According to Noether's theorem, every symmetry implies and conserved quantity. And, from Einstein's equation, every mass have an amount of energy associated.

Can it say that the mass conservation is a consequence of energy conservation associated to homogeneity of time?

PD.: is not duplicated question

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    $\begingroup$ Technically, yes, if you take the relativistic definition of mass, which essentially identifies it with energy. Historically, however, "mass" in "mass conservation" had a more restrictive meaning under which it is not conserved because it can be "converted" into energy, e.g. in nuclear reactions, see Conservation of mass $\endgroup$ – Conifold May 5 '17 at 2:48
  • $\begingroup$ Related: physics.stackexchange.com/q/2690/2451 $\endgroup$ – Qmechanic May 5 '17 at 5:13

In classical mechanics mass is conserved, conservation of mass is not valid in relativistic physics unless you consider the so-called "invariant mass" of a closed system. The sum of the masses of particles can change due to interactions. In classical mechanics this sum of masses is conserved and this follows from conservation of energy.

If we have an initial state of $N$ free particles and a final state of $M$ free particles then the total energy is conserved. We then have:

$$\sum_{j=1}^{N}m_j v_j^2 = \sum_{j=1}^{M}m'_j w_j^{2}$$

where the $v_j$ and $m_j $are the initial velocities and masses, while the $w_j$ and $m'_j$ are the final velocities and masses. The particles being free particles only have a kinetic energy, but this still a completely general statement because in the intermediary stage, arbitrary interactions can have taken place. We're then only neglecting radiation escaping from the system (note that radiation is an extremely relativistic phenomenon).

Then suppose that the exact same process is observed in another frame that moves with velocity $\vec{u}$ relative to the original frame. According to Galilean invariance, we also have conservation of energy in this new frame, therefore:

$$\sum_{j=1}^{N}m_j (\vec{v}_j-\vec{u})^2 = \sum_{j=1}^{M}m'_j (\vec{w_j}-\vec{u})^{2}$$

Expanding out the square of each term:

$$ (\vec{v}_j-\vec{u})^2 = v_j^2 - 2\vec{v}_j\cdot\vec{u} + u^2$$

on both sides, and using that $\vec{u}$ is arbitrary, yields conservation of momentum and conservation of mass.

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  • $\begingroup$ please note that the OP is dragging in general relativity. You say " According to Galilean invariance" . mass is not conserved in GR $\endgroup$ – anna v May 5 '17 at 3:26

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