Analytical Proof that g_{A} → 0 in the Ultra-Relativistic Limit for the Harmonic Oscillator Relativistic Constituent Quark Model ^{∗}

John P. Costella and Bruce H. J. McKellar

School of Physics, The University of Melbourne, Parkville, Victoria 3052, Australia (6 February 1994)

Abstract

We show analytically that g_{A} → 0 in the ultrarelativistic limit for the harmonic oscillator relativistic constituent quark model.

I. PROOF

Our notation essentially follows Berestetskii and Terent’ev [2–4]. Upon application of the Melosh transformation [5] one finds that g_{A} is reduced from its non-relativistic value by a factor

Z ≡ ∫ dΓ |Φ(M_{0}^{2})|^{2} \{1 − 2 \frac{Q_{⊥}^{2}}{Q_{⊥}^{2} + [m + (1 − η)M_{0}]^{2}} \} .

(This same result—without the factor of 2 in the second term—holds for the reduction in electric dipole moment [1] and the contribution of the quark anomalous magnetic moment [6].) We parametrise the harmonic oscillator potential by

Φ(M_{0}^{2}) = A exp(−\frac{M_{0}^{2}}{12α^{2}} ) .

The ultra-relativistic limit, α/m → ∞, can be realised here by taking m = 0 with α fixed. We then have

Z = 1 − 2 \frac{I}{N}  (1)

where

I ≡ \frac{4}{(2π)^{2}} ∫ \frac{d^{2}q_{⊥} dξ}{ξ(1 − ξ)} \frac{d^{2}Q_{⊥} dη}{η(1 − η)} \frac{Q_{⊥}^{2}}{Q_{⊥}^{2} + (1 − η)^{2}(\frac{Q_{⊥}^{2}}{η(1 − η)} + \frac{q_{⊥}^{2}}{ηξ(1 − ξ)})}

× exp \{−\frac{Q_{⊥}^{2}}{12α^{2}η(1 − η)} − \frac{q_{⊥}^{2}}{12α^{2}ηξ(1 − ξ)}\}

∗This paper is taken directly from Appendix D of ref. [1].

and

N ≡ \frac{4}{(2π)^{2}} ∫ \frac{d^{2}q_{⊥} dξ}{ξ(1 − ξ)} \frac{d^{2}Q_{⊥} dη}{η(1 − η)} exp \{−\frac{Q_{⊥}^{2}}{12α^{2}η(1 − η)} − \frac{q_{⊥}^{2}}{12α^{2}ηξ(1 − ξ)}\} .

The quantity N takes care of the normalisation of Φ; the common factors of 4/(2π)^{2} are introduced for later convenience. Converting Q_{⊥} and q_{⊥} to plane polar coördinates, the angular integrations are trivially equal to 2π, and hence

I = 4∫ \frac{qQ dq dQ dξ dη}{ξ(1 − ξ)η(1 − η)} \frac{Q^{2}}{\frac{Q^{2}}{η} + \frac{(1 − η)^{2}q^{2}}{ηξ(1 − ξ)}} exp(−\frac{Q^{2}}{12α^{2}η(1 − η)} − \frac{q^{2}}{12α^{2}ηξ(1 − ξ)}) ,

N = 4∫ \frac{qQ dq dQ dξ dη}{ξ(1 − ξ)η(1 − η)} exp(−\frac{Q^{2}}{12α^{2}η(1 − η)} − \frac{q^{2}}{12α^{2}ηξ(1 − ξ)}) ,

where Q ≡ |Q_{⊥}| and q ≡ |q_{⊥}| are integrated from 0 to ∞. Now change to the variables

X ≡ Q^{2},
Y ≡ q^{2},

and define the quantities

β ≡ 12α^{2}η(1 − η),
γ ≡ 12α^{2}ηξ(1 − ξ),

to obtain

I = ∫_{0}^{1}dξ ∫_{0}^{1}dη ∫_{0}^{∞}dX ∫_{0}^{∞}dY exp(−\frac{X}{β} − \frac{Y}{γ} ) \frac{1}{ξ(1 − ξ)η(1 − η)} (\frac{1}{η} + \frac{(1 − η)^{2}}{ηξ(1 − ξ)} \frac{Y}{X} )^{−1} ,
N = \{∫_{0}^{1}dξ ∫_{0}^{1}dη \frac{1}{ξ(1 − ξ)η(1 − η)}\} · \{∫_{0}^{∞}exp(−\frac{X}{β} )dX\} · \{∫_{0}^{∞}exp(−\frac{Y}{γ} )dY \} .

The X and Y integrals for N can be done immediately, leaving

N = ∫_{0}^{1}dξ ∫_{0}^{1}dη \frac{βγ}{ξ(1 − ξ)η(1 − η)};

substituting back in the values of β and γ gives

N = 144α^{4}∫_{0}^{1}dξ ∫_{0}^{1}dη η.

N can now be finished off completely, with the ξ-integration giving 1, and the η-integration giving 1/2, yielding

N = 72α^{4}.  (2)

Returning, now, to I, concentrate on the X and Y integrations first. To this end, define the quantities

δ ≡ \frac{1}{η},

ε ≡ \frac{(1 − η)^{2}}{ηξ(1 − ξ)},

ϕ ≡ \frac{1}{ξ(1 − ξ)η(1 − η)},

which simplifies the integral to

I = ∫_{0}^{1}dξ ∫_{0}^{1}dη ∫_{0}^{∞}dX ∫_{0}^{∞}dY \frac{ϕ}{δ + ε\frac{Y}{X}} exp(−\frac{X}{β} − \frac{Y}{γ} ) .

Concentrating on the Y -integral,

I_{Y} ≡ ∫_{0}^{∞}dY \frac{exp(\frac{−Y}{γ} )}{δ + ε\frac{Y}{X}},

define a new variable

W ≡ \frac{1}{γ} \{Y + \frac{δX}{ε} \} ,

which gives [7, Eq. 5.1.1]

I_{Y} = \frac{X}{ε} exp(\frac{δX}{γε} ) ∫ ∞_{\frac{δX}{γε}} \frac{e^{−W}}{W} dW
≡ \frac{X}{ε} exp(\frac{δX}{γε} ) E_{1}(\frac{δX}{γε} ) ,

where E_{1}(z) is the exponential integral. Defining another two convenient quantities

ζ ≡ \frac{δ}{γε},

ω ≡ \frac{1}{βζ} − 1,

one can now simplify the format of the X-integral:

I_{X} ≡ \frac{1}{εζ^{2}} ∫_{0}^{∞}dX e^{−ωX}X E_{1}(X).

It will be noted that I_{X} is now in the form of a Laplace transform, i.e.

f̃(s) ≡ L{f(t)} ≡ ∫_{0}^{∞}dt e^{−st}f(t).

Using the fact the Laplace transform of the exponential integral is given by [8, Eq. 19.1]

L\{E_{1}(t)\} = \frac{1}{s} ln(s + 1),

and that

L\{t f(t)\} = −\frac{d}{ds}L\{f(t)\} ,

one then finds that

I_{X} = \frac{ln(ω + 1)}{εζ^{2}ω^{2}} − \frac{1}{ω(ω + 1)}.

Expanding the quantities ω, ζ and ε back out in terms of ξ and η, one has

ε ≡ \frac{(1 − η)^{2}}{ηξ(1 − ξ)},

ζ ≡ \frac{1}{12α^{2}η(1 − η)^{2}} ,

ω ≡ −η,

giving

I_{X} = 144α^{4}ξη^{3}(1 − ξ)(1 − η)^{2} \{\frac{ln(1 − η)}{η^{2}} + \frac{1}{η(1 − η)}\} .

Returning now to the full expression including the ξ and η integrals, we have

I = 144α^{4}∫_{0}^{1}dξ ∫_{0}^{1}dη η^{2}(1 − η) \{\frac{ln(1 − η)}{η^{2}} + \frac{1}{η(1 − η)}\} .

As ξ has now dropped out of our equations, it can be trivially integrated to yield a factor of 1. Expanding out the braces, one finds that

I = 144α^{4}∫_{0}^{1}(1 − η) ln(1 − η) dη + 144α^{4}∫_{0}^{1}η dη.

Noting that

\frac{d}{dx} \{−\frac{1}{2}(1 − η)^{2} ln(1 − η) + \frac{1}{4}(1 − η)^{2}\} = (1 − η) ln(1 − η),

one finally obtains the desired result

I = 36α^{4}.  (3)

Inserting (2) and (3) into (1) gives the final result of

Z = 1 − 2 \frac{36α^{4}}{72α^{4}}
= 0.

(It should be noted that, for the equivalent electric dipole moment and anomalous magnetic moment calculations, the result is 1/2, rather than zero.)

REFERENCES

[1] J. P. Costella, Honours Research Report, The University of Melbourne (November 1990).

[2] V. B. Berestetskii and M. V. Terent’ev, Yad. Fiz. 25 (1977) 653.

[3] M. V. Terent’ev and V. B. Berestetskii, Yad. Fiz. 24 (1976) 1044.

[4] M. V. Terent’ev, Yad. Fiz. 24 (1976) 207.

[5] W. Melosh, Phys. Rev. D 9 (1974) 1095.

[6] I. G. Aznauryan and N. L. Ter-Isaakyan, Yad. Fiz. 31 (1980) 1680.

[7] M. Abramowitz and I. E. Stegun, Handbook of Mathematical Functions (Dover, New York, 1964).

[8] G. E. Roberts and H. Kaufman, Table of Laplace Transforms (W. B. Saunders, 1966).