  It is conventionally assumed that the negative mass squared term in the
linear sigma model version of the pion Lagrangian is $M^2 \sim \Lambda_{\rm
QCD}^2$ in powers of $N_c$. We consider the case where $M^2 \sim \Lambda^2_{\rm
QCD}/N_c$ so that to leading order in $N_c$ this symmetry breaking term
vanishes. We present some arguments why this might be plausible. One might
think that such a radical assumption would contradict lattice Monte Carlo data
on QCD as function of $N_c$. We show that the linear sigma model gives a fair
description of the data of DeGrand and Liu both for $N_c = 3$, and for variable
$N_c$. The values of quark masses considered by DeGrand and Liu, and by Bali
et. al. turn out to be too large to resolve the case we consider from that of
the conventional large $N_c$ limit. We argue that for quark masses $m_{q} \ll
\Lambda_{\rm QCD}/N_c^{3/2}$, both the baryon mass and nucleon size scale as
$\sqrt{N_c}$. For $m_{q} \gg \Lambda_{\rm QCD}/N_c^{3/2}$ the conventional
large-$N_c$ counting holds. The physical values of quark masses for QCD
correspond to the small quark mass limit. We find pion-nucleon coupling
strengths are reduced to order ${\cal O}(1)$ rather than ${\cal O}(N_c)$. Under
the assumption that in the large $N_c$ limit the sigma meson mass is larger
than that of the omega, and that the omega-nucleon coupling constant is larger
than that of the sigma, we argue that the nucleon-nucleon large range potential
is weakly attractive and admits an interaction energy of order $\Lambda_{\rm
QCD}/N_c^{5/2} \sim 10$ MeV. With these assumptions on coupling and masses,
there is no strong long range attractive channel for nucleon-nucleon
interactions, so that nuclear matter at densities much smaller than that where
nucleons strongly interact is a weakly interacting configuration of nucleons
with strongly interacting localized cores.
