Stars as Laboratories for Fundamental Physics - MPP Theory Group

112 Chapter 3
One may also consider the Yukawa coupling of scalar (vector) bosons
to baryons (Grifols and Massó 1986; Grifols, Massó, and Peris 1989b).
In this case one may use the Compton process γ + 4 He → 4 He + ϕ on
a helium nucleus for which the emission rate is given mutatis mutandis
by the same formula as for nonrelativistic electrons. Assuming the
same coupling to protons and neutrons the emission rate is coherently
enhanced by a factor 4 2 . Moreover, in Eq. (3.22) one must replace
Y e with Y He (number of 4 He nuclei per baryon) which is 1 for pure
4
helium, m e is to be replaced by m He ≈ 4m u , and α → 4α to account
for the coherent photon coupling. Pulling these factors together one
finds ϵ = α ϕN 0.7×10 23 erg g −1 s −1 for pure helium. Because the heliumburning
lifetime argument limits this energy-loss rate to 10 erg g −1 s −1
one finds
α ϕN ∼ < 1.5×10 −22 (3.47)
as a limit on the coupling of a scalar boson to a nucleon N. Again, for
vector bosons this limit is a factor of 2 more restrictive.
3.6.3 Long-Range Forces
Scalar or vector particles mediate long-range forces between macroscopic
bodies. For pseudoscalars this is not the case because their CPconserving
coupling to fermions has a pseudoscalar structure, i.e. in the
nonrelativistic limit they couple to the fermion spin. Therefore, even
if the mass of the new particles is very small or exactly zero, they do
not mediate a long-range force between unpolarized bodies. The residual
force caused by the simultaneous exchange of two pseudoscalars is
found to be extremely small (e.g. Grifols and Tortosa 1994).
Consider a scalar of mass m = λ −1 (Compton wave length λ) which
couples to nucleons with a Yukawa strength g. It mediates an attractive
force between two nucleons given in terms of the potential
−(g 2 /4π) r −1 e −r/λ . Two macroscopic test bodies of geometric dimension
much below λ are then attracted by virtue of the total potential
m 1 m 2
V (r) = −G N (1 + βe −r/λ ), (3.48)
r
where G N = m −2
Pl is Newton’s constant, m Pl = 1.22×10 19 GeV is the
Planck mass, m 1 and m 2 are the masses of the bodies, and
β ≡ g2 m 2 Pl
. (3.49)
4π u 1 u 2
Here, u 1,2 = m 1,2 /N 1,2 with N 1,2 the total number of nucleons in each
body. Apart from small binding-energy effects which are different for