =S for the proton with possible Pauli principle non S=½ possibilities for larger mass eigenvalue. Details of Above Solutions for Case 1 Thus besides the ground state (N=0 F = a rn = a r-0=F proton) we have the two groundstate n 0 0 solutions: F =a rn = a- r-1=F ., j= ½, 0., F =a rn =a r–½ =F . For j = ½. 0. N=-1 n 1 1 N=-½ n -½ 2 63 Note the energy eigenvalues (E) can be found from the solution to equation 19.14 and k =1 with H E=1=938MeV. Thus N= 0, j= ½ then 19.14 gives +Nucleon (ground state) mass eigenvalue. Note that for the N=0, (with J= ½ and also J= 0 in section 19.5) ground state that the charge density is uniform (i.e., =Kr0 ) for r<k. N=-½ , j=½ two valued because of the two square root solutions. Equation 19.14 then gives  (charged sigma particle) 1184Mev particles, F eigenfunction(s). Actual 1189Mev 2 N=-1, j= ½ gives one charged  particle. Therefore the energy from equation 19.14 is 1327 Mev (actual 1321), F eigenfunction. 1 Case 2 and case 3 give the neutral hyperons and  respectively (see case 3 below). o 19.5 Nucleon Wavefunction: J=1, q0, N=-1 of Case 1 Here we recall case 1, section 19.3 above and compute energy eigenvalues for J=0 and J=1.again using equation 19.14 in case 1. J=0 N=-1, j=0 E=490 MeV from equation 19.14 case 1. K . Substitute into strangeness equation 19.34 case 1 we obtain strangeness =1. N=0, j=0 then from equation 19.14 E=139.7 MeV case (note again m=1+=1.061 in 19.14 for outside). This is the nontrivial F zero point energy (and so has a fundamental harmonic) for r<k since the square root in equation 20.1 becomes imaginary then. Thus the mass of  is now the vacuum (e.g., note Fro for N=0 here) ‘at rk explaining why this fundamental harmonic result for  is used in all the successful nuclear force theories such as in the Skyrmion Lagrangian for example. Note that: m =139Mev=1.3(105.6MeV) =1.3=.08=‘ (19.22)   N=-1,J=1 case 1. Recall for J=1 we have   rsin Y1 () double lobe * along the z 1 axis: From equation 19.14 we find with these inputs that E=1867Mev implying that (because E~2m and J=1) this eigenstate is responsible for the spin 1 deuteron (state). The L=1, 2P state p solution(s) are symmetric and so of the form (1/2)(  +   ) = and have positive parity 1 2 1 2 s even if the 2P  and  each has negative parity. The Deuteron thus has + parity (Enge, 1966). 1 2 Recall if we include the background metric in eq.4.11  =1+r /r+2’+ and  =1/(1+r /r+). oo H rr H So rescaling rr-’ =r’for r near r allows us to use our above solutions again. So in equation H 18.1 1/ ==1/(1+r /r’+)1/(1+r /r)+(/2). Note if we again rescale our numerator rr H H J=11+(’/2)2 so that we have perturbed our Y spherical harmonic with a (/2)Y giving a 1 2 measure of the oblate, non spherical structure (e.g quadrupolar  and higher. ’/2 .04 from D 19.22 therefore the nonspherical component of  is approximately 4% of the total  and is often called the tensor component of the Deuteron eigenstate (Enge, 1966). This simplest multiparticle state represents the deuteron state and this is then the explanation for the deuteron tensor component of the nuclear force. Also the energy of the Deuteron is given just outside the r boundary (so ’i in 4.11) by H E =Rel 1876/ =Rel 1876/(1+i’)+..=1876(1-i’/2+ (3/8)(i’)2+..). So the added real term D oo due to the ’ is equal to 1876(3/8)’2=1876(3/8)(.08)2=4MeV. In free space ’=0 and just outside the nucleus it gives this contribution to the Deuteron energy. Thus this (3/8)’2 is the binding energy of the Deuteron. 64 Note from the equation 19.15 discussion for N not -1 we can only use J=1/2 and J=3/2 thus are restricted for two particles to S and P states (i.e. ½ + ½ =1) which then gives us the hyperons. For N=-1 we can use other J and can thereby construct large nuclei. The multinucleon nuclei really are the solutions of the indicial equations of 19.15. Recall in the shell model a hard shell nuclear outer wall is assumed with free space oscillations allowed inside this shell. The solutions to the Schrodinger equation are then spherical Bessel functions with corrections for spin orbit interaction, finite well height and tapered wells (Herald Enge, Introduction To Nuclear Physics, P.145). In any case an infinite mean free path for these oscillations is assumed to exist inside this shell. So how can there be an infinite mean free path inside this extremely high mass density region? In that regard the above 2, J=1, N=-1 2P deuteron state can also be viewed as yet another Bogoliubov pairing interaction (such as in the SC section 17.1) giving this infinite mean free path of the electron pairs comprising a pion acting as a Cooper pair, just as in SC In the context of the section 17.1 pairing interaction model A(dv/dt)/v2 is no longer as small but dv/dt becomes very large to due to the ultrarelativistic motion of the electrons inside the nucleons. In any case this infinite mean free path for these oscillations (recall Cooper pairs have an infinite mean free path) is thereby explained here as a new type of superconductivity. Spin Orbit Interaction In Shell Model If equal numbers of Neutrons and Protons gyromagnetic ratios then gy -gy =2.7-1.9 =.8. P N Since more neutrons in heavier elements: (1/1.1)(.8)=.7. R=r ½ Fermi measured from singularity at 1-½ = ½ . H From 2P3/2 at r=r Fitzgerald contraction discussion in section 2.2: rR=½(1-½) = ¼ Fermi  H R (r-r ) so R (r-r )Kr. From chapter 14 V=1/(r-r ). Spin orbit interaction= V H v H H a 2(1/r)(V/r)(sL)= o 1 V .7  1  1 V a2 sL  sL.7(43)sL  0 R rr R rr  R rr R rr 2  r r V H V H V H  V H  1 V 1 V .7(64)sL a2 (sL) 45*E&M spin orbit interaction. r r o r r Thus the a =1Fermi. Thus the nuclear spin-orbit interaction is much larger than the E&M spin o orbit interaction because the nucleons are much closer to r than to r=0 and the Fitzgerald H contraction of the nucleon 2P state is on the order of ½. 3/2 At close range there are higher energies available so the 4mev (=be) in equation 19.3 (if we include r2 contributions) becomes the binding energy for the deuteron in g =1-k/r+be in 18.1 oo particles, F eigenfunction(s). Actual 1189Mev 2 N=-1, j= ½ gives one charged  particle. Therefore the energy from equation 19.14 is 1327 Mev (actual 1321), F eigenfunction. 1 Case 2 and case 3 give the neutral hyperons and  respectively (see main Frobenius series o solution paper). The multinucleon nuclei are the solutions of the indicial equations of 19.15. Recall in the shell model a hard shell nuclear outer wall is assumed with free space oscillations allowed inside this shell. The solutions to the Schrodinger equation are then spherical Bessel functions with corrections for spin orbit interaction, finite well height and tapered wells (Herald Enge, Introduction To Nuclear Physics, P.145). In any case an infinite mean free path for these