According to Noether's theorem, all conservation laws originate from invariance of a system to shifts in a certain space. For example conservation of energy stems from invariance to time translation.

What kind of symmetry creates the conservation of mass?


Noether's theorem says that symmetries lead to conservation laws, not the converse. Conservation of mass doesn't follow from any of the obvious symmetries of nonrelativistic motion. Those symmetries are translations in space (leading to conservation of momentum), translations in time (conservation of energy), rotations (conservation of angular momentum), and boosts (i.e. changes to a frame moving at constant velocity with respect to the original frame, leading to conservation of center-of-mass motion). The algebra of these symmetries, known as Galilean transformations, involves mass as a central charge. Since it commutes with everything in sight, I think it's fair to say that there's no nontrivial associated symmetry.

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    $\begingroup$ By the way, conservation of COM motion with conservation of linear momentum implies conservation of mass. $P$ constant and $P/M$ constant $\rightarrow M$ constant. Though, of course, mass conserves in classical mechanics even when $P$ doesn't. $\endgroup$
    – Malabarba
    Jan 11 '11 at 14:59
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    $\begingroup$ For a superfluid, mass generates a change in the phase. That's because a superfluid is a superposition of states with different masses, i.e. a mass condensate. $\endgroup$
    – QGR
    Jan 13 '11 at 17:03
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    $\begingroup$ This is different from Lubos answer regarding Noether theorm, namely you are saying the converse is not true while he is saying the converse is true. Which one of you is correct? $\endgroup$
    – Revo
    Aug 23 '12 at 1:14
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    $\begingroup$ see also physics.stackexchange.com/q/24596 on the discussion of the converse Noether's theorem $\endgroup$ Jun 18 '14 at 8:44

Mass is only conserved in the low-energy limit of relativistic systems. In relativistic systems, mass can be converted into energy, and you can have processes like massive electron-positron pairs annhillating to form massless photons.

What is conserved (in theories obeying special relativity, at least) is mass energy--this conservation is enforced by the time and space translation invariance of the theory. Since the amount of energy in the mass dominates the amount of energy in kinetic energy ($mc^{2}$ means a lot of energy is stored even in a small mass) for nonrelativistic motion, you get a very good approximation of mass conservation. out of the energy conservation.

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    $\begingroup$ Interesting that this gets up voted without comments. The correctness of the above answer depends on how one defines 'mass'. Many (including yours truly) define mass as the norm of the energy-momentum vector. This vector is conserved, and so is it's norm. Thus defined, mass conservation is rigorous and independent of non-relativistic limits. $\endgroup$
    – Johannes
    Jan 23 '11 at 0:54
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    $\begingroup$ @Johannes, I'm sure you know this, but it bears pointing out that with that definition, "mass" is not additive. You can have system A and system B with their own energy-momentum vectors, and the combined system A+B will have the sum of those vectors, but the norm of the sum is not necessarily the sum of the norms. So the mass of A and B together is not necessarily the sum of the masses. That's certainly quite unlike non-relativistic mass. $\endgroup$ Feb 14 '11 at 12:58
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    $\begingroup$ @Keenan - that's correct. In SR there is two sensible ways to define mass of a multiparticle system: 1) as the sum of the rest energies of the particles, 2) as the center-of-mass energy of the whole system. 1) leads to a notion of mass that unlike the Newtonian concept is frame dependent and not conserved, 2) gives a mass that is covariant and conserved, but unlike the Newtonian concept can not be associated with individual particles. $\endgroup$
    – Johannes
    Feb 14 '11 at 18:11
  • $\begingroup$ @Johannes: Good comments. If we define the mass of a system as the total energy in the frame in which the system has zero momentum, then it's always conserved. One photon has no mass, but two photons moving in opposite directions collectively have mass. Getting back to the original question, the conservation of this sort of mass is just conservation of energy (which should hold in all frames including the rest frame), and is due to time translation invariance. But the possibility of a frame with constant zero momentum is due to momentum conservation (due to spacial translation invariance). $\endgroup$ Jun 5 '11 at 22:46
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    $\begingroup$ @Johannes: all that I would say is that "the law of conservation of mass", as used in a Chemistry class, applies to definition 1), and is not conceptualized according to definition 2), and considering the question, this is the sense in which I wrote the answer. $\endgroup$ Sep 29 '13 at 20:23

The quantity that is conserved is the squared momentum four-momentum $p^2$ of the entire universe (or any isolated system), caused by the Poincaré invariance. It also results in the conservation of total spin by the way. Wigner's classification provides some interesting further reading, too.

  • $\begingroup$ +1: This is the correct answer. Galilean invariance is the limiting case. $\endgroup$
    – Ron Maimon
    Dec 29 '11 at 18:46
  • $\begingroup$ +1: Indeed, this is the correcte answer: your symmetry group for change of reference frames defines the dispersion relation $E(p)$ which can be re-written in the form of $m =\rm{const}$. Works for Gallilean and Lorentz symmetries like charm. $\endgroup$
    – Slaviks
    Mar 10 '12 at 16:34
  • $\begingroup$ I believe this is the answer OP is looking for. Just for the sake of disambiguation, $p$ is 3-momentum, and not 4-momentum, right? $\endgroup$
    – J. Manuel
    Nov 3 '21 at 10:14
  • $\begingroup$ @J.Manuel Poincaré invariance conserves only the four-momentum $p^2 = \vec p^2 + m_0^2c^2$. But of course in many cases $\sum \vec p$ is also conserved due to a translational invariance. $\endgroup$ Nov 3 '21 at 10:18
  • $\begingroup$ @TobiasKienzler. Thanks. This clarifies things now. $\endgroup$
    – J. Manuel
    Nov 3 '21 at 11:10

Noether's relationship implies that for every conserved quantity, there is a symmetry, and vice versa.

In the real world, mass may be converted to energy. For example, uranium power plants convert about 0.1% of the uranium mass to a huge energy, according to the $E=mc^2$ formula. So up to the conventional factor $c^2$, the total energy - including the latent one - and the total mass is the same thing.

Its conservation is linked to the time-translational symmetry of the laws of physics.

In the old world "before Marie Curie", people didn't know any relativity or any other indications of relativistic physics such as radioactivity. So they believed that mass was conserved even if the energy is not included. In their understanding of the world, the total mass of the Universe was the sum of the rest masses of the electrons, protons, and other massive particles.

The conservation law for the mass defined in this way didn't lead to any symmetries because the definition only depends on parameters - masses of all point masses are parameters - and not on dynamical, time-dependence quantities such as the positions or velocities. That's why Noether's theorem didn't apply.

It is not a real flaw of Noether's theorem because, as we know today, the total mass defined in the old way - and neglecting the mass increases from velocity (kinetic energy) and other forms of energy - is actually not conserved. This is no coincidence; a world compatible with general relativity doesn't allow any mass-like quantities to be non-dynamical. All quantities describing particular objects have to be dynamical i.e. changeable, and that's why Noether's theorem always holds in the world.

  • $\begingroup$ In the Classical Mechanics, masses are constant by their definitions as constant parameters. Everything else is calculated from these constants. If in course of your calculation a mass obtains (perturbative) corrections, then it is a wrong calculation: it contradicts definitions. $\endgroup$ Jan 21 '11 at 20:42

For a superfluid, mass generates a change in the phase. That's because a superfluid is a superposition of states with different masses, i.e. a mass condensate. This also applies to supersolids. For other cases, states with different masses lie in different superselection sectors.

You might object that for say, a helium-4 superfluid, a change in phase generates the number of helium-4 atoms, and not the mass per se, and that's true. But going the other way around, mass generates a change in the phase, scaled by the mass of a helium-4 atom. The reason for this asymmetry is the total number of other forms of matter decomposes the state space into superselection sectors.

This entire analysis presupposes Galilean relativity instead of special relativity. For the latter, mass isn't conserved.


The rest mass of a particle in Classical Electrodynamics (as well as the particle charge) is a phenomenological parameter (number) in differential equations of motion. It is defined to be constant. No dynamics can change it, just by definition. It is not expressed via dynamical variables and its conservation is not due to some symmetry. $dm/dt = 0, de/dt = 0$ are experimental facts, if you like. Of course, physical models and their equations should be compatible with such definitions. Some "theoretical" hypothesis are incompatible with it, for example, self-acting particles.

If one obtains "corrections" to the particle mass (or/and charge) in course of perturbative calculations, then the theory formulation is wrong. In some theories they discard such "corrections" and call it "renormalizations". P. Dirac was unhappy about that "renormalization prescription" and invited researchers to change the original equations, not the results.

In Classical Electrodynamics the sum of particle masses $\sum m_i$ is also conserved but it does not define the system mass (the latter is just defined differently).


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