3

Is there any theorem that forbids the bound system of two massive particles to have negative mass?

Qmechanic
  • 201,751
user58480
  • 73
  • Is this even true classically? Cf. http://en.wikipedia.org/wiki/Positive_energy_theorem –  Sep 04 '14 at 02:26
  • Is your system conservative? If it is not, you can have as much "effective negative mass", as you wish, you just have to supply the necessary energy from somewhere. – CuriousOne Sep 04 '14 at 04:08
  • Related: http://physics.stackexchange.com/q/18925/2451 , http://physics.stackexchange.com/q/44934/2451 and links therein. – Qmechanic Sep 04 '14 at 11:43

2 Answers2

4

A negative binding energy would make the vaccum unstable.

For example suppose a virtual electron and positron pop out of the vacuum. This costs energy to create the particles, but if their binding energy could be greater in magnitude than their rest masses then they could bind to form an energy state lower than the vacuum from which they were created. The result is that the vaccum would spontaneously decay into a lower energy state, which would then be the new vacuum state.

So the vacuum is by definition the lowest energy state that can exist, and no bound state can have a total energy lower than this.

John Rennie
  • 355,118
  • This doesn't seem to me like a completely satisfactory answer, because it assumes that we will have a lower bound on the energies of states, so that we have something we can call a ground state. If the mass of a pion was negative, then we would presumably start producing quark-antiquark pairs and forming lots and lots of pions. Wouldn't this be a runaway process? Why would it terminate in a well-defined ground state? –  Nov 10 '19 at 02:19
1

Within special relativity definitions mass is the positive root of the square root of the dot product in four vector space.

mass

With this definition there is no way a mass can be negative by construction. Two particles at rest will have zero momenta and their masses will add linearly for the minimum invariant mass of their system. Once they get momentum the invariant mass goes up. Bound particles have a momentum .

anna v
  • 233,453
  • How does potential energy come into play in this argument? Or it doesn't? – user58480 Sep 04 '14 at 06:15
  • It does not.the sign in a potential can be relative to where the zero is assumed. http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html – anna v Sep 04 '14 at 08:10
  • I meant if the mass of the bound system can be negative relative to the free, unbound, infinitely separated constituent particle's mass, just as when the mass of the hydrogen atom (bound system) is compared relative to the masses of free electron and proton. So I thought it natural to set the potential energy to be zero at infinity. (I should have mentioned that I am not thinking of a confining potential.) So I'm afraid I still don't see why the potential energy doesn't matter in the argument. – user58480 Sep 04 '14 at 09:00
  • 1
    The argument has to do with the constraints that special relativity sets on masses..Mass has to go through the addition of the four momentum vectors. Take a bound state part of the mass goes into the binding energy, which is important in nuclear bindings into nuclei. Then the sum of the masses of the constituents becomes larger than the mass of the nucleus. You are asking whether the mass of the nucleus can become negative? It would have to be possible to go to zero before becoming negative, and quantum mechanical solutions have ground states way above zero. – anna v Sep 04 '14 at 12:16