 Common-pressure melting-point slope

When the solid and liquid have the same pressure, equality of chemical potentials implies $-(s_l-s_s)dT+(1/{\rho }_l-1/{\rho }_s)dp=0$. With latent heat of fusion $L=T_m(s_l-s_s)$, this gives the displayed slope. If solid is less dense than liquid, raising the common pressure depresses the melting temperature. $T_m$ is absolute temperature. This coexistence-curve derivative differs from a thermomolecular pressure difference between equal-density phases.