Tampan, Paminggir people. Lampung region of Sumatra, 19th century, 39 x 40 cm. From the library of Darwin Sjamsudin, Jakarta. Photograph by D Dunlop. |

Let particle P have a rest mass of $m$ and enthalpy marked by $H$. If P is Newtonian then as discussed earlier the enthalpy and mass are approximately related as

$mc^{2} \simeq \left| \, H \, \vphantom{\sum^{2}} \right|$

where $c$ is a constant number. Moreover, we have already shown that the Lorentz factor of a Newtonian particle is always close to one

$\gamma \simeq 1$

and so P is always at rest or in slow motion. But the mechanical energy $E$ of a particle is approximately

$E \simeq \gamma m c^{2}$

Thus for a Newtonian particle, the mechanical energy and the modulus of the enthalpy are just about the same as each other

$E \simeq \left| \, H \, \right|$

The absolute-value signs can usually be ignored because ordinary particles are composed from electrons, neutrons and protons which all have an enthalpy greater than zero (see the WikiMechanics spreadsheet for details). So the enthalpy and the mechanical energy are almost interchangable for any macroscopic particle. That is, if we ignore collisions with anti-particles and processes like annihilation then

$H \simeq E \simeq m c^{2}$

But enthalpy is conserved for *all* particles and conditions, so the mechanical energy and mass must be approximately conserved too. Finally recall that photons have no mass. So for interactions with photons, the mass of a Newtonian particle is usually a dependable constant. This is an important part of the stability that characterizes Newtonian particles.