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Is Geodetic Precession of Massive Neutrinos the solution to the

arXiv:hep-ph/9304260v1 16 Apr 1993UW/PT-93-04Is Geodetic Precession of Massive Neutrinos the solution to theSolar Neutrino Problem?Stamatis Vokos∗Department of Physics, FM-15University of Washington,Seattle, WA 98195(April 20, 2018)AbstractWe examine a recent suggestion by Milburn that slightly massive electronneutrinos, produced left-handed at the core of the sun, suffer geodetic preces-sion adequate to render them right-handed (and therefore sterile) in sufficientnumbers to solve the solar neutrino problem.In that light, we perform acomplete, general-relativistic calculation of the geodetic spin precession of anultrarelativistic particle in the Schwartzschild metric. We conclude that theeffect is negligible, in disagreement with Milburn’s analysis.Typeset using REVTEX1

I. INTRODUCTIONRecently, Milburn [1] suggested a solution to the solar neutrino problem which doesnot require any assumptions beyond the existence of a small, but non-zero, mass for theelectron neutrinos, consistent with experimental limits (mν ≤10 eV). These neutrinos areproduced in the core of the sun with typical values for γ = Eν/mν of the order of 104–106.

Ifproduced eccentrically, they will suffer a small bending due to the gravitational pull of thesun. Milburn argued that these neutrinos undergo a Thomas precession of their spin givenby the same formula relating the spin precession angle of charged particles in an acceleratorto their bending angle [2],θp = −γθb ,(1.1)where in the case at hand θb is the angle of gravitational bending.

The smallness of θbis overcompensated by the magnitude of γ, which for reasonable values of the parameters,results in an adequate repolarization of the originally left-handed neutrinos, turning theminto right-handed, and hence sterile for the purposes of weak interactions.Despite its ingenious simplicity, this suggestion is wrong, inasmuch as it is based on theaforementioned formula. Spin precession calculations have been carried out in the specialcases of circular orbits or arbitrary orbits with small velocities (in the sense of the PPNformalism) [3].

Since both special cases are clearly inapplicable to our current application,we perform, in this letter, the complete general-relativistic calculation of the spin precessionof a particle in the Schwartzschild geometry. We obtain,θp = −12γ θb ,(1.2)which demonstrates that for ultrarelativistic particles, θp ≪θb, and therefore unable toprovide us with a satisfactory rate of left- to right-handed neutrino conversion.2

II. THE CALCULATIONWe consider a test particle approaching a body of mass M. Its original direction is alongφ = 0, while its final direction is along the asymptote φ = π+θb.

The Schwartzschild metricisds2 = B(r)dt2 −A(r)dr2 −r2dθ2 −r2 sin2 θdφ2 ,(2.1)whereB(r) = A(r)−1 = 1 −2m/r ,(2.2)with m = GM/c2. In the remaining analysis we use units with G = 1 = c. Instead ofconsidering this coordinate frame, it is useful to construct an orthonormal frame (vierbein),eα µ, and its inverse eα µ, wheregµν = eαµeβνηαβ .

(2.3)This frame is given bye0t =√B,e1r = 1/√B,e2θ = r,e3φ = r sin θ . (2.4)To separate orthonormal indices from coordinate indices, we will use letters from the begin-ning of the greek alphabet (α, β, γ, .

. .) for the former, while the latter shall be denoted bygreek letters from the middle of the alphabet (µ, ν, .

. .

). Then the components of any vectorV can be expressed asV µ = eαµV α ,V α = eαµV µ .

(2.5)In particular, the four-velocity of the particle can be expressed asuµ = (˙t, ˙r, ˙θ, ˙φ) ,uα = (√B ˙t, ˙r/√B, r ˙θ, r sin θ ˙φ) ,(2.6)where ˙ =dds is the derivative with respect to the proper time of the particle. Covariantderivatives are given in terms of the connection coefficients ωα β γ.

The only non-vanishingsuch coefficients are3

ω001 = ω010 = ddr√B ,(2.7)ω212 = ω313 = −ω221 = −ω331 = −1r√B ,(2.8)ω332 = −ω323 = r−1 cot θ . (2.9)The spin vector of the particle will in general be Fermi-Walker transported along an arbitrarypath xβ(s) under external non-gravitational forces [3],uα∇αSβ = −Sγaγuβ ,(2.10)where aβ = Duβds is the four-acceleration.

However, in the absence of non-gravitational forcesthe four-acceleration is zero and the particle will move along a geodesic, while its spin willbe parallel-transported,uα∇αSβ = 0 . (2.11)Thomas precession is a purely kinematical effect which comes about when there is a non-zerofour-acceleration, for instance in an electron moving around a nucleus or in an accelerator.For a particle moving along a geodesic, the Thomas precession is identically zero.

Therefore,no direct analogy with special relativistic electrodynamics can be drawn and the reasoningleading to (1.1) is incomplete.However, the spin will still experience geodetic precession. Eq.

(2.11) above becomes incomponents,dSαds + ωβαγuβSγ = 0 . (2.12)Explicitly,dS0ds = −(√B)′u0S1 ,(2.13)dS1ds = −(√B)′u0S0 +√Br u2S2 +√Br u3S3 ,(2.14)dS2ds = −√Br u2S1 + cot θru3S3 ,(2.15)dS3ds = −√Br u3S1 −cot θru3S2 .

(2.16)4

For motion in the equatorial plane, θ = π/2 and u2 = 0. Furthermore the spin is a spacelikevector, which in the rest frame of the particle has vanishing zeroth component.

Given thatthe four-velocity is a timelike vector with vanishing spatial components in the rest-frame ofthe particle implies that the spin is orthogonal to the four-velocity. This allows us to expressS0 as a linear combination of the spatial components of S, namelyS0 = 1u0(u1S1 + u3S3) .

(2.17)Substituting into Eq. (2.14) gives three independent equationsdS1ds = −(√B)′u1S1 + √Br−(√B)′!u3S3 ,dS2ds = 0 ,(2.18)dS3ds = −√Br u3S1 .We can simplify further the equations of motion of the spin components, in view of thenon-uniqueness of this orthonormal basis {eα µ}.

We can use the freedom of choosing anappropriate basis, in which the spin precession is evident. This is achieved by going to therest frame of the particle.

The basis corresponding to the rest-frame {¯eα µ} is constructedthrough a local Lorentz transformation with the following properties: (a) the resulting zerothcomponent of the spin vanishes and (b) the four velocity takes on the form (1,⃗0). Explicitly,¯eβ = eαLαβ ,(2.19)withuαLαi = 0 ,uαLα0 = 1 ,(2.20)Then, ¯ei (i = 1, 2, 3) are orthogonal to the four-velocity, and we are guaranteed to get ¯S0 = 0when we resolve S along the new basis,S = Sαeα = ¯Si¯ei .

(2.21)To construct the requisite Lorentz transformation we note that the vanishing of the α = 0component of5

¯Sα = SγLγα ,(2.22)together with S · u = 0, givesLi 0L0 0= uiu0 . (2.23)Keeping in mind that a Lorentz transformation matrix satisfiesηαδLαβLδγ = ηβγ ,(2.24)we obtainLα0 = uα .

(2.25)One can similarly obtain the remaining entries to arrive atLαβ =u0u10u3u1 1 + (u1)21+u0 0u1u31+u00010u3u1u31+u00 1 + (u3)21+u0. (2.26)The equations of motion for the spin components in the rest frame can be inferred from(2.18) with the aid of the equation for the vanishing of the four-acceleration,duαds + ωβαγuβuγ = 0 .

(2.27)We obtaind ¯S1ds = ¯S3(r −2m + (r + 3m)u0)u3r2√B(1 + u0),d ¯S2ds = 0 ,(2.28)d ¯S3ds = −¯S1(r −2m + (r + 3m)u0)u3r2√B(1 + u0).Next we replace the derivatives with respect to proper time with derivatives with respect tocoordinate time. Recalling that6

u0 =√B ˙t ,u3 = r ˙φ ,(2.29)Eq. (2.28) obtainsd⃗Sdt = ⃗Ω× ⃗S ,(2.30)where we dropped the bars for notational simplicity, and Ω1 = 0 = Ω3, whileΩ2 ≡Ω= r −3mr+mr(1 + u0)!1q1 −2mrdφdt .

(2.31)How can we turn the knowledge of ⃗Ωinto an expression for the geodetic precession angleover the whole trajectory of the particle? First, we recall that the Schwartzschild geometryadmits two Killing vectors, namely∂∂t and∂∂φ.These two isometries correspond to twoconstants of the motion,E ≡(1 −2mr )˙t ,L ≡r2 ˙φ .

(2.32)For a particle coming in from infinity,E = ˙t|r→∞. (2.33)Asymptotically, the Schwartzschild metric becomes Minkowskian, ds2 = dt2 −d⃗x2, henceE = (1 −d⃗x2dt2 )−12r→∞≡γ .

(2.34)We therefore obtain,u0 =γq1 −2mr. (2.35)DenotingS± = S1 ± iS3 ,(2.36)e(φ) = Ω/dφdt ,(2.37)the spin precession equations may be recast in the form7

dS±dφ = ∓ie(φ)S± ,(2.38)with solutionS±(φf) = S±(φi)e∓iR φfφi dφ e(φ) . (2.39)In the absence of gravity e(φ) = 1, and if φi = 0, then φf = π, whereas in the presence ofgravity e(φ) ̸= 1 and φf = π + θb.

The total geodetic deviation is∆S± = S±(π + θb) −S(0)± (π)= S±(π + θb) + S±(0) . (2.40)An ultrarelativistic particle follows approximately the same geodesic as light, which to firstorder in m/b is given byu ≡1r = 1b sin φ + 3m2b2 (1 + 13 cos 2φ) ,(2.41)where b is the impact parameter.

With the aid of Eqs. (2.35) and (2.41),e(φ) = 1 −3mu√1 −2mu +muγ + √1 −2mu ,(2.42)= 1 −mu(2 −11 + γ) + · · · .

(2.43)Substituting into Eq. (2.40) gives∆S±S±(0) = e∓i(π+ 12θb/γ) + 1(2.44)= ±i θb2γ + · · ·(2.45)This clearly agrees with the expectation that for light (γ →∞), there is no excess spinprecession.III.

DISCUSSIONWe have shown that geodetic precession of massive, initially left-handed, neutrinos mov-ing along geodesics in the Schwartzschild geometry, cannot produce enough sterile neutrinos8

to solve the solar neutrino puzzle. In fact, the answer is O(∞/γ) = O(⇕/E), and conformswith conventional field-theoretic experience, in which helicity flips vanish in the masslessfermion (or ultrarelativistic) limit.

However, one may argue that our analysis is steeped inthe spirit of general relativity and may rely on some special features which may not be sharedby some other interaction (i.e. one which does not satisfy the principle of equivalence).

Insuch a case, it could be imagined that a small deflection due to a non-gravitational “fifthforce” interaction could be amplified to yield the requisite sterilizing effect suggested byMilburn. But how can an interaction produce counter-intuitive O(E/⇕) helicity-flip proba-bilities?

Milburn’s argument suggests that kinematics, not dynamics, can yield this effect.After all, in the case of electrons deflected at SLAC [2], the vector couplings of the QED la-grangian do not even connect left- and right-handed spinors. But if such kinematical effectsheld for other interactions, why would gravity (which, in many respects, can be modelledby graviton exchange amplitudes) yield such a dramatically different answer?

It is, there-fore, worth asking whether our result carries over to other interactions, within the scope ofMilburn’s suggestion.ACKNOWLEDGMENTSI am indebted to Cosmas Zachos for bringing this issue to my attention and for usefulconversations. I am also grateful to David Boulware and Lowell Brown for several invaluablesuggestions.9

REFERENCES∗This work was supported by DOE grant DE-FG06-91ER40614. [1] R. H. Milburn, “Gravitational Sterilization of Neutrinos and the Solar Puzzle”, TuftsUniversity preprint TUHEP-93-01.

[2] P. S. Cooper et al, Phys. Rev.

Lett. 42, 1386 (1979) and references therein.

[3] See any book on general relativity, for instance N. Straumann, General Relativity andRelativistic Astrophysics (Springer-Verlag, New York, 1984).10

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