= Common-pressure melting-point slope
{title2=$dT_m/dp=-T_m(1/\rho_s-1/\rho_l)/L$}
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.
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