Volume 40B, number 3
PHYSICS LETTERS
10 July 1972
DYNAMIC INTERPRETATION OF BARYON AND MESON MASS FORMULAE L.M. LIBBY*
Engineering School, UCLA, Los Angeles, Cahfornia. USA Recieved 14 May 1972
A dynamical interpretation of baryon and meson mass formulae offers an explanation for associated production, for parity non-conservation, and for the T = 1/2 rule.
The mass formulae [1], for the baryon octet which may be written,
MB=(925+I85S+40[T(T+l)-¼S2])MeV
(I)
and for the meson octet which may be written mm =(550-
2 0 5 [ T ( T + l ) - ½ S 2 ] ) MeV
(2)
have an appearance which suggests that the bracket may be interpreted as the rotational energy o f a symmetrical top and the linear term in M B as an energy of vibration. The reason for studying a dynamical interpretation is that it offers to explain existence of (1) several particles of the same mass, (2) associated production, (3) the A T = 1/2 rule, and (4) parity non-conservation in particle decay. Each energy level of a symmetric top is doubly degenerate (for S g: 0) corresponding to two directions of rotation about the third axis (the figure axis) of the system [2, 3]. (In a symmetric t o p , I x =Iy ¢1z, where Ix, ly, I e are the moments of inertia about the x, y, z axes; T and S are the quantum numbers for angular momenta normal to and parallel to the figure axis respectively.) In a non-planar system, reflection at the center o f mass produces a configuration that cannot be obtained by rotation of the system; thus, for each value o f T there exist to modifications corresponding to +-S, namely, both a left hand and a right hand form as in the case of optically active isomers [3, p. 26], If the system is a dumbbell with a flywheel along its axis, then two energy-degenerate modes result. * On leave from the University of Colorado, Boulder, Colorado.
That is, while the dumbbell rotates about an axis normal to its figure axis, the flywheel on the figure axis may rotate either right or left (R, L). A more complicated system resembles a helicopter, with a rotor able to rotate either right or left (R, L) and a propeller able to rotate either right or !eft (r, 1). Four possible dynamic configurations result corresponding to (R, r), (R, 1), (L, r) and (L, 1). A third even more complicated situation occurs if an additional flywheel is added along the third axis of the helicopter, with right hand and left hand rotation, so that eight possible energydegenerate dynamic modes result for the three flywheels on the mutually orthogonal axes. Every single one of these dynamic situations described above is "optically active". If "optically active" particles described by such systems undergo decay, the decay products must carry the "optical activity" or decay in an "optically active" way, that is with violation of parity conservation t. Starting from a non-"optically active" system, "optically active" particles must be made in ( L + R ) pairs in order that angular momentum may be con-
t Among baryons, sets of mass-degenerate particles such as (p, n, fi, ~) may correspond to the four-fold degenerate system. Among mesons, four-fold mass-degenerate systems are the kaons (K-, K°, K°, K-). Of eight-fold mass-degenerate baryons, examples are (A++, A+, A°, A-, ~+, ~°, A-, A-). Three-fold mass-degenerate sets such as (n +, ~r, rr-) may be sub-sets of four-fold degeneracy with an internal symmetry such that (RI) and (Lr) rotate into each other. Mass splittings within a mass-degenerate set of particles may now derive from an interaction analogous to "A-type doubling" in molecules in that the energy may depend on the direction of rotation of the meson-cloud fly-wheel about the figure axis of a rotating particle. 307
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served. If an "optically activ~ system undergoes an interaction, odd numbers o f "optically active" particles must result,
where H v is an Hermite polynomial, and r is the dimension of the system. The optical model relates r to momentum transfer t according to [6],
L ~ L + ( L + R ) , L + 2 ( L + R ) ....
r = rotl/Z/2MpC2 sin ½0
R ~ R + ( L + R ) , R + 2 ( L + R ) .... According to this dynamic hypothesis, associated production is a manifestation o f conservation of total angular momentum. Parity non-conservation in particle decay, indicating the "optically active", i.e. right hand or left hand, nature of elementary particles, is caused by internal flywheel rotational motions about orthogonal axes in real space being carried away by the decay particles, as the system breaks up. Some consequences of this interpretation are the following: l) If the vibrational energy is given (in MeV) by the linear term in eq. (1), then it follows that, E v = 185 [SI = hCOv ISI
(3)
where COv = 0.3 X 1023 sec -1 is the frequency of vibration. From it the curvature of the interaction energy E(r) may be computed according to [e.g. 2, p. 78], E(r) = ½/.tCO2(r - re)2;
d2E/dr2 =/~CO2 .
(4)
where r o is the "black disc radius", r o = 0.8 fm for proton-proton scattering [6], 0 is the scattering angle in the center o f mass. The scattering cross section is proportional to the square o f the matrix element and hence to I'I'14 , d o / d t ~ IHdr[ 2 ~ qz exp(_2~a. 2)
d~/dt ~ exp(
bt)
b = - 2a(r2/2MpC 2 sin( 1 0 ) 2 ) .
Letting 0 = 90 °,/~ ~ Mp, it follows that b ~ 11 (GeV/c) 2, in good agreement with slopes found for elastic scattering at small values of t. 3) The moments o f inertia are of reasonable size. The energy of a symmetrical top is [2, p. 125] Ero t = B T ( T + I )+(A -- B ) S 2 B = hZ/2IB;
A =/i2/21 A
qJv=NHv(x/~r)exp(
L =/COro t =/~ "
ct = COtt/21rh = 0.8 X 1026 cm - 2
308
i
(8)
Setting B = 40 MeV and (A - B) = I 0 MeV or A = 30 MeV from eq. (1), it follows that the m o m e n t o f inertia for the angular momentum normal to the figure axis is,
I A = pA r 2 = 320 MeV fm 2 .
(5)
(7)
and, after substitution from eq. (6) we find that
To compute a number we may insert, for example, the proton mass in fomula (4), tt ~ 1000 MeV, and find d2E/dr 2 = 1000 MeV/fm 2, in reasonable agreement with d2E/dr 2 ~ 3000 MeV/fm 2 computed from the proton-proton cross section [4]. In fact one expects to compute a slightly smaller number from eq. (4) because the vibrational term in eq. (1) represents an average curvature over two vibrational states of the octet (ISI = 1,0) whereas the value of 3000 MeV/fm 2 from ref. [4] is specifically for the minimum of the interaction energy E(r) where the curvature is expected to be maximal. 2) The wave function for vibration products exponential dependence of scattering cross-section on momentum transfer t and predicts a reasonable slope for it. Namely, the wave function for vibration [2, p. 83] is, ½ c z r 2 ) ~ e x p ( - ½ ~ r 2)
(6)
I B = tzBr2 = 430 MeV fm 2
(9a)
and along the figure axis is, (9b)
Taking r A ~ r B ~ 0.8 fin for th'e effective separation distance in proton-proton scattering we find the reduced mass associated with rotation to be about 600 MeV, which seems reasonable. The moments of inertia for the meson system, from eq. (2), are about 1/5 th those of the baryon system, also reasonable. We may compute a representative rotation frequency according to, I ~ 400 MeV fin 2
COrot ~ Ii/I= 1.4 X 10 +23 s e c - I .
(1,0)
Volume 40B, number 3
PHYSICS LETTERS
The near equality of COrot and COve, eq. (3), suggests that there may be strong resonant coupling between rotation and vibration, and this would explain why S appears both as a rotational and as vibrational quantum number. We may expect that such resonances occur both for harmonics and subharmonics of vibration, and affect both iso-rotation (quantum number 7") and hyper-rotation (quantum nunther S) to be point that they are uncoupled from each other, and instead are coupled to vibration simultaneously, if they are uncoupled, the rotational energy may be written, Ero t =
(h2/2IT) T(T+ 1)+(h2/21s)ISI[ISI + 11
= e~T(T+ I)+NSI[[S[
+ 1]
= ~lSl+ ~ r ( T + :) + # s 2 .
(: 1)
The linear term in IS[ then combines with the vibrational term, and so the situation of T a n d S uncoupled may not be clearly distinguishable form the coupled situation. 4) If we assume that the hyper-rotation has a frequency twice that o f vibration, namely coS = 2covib, then it seems that the A T = 1/2 rule may be clarified. Let us assume that the proper hy~per-rotation quantum number is half integral, S = S/2, S to be called here the hyper-spin. Let us rewrite the rotational enerogy of a symmetrical top, eq. (8), in terms o f hyper-spin S, as, o j_ /:'rot =BT(T+I) + (A - B ) S 2 ; S = 2 S . (12) We set now B = 40 MeV and (,4 - B) = 40(1 +e) MeV from eq. (1), so that A = 40 MeV, and e is small but not zero. As a consequence o f the hyper-spin taking half integral as well as integral values, the rotational eigenfunction becomes double valued. Namely, in terms of Euler angles 0, SOand ?( it may be written as [71, O
*rot =
O(T, S,
o
Tz) (0) exp(iSso) exp(iTzX )
(13)
where T z is the projection o f the total roational momentum T o n a space axis. The-double valued nature now becomes understandable in that so goed from 0 to 2rr while the vibrating system expands, and from 2rr to 4rr while the system contracts. The condition coS = 2cov~ may be implemented by
10 July 1972
rewriting eq. (1) in approximately equivalent form, M = 8 9 0 + [ 1 8 5 ( I S 1 + I / 2 ) - 70(1S1+1/2) 2] 02 + 4 0 [ T ( T + I ) + 6S I •
(14)
Then, from the form of eq. (12), it follows that, COvib = 2(185 MeV)/h = 2 × 1023 sec -1 ,
A=h2/2Is=hcoS=16OMeV;
(is)
COS = 4.0 X 1023 sec -1
The first bracket may now be interpreted as the energy of an anharmonic oscillator. The coefficients of first and second terms in the bracket provide an estimate of the well depth D of the curve of interaction energy versus separation distance [2, p. 114] according to, E = hcovib [ISI + I/2] - (h2co2N),(ISI + 1/2)2/4D, (16) D = 490 MeV for [SI = 1 . This well depth is in agreement with that found from p-p elastic scattering cross-sections [6, eq. (11)]. Such a shallow well cannot contain more than two vibrational levels (ISP ~< 2), which explains why the first baryon octet is complete at ISt = 2, or ISI o 1. 5) As a consequence of the hypothesis S = ½S, we may now be able to explain the so-called A T = 1/2 rule, and to dismiss it in favor of conservation of total rotational momentum. Consider for example decay of the A ° (T = 0, S = - 1 ) , according to, A ° -+ rr- +p where the A °, with T = 0, decays to a pion and proton with total isospin 3/2 or 1/2. Apparently, A T = 1/2 in this decay. Instead, the present interpretatioon allows us to write the total rotational momentum J of the A ° as J = S = 1]2, and that of the pion-proton system as Jo = 3/2 or 1/2. If the decay occurs f r o m J = 1/2 to J = 1/2, total rotational momentum is conserved. If Pauli spin angular momentum (s = 1/2 for both A ° and p), flips with an accompanyinog change of one unit of Pauli spin, then (J = 1/2)--* (9' = 3/2) is allowed with simultaneous conservation of total rotational angular momentum. The long lifetime of A ° decay may be attributed to difficulty in converting hyper-spin into isospin. In short, we are converting the A ° ' s hyperangular m o m e n t u m of (1/2)h into the (p~r)'s unit o f ( l / 2 ) h isospin, and conservation of angular mon:entum results. 300
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Consider for a second example the decay of the K meson according to, K ~ rr+zr. The K ° has T = 1/2, [S[ = 1/2. The total rotational angular m o m e n t u m may be written, for uncoupled hyper-spin and iso-spin, as ([Tz[ -+ [S[) equal to 1 or 0. The two decay pions together have T = 2, 1 or 0. Let the total (iso + hyper) angular momentum o f the kaon be converted into iso-angular m o m e n t u m of the two pions, and again angular momentum is able to be conserved. As with the A, the long life of the kaon against decay may result from difficulty of converting hyperspin into iso-spin. Considering that beta decay o f the neutron violates parity conservation, indicating a right or left handedhess for the neutron, symmetry between neutron and proton requires the proton to have a corresponding left or right handedness *. However, the (p, n) mass is computed from formula (1) be setting S = 0; and S = I S is the quantum number o f fly-wheel rotation which is the source of "handedness". Thus, we are led to suggest that there exists a "zero-point handedness" for S = 0 which can never be lost from baryon and anti-baryon ground states except by annihilation. Finally, we take up the curious question of why the (T = 0, S = 0) meson level lies above meson levels of non-zeoro spin (see formula (2)). Namely, the r~(T = 0, So= 0) has higher mass than the kaon (T = 1/2, S = 1/2) and the pion ( r = 1, ~ = 0). We suggest that the zt has three symmetrically oriented iso-spin axes, each with one unit of iso-spin, so that
310
10 July 1972
the overall spin is zero. According to this picture, the r/is an excited state; the ground state, of no internal spin, is missing, presumably the as-yet-undiscovered o meson. In fact, it might have the same mass as the pion (because the pion has no preferred space axis to orient its iso-spin vector) and so have been until now not separately identified. Identification as a particle separate from the pion could be made by finding greater ratios of 7r°/Tr- , zr°/Tr+ than would be predicted for a particle of T = 1. In this case, formula (1) would have different numerical values for the constants. * See footnote on page 307.
References [ll M. Gell-Mann, Phys. Rev. 125 (1962) 1067; S. Okubo, Progr. Theor. Phys. (Kyoto) 27 {1963) 949; J. Schwinger, Phys. Rev. Letters 12 (1964) 237; F. Gurscy and L.Radicati, Phys Rev. Letters 13 (1964) 173; M.M. Beg and V. Singh, Phys. Rev. Letters 13 (1964) 418; A.O. Barut and A. Bohm, Phys. Rev. 139 (1965) B1107. [2] G.Herzberg, Molecular spectra and molecular structure I, Diatomic molecules (Prentice Hall, Inc., New York, 1930). [3] G.Herzberg, Infra-red and Raman spectra of polyatomic molecules (D. Can Nostrand Company, Inc., New York, 1945). [4] L.M. Libby, Phys. Letters 29B (1969) 348. [51 L.M. Libby, Phys. Letters 29B (1969) 345. [6] S. Miyashita and L.M. Libby, Phys. Rev. 168 (1968) 1779. [7] H. Margenau and G.M. Murphy, Mathematics of physics and chemistry (D. Van Nostrand Company, Inc., New York, 2nd ed. 1956)p. 370.