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It is conventionally assumed that the negative mass squared term in thelinear sigma model version of the pion Lagrangian is $M^2 \sim \Lambda_{\rmQCD}^2$ in powers of $N_c$. We consider the case where $M^2 \sim \Lambda^2_{\rmQCD}/N_c$ so that to leading order in $N_c$ this symmetry breaking termvanishes. We present some arguments why this might be plausible. One mightthink that such a radical assumption would contradict lattice Monte Carlo dataon QCD as function of $N_c$. We show that the linear sigma model gives a fairdescription 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 Baliet. al. turn out to be too large to resolve the case we consider from that ofthe 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 conventionallarge-$N_c$ counting holds. The physical values of quark masses for QCDcorrespond to the small quark mass limit. We find pion-nucleon couplingstrengths are reduced to order ${\cal O}(1)$ rather than ${\cal O}(N_c)$. Underthe assumption that in the large $N_c$ limit the sigma meson mass is largerthan that of the omega, and that the omega-nucleon coupling constant is largerthan that of the sigma, we argue that the nucleon-nucleon large range potentialis weakly attractive and admits an interaction energy of order $\Lambda_{\rmQCD}/N_c^{5/2} \sim 10$ MeV. With these assumptions on coupling and masses,there is no strong long range attractive channel for nucleon-nucleoninteractions, so that nuclear matter at densities much smaller than that wherenucleons strongly interact is a weakly interacting configuration of nucleonswith strongly interacting localized cores.

The Large N Limit with Vanishing Leading Order Condensate for Zero Pion Mass