– page 2, middle of paragraph 2: Op(H) is sending functions to functions right?

– page 3, line 4: typo in “describes”.

]]>I think the confusion may be caused by (a) mistakenly equating mass-energy with the invariant mass ; and (b) mistakenly assuming that invariant mass is additive (i.e. that the invariant mass of a system equals the sum of the invariant mass of the components). If one makes both of these mistakes, it may seem that mass can “disappear”. For instance in Einstein’s original paper (a beautiful and short paper, by the way, and well worth reading), he considers a stationary body of mass m emitting two photons in opposite directions with total energy E, and shows that the remaining body now has a mass of . Since photons have zero *invariant* mass (but positive mass-energy), it does look like mass has been converted into energy here, but that is only because of the false assumptions (a) and (b).

This Wikipedia article gives a fairly accurate discussion of these topics:

http://en.wikipedia.org/wiki/Mass-energy_equivalence

Dear John B.: I think that the symmetry associated with the invariant mass of a system is the action of translating the system in spacetime along the four-velocity vector of that system. Note that when one chooses an inertial frame in which the system has zero total momentum, then the invariant mass and mass-energy agree to second order and so must yield the same symmetry, and indeed in this case translation along the four-velocity vector is just time translation.

]]>Mass and energy are (approximately) separately conserved, so each one gets a symmetry. We’ll describe the mass one by an “infinitesimal symmetry” — an operator — and the energy one by a similar operator . Saying that these two operators are infinitesimal symmetries means that physically realistic states are in the kernels of each operator.

Now here’s where I’m sort of guessing. Since the combination of mass and energy is conserved exactly, the sum of these two infinitesimal symmetries is exact. That is, physically realistic states are in the kernel of the sum. In particular, the intersection of the kernels is in the kernel, but there are states which and don’t kill, but their errors exacctly cancel each other out.

]]>If “mass” really means “invariant mass”, as Terry T. suggests – that is, the magnitude of the energy-momentum 4-vector – then it’s not so clear what the corresponding symmetry is supposed to mean, physically. I’ll try to come up with a snappy answer, but I’m not sure I’ll succeed.

(The debate over what “mass” really means is endless and unproductive, but modern particle physicists always use it to mean “invariant mass”, since the other sort might as well be called “energy” – especially if you use units where c = 1.)

All this is for relativistic physics. For nonrelativistic quantum mechanics, governed by the Galilei group instead of the Poincare group, mass is a cocycle! In other words, a quantum particle satisfying Schrodinger’s equation gives a representation of the Galilei group *only up to phase* – and this phase fudge factor depends on the particle’s mass.

There’s a nice (but advanced) discussion of this near the end of Guillemin and Sternberg’s *Symplectic Techniques in Physics*.

First, remember that since Einstein we know that mass-energy is what’s really conserved. That is, “mass” and “energy” are really different ways of looking at the same thing, and we convert from units of mass to units of energy by multiplying a mass by the square of the speed of light: .

Now, what’s the quantity that’s Noether-conjugate to mass-energy? It’s time! That is, conservation of mass-energy is equivalent to the fact that the laws of physics stay the same as time goes by. In fact, if mass-energy was *not* conserved, then the laws of physics would change over time, we wouldn’t be able to replicate experiments, and science couldn’t ever get off the ground in the first place!

It’s an interesting question. It seems that for the Hamiltonian systems that arise in non-relativistic physics, the flow (or symmetry) corresponding to the total mass is always trivial (the Hamiltonian vector field associated to the mass is zero). In many cases it is because the mass is already constant on the whole phase space, but in other cases (e.g. KdV or compressible Euler) it seems that the symplectic form happens to degenerate to the constant-mass surfaces anyway. I don’t really have a good explanation of why this should be the case, though.

In special relativity, of course, the total mass is the magnitude of the total energy-momentum, and the latter is associated to spacetime translation invariance. (Somewhat confusingly, even though energy-momentum is additive – the total energy-momentum of a system is the sum of the energy-momenta of its components – mass is not additive in relativity. But total mass is still conserved, even though it is not the sum of the masses of the components.)

Dear Peter: thanks for the corrections!

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